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ELECTROCHEMISTRYAND

CORROSION SCIENCE— SOLUTION MANUAL –

NESTOR PEREZUniversity of Puerto Rico

Department of Mechanical EngineeringMayaguez, PR 00680

Contents

1 FORMS OF CORROSION 11.1 QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 ELECTROCHEMISTRY 52.1 PROBLEMS/QUESTIONS . . . . . . . . . . . . . . . . . . . . . 5

3 THERMODYNAMICS OF AN ELECTROCHEMICAL CELL 93.1 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

4 NANO ELECTROCHEMISTRY 254.1 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

5 KINETICS OF ACTIVATION POLARIZATION 415.1 PROBLEMS/QUESTIONS . . . . . . . . . . . . . . . . . . . . . 41

6 MASS TRANSPORT BY DIFFUSION AND MIGRATION 716.1 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

7 CORROSIVITY AND PASSIVITY 977.1 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

8 DESIGN AGAINST CORROSION 1078.1 PROBLEMS/QUESTIONS . . . . . . . . . . . . . . . . . . . . . 107

9 ELECTRODEPOSITION 1259.1 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

10 HIGH-TEMPERATURE CORROSION 153

iii

iv CONTENTS

CHAPTERS:

v

vi CONTENTS

Chapter 1

FORMS OF CORROSION

1.1 QUESTIONS

1.1 During metallic corrosion there is loss of weight of the metal. Why metalsundergo corrosion in a suitable environment?

Answer: Metals are thermodynamically unstable in their free state andconsequently, they corrode forming a metallic compound due to their reactivityin the surrounding environment.

1.2 Is copper corrosion an oxidation process?Answer: Corrosion is an oxidation process in which a pure metal surface

reacts with its environment by loosing electrons, Cu ! Cu2++2e¡ , and forminga green …lm on the surface of copper.

1.3 It is known that corrosion on the surface of a metal is due to a directreaction of atmospheric gases. Which gas is mainly responsible for the corrosionof most metallic iron and steel surfaces?

Answer: Oxygen when compared to halogens, oxides of sulphur, oxides ofnitrogen, hydrogen sulphide.

1.4 What are the two mechanisms for oxidation of iron? Write down thesuitable electrochemical reactions.

Answer: (a) In the absence of oxygen, the evolution of hydrogen in acidsolutions (H2) is represented by the following electrochemical reactions

Fe ! Fe2+ + 2e¡ (anode)

2H+ + 2e¡ ! H2 (cathode)

Fe + 2H+ ! Fe2+ + H2 (redox)

(b) Absorption of oxygen (O2) in neutral or basic medium.

1

2 CHAPTER 1. FORMS OF CORROSION

Fe ! Fe2+ + 2e¡ (anode)

O2 + H2O + 2e¡ ! 2OH ¡ (cathode)

Fe + O2 + H2O + 2e¡ ! Fe2+ + 2OH ¡ (redox)

Fe2+ + 2OH ¡ ! Fe(OH)2

1.5 Investigate the di¤erences between dry and wet corrosion Write downat least three di¤erences per type of corrosion.

Solution:

Dry Corrosion1. Corrosion occurs in the absence of moisture2. Chemicals in the dry environment attack the metal surface3. It is a slow process

Wet Corrosion1. Corrosion occurs in presence of moisture2. It involves formation of electrochemical cells3. It is a fast process.

1.6 What is bimetallic corrosion?Answer: It is an electrochemical corrosion due to dissimilar metals being

connected and exposed to an electrolyte, forming a galvanic cell.

1.7 What form of corrosion will cause sand grains, dust particles and anoxide scale on the surface of metals exposed to a corrosive medium?

Answer: Pitting corrosion.

1.8 Why stress corrosion is not uniform?Answer: Stress corrosion not uniform because a mechanically stressed

structure is highly strained at certain locations, which become the localizedanodic sites for corrosion to take place there.

1.9 What form of corrosion develops at rivets and bolts?Answer: Crevice corrosion

1.10 Corrosion is a process of destruction of a metal surface by the sur-rounding environment. Environmental parameters like temperature, humidity,gases (CO2, SO2, NOx) and pH. play an important role in studying corrosion.What are the two factors that govern the corrosion process?

Answer: Metallic and environmental

1.11 What are the main e¤ects or consequences of corrosion of structures?Answer: To name a few

1.1. QUESTIONS 3

² Corrosion reduces the thickness of metal parts of a structure and as result,loss of mechanical strength occurs, leading failure of the structure.

² E¢ciency of machines is reduced due to corrosion.

² Because of corrosion, pipes, boilers and pumps are blocked by the corrosioncompound and therefore, they are di¢cult to operate e¢ciently.

1.12 The principle constituent of wood is cellulose, which is a polysaccha-ride. This is a polymer made of sugar molecules joined in long chains. Willthese sugar molecules corrode a galvanized steel nail and weaken a wood beamin a hypothetical …shing vessel?

Answer: Anodic coating is used as a sacri…cial material during a certainperiod of time and cathodic coating is used as a cathodic protection technique.

1.13 What method is used to prevent corrosion of iron by zinc coating?Answer: Galvanisation

1.14 The principle constituent of wood is cellulose, which is a polysaccha-ride. This is a polymer made of sugar molecules joined in long chains. Willthese sugar molecules corrode a galvanized steel nail and weaken a wood beamin a hypothetical …shing vessel?

Answer: The galvanized steel nail will corrode and the corrosion productwill soften the wood beam at contact

with the nail. Consequently, the vessel will begin to lose structural stability.

4 CHAPTER 1. FORMS OF CORROSION

Chapter 2

ELECTROCHEMISTRY

2.1 PROBLEMS/QUESTIONS

2.1 De…ne the galvanic electrode potential for the cell shown in Figure 2.4using the interfacial potentials involved in the electrochemical process.

Solution:

E =£ÁCu ¡ ÁCuSO4

¤Half ¡ cell

+£ÁZn SO4

¡ ÁZn

¤Half ¡ cell

+£ÁCu SO4

¡ ÁZnSO4

¤+

£ÁZn ¡ Á0Cu

¤Wire

+£ÁCu ¡ Á00Cu

¤Wire

+£ÁCu SO4

¡ ÁZnSO4

¤Bridge

E = ÁCu + ÁCu SO4¡ ÁCu SO4

+ ÁZnSO4¡ ÁZnSO4

+ ÁZn ¡ ÁZn ¡ Á0Cu

+ÁCu ¡ Á00Cu

E = ÁCu ¡ Á0Cu

since ÁCu = Á00Cu at the copper electrode and copper wire junction. In addi-tion, let Á0Cu and Á00Cube the potential at the copper wire-zinc electrode and thepotential at copper wire-copper electrode junctions, respectively.

2.2 Described what happens in the galvanic cell shown in Figure 2.4 whenelectrons leave the Cu terminal at the Cu-Zn interface.

Explanation: The Cu¡Zn interface is disturbed causing electrons to ‡owout of Zn into Cu terminal and consequently, Zn oxidizes according to theanodic reaction Zn = Zn+2 + 2e. This reaction occurs at the Zn-ZnSO4(aq:)

interface where Zn+2 cations go into solution loosing electrons. Then, theseelectrons react with Cu+2 cations according to the cathodic reaction Cu+2+2e =Cu.

5

6 CHAPTER 2. ELECTROCHEMISTRY

2.3 Explain why an electrode potential di¤erence occurs in Figure 2.4.

Explanation: The CuSO4 solution is depleted of Cu+2 cations and theZnSO4 is enriched in Zn+2 cations. This causes a ‡ow of positive ions (cations)through the solutions from the Zn electrode to the Cu electrode and a ‡ow ofnegative SO¡ 2

4 ions (anions) towards the Zn electrode. Hence, the current iscarried through the solution by these ions.

2.4 Why are Cu+2 cations electroplated on the Cu electrode surface inFigure 2.4?

Answer: During the operation of the galvanic cell both Zn ! Zn+2 + 2eand Cu+2+2e ! Cu reactions occur since the electron ‡ow process is 2e (Zn) !2e (Cu) and Cu+2 is reduced on the Cu electrode surface as a metallic element.

2.5 Calculate the standard potential for the formation of ferric hydroxideFe(OH)3 (brown rust).

Solution: The possible half-cell reactions and their respective standard po-tential are

4Fe + 12OH ¡ ! 4Fe (OH)3 + 12e =) Eoa = 0:771 V

3O2 + 6H2O + 12e ! 12 (OH ¡ ) =) Eoc = 0:401 V

4Fe + 3O2 + 6H2O ! 4Fe (OH)3 =) Eo = 1:172 V

The standard potentials were taken from Table 2.3. Thus, Eo = Eoa + Eo

c =1:172 V:

2.6 Explain why care must be exercised in using the galvanic series illus-trated in Table 2.2.

Explanation: The electrode anodic and cathodic reactions dissipate energyand consequently, the electrode potentials and activities may be altered.

2.7 If zinc and copper rods are placed in salt water, a direct chemicalreaction may slightly corrode zinc. Why?

Answer: Because Zn is more active than Cu and therefore, Zn is anodicto Cu.

2.8 Is the cell potential a surface potential? If so, why?

Answer: Ecorr must be a surface potential because it arises due to simulta-neous anodic and cathodic reactions on the metal surface.

2.1. PROBLEMS/QUESTIONS 7

2.9 If a monopole contains a net charge (Q) due to 10¡ 12 moles of a speciesj in a electrically neutral system, then calculate the electric potential (Áx), theelectric potential strength (Ex) and the electric force (Fx) acting on a particlewithin a distance of 9 cm from another charged particle in the x-direction.

Solution:

From eq. (2.4),

Q = zFX = (2)(96; 500 C=mol)¡10¡ 12 mol

¢

Q = 1:93x10¡ 7 C = 0:193 ¹C

But,

¸ =1

4¼²o= 9x109 N:m2

C2 = 9x109 J:m

C2

r = 0:09m

Inserting these values into eq. (2.8) give the magnitude of the electric poten-tial on a monopole system. Thus,

Áx =¸Q

r=

¡9 £ 109 J:m= C2

¢ ¡1:93x10¡ 7 C

¢

0:09 mÁx = 19; 300 J/C = 19; 300 N:m/C

Áx = 19; 300 J/(J=V ) = 19; 300 V = 19:30 kV

which is a high value for the isolated monopole case. The magnitude of theelectric potential strength is expected to have a very high magnitude given by eq.(2.17). Thus,

Ex =¸Q

r2=

Á

r=

19; 300 N:m=C

0:09 m

Ex = 2:14 £ 105 N=C = 2:14 £ 105 V=m

Thus, the electric force acting on the particle becomes

Fx = QEx =¡1:93x10¡ 7 C

¢ ¡2:14 £ 105 N=C

¢

Fx = 0:04 N

2.10 A battery (Example 2.2) containing 0:4 moles of MnO2 delivers 1:5 V .For 4-hour operation, calculate the electric current and the power (in watts).

Solution:

8 CHAPTER 2. ELECTROCHEMISTRY

From example 2.2, MnO2 + H+ + e ! MnOOH and

Q = zFXMnO2 = (1) (96500 A:s=mol) (0:4 mol)

Q = 38; 600 A:s = 38; 600 C = 38:60 kJ=V

Then,

I =Q

t=

38; 600 A:s

4 £ 60 £ 60 s= 2:68 A

P = EI = (1:5 V ) (2:68 A) = 4:02 V:A = 4:02 watts

2.11 Write down the reversible hydrogen reaction for a standard electrodeand determine its standard electric potential.

Solution:

H2 = 2H+ + 2e¡

The standard electric potential for the above reaction is taken a zero becauseit is a standard reference. Thus, Eo

H2=H+ = 0.

2.12 Calculate the Eo values for each of the following reactions as writtenand determine which one is the cathode and anode.

2Fe3+ + Cd ! Cd2+ + 2Fe2+

Solution:From Table 2.2,

Eo = EoFe3+=Fe2+ + Eo

Cd=Cd2+ = 0:77 VSHE + 0:403 VSHE

Eo = 1:173 VSHE

Fe+3 = Cathode because it reduces to Fe+2

Cd = Anode because it oxidizes to Cd2+

Recall that the reactions illustrated in Table 2.2 are for reduction processes.In this problem cadmium (Cd) oxidizes and therefore, the standard potentialchanges its sign from negative to positive. That is,

Cd2+ + 2e¡ ! Cd EoCd2+=Cd = ¡0:403VSHE

Cd ! Cd2+ + 2e¡ EoCd=Cd2+ = +0:403VSHE

Chapter 3

THERMODYNAMICS OFAN ELECTROCHEMICALCELL

3.1 PROBLEMS

3.1 Consider the following electrochemical cell¯¯Cu+2; Cu

¯¯ jH2;H

+jPt to cal-culate the maximum activity of copper ions Cu+2 in solution due to oxidation ofa copper strip immersed in sulfuric acid H2SO4 at 25 ± C, 101 kPa and pH = 2.

Solution: The possible half-cell reactions and their respective standard po-tentials (Table 2.2) are

Cu+2 + 2e ! Cu =) Eoc = 0:337 V

H2 ! 2H+ + 2e =) Eoa = 0 V

Cu+2 + H2 ! Cu + 2H+ =) Eo = 0:337 V

Then,

¢Go = ¡zFEo = (2)

µ

96; 500J

mol:V

¶

(0:337 V )

¢Go = ¡65kJ

mol

Therefore, the reactions will occur as written since ¢Go < 0. The maxi-mum activity of Cu+2 cations can be calculated when the potential and currentdensities are ECu = EH at icorr = imax. Using the Nernst equation yields

9

10CHAPTER 3. THERMODYNAMICS OF AN ELECTROCHEMICAL CELL

ECu = EoCu ¡

RT

zFln

[Cu]

[Cu+2]

EH = EoH ¡

RT

zFln

[H+]2

[H2]= ¡0:059pH

EH = ¡0:059 (2) = ¡0:118 V

Letting ECu = EH = ¡0:118 V yields

EH = EoCu ¡

RT

zFln

[Cu]

[Cu+2]= ¡0:118 V

ln[Cu]

[Cu+2]= (Eo

Cu ¡ EH)

µzF

RT

¶

Cu

(EoCu ¡ EH)

¡zFRT

¢Cu

= (0:337 V + 0:118 V)

·(2)(96500 J

m o l .V )(8:314 J

m o l .oK )(298 oK)

¸

(EH ¡ EoCu)

µzF

RT

¶

Cu

= 35:44

and

ln[Cu]

[Cu+2]= 35:44

[Cu]

[Cu+2]= 2:46x1015

Thus,

£Cu+2

¤=

[Cu]

2:46x1015=

1 mol=l

2:46x1015

aCu+2 =£Cu+2

¤= 4:06x10¡ 16 mol=l

3.2 Consider a galvanic cell j Zn j0:10MZnSO4j j0:10 M NiSO4jNi j.Calculate the cell potential using the Nernst equation at T = 25 ± C.

Solution: The possible half-cell reactions and their respective standard po-tential (Table 2.2) are

Ni+2 + 2e ! Ni =) EoNi = ¡0:250 V

Zn ! Zn+2 + 2e =) EoZn = +0:763 V

Ni+2 + Zn ! Ni + Zn+2 =) E = +0:513 V

3.1. PROBLEMS 11

Using the Nernst equation along with£Ni+2

¤=

£Zn+2

¤= 0:10 mol=l and

[Ni] = [Zn] = 1 mol/l gives

ENi = EoNi ¡

RT

zFln

[Ni]

[Ni+2]= ¡0:280 V

EZn = EoZn ¡

RT

zFln

£Zn+2

¤

[Zn]= 0:793 V

E = ENi + EZn = 0:513 V

3.3 The dissociation constant of silver hydroxide, AgOH, is 1:10x10¡ 4 at25 ± C in an aqueous solution. (a) Write down the chemical reaction and (b)determine the Gibbs free energy change ¢Go.

Solution:

(a) The dissociation chemical reaction is

AgOH = Ag+ + OH¡

Ke =[Ag+] [OH ¡ ]

[AgOH]= 1:10x10¡ 4

(b) The free energy change is

¢Go = ¡RT ln (Ke) = ¡

µ

8:314J

mol:oK

¶

(298oK) ln¡1:1x10¡ 4

¢

¢Go ¼ ¡22:58 kJ=mol

Therefore, the reaction will proceed as written since ¢Go < 0.

3.4 Calculate (a) the concentration of AgOH and Ag+ in g=l at 25 ± C and(b) the pH if [OH ¡ ] = 4:03x10¡ 3 mol=l and Ke = 1:10x10¡ 4.

Solution:

(a) AgOH = Ag+ + OH¡ and Ke = 1:10x10¡ 4. Let [Ag+] = [OH ¡ ] forcharge balance so that

Ke =[Ag+] [OH¡ ]

[AgOH]=

[OH¡ ]2

[AgOH]

[AgOH] =[OH¡ ]

2

Ke=

¡4:03 ¤ 10¡ 3

¢2

1:10 ¤ 10¡ 4= 0:15 mol=l

But,


CAgOH = [AgOH]Aw;AgOH

CAgOH = (0:15 mol=l)³107:87

g

l

´= 18:73 g=l

£Ag+

¤=

£OH ¡

¤= 4:03x10¡ 3 mol=l

CAg+ =£Ag+

¤Aw;Ag =

µ

4:03x10¡ 3 mol

l

¶³107:87

g

l

´= 0:43 g=l

COH¡ =£OH ¡

¤Aw;OH =

µ

4:03x10¡ 3 mol

l

¶³17

g

mol

´= 6:85x10¡ 2 g=l

(b) The dissociation of water ant its rate constant are

H2O = H+ + OH ¡

Kw =[H+] [OH¡ ]

[H2O]= 10¡ 14

where [H2O] = 1 mol=l and

£H+

¤=

Kw

[OH ¡ ]=

10¡ 14

4:03x10¡ 3 mol=l= 2:48x10¡ 12 mol=l

pH = ¡ log£H+

¤= ¡ log

¡2:48x10¡ 12

¢= 11:61

3.5 For an electrochemical copper reduction reaction at 25 ± C and 101 kPa,calculate (a) the Gibbs free energy change ¢Go as the driving force, (b) theequilibrium constant Ke and (c) the copper ion concentration

£Cu+2

¤in g=l.

Solution:

(a) The reduction reaction and its standard potential are

Cu+2 + 2e = Cu Eo = 0:337 V

¢Go = ¡zFEo = ¡ (2)

µ

96; 500J

mol:V

¶

(0:337 V )

¢Go = ¡65; 041J

mol

(b) The equilibrium constant is

lnKe = ¡¢Go

RT= ¡

¡¡65; 041 J

mol

¢

¡8:314 J

mol:oK

¢(298 oK)

= 26:25

Ke = 2:52x1011

3.1. PROBLEMS 13

(c) The concentration of copper cations is

Ke =[Cu]

[Cu+2]

£Cu+2

¤=

[Cu]

Ke=

1 mol=l

2:52x1011= 3:97x10¡ 12 mol=l

CCu+2 =£Cu+2

¤Aw;Cu =

¡3:97x10¡ 12 mol=l

¢(63:55 g=mol)

CCu+2 = 2:52x10¡ 10g=l

3.6 Use the electrochemical cell j Cd¯¯0:05 M Cd+2

¯¯¯¯0:25 M Cu+2

¯¯Cu j to

calculate a) the temperature for an electric potential of 0:762 VSHE, and b) ¢S,¢G; ¢H, and Q.

Solution:

a) The reactions and the required temperature are

Cu+2 + 2e = Cu =) EoCu = 0:337 V

Cd = Cd+2 + 2e =) EoCd = 0:403 V

Cu+2 + Cd = Cu + Cd+2 =) Eo = +0:740 V

E = Eo ¡RT

zFln (Ke)

Ke =[Cu]

£Cd+2

¤

[Cd] [Cu+2]=

[1] [0:05]

[1] [0:25]= 0:2

T = ¡zF (E ¡ Eo)

R ln (Ke)

T = ¡(2)

¡96; 500 J

mol:V

¢(0:762 V ¡ 0:740 V )

¡8:314 J

mol:oK

¢ln (0:2)

T = 317:32oK = 44:32oC

b) Using eqs. (2.32) and (2.35) at constant pressure yields

E = Eo ¡RT

zFln (Ke)

@E

@T= ¡

R

zFln (Ke)

¢S = zF@E

@T= ¡R ln (Ke)

¢S = ¡

µ

8:314J

mol:oK

¶

ln (0:2) = 13:38J

mol:oK

From eq. (2.26),


¢G = ¡zFE = (2)

µ

96; 500J

mol:V

¶

(0:762 V )

¢G = ¡147 kJ=mol

From eq. (2.34),

¢H = ¢G + T¢S

¢H = ¡147; 000 J=mol + (317:32oK)

µ

13:38J

mol:oK

¶

¢H = 151:25 J=mol

From eq. (236),

Q = ¡T¢S = ¡ (317:32oK)

µ

13:38J

mol:oK

¶

Q = ¡4:25 kJ=mol

3.7 Pure copper and pure cobalt electrodes are separately immersed insolutions of their respective divalent ions, making up a galvanic cell that yieldsa cell potential of 0:65 VSHE . Calculate a) the cobalt activity if the activity ofcopper ions is 0:85 mol=l at 25 ± C and b) the change in entropy ¢So and Gibbsfree energy ¢Go.

Solutions:

a) The reactions and the required cobalt activity are

Cu+2 + 2e = Cu =) EoCu = 0:337 V

Co = Co+2 + 2e =) EoCo = 0:227 V

Cu+2 + Co = Cu + Co+2 =) Eo = +0:614 V

Using eq. (2.32) yields

E = Eo ¡RT

zFln (Ke)

ln (Ke) = ¡zF (E ¡ Eo)

RT

ln (Ke) = ¡(2)

¡96; 500 J

mol:V

¢(0:65 V ¡ 0:614 V )

¡8:314 J

mol:oK

¢(298oK)

= ¡2:8044

Ke = 0:0605

3.1. PROBLEMS 15

Thus,

Ke =[Cu]

£Co+2

¤

[Cu+2] [Co]

£Co+2

¤=

K£Cu+2

¤[Co]

[Cu]=

(0:0605) [0:85 mol=l] [1 mol=l]

[1 mol=l]

aCo+2 =£Co+2

¤= 0:0515 mol=l

b) Using eqs. (2.32) and (2.35) at constant pressure yields

E = Eo ¡RT

zFln (Ke)

@E

@T= ¡

R

zFln (Ke)

¢S = zF@E

@T= ¡R ln (Ke)

¢S = ¡

µ

8:314J

mol:oK

¶

ln (0:0605) = 23:32J

mol:oK

From eq. (2.26),

¢G = ¡zFE = ¡ (2)

µ

96; 500J

mol:V

¶

(0:65 V )

¢G = ¡125:45 kJ=mol

3.8 Show that the activities are£M+2

1

¤=

£M+2

2

¤in a galvanic cell, provided

that the potentials is E = Eo.

Solution:

The reactions and the required activities are

M1 = M+21 + 2e =) Eo

M1

M2+2 + 2e = M2 =) EoM2

M1 + M+22 = M+2

1 + M2 =) Eo = EoM1

+ EoM2



E = Eo ¡RT

zFln (Ke)

dE

dT=

R ln (Ke)

zF= 0

ln (Ke) = ln

£M+2

1

¤

£M+2

2

¤ = 0

ln£M+2

1

¤¡ ln

£M+2

2

¤= 0

ln£M+2

1

¤= ln

£M+2

2

¤

£M+2

1

¤=

£M+2

2

¤

3.9 Show that ¢S = ¡R lnKe in a galvanic cell at constant pressure.

Solution:

Using eqs. (2.32) and (2.35) at constant pressure yields

E = Eo ¡RT

zFln (Ke)

@E

@T= ¡

R

zFln (Ke)

¢S = zF@E

@T= ¡R ln (Ke)

3.10 Derive an expression for ln£Cu+2

¤= f(T ) for the given cells below.

Assume that the current ceases to ‡ow and that the cells are not replenishedwith fresh solutions. Plot the expressions for a temperature range of 0 ± C ·T · 60 ± C, and draw some suitable conclusions to explain the electrochemicalbehavior based on the resultant trend.

Cu

CuSO4 (aq)a(Zn+2) = 1 mol/l a(H+) = 1 mol/l

Zn

Cu

Pb

e- e-

V+ - V+ -

Solution:a) The reactions and the required copper activity are

3.1. PROBLEMS 17

Cu+2 + 2e = Cu =) EoCu = 0:337 V

Zn = Zn+2 + 2e =) EoZn = 0:763 V

Cu+2 + Zn = Cu + Zn+2 =) Eo = +1:10 V

Ke =[Cu]

£Zn+2

¤

[Cu+2] [Zn]=

[1] [1]

[Cu+2] [1]=

1

[Cu+2]

E = Eo ¡RT

zFln (Ke) = 0

ln (Ke) =zFEo

RT=

ln1

[Cu+2]=

zFEo

RT

ln£Cu+2

¤= ¡

zFEo

RT¡

(2)¡96; 500 J

mol:V

¢(1:10 V )

¡8:314 J

mol:oK

¢T

ln£Cu+2

¤= ¡

2:55x104

T= y (Solid curve)

b) Similarly, the Pb-Cu Cell gives

H2 = 2H+ + 2e =) EoCu = 0 V

Cu+2 + 2e = Cu =) EoH = 0:337 V

Cu+2 + H2 = Cu + 2H+ =) Eo = +0:337 V

Ke =[Cu]

£Pb+2

¤

[Cu+2] [Pb]=

[1] [1]

[Cu+2] [1]=

1

[Cu+2]

E = Eo ¡RT

zFln (Ke) = 0

ln (Ke) =zFEo

RT

ln1

[Cu+2]=

zFEo

RT

ln£Cu+2

¤= ¡

zFEo

RT¡

(2)¡96; 500 J

mol:V

¢(0:337 V )

¡8:314 J

mol:oK

¢T

ln£Cu+2

¤= ¡

7823: 1

T= y (Dash curve)

The required plots are y = ln£Cu+2

¤= f (T )


337.5325312.5300287.5275

0

-25

-50

-75

-100

T

y

The plot below shows the trend of the derived expressions at a larger range.

350300250200150100500

0

-500

-1000

-1500

-2000

T

y

Besides the di¤erences in activity behavior, either trend means that the activ-ity, solution conductivity and ionic di¤usivity increase with increasing tempera-ture. In addition, the reduction of copper ions is faster at higher temperature.

EXTRA:

EX1. From Levine [2].

The entropy change (dS), the electrical energy change (dE), and the changeof charge (dq) for a ®-² coupling are, respectively

3.1. PROBLEMS 19

dS = dS® + dS² (3.1a)

T®dS® = dU® + P®dV ® ¡ [(¹®d´®)M + (¹®d´®)e] (3.1b)

T ²dS² = dU² + P ²dV ² ¡ (¹²d´²)Mz+ (3.1c)

dE = dU® + dU ² + Á®dq® + Á²dq² (3.1d)

dq = dq® + dq² = zeF (d´®)e + zF (d´²)Mz+ (3.1e)

where ¹ = Chemical potential P = Pressure E = Total energyU = Internal energy V = VolumeS = Entropy T = Temperatureq = Electron charge ´ = Number of molesÁ = Electric potential F = Faraday’s constant

Figure 3.1 Galvanic cell and its phase components.

For an isolated ®-² thermodynamic system, the following constraints can beused to develop the equilibrium conditions and the Gibbs free energy change(¢G) [2]

(d´®)Mz+ = ¡ (d´®)M (3.2a)

dV ² = ¡dV ® (3.2b)

dV = 0 (3.2c)

dE = 0 (3.2d)

dq = 0 (3.2e)


Hence, combining eq. (3.1e) and (3.2e) and the resultant expression witheq. (3.2a) yields

(d´®)e = +(d´²)e since ze = ¡1 (3.3a)

(d´®)e = ¡z (d´®)M since z ¸ 1 (3.3b)

Adding (d´®)e and (d´²)Mz+ moles to the ® phase and the ² phase, respec-tively, the change in electron charges becomes [2]

dq® = ¡F (d´®)e (3.4a)

dq² = +zF (d´²)Mz+ (3.4b)

Substituting eqs. (3.2a) and (3.3a) into (3.1e) and (3.1e) into (3.1d) yields

dU® + dU ² + zF (Á® ¡ Á²) (d´®)M = 0 (3.5a)

dU ² = ¡ [dU® + zF (Á® ¡ Á²) (d´®)M ] (3.5b)

Substituting eqs. (3.2a) through (3.5b) into (3.2b) and (3.2c) and the re-sultant expressions into (3.2a), and subsequently, collecting terms yields theentropy change of the ®-² system

dS =

·1

T®¡

1

T ²

¸

dU® +

·P®

T®¡

P ²

T ²

¸

dV ® (3.6)

+

·

¡(¹®)M

T®+

z (¹®)e

T®¡

(¹²)Mz+

T ²+

(zF ) (Á² ¡ Á®)

T ²

¸

(d´®)M

At equilibrium, dS = 0 and the following is deduced from eq. (3.6) [2]

1) For thermal equilibrium,

·1

T®¡

1

T ²

¸

= 0 since T® = T ² (3.7)

2) For mechanical equilibrium,

·P®

T®¡

P ²

T ²

¸

= 0 since P® = P ² (3.8)

3) For electrochemical equilibrium, T® = T ² and

(¹®)M ¡ [z (¹®)e ¡ (¹²)Mz+ ] + zF (Á® ¡ Á²) = 0 (3.9a)

(¹®)M ¡ [z (¹®)e ¡ (¹²)Mz+ ] = ¡zF (Á® ¡ Á²) (3.9b)

3.1. PROBLEMS 21

and

¢G® = (¹®)M ¡ [z (¹®)e ¡ (¹²)Mz+ ] (3.9c)

¢G® = ¡zF (Á® ¡ Á²) (3.9d)

Thus, (¹®)M ¡ [z (¹®)e ¡ (¹²)Mz+ ] is the a¢nity for M® = (Mz+)²+(ze¡ )

²

and ¢G® is the Gibbs free energy change for the ®-² subsystem.In addition, the potential di¤erence (Á® ¡ Á²) between two phases in a half-

cell at equilibrium cannot be measured. Instead, two electrodes are connectedas shown in Figure 3.1 to measure the electric potential di¤erence between theelectrodes. This can be done by connecting, say, ¯-± system to the previous ®-²system using a salt bridge or porous membrane. Thus, the electrical potential

di¤erence between the ¯ and ® electrodes is E =³Á® ¡ Á¯

´. Recall that the

salt bridge is a glass tube …lled with an electrolyte that allows charge transferbetween the cell electrolytes without disturbing the equilibrium state.

Additionally, the anodic and cathodic reactions, the Gibbs free energy changesand the electron balance expressions for the electrochemical system schemati-cally shown in Figure 3.1 are given next.

Case 1: The electron balance in the reactions is (z1e¡ )

®= (z2e

¡ )¯

= ze¡

and Á² = Á±. Case 2: If (z1e¡ )

®6= (z2e

¡ )¯

and Á² = Á±, then z = x (z1e¡ )

®=

y (z2e¡ )

¯, where x and y are integer numbers. Hence, the reactions are

M®1 =

¡Mz+

1

¢²+

¡z1e

¡¢®

¢G® = ¡zFE® (3.10)¡Mz+

2

¢±+

¡z2e

¡¢¯

= M¯2 ¢G¯ = ¡zFE¯ (3.11)

M®1 +

¡Mz+

2

¢±=

¡Mz+

1

¢²+ M¯

2 ¢G = ¡zFE (3.12)

where E® = Á® ¡ Á², E¯ = Á± ¡ Á¯, E = Á® ¡ Á¯,¡¹¯

¢e

= (¹®)e and¢G = ¢G® + ¢G¯. In general, the standard and non-standard Gibbs freeenergy change expressions are, respectively

¢Go = ¡zFEo (standard) (3.13)

¢G = ¡zFE (non-standard) (3.14)

EX2. From thermodynamics,The …rst law for a closed system is

dQ ¡ dW = dU (1)

where dV is the di¤erential volume of the electrolyte and the total di¤erentialwork due to pressure (P) and electrical potential (E) is

dW = dWP + dWe = PdV + zFdE (2)


Combining eqs. (1) and (2) yields

dQ = dU + dW (3)

dQ = dU + PdV + zFdE (4)

From the second law of thermodynamics,

dQ = TdS (5)

Then, eq. (4) becomes

TdS = dU + PdV + zFdE (6)

TdS = dH + zFdE (7)

where dH is the di¤erential enthalpy.

By de…nition, the di¤erential Gibbs free energy is

dG = dH ¡ TdS (8)

Combining eqs. (7) and (8) gives

dG = ¡zFdE (9)

Integrating

¢Go = ¡zFEo (standard) (10)

¢G = ¡zFE (non-standard) (11)

EX3 3.3 Use the data given in Table 3.1 to determine the half-cells and theredox standard potentials for the galvanic cell shown in Figure 3.1 if ® = Znand ¯ = Cu. Compare the results with the data given in Table 2.2.

Solution:The redox reaction is

Zn + Cu2+ = Zn2+ + Cu

From eq. (3.16) and Table 3.1,

EoCu2+=Cu = ¡

(º¹o)Cu ¡ (º¹o)Cu2+

zF

EoCu2+=Cu = ¡

·(1)(0) ¡ (1)(+64:98 kJ=mol)

(2)(96:50 kJ=mol ¢ V )

¸

= +0:337 V

3.1. PROBLEMS 23

EoZn=Zn2+ = ¡

(º¹o)Zn2+ ¡ (º¹o)Zn

zF

EoZn=Zn2+ = ¡

·(1)(¡147:21 kJ=mol) ¡ (1)(0)

(2)(96:50 kJ=mol ¢ V )

¸

= +0:763 V

These results are exactly the same given in Table 2.2. Thus, the standardcell potential is

Eo = EoCu2+=Cu + Eo

Zn=Zn2+ = +1:10 V

Chapter 4

NANOELECTROCHEMISTRY

4.1 PROBLEMS

4.1 In an electrochemical cell, the bulk electrolyte has uniform and constantion densities, but the electrical-double layer is an inhomogeneous ‡uid. Why?

Answer: The ion densities in the electrical-double layer vary in space inthe vicinity of the electrode surface due to the charged particle forming the in-homogeneous structure.

4.2 The interface region between two bulk phases may contain a complexdistribution of electric charge. Name at least two causes for this phenomenon.

Answer: (a) The charge transfer between phases and (b) unequal adsorptionof positive and negative ions.

4.3 Explain how it is possible for a scanning tunneling microscope (STM)to image atoms on the surface of a sample. Recall that one of the metals is thesample and the other is the probe.

Answer: Move the tip up and down in the z-direction to obtain a constantcurrent across the contour of the substrate (sample) surface. Then, record theposition of the tip as it scans the substrate surface along the x and y directions.The tip will trace out a contour of constant probability for …nding an atom onthe surface. The constant tunneling current for this process can be approximatedby

It = Io exp (¡¾z)

25

26 CHAPTER 4. NANO ELECTROCHEMISTRY

where Io and ¾ is a constant. Nonetheless, when the tip is scanned over thesurface, the tip height (z) is determined by the local geometric and electronicstructure of the surface and thus it produces a surface map in real space.

4.4 For a …nite depth energy well, the wave function Ã (x) within the barrierof width w, the tunneling current It through the barrier and the decay lengthare related as shown below:

Ã (x) = Ão exp (¡k2x)

It / jÃoj2exp (¡k2x)

k2 =2¼

h

p2m (Uo ¡ E) =

2¼

h

p2mÂ

If the work function and the barrier width are Â = 4:32 eV and w = 0:151nm in a STM experiment, then determine It for a barrier of width 2w. Explain.

Solution:

k2 =³5:1268 nm¡ 1:eV ¡ 1=2

´(Uo ¡ E)

1=2

k2 =³5:1268 nm¡ 1:eV ¡ 1=2

´Â1=2 =

³5:1268 nm¡ 1:eV ¡ 1=2

´(4:32 eV )1=2

k2 = 10:656 nm¡ 1

Then,

It;w

It;2w=

jÃoj2exp (¡k2w)

jÃoj2 exp (¡2k2w)

It;w

It;2w= exp (k2w) = exp

£¡10:656 nm¡ 1

¢(0:20 nm)

¤

It;w

It;2w= 5 or It;w = 5It;2w or It;2w = 0:2It;w

Denote the signi…cant change in It;2w as w ! 2w. Therefore, the STM cellis very sensitive to changes in the barrier size.

4.5 Determine (a) the Fermi energy and (b) the velocity of the electron fora copper (Cu) substrate and a tungsten (W) tip in a ultra high vacuum STMcell at temperature T .

Solution:

(a) The Fermi energy: For gold, Cu ! Cu2+ + 2e¡ , with two free electronper reaction. The electron density, eq. (4.11), is

4.1. PROBLEMS 27

½e;Cu = [ne½NA=Aw]Cu

½e;Cu = (2)

µ8:94 £ 103 Kg=m3

63:55 Kg=Kmol

¶µ6:02 £ 1023 atom

10¡ 3 Kmol

¶

½e;Cu = 1:6937 £ 1029 1=m3

and

UF;Cu =h2

2me

µ3

8¼½e

¶2=3

UF;Cu =

¡6:63 £ 10¡ 34 J:s

¢2

2 ¤ 9:11 £ 10¡ 31 Kg

·µ3

8¼

¶¡7:6514 £ 1028 1=m3

¢¸2=3

UF;Cu = 1:0541 £ 10¡ 18 J2.s2

Kg:m2= 1:0541 £ 10¡ 18 J

UF;Cu =1:0541 £ 10¡ 18 J

1:60 £ 10¡ 19 J=eV= 6:59 eV

The units: J2.s2=¡Kg:m2

¢= J2= (N:m) = J . Similarly, for tungsten,

W ! W 4+ + 4e¡ , with 4 free electrons, eqs. (4.7) and (4.6) give

½e;W = [ne½NA=Aw]W

½e;W = (4)

µ19:3 £ 103

183:85

¶µ6:02 £ 1023

10¡ 3

¶

= 2:5278 £ 1029 1=m¡ 3

UF;W =

¡6:63 £ 10¡ 34 J:s

¢2

2 ¤ 9:11 £ 10¡ 31 Kg

µ3 ¤ 1:694 £ 1029

8¼

1

m3

¶2=3

= 2:3383 £ 10¡ 18 J

UF;W =2:3383 £ 10¡ 18 J

1:60 £ 10¡ 19 J=eV= 14:61 eV

Obviously, UF;W > UF;Cu due to the di¤erences in the amount of free elec-trons and the electron densities between gold and tungsten metals.

(b) The velocity of electrons: Using eq. (4.12) yields

ve =

r2UF

me

ve;Cu =

s2 (1:0541 £ 10¡ 18 J)

9:11 £ 10¡ 31Kg= 1:52 £ 106

sN:m

Kg

ve;Cu = 1:52 £ 106

sKg:m

s2

m

Kg= 1:52 £ 106 m=s


and for tungsten

ve;W =

s(2) (2:3383 £ 10¡ 18 J)

9:11 £ 10¡ 31Kg

ve;W = 2:27 £ 106 m=s = 2; 270 Km=s

Therefore, an electron travels very fast in the vacuum gap. Let’s determinethe atom density using eq. (4.9)

na;Cu =½NA

Aw=

¡8:94 £ 103 Kg=m3

¢

63:55 Kg=Kmol

µ6:02 £ 1023 atoms

10¡ 3 Kmol

¶

na;Cu = 8:4687 £ 1028 atoms=m3

na;W =½NA

Aw=

¡19:30 ¤ 103 Kg=m3

¢

183:85 Kg=Kmol

µ6:02 ¤ 1023 atoms

10¡ 3 Kmol

¶

na;W = 6:3196 £ 1028 atoms=m3

Verify the number of free electrons being used. Thus,

ne;Cu =

µ½e

na

¶

Cu

=1:6937 £ 1029 1=m3

1:8302 £ 1029 atoms=m3= 2

ne;W =

µ½e

na

¶

W

=2:5278 £ 1029

6:3196 £ 1028= 4

According to the above calculations, the amount of energy (o¤set) being givento the electrons during tunneling is

Â = ESF;Cu ¡ ET

F;W = 6:59 eV ¡ 14:61 eV = ¡8:02 eV

This is the energy that causes the electrons to acquire a speci…c velocity inthe tunneling gap (electrical conduction region).

4.6 For the reaction O + e¡ À R in an electrochemical cell, the cyclicvoltammetry method provided the formal potential (standard potential) Áo =133 mV against a reference electrode. If the activity ratio is [O] = [R] = 0:2,then calculate (a) the applied potential Á at 25oC and (b) the required energyfor the reaction to proceed to the right as written.

Solutions:

(a) Using the Nernst equation yields

4.1. PROBLEMS 29

Á = Áo ¡RT

zFln

[R]

[O]

Á = Áo +RT

zFln

[O]

[R]= 0:133 V +

(8:314 J=mol:K) (298 K)

(1) (96; 500 J=mol:V )ln (0:2)

Á = 0:09168 V = 91:68 mV

(b) The reaction energy:

U = qeÁ =¡1:602 £ 10¡ 19 J=V

¢(0:09168 V ) = 1:4687 £ 10¡ 20 J

U = 0:092 eV

4.7 Show that PT = exp (¡2k2x), eq. (4.48), using the following equations

Ã = A exp (iµ)

d2Ã

dx2= ¡

p2m [U ¡ E]

}2Ã = ¡

P 2

}2Ã

where P is the moment, x = w and } = h=2¼. Hint: extract the resultantreal and imaginary parts for solving this problem.

Solution:

Derivatives:

Ã = A exp (iµ) (a)

dÃ

dx=

·dA

dx+ iA

dµ

dx

¸

exp (iµ)

d2Ã

dx2=

"d2A

dx2+ iA

d2µ

dx2+ 2i

dA

dx

dµ

dx¡ A

µdµ

dx

¶2#

exp (iµ) (b)

d2Ã

dx2=

"d2A

dx2+ iA

d2µ

dx2+ 2i

dA

dx

dµ

dx¡ A

µdµ

dx

¶2#

Ã

A

Equate d2Ã=dx2 terms:

¡P 2

}2Ã =

"d2A

dx2+ iA

d2µ

dx2+ 2i

dA

dx

dµ

dx¡ A

µdµ

dx

¶2#

Ã

A

¡AP 2

}2=

"d2A

dx2+ iA

d2µ

dx2+ 2i

dA

dx

dµ

dx¡ A

µdµ

dx

¶2#

(c)


Imaginary parts:of eq. (c) and multiply by A and eliminate i:

0 = iAd2µ

dx2+ 2i

dA

dx

dµ

dx

0 = A2 d2µ

dx2+ 2A

dA

dx

dµ

dx(d)

Eq. (d) becomes

·

A2 d2µ

dx2

¸

=dx = 0 (e)

Integrating eq. (e) yields the wave amplitude A as

A2 dµ

dx= C1

A =

pC1p

dµ=dx=

Cp

dµ=dx(f)

Real Parts of eq. (c):

¡AP 2

}2=

d2A

dx2¡ A

µdµ

dx

¶2

¡P 2

}2=

1

A

d2A

dx2¡

µdµ

dx

¶2

and

1

A

d2A

dx2=

µdµ

dx

¶2

¡P 2

}2

If

1

A

d2A

dx2< <

µdµ

dx

¶2

¡P 2

}2

·1

A

d2A

dx2

¸

! 0

then

µdµ

dx

¶2

=P 2

}2

dµ

dx=

P

}

µ =1

}

Z

Pdx (g)

4.1. PROBLEMS 31

and

A =

pC1p

dµ=dx=

Cp

dµ=dx=

}Cp

P

A =D

pP

Thus,

Ã = A exp (iµ)

Ã =D

pP

exp

·i

}

Z

Pdx

¸

(h)

But,

P =p

2m (E ¡ U)

P = ip

2m (U ¡ E)

and eq. (h) along with i2 =¡p

¡1¢2

= ¡1 and …xed U = Uo

Ã =D

pP

exp

·i2

}

Z p2m (U ¡ E)dx

¸

Ã =D

pP

exp

"

¡

Z p2m (U ¡ E)

}dx

#

Ã =D

pP

exp

·

¡2¼

h

p2m [U (x) ¡ E]

Z

dx

¸

Ã =D

pP

exp

·

¡

Z

k2dx

¸

Ã =D

pP

exp

·

¡

Z x

0

k2dx

¸

=D

pP

exp (¡k2x)

Ãp

P

D= exp (¡k2x)

where

k2 =2¼

h

p2m [U (x) ¡ E]

Then,

PT =

ÃÃ

pP

D

!2

PT = exp (¡2k2x)


where x is the barrier width and according to Figure 4.10, x = x2 ¡ x1 = w

4.8 Below is a sketch for the principle of the STM technique. If the workfunction (Â) and the tunneling gap column (z) are 4 eV and 0:3 nm, respectively,calculate the probability for an electron to tunnel from the probe tip to the metalsample.

Solution:

k2 =³5:1268 nm¡ 1:eV ¡ 1=2

´(Uo ¡ E)

1=2=

³5:1268 nm¡ 1:eV ¡ 1=2

´(4 eV )

1=2

k2 = 10:254nm¡ 1

and

PT = exp (¡2k2z) = exp£¡2

¡10:254 nm¡ 1

¢(0:3 nm)

¤= 2:13 £ 10¡ 3

PT ' 0:21%

4.9 Calculate the electron energy (E) if the tunneling probability (PT ) is0:106 and the barrier height (Uo) 4:5 eV . Assume a barrier width of 0:4 nm.

Solution:

PT = exp (¡2k2w)

k2 = ¡ ln (PT ) =2w = ¡ ln (0:106) = (2x0:4) = 2:81 nm¡ 1

Then,

k2 =³5:1268 nm¡ 1:eV ¡ 1=2

´(Uo ¡ E)

1=2

E = Uo ¡ (2:81=5:1268)2

= 4:5 eV ¡ (2:81=5:1268)2

eV

E = 4:2 eV

4.1. PROBLEMS 33

4.10 Assuming that there is no wave re‡ection, eq. (4.20), at the boundary

x = 0, derive PR = jB=Aj2

and PT for E > Uo. Use the continuity of the wavefunction and its derivative at the origin.

Solution:From eq. (4.20),

Ã1 (x) = A exp (ik1x) + B exp (¡ik1x)

dÃ1 (x)

dx= ik1A exp (ik1x) ¡ ik1B exp (¡ik1x)

Ã2 (x) = C exp (ik2x) + D exp (¡ik2x)

Ã2 (x) = C exp (ik2x) since D = 0

dÃ2 (x)

dx= ik2C exp (ik2x) ¡ ik2D exp (¡ik2x)

dÃ2 (x)

dx= ik2C exp (ik2x) since D = 0

At x = 0;

Ã (0)1 = Ã (0)2A + B = C (1)

dÃ1 (0)

dx=

dÃ2 (0)

dx

A ¡ B =ik2

ik1C (2)

or

1 +B

A=

C

A

1 +B

A=

k2

k1

C

A

Combining eqs. (1) and (2) gives

1 +B

A=

k2

k1

µ

1 +B

A

¶

=k2

k1+

k2

k1

B

Aµ

1 ¡k2

k1

¶B

A= 1 +

k2

k1

B

A=

1 + k2=k1

1 ¡ k2=k1


Then,

PR =

¯¯¯¯B

A

¯¯¯¯

2

=

¯¯¯¯1 + k2=k1

1 ¡ k2=k1

¯¯¯¯

2

and

PT + PR = 1

PT = 1 ¡ PR

PT = 1 ¡

¯¯¯¯1 + k2=k1

1 ¡ k2=k1

¯¯¯¯

2

4.11 Assuming that the wave amplitude, eq. (4.23), C = 0 when x ! 1,

derive PR = jB=Aj2

and PT for E < Uo. Use the continuity of the wave functionand its derivative at the origin. Explain.

Solution:

From eq. (4.31) through ) (4.34),

A + B = C + D (4.31)

ik1 (A + B) = k2 (C + D) (4.32)

C exp (k2w) + D exp (¡k2w) = F exp (ik1w) (4.33)

k2 [C exp (k2w) ¡ D exp (¡k2w)] = ik1F exp (ik1w) (4.34)

Divide these eqs. by A so that

A + B = C + D (a)

ik1 (A + B) = k2 (C + D) (b)

C exp (k2w) + D exp (¡k2w) = F exp (ik1w) (c)

k2 [C exp (k2w) ¡ D exp (¡k2w)] = ik1F exp (ik1w) (d)

and

Ã1 (x) = A exp (ik1x) + B exp (¡ik1x)

dÃ1 (x)

dx= ik1A exp (ik1x) ¡ ik1B exp (¡ik1x)

Ã2 (x) = C exp (ik2x) + D exp (¡ik2x)

Ã2 (x) = D exp (¡ik2x) since C = 0

dÃ2 (x)

dx= ¡ik2D exp (¡ik2x)

4.1. PROBLEMS 35

At x = 0;

Ã (0)1 = Ã (0)2A + B = D (1)

dÃ1 (0)

dx=

dÃ2 (0)

dx

A ¡ B = ¡ik2

ik1D (2)

Divide eqs. (1) and (2) by A so that

1 +B

A=

D

A(3)

1 ¡B

A= ¡

k2

k1

D

A(4)

Eliminate D=A

1 +B

A= ¡

k2

k1

µ

1 +B

A

¶

= ¡ik2

ik1¡

k2

k1

B

A

B

A

µ

1 +k2

k1

¶

= ¡k2

k1¡ 1

B

A= ¡

1 + k2=k1

1 + k2=k1

Then,

PR =

¯¯¯¯B

A

¯¯¯¯

2

=

¯¯¯¯¡

1 + k2=k1

1 + k2=k1

¯¯¯¯

2

= 1

PT + PR = 1

PT = 1 ¡ 1 = 0

The particle will always be re‡ected at x = 0. This means that the particlecan not penetrate into the classically forbidden barrier (region 2, Figure 4.10 )and therefore, there is no particle transmission.

4.12 ForE < Uo, use eq. (4.23) to show that

F

A=

4ik1k2 exp (¡ik1w)

(k2 + ik1)2exp (¡k2w) ¡ (k2 ¡ ik1)

2exp (k2w)

PT =16k2

1k22h

(k1 + k2)2e¡ 2k2w + (k1 + k2)

2e2k2w

i+ [12k2

1k22 ¡ 2 (k4

2 + k41)]


Solution:

Ã (w)2 = Ã (w)3C exp (k2w) + D exp (¡k2w) = F exp (k1w) (3)

dÃ2 (w)

dx=

dÃ3 (w)

dxk2C exp (k2w) ¡ k2D exp (¡k2w) = ik1F exp (ik1w) (4)

Divide eqs. (a) through (d) by A and group them to be algebraically manip-ulated.

A + B = C + D (5)

A ¡ B =k2

ik1C ¡

k2

ik1D (6)

C exp (k2w) + D exp (¡k2w) = F exp (k1w) (7)

C exp (k2w) ¡ D exp (¡k2w) =ik1

k2F exp (ik1w) (8)

The tedious work: Divide by A

1 +B

A=

C

A+

D

A(9)

ik1

k2¡

ik1

k2

B

A=

C

A¡

D

A(10)

C

Aexp (k2w) +

D

Aexp (¡k2w) =

F

Aexp (k1w) (11)

C

Aexp (k2w) ¡

D

Aexp (¡k2w) =

ik1

k2

F

Aexp (ik1w) (12)

Add eqs. (11) and (12) to get

2C

Aexp (k2w) =

µik1

k2+ 1

¶F

Aexp (ik1w)

C

A=

F

A

µk2 + ik1

2k2

¶

exp (ik1w) exp (¡k2w) (13)

Next, subtract eqs. (11) ¡ (12)

2D

Aexp (¡k2w) =

µ

1 ¡ik1

k2

¶F

Aexp (ik1w)

D

A=

F

A

µk2 ¡ ik1

2k2

¶

exp (ik1w) exp (k2w) (14)

4.1. PROBLEMS 37

Add eqs. (10) + (9) to get

2C

A=

ik1

k2¡

ik1

k2

B

A+ 1 +

B

A

C

A=

k2 + ik1

2k2+

µk2 ¡ ik1

2k2

¶B

A(15)

Subtract eqs. (9) ¡ (10) so that

2D

A=

µ

1 ¡ik1

k2

¶

+

µ

1 +ik1

k2

¶

D

A=

k2 ¡ ik1

2k2+

µk2 + ik1

2k2

¶B

A(16)

Now, equate eqs. (13) = (15)

F

A

µk2 + ik1

2k2

¶

exp (ik1w) exp (¡k2w) =k2 + ik1

2k2+

µk2 ¡ ik1

2k2

¶B

A

F

A(k2 + ik1) exp (ik1w) exp (¡k2w) = (k2 + ik1) + (k2 ¡ ik1)

B

A(17)

Equate eqs. (14) = (16)

F

A

µk2 ¡ ik1

2k2

¶

exp (ik1w) exp (k2w) =k2 ¡ ik1

2k2+

µk2 + ik1

2k2

¶B

A

F

A(k2 ¡ ik1) exp (ik1w) exp (k2w) = (k2 ¡ ik1) + (k2 + ik1)

B

A(18)

Rearrange eqs. (17) and (18) so that

F

A(k2 + ik1) exp (ik1w) exp (¡k2w) ¡ (k2 + ik1) = (k2 ¡ ik1)

B

A(19)

F

A(k2 + ik1) exp (ik1w) exp (¡k2w) ¡ (k2 ¡ ik1) = (k2 + ik1)

B

A(20)

Now divide eqs. (19)=(20) to eliminate B=A

FA (k2 + ik1) exp (ik1w) exp (¡k2w) ¡ (k2 + ik1)FA (k2 + ik1) exp (ik1w) exp (¡k2w) ¡ (k2 ¡ ik1)

=k2 ¡ ik1

k2 + ik1(21)

Solve for F=A


PT =

¯¯¯¯F

A

¯¯¯¯

2

=F ¤ F

A¤A

F

A=

4k1k2eik1w

(k2 + ik1)2 e¡ k2w ¡ (k2 ¡ ik1)

2 ek2w

F ¤

A¤=

¡4k1k2e¡ ik1w

(k2 ¡ ik1)2e¡ k2w ¡ (k2 + ik1)

2ek2w

(Conjugate)

F ¤ F

A¤A=

¡4k1k2e¡ ik1w

h(k2 ¡ ik1)

2e¡ k2w ¡ (k2 + ik1)

2ek2w

i

x4k1k2e

ik1w

h(k2 + ik1)

2 e¡ k2w ¡ (k2 ¡ ik1)2 ek2w

i

Algebiaic work for the denominator:

[(k2 ¡ ik1) (k2 + ik1)]2e¡ k2we¡ k2w =

¡k21 + k2

2

¢2e¡ 2k2w

¡ [(k2 ¡ ik1) (k2 ¡ ik1)]2 e¡ k2wek2w = 4ik1k

32 ¡ k4

2 ¡ k41 ¡ 4ik3

1k2 + 6k21k

22

¡ (k2 + ik1)2(k2 + ik1)

22ek2w2e¡ k2w = 4ik3

1k2 ¡ k42 ¡ 4ik1k

32 ¡ k4

1 + 6k21k

22

[(k2 + ik1) (k2 ¡ ik1)]2 =

¡k21 + k2

2

¢2e2k2w

X = 4ik1k32 ¡ k4

2 ¡ k41 ¡ 4ik3

1k2 + 6k21k

22 + 4ik3

1k2 ¡ k42 ¡ 4ik1k

32 ¡ k4

1 + 6k21k

22

Y = 12k21k

22 ¡ 2k4

2 ¡ 2k41

X = Y

Then, the full-blown tunneling probability equation becomes

PT =

¯¯¯¯F

A

¯¯¯¯

2

=F ¤ F

A¤A

PT =16k2

1k22

(k1 + k2)2e¡ 2k2w + (k1 + k2)

2e2k2w + 12k2

1k22 ¡ 2k4

2 ¡ 2k41

PT =16k2

1k22h

(k1 + k2)2e¡ 2k2w + (k1 + k2)

2e2k2w

i+ [12k2

1k22 ¡ 2 (k4

2 + k41)]

4.1. PROBLEMS 39

Test:If k1 = 14; k2 = 7 and w = 1:2, then

PT =16k2

1k22

(k1 + k2)2e¡ 2k2w + (k1 + k2)

2e2k2w + 12k2

1k22 ¡ 2 (k4

2 + k41)

PT = 1:761 9 £ 10¡ 5

(k1 + k2)2e¡ 2k2w + (k1 + k2)

2e2k2w = 8:721 4 £ 109

12k21k

22 ¡ 2

¡k42 + k4

1

¢= 33; 614

h(k1 + k2)

2e¡ 2k2w + (k1 + k2)

2e2k2w

i> >

£12k2

1k22 ¡ 2

¡k42 + k4

1

¢¤

8:721 4 £ 109 > > 33; 614

According to these results, the full-blown equation can be simpli…ed as

PT =16k2

1k22

(k1 + k2)2e¡ 2k2w + (k1 + k2)

2e2k2w

= 1:761 9 £ 10¡ 5

If e2k2w >> e¡ 2k2w, then a further simpli…cation of the PT equations on-tained as

PT =16k2

1k22

(k1 + k2) e2k2w= 1:761 9 £ 10¡ 5

Thus,

PT =

¯¯¯¯F

A

¯¯¯¯

2

=F ¤ F

A¤A=

16k21k

22

(k22 ¡ k2

1)2 exp (¡k2w)

Chapter 5

KINETICS OFACTIVATIONPOLARIZATION


5.1 What are the three conditions for galvanic corrosion to occur?

Answer: Dissimilar metals, corrosive media, and electrical contact.

5.2 If the state of equilibrium of an electrochemical cell is disturbed by anapplied current density ix, then ia = ¡ic = icorr no longer holds. Let ix be acathodic current density and the slope of the corresponding polarization curvebe dE=d log (ix), which increases approaching the Tafel constant ¯c. Determinea) the value of ix when ic = 10¡ 3 A=cm2 and ia = 10¡ 9 A=cm2 and b) thevalue of ¯c when icorr = 10¡ 5 A=cm2 and ´ = ¡0:20 V .

Solution:

a) ix = ic ¡ ia ¼ ic = 10¡ 3 A=cm2

b) The cathodic overpotential is given by

´c = ¡¯c logix

icorr

¯c = ¡´c

log (ix=icorr)=

0:20

log (10¡ 3=105)

¯c = 0:10 V

5.3 According to the Stockholm Convention Cell Fe¯¯Fe+2

¯¯ jH+;H2jPt,

the corrosion potential of iron (Fe) is ¡0:70 VSCE at pH = 4:4 and 25oC in a

41

42 CHAPTER 5. KINETICS OF ACTIVATION POLARIZATION

deaerated (no oxygen is involved) acid solution. Calculate a) the corrosion ratein mm=y when

io;H = 10¡ 6 A=cm2 ¯c = ¡0:10 VSHE

io;Fe = 10¡ 8 A=cm2 EFe = ¡0:50 VSHE

b) the Tafel anodic slope ¯a and c) draw the kinetic diagram E vs: log i.

Solutions:

The possible reactions are

Fe ! Fe+2 + 2e ) EFe = ¡0:50 V2H+ + 2e ! H2 ) EH =?

Fe + 2H+ ! Fe+2 + H2 ) E = EFe + EH

a) From Table 2.7 and Nernst equation,

Ecorr = ¡0:70 + 0:241 ¼ ¡0:46 VSHE

EH = EoH ¡

RT

zFln

[H2]

[H+]2 = Eo

H +4:606RT

zFlog

£H+

¤

EH = EoH ¡

4:606RT

zF(pH)

EH = 0 ¡(4:606) (8:314 J=mol:K) (298K)

(2) (96; 500 J=mol:V )(pH)

EH = ¡ (0:0591) (4:4) = ¡0:26 V

Thus, the cell potential and the overpotentials becomes

E = EFe + EH = ¡0:50 V ¡ 0:26 V = ¡0:76 V

á = EFe ¡ Ecorr = ¡0:46 V + 0:50 V = ¡0:04 V

ć = EH ¡ Ecorr = ¡0:26 V + 0:46 V = ¡0:20 V

Using eq. (5.35) yields the corrosion current density

ć = ¡¯c log

µio;H

icorr

¶

icorr = io;H10ć=¯c

icorr =¡10¡ 6 A=cm2

¢100:2=0:10

icorr =¡10¡ 6 A=cm2

¢102

icorr = 10¡ 4 A=cm2

Using eq. (5.48) gives the iron corrosion rate

CR =icorrAw

zF½=

¡10¡ 4 A=cm2

¢(55:85 g=mol)

(2) (96; 500 A:s=mol) (7:86 g=cm3)

CR = 3:68x10¡ 9 cm=s = 1:16 mm=y


b) From eq. (5.32),

´a = ¯a log

µicorr

io;Fe

¶

¯a = ´a= log

µicorr

io;Fe

¶

¯a = (0:04 V ) = log¡10¡ 4A=cm2=10¡ 8A=cm2

¢

¯a = 0:01 V

c) The required diagram is

ηa

10-10 10-8 10-6 10-4 10-2

-0.30

i (A/cm2)

-0.20

-0.60

-0.50

-0.40ηc

βc

βa

E (V)

(io,E)H

(io,E)Fe

(i,E)coorr

5.4 Let the following anodic and cathodic reactions be, respectively

M = M+2 + 2e¡ EM = ¡0:941 VSCE io;M = 10¡ 2 ¹A=cm2

2H+ + 2e¡ = H2 EH = ¡0:200 VSHE io;H = 0:10 ¹A=cm2

where icorr = 102 ¹A=cm2 and Ecorr = ¡0:741 VSCE. a) Construct the cor-responding kinetic diagram and b) determine both ¯a and ¯c from the diagramand from the de…nition of overpotential equations.

Solutions:

From Table 2.7,

EM = ¡0:941 + 0:241 = ¡0:70 V

Ecorr = ¡0:741 + 0:241 = ¡0:50 V

Thus, the required diagram is


ηa

10-4 10-2 100 102 104

-0.20

i (μA/cm2)

0

-0.70

-0.60

-0.40ηc

βc

βa

E (V)

(io,E)H

(io,E)M

(i,E)coorr

From the diagram,

¯a =Ecorr ¡ EM

4 decades=

(¡0:50 V ) ¡ (¡0:70 V )

4¯a = 0:05 V

¯c =Ecorr ¡ EH

3 decades=

(¡0:20 V ) ¡ (¡0:50 V )

3¯c = 0:10 V

From the overpotential eq. (5.32),

á = Ecorr ¡ EM = 0:20 V

ć = Ecorr ¡ EH = ¡0:30 V

á = ¯a log

µicorr

io;M

¶

ć = ¡¯c log

µicorr

io;H

¶

Then,

¯a = á= log

µicorr

io;M

¶

¯a = (0:20 V ) = log£¡

102¹A=cm2¢=10¡ 2¹A=cm2

¤

¯a = 0:05 V

and


¯c = ¡ć= log

µicorr

io;H

¶

¯c = (¡0:30 V ) = log£¡

102¹A=cm2¢=¡0:10¹A=cm2

¢¤

¯c = 0:10 V

5.5 A steel tank is hot dipped in a deaerated acid solution of 5x10¡ 4

mol=cm3 molality zinc chloride (ZnCl2) so that a 0:15-mm zinc coating is de-posited on the steel surface. This process produces a galvanized steel tank. Cal-culate the time it takes for the zinc coating to corrode completely at a pH = 4:Data: EZn = ¡0:8 V; io;Zn = 10 ¹A=cm2, io;H = 10¡ 3 ¹A=cm2, ¯a = 0:08 V ,¯c = 0:12 V; T = 25oC.

Solution:

The possible half-cell reactions and their respective standard potential are

Zn ! Zn+2 + 2e =) EZn = ¡0:80 V2H+ + 2e ! H2 =) EH =?

Zn + 2H+ ! Zn+2 + H2 =) E = EZn + EH

The activity of hydrogen ions and the hydrogen potential are, respectivily

pH = ¡ log£H+

¤

£H+

¤= 10¡ pH = 10¡ 4 mol/l

EH = 0:059pH = 0:059 (4) = 0:24 V

Using the de…nition of overpotential along with the condition á = ć yields

á = E ¡ EZn = ¯a log

µicorr

io;Zn

¶

ć = E ¡ EH ¡ ¯c log

µicorr

io;H

¶

EZn + ¯a log

µicorr

io;Zn

¶

= EH ¡ ¯c log

µicorr

io;H

¶

which leads to

log (icorr) =1

¯a + ¯c

[¡EZn + EH + ¯a log (io;Zn) + ¯c log (io;H)]

log (icorr) = 10:08+0:12

£¡ (¡0:80) + 0:236 + 0:08 log (10) + 0:12 log

¡10¡ 3

¢¤


log (icorr) = 1:40

icorr = 101:40 ¹A=cm2 = 25:12 ¹A=cm2

From eq. (5.48),

CR =icorrAw

zF½=

¡25:12x10¡ 6 A=cm2

¢(65:37 g=mol)

(2) (96; 500 A:s=mol) (7:14 g=cm3)

CR = 1:192x10¡ 9 cm=s = 0:38 mm=y

CR =x

t

t =x

CR=

0:15 mm

0:38 mm=y= 0:40 years = 146 days

5.6 Calculate the activity and the corrosion rate of iron (Fe) immersed inan aerated aqueous solution of pH = 9: The dissociation constant for ferroushydroxide, Fe (OH)2, is 1:64x10¡ 14. Given data:

io;Fe = 10¡ 2 ¹A=cm2 Ecorr = ¡0:30 V ¯a = 0:10 VAw;Fe = 55:85 g=mol ½Fe = 7:86 g=cm3

Solution:

The possible half-cell reactions and standard potentials are

Fe = Fe+2 + 2e =) EoFe = 0:440 V

12O2 + H2O + 4e = 2OH¡ =) Eo

H2O = ¡0:401 V

Fe + 12O2 + H2O = Fe+2 + 2OH ¡ =) Eo = Eo

Fe + EoH2O = 0:039 V

Fe+2 + 2OH ¡ = Fe (OH)2 EoFe(OH)2

= 0:039 V

Thus, the oxidation of iron may be represented by the following reaction

Fe (OH)2 = Fe+2 + 2OH ¡

From Table 2.6, [OH ¡ ] = 10pH¡ 14 = 10¡ 5 mol=l so that the activity offerrous iron becomes

KFe(OH)2=

£Fe+2

¤[OH ¡ ]

2

[Fe (OH)2]

£Fe+2

¤=

[Fe (OH)2]KFe(OH)2

[OH¡ ]2 =

[1]¡1:64x10¡ 14

¢

[10¡ 5]2

£Fe+2

¤= 1:64x10¡ 4 mol=l


Using eq. (5.32) gives the corrosion density

á = Ecorr ¡ EoFe(OH)2

= ¡0:30 V ¡ 0:039 V

á = ¡0:339 V

á = ¯a log

µicorr

io;Fe

¶

log

µicorr

io;Fe

¶

= ¡0:339 V

0:10 V= ¡3:39

icorr = io;Fe10¡ 3:39 =¡10¡ 8 A=cm2

¢ ¡10¡ 3:39

¢

icorr = 4:07x10¡ 12 A=cm2

From eq. (5.48), the corrosion rate is

CR =icorrAw

zF½=

¡4:07x10¡ 12 A=cm2

¢(55:85 g=mol)

(2) (96; 500 A:s=mol) (7:86 g=cm3)

CR = 1:50x10¡ 16 cm=s = 4:73x10¡ 8 mm=y = 0:0473 ¹m=y

5.7 Plot the anodic data given below and determine the polarization re-sistance (Rp) and the anodic Tafel slope (¯a) for a metal M . Use ¯c = 0:07 Vand icorr = 0:019 A=cm2.

M = M+2 + 2e¡ i¡¹A=cm2

¢E (VSHE)

2H+ + 2e¡ = H2 0:08 ¡ 0:32

0:10 ¡ 0:30

0:15 ¡ 0:25

0:18 ¡ 0:22

0:20 ¡ 0:20

Solution:

Linear least squares on the given data gives E = ¡0:4 V +¡1:0V:cm2=A

¢i


Slope = 1.0 V.cm2/A

i (A/cm2)

E (V)

-0.4

-0.35

-0.3

-0.25

-0.2

-0.15

-0.1

0 0.05 0.1 0.15 0.2 0.25

The polarization resistance and the anodic Tafel slope are

Rp = slope = 1:0 ohm:cm2

Rp =¯

icorr

¯ =¯a¯c

2:303 (¯a + ¯c)= 0:019 V

Thus,¯a = 0:12 V

5.8 Why does a pearlitic steel corrode rapidly in an acidic solution?

Answer: Because of the galvanic e¤ects between the body-centered cubicferrite (BCC ®-Fe) and tetragonal cementite (Fe3C) crystal structures. Infact, this type of corrosion is bene…cial for revealing the steel microstructure.

5.9 Why will the tip and the head of an iron nail behave as anodes relativeto the shank? See Figure 1.10a.

Answer: Because the tip and the head of the nail are highly strained areascontaining large amounts crystal defects, such as dislocations, voids, and possiblymicrocracks. These areas contain large amounts of stored strain energy.

5.10 It is known that the standard electrode potential (Eo) for pure crys-talline zinc is ¡0:763 V . Will this value change by cold working and impurities?Explain.

Answer: Cold working causes microstructural changes and crystal defects,which in turn change the mechanical properties and increases corrosion. On


the other hand, impurities also change properties due to segregation along grainboundaries. Therefore, Eo must change and it is di¤erent at grain boundaries(atomic mismatch) due to the straining e¤ects induced during cold working thanthe potential within grains. Consequently, the grain boundaries and possiblysecondary phases become anodic sites to grains.

5.11 An electrolyte contains a very low activity¡8x10¡ 9 mol=l

¢of silver

cations (Ag+) and an unknown concentration of copper cations. If the cellpotential di¤erence between the copper anode and the silver cathode is ¡0:04V , determine a) which cation will be reduced (electroplated) on the cathodeand b) the concentration of

£Cu+2

¤in g=l at 40oC. Neglect the e¤ects of other

ions that might react with silver.

Solution:

The possible half-cell reactions and standard potentials are

2Ag+ + 2e = 2Ag =) EoAg = 0:799 V

Cu = Cu+2 + 2e =) EoCu = ¡0:337 V

2Ag+ + Cu = 2Ag + Cu+2 =) Eo = EoAg + Eo

Cu = 0:462 V

The free energy change is

¢Go = ¡zFEo = ¡ (2) (96; 500 J=mol:V ) (0:462 V )

¢Go = ¡89; 166 J=mol

Therefore, silver (Ag) will be reduced since ¢Go < 0: Using the Nernstequation yields the copper ion concentration

E = Eo ¡RT

zFln (K)

ln (K) =zF (Eo ¡ E)

RT=

(2)¡96; 500 J

mol:V

¢(0:462 V + 0:04 V )

¡8:314 J

mol:oK

¢(313oK)

ln (K) = 37:23

K = 1:48x1016


Then,

K =[Ag]

2 £Cu+2

¤

[Ag+]2[Cu]

£Cu+2

¤=

K [Ag+]2[Cu]

[Ag]2

£Cu+2

¤=

¡1:48x1016

¢ £8x10¡ 9 mol=l

¤2[1 mol=l]

[1 mol=l]2

£Cu+2

¤= 0:95 mol=l

CCu+2 =£Cu+2

¤Aw;Cu = (0:95 mol=l) (63:55 g=mol)

CCu+2 ' 60 g=l

5.12 Suppose that a cold worked copper electrode has 8; 000 J=mol storedenergy and dissolves in an aqueous electrolyte. Calculate the overpotential.

Solution:

Cu = Cu+2 + 2e =) Eo = ¡0:337 V

From eq. (2.24),

¢G = ¡zFE

E = ¡¢G

zF= ¡

8; 000 J=mol

(2)¡96; 500 J

mol:V

¢

E = ¡0:042 V > EoCu

á = E ¡ Eo = ¡0:042 V ¡ 0:337V

á ' ¡0:38 V

5.13 What is the signi…cant di¤erences between the overpotential (´) andthe Ohmic potential (E)?

Answer: Mathematically, both potentials have di¤erent physical trend. Thus,the overpotential is nonlinear with respect to the current or current density since´ = ¯ log (I=Icorr) : If ´ 6= 0, the anode electrode is polarized and ´ is a measureof the energy required for polarization since ¢G = ¡zF´. On the other hand,Ohm’s law, E = IR is linear and it may include other terms, such as overpo-tential and solution (electrolyte) potential, Ás = IRs. Equation (3.25) describesthe Ohm’s potential in details.

E = Ecorr ¡ á ¡ ć ¡ Ás = IR


5.14 If a sheet of zinc (Zn) is placed in hydrochloric acid (HCl), Zn goesinto solution. Why?

Answer: Localized anodic and cathodic areas must exist on the sheet surfaceso that the following reactions occur

Zn = Zn+2 + 2e =) EoZn = 0:763 V

2H+ + 2e = H2 =) EoH = 0

Zn + 2H+ = Zn+2 + H2 =) Eo = EoZn + Eo

H = 0:763 V

The standard free energy change must be ¢Go < 0 in order for the abovereactions to occur. Using eq. (2.24) yields

¢Go = ¡zFEo = ¡ (2) (96; 500 J=mol:V ) (0:763 V )

¢Go = ¡147:26 kJ=mol

5.15 Calculate the equilibrium constant at 25oC for the electrochemical cellshown in Figure 2.4.

Solution:

Combining eqs. (2.24) and (2.31b) along with Eo = 1:10 V gives

¡zFEo = ¡RT ln (K)

ln (K) =zFEo

RT

ln (K) =(2) (96; 500 J=mol:V ) (1:10V )

(8:314 J=mol:oK) (298oK)= 85:69

K = 1:64x1037

5.16 Show that ¢G = zF¯a log (ia=icorr) where the local potential E canbe de…ned by the Nernst equation.

Solution:

Using eqs. (3.31) and (2.24) yields

E = Eo ¡RT

zFln (K)

ln (K) =zF (Eo ¡ E)

RT


But,

¢G = ¡RT ln (K)

¢G = ¡RT:zF (Eo ¡ E)

RT¢G = zF (E ¡ Eo) = zF´a

But,

´a = ¯a log

µia

icorr

¶

¢G = zF¯a log

µia

icorr

¶

5.17 If an electrochemical cell operates at 10 amps, 25oC; and 101 kPa for30 minutes, calculate a) the number of moles and b) the weight of copper wouldbe produced.

Solution:

The reduction reaction is Cu+2 + 2e = Cu. From eq. (5.12b),

N =It

zF=

(10 A) (1; 800 s)

(2) (96; 500 A:s=mol)= 0:093moles

From eq. (5.12c),

W =ItAw

zF= NAw = (0:093mol) (63:55 g=mol)

W = 5:93 grams

5.18 (a) Derive the Arrhenius equation from the general de…nition of theactivation energy of any rate reaction process, (b) determine the Arrheniusconstants B and Ae using the derived equation and the following data for thedissociation constants: Ksp@300 = 1011, Ksp@350 = 3Ksp@300 between 300oKand 350oK. It is assumed that the activity coe¢cients approach 1 in dilutedelectrolyte solution and that Ksp follows an Arrhenius type trend. Recall thatKsp can be expressed in terms of concentrations. (c) Plot the given data belowand draw the trend line in the close interval [200,400] for the temperature axisand [0,4]x1011 for the reaction constant Ksp axis.

T (oK) Dissociation Constant Ksp

200 0:02x1011

250 0:22x1011

280 0:50x1011

300 1:00x1011

330 1:95x1011

350 3:00x1011

360 3:65x1011


Solution:

(a) From eq. (5.20),

¢G¤ = RT 2 d (lnKsp)

dTdT

T 2=

Rd (lnKsp)

¢G¤Z T

o

T ¡ 2dT =R

¢G¤

Z

d (lnKsp)

¡T ¡ 1 =R

¢G¤(lnKsp ¡ lnB)

lnKsp

B= ¡

¢G¤

RT

Ksp = B exp

µ

¡¢G¤

RT

¶

(b) Using the derived expression for two di¤erent rates yields

Ksp@300 = B exp

µ

¡¢G¤

RT300

¶

Ksp@350 = B exp

µ

¡¢G¤

RT350

¶

Ksp@300

Ksp@350= exp

·¢G¤

RT

µ1

T350¡

1

T300

¶¸

Thus,

¢G¤ =R ln (Ksp@300=Ksp@350)

1=T350 ¡ 1=T300

¢G¤ =(8:314) ln (1=3)

1=350 ¡ 1=300= 19; 181 J/mol

and

B = Ksp@300 exp

µ¢G¤

RT300

¶

B =¡1011

¢exp

½

¡19181 J/mol

[8:314 J/(mol.K)] (300 K)

¾

B = 2:1869 £ 1014

Check:


Ksp@300 = B exp

µ

¡¢G¤

RT300

¶

Ksp@300 =¡2:1869 £ 1014

¢exp

µ

¡19181

(8:314) (300)

¶

= 1 £ 1011

Ksp@350 =¡2:1869 £ 103

¢exp

µ

¡19181

(8:314) (350)

¶

= 3 £ 1011

Ksp@350

Ksp@300=

3 £ 1011

1 £ 1011= 3

(c) Plot Ksp =¡2:1869 £ 103

¢exp

¡¡ 19181

8:314T

¢= 2186:9 exp

¡¡2307:1

T

¢

400350300250200

4

3

2

1

0

T (Degree Kelvin)

Ksp

5.19 Plot the given conductivity data vs: the aqueous Cu+2SO¡4 concen-tration at 25oC and 1 atm. (The data was taken from Ref. [5]). Conduct anonlinear regression analysis on these data and explain the resultant curve.


C Kc¡mol=cm3

¢ ¡ohm¡ 1 ¢ cm¡ 1

¢

x10¡ 6 x10¡ 3

025:0050:0075:00100:00102:50200:00202:50

03:005:107:408:5010:0015:0016:00

Linear least squares on the given data yields

Kc =¡1:1922x10¡ 3 ohm¡ 1 ¢ cm¡ 1

¢+

¡73:46 mol=cm3

¢C

C (x10-6 mol/cm3)

Kc

(x1

0-3

oh

m-1

.cm

-1)

0

5

10

15

20

25

50 100 150 200 250

In fact, Cu+2SO¡4 solution is a weak electrolyte due to the interactionsamong Cu+2 and SO¡4 ions. However, Kc increases because the number ofcharge carriers per unit volume increase with the electrolyte concentration.

5.20 An electrochemical cell operates at a small overpotential and the cor-roding metal is exposed to a H+ ion-containing electrolyte. Use the given data

´ = 0:005 V T = 25oCi = 3:895 £ 10¡ 8 A=cm2 ½ = 7:14 g=cm3

io = 10¡ 7 A=cm2 D = 10¡ 5 cm2=sCR = 5:8277x10¡ 4 mm=y


in order to determine a) the corroding metal by calculating the atomic weightAw and determining its anodic reaction. Assume that there is a linear relation-ship between the current density and the overpotential. b) The activity of thecorroding metal if aH+ = 1 mol=l, c) the free energy change ¢G. Will thereaction you have chosen occur?, d) the ionic velocity of the corroding metal.As a crude approximation, neglect the ionic velocity of other ions in solution.e) The ionic mobility B and f) electrolyte conductivity by neglecting other ionspresent in the electrolyte.

Solution:

a) From eq. (5.16c),

i = io

µzF´

RT

¶

(3.16c)

z =iRT

ioF´=

¡3:895x10¡ 8 A=cm2

¢(8:314 J=mol:oK) (298oK)

(10¡ 7 A=cm2) (96; 500 J=mol:V ) (0:005 V )

z = 2


CR =iAw

zF½= 5:8277x10¡ 4mm=y = 1:848x10¡ 12 cm=s (3.48)

Aw =zF½CR

i=

(2) (96; 500 A:s=mol)¡7:139 g=cm3

¢ ¡1:848x10¡ 12 cm=s

¢

(3:895 £ 10¡ 8 A=cm2)

Aw = 65:38 g=mol

Therefore, the corroding metal is zinc and its anodic reaction is

Zn = Zn+2 + 2e

b) According to the results obtained in part a), the reactions and potentialstaking place at 25oC are

Zn = Zn+2 + 2e =) EoZn = 0:763 V

2H+ + 2e = H2 =) EoH = 0

Zn + 2H+ = Zn+2 + H2 =) Eo = EoZn + Eo

H = 0:763 V

Using the Nernst equation, eq. (2.32), along with ´ = E ¡ Eo = 0:005 Vgives the activity of Zn+2 ions


E = Eo ¡RT

zFln (K) (2.32)

´ = ¡RT

zFln

£Zn+2

¤[H2]

[Zn] [H+]2

£Zn+2

¤=

[Zn] [H+]2

[H2]exp

·

¡zF´

RT

¸

£Zn+2

¤=

[1 mol=l] [1 mol=l]2

[1 mol=l]exp

·

¡(2) (96; 500 J=mol:V ) (0:005 V )

(8:314 J=mol:oK) (298oK)

¸

£Zn+2

¤= 0:677 mol=l = 6:77x10¡ 4 mol=cm3

c) From eq. (2.24),

¢G = ¡zFE (2.24)

E = ´ + Eo = 0:005 V + 0:763V = 0:768 V

¢G = ¡ (2) (96; 500 J=mol:V ) (0:768 V )

¢G = ¡148:224 kJ=mol

Therefore, the above reactions occur because ¢G < 0.

d) Using eq. (5.95) yields the zinc ionic velocity

i = zFvjCj = zFvZn+2CZn+2 = zFvZn+2

£Zn+2

¤(3.95)

vZn+2 = vj =i

zFCZn+2

vZn+2 =

¡3:895x10¡ 8 A=cm2

¢

(2) (96; 500 A:s=mol) (6:77x10¡ 4 mol=cm3)

vZn+2 = 2:98x10¡ 10 cm=s

e) Using eq. (5.97) yields the zinc ionic mobility

DZn+2 = kTBZn+2 (3.97)

BZn+2 =DZn+2

kT=

10¡ 5 cm2=s

(1:38x10¡ 23 J=mol:oK) (298oK)

BZn+2 = 2:43x1015 cm2:mol

J:s= 2:43x1015 cm2

V:s

f) Using eq. (5.93) gives the zinc ionic conductivity

Kc;Zn+2 = zFBZn+2CZn+2

Kc;Zn+2 = (2) (96; 500 A:s=mol)

µ

2:43x1015 cm2

V:s

¶¡6:77x10¡ 4 mol=cm3

¢

Kc;Zn+2 = 3:18x1017 ohm¡ 1:cm¡ 1


5.21 If an electrochemical copper reduction process is carried out at 5amperes. for 20 minutes, determine a) the electric charge (amount of coulombsof electricity), b) the number of electrons if there are 1=

¡1:6022x10¡ 19

¢=

6:24x1018 electron=C, c) the number of moles, d) Faraday’s weight reducedon a cathode, and the reduction rate (production rate in Chapter 7). Data:Aw = 63:55 g=mol and T = 35oC:

Solution:

a) The reactions and potentials taking place at 35oC are

Cu = Cu+2 + 2e =) EoCu = 0:337 V

2H+ + 2e = H2 =) EoH = 0

Cu + 2H+ = Cu+2 + H2 =) Eo = EoCu + Eo

H = 0:763 V

From eq. (5.12a),

Q = It = (5 A) (1; 200 s) = 6; 000 C

b)

Ne = (6; 000 C)¡6:24x1018electron=C

¢

Ne ' 4x1022 electrons

c) From eq. (5.12b),

N =Q

zF=

6; 000 C

(2) (96; 500 C=mol)= 0:0311 moles

d) From eq. (5.12c),

W = NAw = (0:0311 moles) (63:55 g=mol)

W = 1:98 grams

e)

PR =W

t=

1:98 g

1; 200 s= 1:65x10¡ 3 g=s = 5:94 g=h

5.22 Calculate the overpotential that causes hydrogen evolution on a ‡atplatinum (Pt) electrode surface immersed in an acid solution, when the appliedcathodic and exchange current densities are 6; 000 ¹A=cm2 and 100 ¹A=cm2,respectively. Assume a symmetry factor of 0:50 at room temperature (25oC).

Solution:

2H+ + 2e = H2 =) EoH = 0


From eq. (5.20b),

ć = ¡RT

zF®ln

µicio

¶

ć = ¡

¡8:314 J

mol:oK

¢(298oK)

(1) (96; 500 J=mol:V ) (0:5)ln

µ6; 000

100

¶

ć = ¡0:21V

5.23 Consider a discharge (chemical desorption) mechanism as the ratedetermining for Ni+2 + 2e¡ = Ni at 25oC in a nickel battery. Calculate thecathodic Tafel slope per decade if the symmetry factor is 0:50, the exchangecurrent density is constant, and the cathodic overpotential is ć < 0.

Solution:

From eq. (5.20b),

ć = ¡RT

zF®ln

µicio

¶

ć = ¡2:303RT

zF®log

µicio

¶

dć

d log (ic)= ¡

2:303RT

zF®

Thus,

¯c =dć

d log (ic)= ¡

(2:303)¡8:314 J

mol:oK

¢(298oK)

(2) (96; 500 J=mol:V ) (0:5)

¯c = ¡0:06 V=decade

5.24 Plot the normalized current pro…le as a function of both symmetryfactor (®) and overpotential; that is, i=io = f (®; ´) when the oxidation stateis de…ned is z = 2 and ® = 0:35; 0:50 and 0:75 at room temperature (25oC).Explain the polarization behavior very succinctly.

Solution:


i

io= exp

·®zF´

RT

¸

¡ exp

·

¡(1 ¡ ®) zF´

RT

¸

i

io= exp [77:90®´] ¡ exp [¡ (1 ¡ ®) 77:90´]


-15

-10

-5

0

5

10

15

-0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08

α = 0.75 0.50

0.35

Tafel Region

CathodicRegion

AnodicRegion

i/io

η (V)

The normalized current density (i=io) shows the expected linearized Tafel re-gion at small values of the anodic and cathodic overpotentials (´). However, thedeparture from this linearity is obvious at large ´ and the shape of the curveschange signi…cantly. Therefore, variations of the current density (i) stronglydepend on the overpotential (´) and the shape of the curves depend on the sym-metry factor (®). The slopes of the curves at large ´ are characteristics of ®since it arises due to changes in the electric double layer structure.

5.25 Calculate the mass and number of moles a) of a battery zinc (Zn)casing and b) of the manganese dioxide (MnO2) in the electrolyte if the batteryhas a stored energy of 36 kJ=V and a power of 3 Watts. c) Find the time ittakes to consume the stored energy if the battery operates at a current of 2 Aand the potential (voltage). The thickness of the cell is x = 1 mm and otherdimensions are indicated below. The discharging reaction is

MnO2 + H+ + e ! MnOOH


60 mm

2 mm

MnO2

Carbon( )15 mm

0.3 mm

Solution:Q = 36 kJ=V = 36; 000 A:s = 36; 000 CoulombsAs = 2¼Lr = (2¼) (6 cm) (1:5 cm) = 56:55 cm2

V = xAs = (0:1 cm)¡5:6549 cm2

¢= 5:65 cm3

a) The mass and moles of Zn

MZn = V ½ =¡5:65 cm3

¢ ¡7:14 g=cm3

¢= 40:34 g

XZn =MZn

Aw;Zn=

40:34 g

65:37 g=mol= 0:62 mol

b) The mass and moles of MnO2 are

XMnO2 =Q

zF=

36; 000 A:s

(1) (96; 500 A:s=mol)

XMnO2= 0:37 mol

MMnO2 = XMnO2Aw;MnO2 = (0:37 mol) (86:94 g=mol)

MMnO2= 32:17 g

c) The time for discharging the stored energy is

t =Q

I=

36; 000 A:s

2 A= 5 hours

E = P=I = (3 V:A) = (2 A) = 1:5 V

5.26 A plate of pure nickel (Ni) oxidizes in an electrochemical cell contain-ing an acid solution at 25oC. The total surface area of the nickel plate is 100cm2. If 2x1016 electrons per second (e=s) are relieved on the plate surface, thencalculate a) the corrosion rate in mm=y and b) the mass of nickel being lost ina year..

Solution:


Ni = Ni+2 + 2e z = 2 ½ = 8:90 g=cm3

r = 2x1016 e=s Aw = 58:71 g=molqe = 1:6022x10¡ 19 A:s=e As = 100 cm2

a) From eq. (5.50),

I = rqe =¡2x1016 e=s

¢ ¡1:6022x10¡ 19A:s=e

¢

I = 3:20x10¡ 3 A

i =I

As=

3:20x10¡ 3 A

100cm2= 32 ¹A=cm2

Then,

CR =iAw

zF½=

¡32x10¡ 6 A=cm2

¢(58:71g=mol)

(z) (96; 500 A:s=mol) (8:90g=cm3)

CR = 1:10x¡ 9 cm=s = 0:35 mm=y

b)

i =I

As=

zFm

AstAw

zFm

AstAw=

zF½CR

Aw

m = ½CRAst =¡8:90 g=cm3

¢(0:035 cm=s)

¡100 cm2

¢(1 y)

m = 31:15 grams

5.27 Use the data listed in Table 3.3 to perform a least squares analysisand subsequently, determine the polarization proportionality constant ¯. Letthe atomic weight and the density of the steel be Aw;steel = Aw;Fe = 55:85g=mol and ½ = 7:85 g=cm3, respectively.

Solution:

Curve …tting:

Rp = 3:8502 £ 10¡ 2 ohm:cm2 +

µ

30:147ohm:cm3

y

¶1

CR


1/CR (y/cm)

Rp

(oh

m.c

m2 )

0

20

40

60

80

100

120

140

1 2 3 4

S = 30.147 ohm.cm3/y

Combining eqs. (3.25a) and (3.48) yields

Rp =

µ¯Aw

zF½

¶1

CR

S =

µ¯Aw

zF½

¶

¯ =zF½S

Aw=

(2)¡3:06x10¡ 3 A:y=mol

¢ ¡7:85 g=cm3

¢ ¡30:147 ohm:cm3=y

¢

55:85 g=mol

¯ = 0:03 V

5.28 Equal amounts of CuSO4 and NiSO4 are dissolved with water tomake up an electrolyte. Hypothetically, the ion velocities and concentrationsare

vCu+2 = 0:22 cm=s CNi+2 = 10¡ 5 mol=cm3

vSO¡24

= 0:1 cm=s CSO¡2 = 10¡ 5 mol=cm3

CCu+2 = 10¡ 5 mol=cm3

If the current density is 1 A=cm2, calculate the velocity of the nickel ions¡Ni+2

¢.

Solution:


From eq. (5.86),

i =X

z+j Fv+

j C+j +

X¯¯z¡j

¯¯Fv¡j C¡j

i = Fh(zvC)Cu+2 + (zvC)Ni+2 + (jzj vC)SO¡2

4

i

i = 2Fh(vC)Cu+2 + (vC)Ni+2 + (vC)SO¡2

4

i

If zCu+2 = zNi+2 =¯¯¯zSO¡2

4

¯¯¯ = 2 and C = CCu+2 = CNi+2 = CSO¡2

4= 10¡ 5

mol=cm3, then

i = 2FChvCu+2 + vNi+2 + vSO¡2

4

i

Solving for vNi+2 yields

vNi+2 =i

2CF¡ vCu+2 ¡ vSO¡2

4

vNi+2 =1

(2)¡10¡ 5 mol

cm3

¢ ¡96; 500 A:s

mol

¢ ¡ 0:22cm

s¡ 0:1

cm

s

vNi+2 = 0:20 cm=s

5.29 An electrochemical cell operates at 10 A, Rx = 0:25 ohm; and ! = 50Hz. Determine a) the electrolyte resistance (Rs), b) the potential Ex when theexternal resistance and the capacitance are Rx = 0:25 ohm and Cx = 20 A:s=Vat 300C and c) the electrolyte conductivity Kc: The distance between electrodesis L = 15 cm and the e¤ective electrode surface is A = 8; 000 cm2:

Solution:

a) From eq. (5.103),

Z (w)x =

s

R2x +

µ1

!Cx

¶2

Z (w)x =

s

(0:25 ohm)2 +

·1

(50 s¡ 1) (20A:s=V )

¸2

Z (w)x = 0:25 ohm

Therefore, Rs = Z (w)x = 0:25 ohm:

b) From eq. (5.104),

Ex = IxZ (w)x

Ex = (10 A) (0:25 ohm) = 2:5 V


c) From eq. (5.99),

¸ = L=A = (15 cm) =¡8; 000 cm2

¢

¸ = 1:88x10¡ 3 cm¡ 1

Kc =¸

Rs=

¸

Z (w)x

Kc =1:88x10¡ 3 cm¡ 1

0:25 ohm= 7:50x10¡ 3 ohm¡ 1:cm¡ 1

5.30 Determine and analyze the impedance pro…le by varying the angularfrequency for …xed Rx = 0:25 ohm and Cx = 20 A:s=V .

Solution:

This problem requires use of eq. (5.103). The intent here is to show how theimpedance depends on the angular frequency at very low values.

Z (w)x =

s

(0:25)2

+

·1

20!

¸2

0.30.250.20.150.10.050

5

3.75

2.5

1.25

0

w

Z(w)

5.31 Assume that an electrolytic cell is used for recovering magnesium froma solution containing 10¡ 4 mol=cm3 of Mg+2 at 350C: The nickel ionic mobilityand the electric-…eld strength are B = 55x10¡ 5 cm2V ¡ 1s¡ 1 and Fx = 10 V=cm[Taken from reference 5], respectively, calculate (a) the ionic velocity (v), (b)the solution electric conductivity (Kc) and the electric resistivity (½c). [Solution:(a) v = 0:0055 cm=s, (b) Kc = 0:0106 ohm¡ 1:cm¡ 1 and (c) ½c = 94:21 ohm:cm].

Solution:


a) From eq. (5.92),

v = BFx =

µ

55x10¡ 5 cm2

V:s

¶µ

10V

cm

¶

= 0:0055 cm=s

b) Using eq. (5.93) yields

Kc = zFBC

Kc = (2)

µ

96; 500A:s

mol

¶µ

55x10¡ 5 cm2

V:s

¶µ

10¡ 4 mol

cm3

¶

Kc = 0:0106 ohm¡ 1:cm¡ 1

and from eq. (5.89)

½c = 1=Kc = 94:21 ohm:cm

5.32 It is known that current ‡ows when there exists a gradient of electricpotential (dÁ=dx) within an electric conductor, such as an electrolyte. Considera current-carrying homogeneous conductor with constant cross-sectional area(Ac) so that the electric-…eld strength (Ex) is constant at every point in theconductor. Derive an expression for the current as a function of the gradientof electric potential. In this particular problem "x" stands for direction as wellas length of the electric conductor. Start with the following current densityde…nition i = KcEx, where Kc is the electric conductivity (ohm¡ 1:cm¡ 1).

Solution:

Using eq. (2.2a) and the given current density equation yields

Ex = ¡dÁ

dx(a)

i = KcEx = ¡KcdÁ

dx(b)

I

Ac= ¡Kc

dÁ

dx(c)

I = ¡AcKcdÁ

dx(d)

This expression agrees with the above statement because I = f (dÁ=dx). Fora homogeneous conductor, eq. (d) becomes

I = ¡AcKc¢Á

¢x

where ¢x becomes the length of the conductor.


EXTRA5.EX1 Determine the atomic weight of 82Mg-10Al-8Zn alloy.

Solution:The state of oxidation and the atomic weight of each element are

zMg = 2 & Aw;Mg = 24:31 g=mol

zAl = 3 & Aw;Al = 26:98 g=mol

zZn = 2 & Aw;Zn = 65:37 g=mol

From the given weight percent of the alloy, the weight fraction for Mg, Aland Zn are

fMg = 0:82 fAl = 0:10 fZn = 0:08

Then the atomic weight of the alloy is

Aw;alloy =X fjAw;j

zj=

fMgAw;Mg + fAlAw;Al + fZnAw;Zn

zMg + zAl + zZn

Aw;alloy = 13:48 g=mol

5.EX2 An electrochemical ionic concentration cell shown below was usedas a concentration cell in order to measure the current due to the di¤erence inionic concentration between the anodic half-cell and the cathodic-half cell. Deter-mine the corrosion rate in terms of electrons/second and the number of anodicreactions/second if the measured current is 3:23x10¡ 7 ¹A = 3:23x10¡ 7C=s.


Solution:

The oxidation of copper (corrosion) and the reduction of copper (electroplat-ing) are, respectively

Cu ¡! Cu2+ + 2e¡

Cu2+ + 2e¡ ¡! Cu

Clearly, the oxidation state is z = 2: Thus, the corrosion rate is

r =I

qe=

3:23x10¡ 7C=s

(1:6022x10¡ 19 C=electrons)¼ 2x1012 electrons=s

Also,

r =I

zqe=

3:23x10¡ 7C=s¡2 electrons

reactions

¢(1:6022x10¡ 19 C=electrons)

r = 1012 reactions=s

5.EX3 Use the Evans diagram given below to determine (a) the linear cor-rosion rate pro…le, CR = f (ix), at 10¡ 8 A=cm2 · ix · 10¡ 3 A=cm2 and (b) thenon-linear counterpart, CR = f (Ea), at ¡0:45 V · Ea · ¡0:25 V for iron (Fe)immersed in a acid solution. Use CR in units mm=yr and let ix be the appliedcurrent density. The atomic weight and density of iron are Aw = 55:85 g=moland 7:87 g=cm3, respectively. Explain.


Solutions:(a) From eq. (5.48),

CR =iaAw

zF½= ¸ix (a)

¸ =Aw

zF½=

55:85

2 ¤ 96500 ¤ 7:87= 3:6770 £ 10¡ 5 V:cm3

A:s

CR =

µ

3:6770 £ 10¡ 5 V:cm3

A:s

¶

(60 ¤ 60 ¤ 24 ¤ 365) (10) ix

CR = 11; 596ix (inmm=yr)

The slope of this plot represents the volume of Fe2+ ions per electric charge.Hence,

dCR

dix= ¸ = 3:6770 £ 10¡ 5 cm3

A:s(slope)

dCR

dix' 0:038

mm3

A:s

The upper and lower CR values are

CR = 11:60 mm=yr @Ecorr (upper limit)

CR = 1:16 £ 10¡ 4 mm=yr @EFe (lower limit)


Therefore, corrosion can be minimized signi…cantly as ix ! iFe = 10¡ 8

A=cm2. Upon decreasing ix into the cathodic region at E < EFe, reduction ofiron, Fe2+ + 2e¡ ! Fe, occurs and CR = 0. This electrochemical process leadsto the principles of cathodic protection (CP ) large structures. The theory of CPis dealt with in a later chapter.

(b) Combining eq. (a) and (5.37a) gives

CR =iaAw

zF½= ¸ia

ia = icorr exp

µ2:303 (Ea ¡ Ecorr)

¯a

¶

CR = ¸icorr exp

µ2:303 (Ea ¡ Ecorr)

¯a

¶

(b)

CR = (11:60 mm=yr) exp (57:58Ea + 14:39) (c)

The non-linear corrosion rate pro…le is given by eq. (c) and it is depicted inthe …gure below.

The upper and lower CR values in this case are

CR = 11:54 mm=yr @Ecorr (upper limit)

CR = 1:15 £ 10¡ 4 mm=yr @EFe (lower limit)

Basically, these results are the same as in part (a).

Chapter 6

MASS TRANSPORT BYDIFFUSION ANDMIGRATION

6.1 PROBLEMS

6.1 Why convective mass transfer is independent of concentration gradient@C=@x in eq. (6.2)?

Answer: Because it arises due to ‡uid motion in which @C=@x doe notplay an important role. Only the bulk concentration and ‡uid velocity are thecontrolling parameters at a plane for convective motion.

6.2 Use the information given in Example 6.3, part b, to determine theconcentration rate in the x-direction at 1 mm from the columns interface. Givendata: D = 10¡ 5 cm2=s, Co = 10¡ 4 mol=cm3and t = 10 sec : [Solution: @C=dt =1:96x10¡ 16 mol=cm3:s].

Solution:

From Example 6.2,

@C

@x= ¡

Cop4¼Dt

exp

µ

¡x2

4Dt

¶

@2C

@x2=

xCo

4Dtp

¼Dtexp

µ

¡x2

4Dt

¶

71

72 CHAPTER 6. MASS TRANSPORT BY DIFFUSION AND MIGRATION

Thus, Fick’s second law becomes

@C

@t= D

@2C

@x2

@C

@t=

xCo

4tp

¼Dtexp

µ

¡x2

4Dt

¶

@C

@t=

(0:1 cm)¡10¡ 4mol=cm3

¢

(4) (10 s)p

(¼) (10¡ 5 cm2=s) (10 s)exp

"

¡(0:1 cm)

2

(4) (10¡ 5 cm2=s) (10 s)

#

@C

@t= 1:96x10¡ 16 mol

cm3:s

6.3 What’s the action of hydrogen on a steel blade exposed to an aqueoussolution of hydrogen sul…de if the following reaction takes place Fe + H2S =FeS + 2H? Assume that the steel blade gets damaged during service.

Answer: Evolution of atomic hydrogen (H) on the steel surface may occuraccording to the reactions

Fe ! F+2 + 2eH2S ! 2H+ + S¡ 2 + 2e

Fe + H2S ! Fe+2 + S¡ 2 + 2H+ + 2e

and

Fe+2 + S¡ 2 + 2H+ + 2e ! FeS + 2H

Atomic hydrogen readily di¤uses into the steel (iron matrix) and locates therein voids, vacancies, and dislocations where H + H ! H2 occurs within thelattice. The formation of molecular hydrogen (H2) in voids combine with otherhydrogen molecules, building up a high pressure (P ) that may exceed the yieldstrength (¾ys) of the steel blade. If P > ¾ys, then blistering and …ssures mayoccur and the steel strength and ductility decrease. This phenomenon is knownas hydrogen embrittlement.

6.4 An aerated acid solution containing 10¡ 2 mol=l of dissolved oxygen (O2)moves at 2x10¡ 4 cm=s in a stainless steel pipe when a critical current densityof 103 ¹A=cm2 passivates the pipe. Calculate a) the thickness of the Helmholtzionic structural layer (±), b) the limiting current density (iL) if the three modesof ‡ux are equal in magnitude, and c) Will the pipe corrode under the currentconditions? Why? or Why not? Data:

O2 + 2H2O + 4e¡ = 4OH ¡ (Cathode)

DO2 = 10¡ 5 cm2=s T = 25oC F = 96; 500 A:s=mol

6.1. PROBLEMS 73

Solutions:

a) From eq. (6.1) along with Cb = 10¡ 2 mol=l = 10¡ 5 mol=cm3, the molar‡uxes are

J (x; t)c = Cbv (x; t) =¡10¡ 5 mol=cm3

¢ ¡2x10¡ 4 cm=s

¢

J (x; t)c = 2x10¡ 9 mol

cm2:s

J (x; t)d = 2x10¡ 9 mol

cm2:s= D

@C

@x=

DCb

±

J (x; t)m = 2x10¡ 9 mol

cm2:s=

zFDCb

RT

dÁ

dx

Thus, the total molar ‡ux becomes

J (x; t) = J (x; t)d + J (x; t)m + J (x; t)c

J (x; t) = 3J (x; t)c

J (x; t) = 3Cbv (x; t) = 6x10¡ 9 mol

cm2:s

But,

DCb

±= Cbv (x; t) = J (x; t)c

± =D

v (x; t)=

10¡ 5 cm2=s

2x10¡ 4 cm=s

± = 0:05 cm = 0:5 mm

b) From eq. (6.8)

i = zFJ (x; t)

i = (4)

µ

96; 500A:s

mol

¶µ

6x10¡ 9 mol

cm2:s

¶

i = iL = 2:32x10¡ 3 A

cm2= 2:32x103 ¹A

cm2

c) Corrosion will not occur because iL = 2:32x103 ¹A=cm2 > icrit = 103

¹A=cm2.

6.5 a) Determine the oxygen concentration for Problem 6.4 that will pro-mote corrosion of the stainless steel pipe. b) How long will it take for corrosionto occur?

Solutions:


a) For a di¤usional process only, From eq. (6.9a) yields along with @C=@x 'DCb=± and i = ic,

Cb =ic±

zFD=

¡10¡ 3 A=cm2

¢(0:05 cm)

(4) (96; 500 A:s=mol) (10¡ 5 cm2=s)

Cb = 1:30x10¡ 2 mol=l = 1:30x10¡ 5 mol=cm3

This result indicates that approximately 100%¡1:30x10¡ 2 ¡ 10¡ 2

¢=1:30x10¡ 2 =

23% increase in oxygen content will cause corrosion.

b) From eq. (6.52),

ic =zFCbp

¼Dt

t =D

¼

·zFCb

ic

¸2

t =

µ10¡ 5 cm2=s

¼

¶"(4) (96; 500 A:s=mol)

¡1:30x10¡ 5 mol=cm3

¢

10¡ 3 A=cm2

#2

t = 80:15 s = 1:34 min:

6.6 Use the given data to calculate a) the di¤usivity of copper ions in acathodic process at 25oC and b) the valence z. The original concentration inan acidic solution is 60 g=l.

t(sec)

i¡A=cm2

¢ @C=@x¡mol=cm4

¢

0 0 05 0:1455 0:07532

10 0:1029 0:0532615 0:0840 0:0434920 0:0728 0:03766

Solution:

Cb = 60 g=l = (60g=l) (63:55 g=mol) = 0:9443 mol=l

Cb = 9:443x10¡ 4 mol=l

a) Linear regression analysis:From eq. (6.48),

@C

@x= 1:801 5 £ 10¡ 6 mol

cm4+

µ

0:16842mol:s1=2

cm4

¶

t¡ 1=2

6.1. PROBLEMS 75

0.03

0.04

0.05

0.06

0.07

0.08

0.2 0.25 0.3 0.35 0.4 0.45

t-1/2 (sec-1/2)

dC

/dx

(mo

l/cm

4)

Slope = 0.1684 mol.s1/2.cm-4

Eq. (4.48)

From eq. (6.48),

@C

@x=

Cbp¼Dt

=

µCbp¼D

¶

t¡ 1=2

S =Cbp¼D

= 0:16842mol:s1=2

cm4

D =1

¼

µCb

S

¶2

=1

¼

µ9:443x10¡ 4mol=cm3

0:16842 mol:s1=2=cm4

¶2

D = 10¡ 5 cm2=s

b) Polynomial …t:

i = 4:8903 £ 10¡ 3 A=cm2 +³0:30648 A:s1=2=cm2

´t¡ 1=2


0.06

0.08

0.1

0.12

0.14

0.2 0.25 0.3 0.35 0.4 0.45

t-1/2 (sec-1/2)

i(A

/cm

2)

S = 0.30648 A.s1/2/cm2

Eq. (4.52)

From eq. (6.52),

i =zFDCbp

¼Dt

i =

µzFDCbp

¼D

¶

t¡ 1=2

S =zFDCbp

¼D= 0:30648 A:s1=2=cm2

z =S

p¼=D

FCb=

¡0:30648 A:s1=2=cm2

¢p¼=10¡ 5 cm2=s

(96; 500 A:s=mol) (9:443x10¡ 4mol=cm3)

z = 2

6.7 It is desired to reduce copper, Cu+2 + 2e¡ = Cu, at 35oC from anelectrolyte containing 60 g=l of Cu+2 ions. Calculate a) the total molar ‡ux thatarises from equal amounts of di¤usion and migration mass transfer under steady-state conditions, b) the gradients dc=dx; dÁ=dx; and d¹=dx, and c) approximatethe thickness of the di¤usion layer at the cathode electrode surface. Operatethe electrowinning cell for 10 minutes and let the di¤usivity for Cu+2 ions be10¡ 5 cm2=s:

Solutions:

Cb = 60 g=l = (60 g=l) (63:55 g=nol) t = 10 min = 600 secCb = 0:944 mol=l z = 2Cb = 9:44x10¡ 4 mol=cm3 F = 96; 500 A:s=molD = 10¡ 5 cm2=s k = 1:38x10¡ 23 J=oK

6.1. PROBLEMS 77

a) From eq. (6.52),

i =zFDCbp

¼Dt= zFCb

rD

¼t

i = (2) (96; 500A:s=mol)¡9:44x10¡ 4 mol=cm3

¢s

(10¡ 5 cm2=s)

(¼) (600 sec)

i = 1:33x10¡ 2 A=cm2

From eq. (6.8),

i = zFJ

J =i

zF=

1:33x10¡ 2 A=cm2

(2) (96; 500 A:s=mol)

J = 6:88x10¡ 8 mol

cm2:s

b) Both di¤usion and migration ‡uxes are equal. Thus,

J = Jd + Jm

Jd = Jm = J=2 = 3:44x10¡ 8 mol

cm2:s

From eq. (6.48),

@C

@x

¯¯¯¯x=0

=Cbp¼Dt

=9:44x10¡ 4 mol=cm3

p(¼) (10¡ 5 cm2=s) (600 sec)

@C

@x

¯¯¯¯x=0

= 3:88x10¡ 3 moil

cm4

From eq. (6.15), the ionic mobility is

B =D

kT=

10¡ 5 cm2=s

(1:38x10¡ 23 J=oK) (308oK)

B = 2:35x1015 cm2

J:s= 2:35x1015 V:cm2

C:s

Now, using eq. (6.12) yields

dÁ

dx= ¡

Jm

zqeCbB

dÁ

dx= ¡

3:44x10¡ 8 molcm2:s

(2) (1:602x10¡ 19 C)¡9:44x10¡ 4 mol

cm3

¢ ¡2:35x1015 V:cm2

C:s

¢

dÁ

dx= ¡4:83x10¡ 2 V=cm


From eq. (6.11),

d¹

dx= zqe

dÁ

dxd¹

dx= (2)

¡1:602x10¡ 19 J=V

¢ ¡¡4:83x10¡ 2 V=cm

¢

d¹

dx= ¡1:55x10¡ 20 J=cm

c) From eq. (6.53),

± =p

¼Dt

± =p

(¼) (10¡ 5 cm2=s) (600 s)

± = 0:137 cm = 1:37 mm

6.8 Show that Jx=o = DCo=±:

Solution:

Using eq. (6.40) for activation polarization (Co > Cb = C1 ) yields

Cx = Co ¡ (Co ¡ Cb) erf (y) (4.40)

y =x

p4Dt

dy =dx

2p

Dt(a)

y2 =x2

4Dt(b)

erf (y) =2

p¼

Z y

o

exp¡¡y2

¢dy (c)

Rearranging eq. (6.40) along with C = Cx and Co > Cb = C1 gives

C = Co ¡ Co

·2

p¼

Z y

o

exp¡¡y2

¢dy

¸

C = Co ¡ Co

·2

p¼

Z y

o

exp

µ

¡x2

4Dt

¶dx

2p

Dt

¸

C = Co ¡Cop¼Dt

·Z y

o

exp

µ

¡x2

4Dt

¶

dx

¸

(d)

Thus,

@C

@x= ¡

Cop¼Dt

@

@x

·Z y

o

exp

µ

¡x2

4Dt

¶

dx

¸

@C

@x= ¡

Cop¼Dt

exp

µ

¡x2

4Dt

¶

@C

@x

¯¯¯¯x=o

= ¡Cop¼Dt

(e)

6.1. PROBLEMS 79

Combining eqs. (6.3) and (e) yields

Jx=o = ¡D@C

@x

¯¯¯¯x=o

Jx=o =DCop¼Dt

=DCo

±((4.51))

6.9 Show that dC=dx = Cb=p

¼Dt:

Solution:

Using eq. (6.38) for activation polarization (Cb = C1 >> Co) yields

Cx = Co + (Cb ¡ Co) erf (y) (4.38)

Cx = Co + Cb erf (y) (a)

y =x

p4Dt

dy =dx

2p

Dt(b)

y2 =x2

4Dt(c)

erf (y) =2

p¼

Z y

o

exp¡¡y2

¢dy (d)

Rearranging eq. (6.40) along with C = Cx gives

C = Co + Cb

·2

p¼

Z y

o

exp¡¡y2

¢dy

¸

C = Co + Cb

·2

p¼

Z y

o

exp

µ

¡x2

4Dt

¶dx

2p

Dt

¸

C = Co +Cbp¼Dt

·Z y

o

exp

µ

¡x2

4Dt

¶

dx

¸

(e)

Thus,

@C

@x=

Cbp¼Dt

@

@x

·Z y

o

exp

µ

¡x2

4Dt

¶

dx

¸

@C

@x=

Cbp¼Dt

exp

µ

¡x2

4Dt

¶

@C

@x

¯¯¯¯x=o

=Cbp¼Dt

(e)

6.10 Prove that the di¤usivity D is constant in the Fick’s second law, eq.(6.17).


Solution:

From eq. (6.39),

Cx ¡ Co

Cb ¡ Co= erf (y)

Cx ¡ Co

Cb ¡ Co= erf

µx

p4Dt

¶

For …xed values of distance x, time t, temperature T and concentration Cx;

Cx ¡ Co

Cb ¡ Co= Constant = erf

µx

p4Dt

¶

Therefore, the di¤usivity is

D = Do exp

µ

¡E¤

RT

¶

= Constant

D 6= f(Cx)

6.11 Assume that the total molar ‡ux of a specie j is due to di¤usion andconvection. The convective force acting on the specie is Fx = ¡ (1=NA) (d¢G=dx),where NA is Avogadro’s number and d¢G=dx is the molar free energy gradient.Recall that the volume fraction is equals to the mole fraction divided by molarconcentration; that is, V = X=C. Based on this information, show that

dC

dx= C

·d ln (C=Co)

dx

¸

where Jd = Jc, K = C=Co < 1 and Co = Constant.

Solution:

Combining eqs. (6.5) and (6.13) yields

Jc = Cv = CBFx

Jc = ¡CB

NA

d¢G

dx

Thermodynamically, the free energy change is de…ned by eq. (2.31a)

¢G = ¢Go + RT ln (K)

¢G = ¢Go + RT ln (C=Co)

d¢G

dx= RT

d [ln (K)]

dx

6.1. PROBLEMS 81

where K is the rate constant. Thus,

Jc = ¡CBRT

NA

d [ln (C=Co)]

dx

Jc = ¡CkBT

·d [ln (C=Co)]

dx

¸

since k = R=NA = Boltzmann constant. Fick’s …rst law gives

Jd = ¡D@C

@x

If Jd = Jc, then

¡D@C

@x= ¡CkBT

·d [ln (C=Co)]

dx

¸

kBT@C

@x= CkBT

·d [ln (C=Co)]

dx

¸

@C

@x= C

·d [ln (C=Co)]

dx

¸

since D = kBT as in eq. (6.15). This concludes the solution of the problem.

6.12 Use the information given in Problem 6.11 to show that

d¢G

dx=

RT

C

dC

C

Solution:




d¢G

dx= RT

d ln (C=Co)

dx

But,

1

C

dC

dx=

d ln (C=Co)

dx

Then,d¢G

dx=

RT

C

dC

dx

6.13 Use the information given in Problem 6.12 to show that

vx = ¡D

C

dC

dx


Solution:

From eqs. (6.13) and Problem 6.11,

v = BFx (4.13)

Fx = ¡1

NA

d¢G

dx




d¢G

dx= RT

d ln (C=Co)

dx

But,

1

C

dC

dx=

d ln (C=Co)

dx

Then,d¢G

dx=

RT

C

dC

dx

Thus, eq. (6.13) yields the drift velocity as

v = ¡BkT

C

dC

dx

v = ¡D

C

dC

dx

where D = BkT:

6.14 If the migration ‡ux is neglected in eq. (6.2), approximate the total‡ux at low and high temperatures. Assume that Cx >> Co at a distance x froman electrode surface.

Solution:

J = ¡DdC

dx+ Cv

J = ¡DCo ¡ C

x+ Cv

If C >> Co, then

J = ¡DC

x+ Cv

6.1. PROBLEMS 83

In addition,

D = Do exp

µ

¡E¤

RT

¶

and

J = ¡CDo

xexp

µ

¡E¤

RT

¶

+ Cv

Mathematically,

² @ If T ! 0, then D ! 0 andJ = Jc ' Cv

² @ If T ! 1, then D ! Do and

J ' ¡CDo

x+ Cv

This analysis indicates that the di¤usivity D and the molar ‡ux J increasewith increasing temperature.

6.15 Consider the concentration plane shown below. This is a transientelectrochemical system having …nite dimensions. Derive a solution for Fick’ssecond law, @C=@t = D@2C=@x2; based on this information

Solution:

Following the analytical procedure in Appendix A yields

y =x

2p

Dty1 =

a ¡ x

2p

Dty2 =

b ¡ x

2p

Dt

Start with eq. (A6) and (A8):

Z Cb

o

dC = B

Z 1

¡ 1

exp¡¡y2

¢dy =

p¼B

Cb =p

¼B

B =Cbp

¼


Then,

Z C

o

dC =Cbp

¼

Z y2

y1

exp¡¡y2

¢dy

Z C

o

dC =Cb

2

·2

p¼

Z y2

o

exp¡¡y2

¢dy ¡

2p

¼

Z y2

o

exp¡¡y2

¢dy

¸

C

Cb=

1

2[erf (y2) ¡ erf (y1)]

6.16 Derive Fick’s second law if the volume element in Figure 6.2 has a unitcross-sectional area and a thickness x.

Solution:

Jx;in = D

µ@C

@x

¶

x

Jx;out = ¡D

µ@C

@x

¶

x+dx

= ¡D

µ@C

@x

¶

x

¡ D

µ@C

@x

¶

dx

The concentration rate is

@C

@t= ¡

@Jx

@x= ¡

@Jx;in

@x¡

@Jx;out

@x@C

@t= ¡D

µ@2C

@x2

¶

x

+ D

µ@2C

@x2

¶

x

+ D

µ@2C

@x2

¶

dx

@C

@t= D

µ@2C

@x2

¶

dx

@C

@t= D

@2C

@x2

6.17 What does Fick’s …rst law mean in terms of atoms or ions of a singlephase?

Answer: Fick’s …rst law, Jx = D¡

@C@x

¢, means that atoms or ions di¤use

from high to low concentration regions.

6.18 What will Fick’s …rst law mean if D does not vary with x or C?

Answer: Fick’s …rst law will mean that Jx = D @C@x becomes a linear re-

lationship under steady-state condition¡

@C@t = 0

¢and hat atoms or ions di¤use

from high to low concentration regions.

6.1. PROBLEMS 85

6.19 Derive Fick’s second law if the volume element in Figure 6.2 has a unitcross-sectional area and the di¤using plane is located between x and x + dx,where Jx is the entering molar ‡ux at x and Jx+dx is that leaving at x + dx.

Solution:

Using Fick’s …rst law and the continuity equation yields

@C

@t= ¡

@Jx

@x

Jx = ¡D@C

@x@C

@t=

1

@x

·

D@C

@x

¸

@C

@t=

@D

@x

@C

@x+ D

@2C

@x2

If D is constant, then Fick’s second law becomes

@C

@t= D

@2C

@x2

6.20 Find an expression for x when

Cx ¡ Cb

Co ¡ Cb= 0:5205 = erf

µx

p4Dt

¶

Solution:

Cx ¡ Cb

Co ¡ Cb= 0:5205 = erf

µx

p4Dt

¶

Cx ¡ Cb

Co ¡ Cb= 0:5205 = erf (y)

0:5205 = erf (y)

From Table 6.2,

y =x

p4Dt

1

2=

x

2p

Dt

x =p

Dt

6.21 Derive eq. (6.85).

Solution:


Using eq. (6.82) for oxidation yields

i = zFD

·dCx

dx+

Cx

RT

dÁ

dx

¸

di

dx= zFD

·d2C

dx2+

C

RT

d2Á

dx2+

1

RT

dC

dx

dÁ

dx

¸

= 0 (a)

From eq. (6.48) along with constant D at …xed time t and with C = Cb =constant at x ! 1 in the bulk electrolyte,

dC

dx=

Cbp¼Dt

(b)

d2C

dx2= 0 (c)

Inserting eqs. (b) and (c) into (a) yields

d2Á

dx2+

1p

¼Dt

dÁ

dx= 0 (4.62)

6.22 Show that dÁ=dx = ¡7Áo=³19

p¼Dt

´in a coupled di¤usion and

migration mass transfer. let x =p

¼Dt:

Solution:


Á = Áo exp

µ

¡x

p¼Dt

¶

dÁ

dx= ¡

Áop¼Dt

exp

µ

¡x

p¼Dt

¶

dÁ

dx= ¡

Áop¼Dt

exp

Ã

¡

p¼Dt

p¼Dt

!

dÁ

dx= ¡

Áop¼Dt

exp (¡1)

dÁ

dx= ¡

0:368Áop¼Dt

dÁ

dx= ¡

7Áo

19p

¼Dt

6.23 For activation polarization, the concentration gradient is given by eq.(6.50). If Co >> Cb, then show that

6.1. PROBLEMS 87

C

Co=

1p

4¼Dt

µx3

12Dt¡ x

¶

Solution:

From eq. (6.49) along with Co >> Cb,

dC

dx= ¡

Cop

4¼Dtexp

µ

¡x2

4Dt

¶

a) Taylor’s series: Letting y = x2= (4Dt) and expanding the exponential

function, exp(¡y) =P

(¡y)k =k! ' 1 ¡ y, yields

dC

dx= ¡

Cop4¼Dt

(1 ¡ y)

dC

dx= ¡

Cop4¼Dt

µ

1 ¡x2

4Dt

¶

dC

dx=

Cop4¼Dt

µx2

4Dt¡ 1

¶

b) Using the …rst three terms of McClaurin series yields

f (x) = f (0) +f0(0)

1!x +

f00(0)

2!x2 +

f000(0)

3!x3 + ::::

f (x) = exp

µ

¡x2

4Dt

¶

f (x = 0) = exp

µ

¡x2

4Dt

¶

x=0

= 1

f0(x = 0) = ¡2x

4Dtexp

µ

¡x2

4Dt

¶

x=0

= 0

f00(x = 0) = ¡1

2Dtexp

µ

¡x2

4Dt

¶

x=0

+x2

4D2t2exp

µ

¡x2

4Dt

¶

x=0

f00(x = 0) = ¡1

2Dt

f (x) = exp

µ

¡x2

4Dt

¶

= 1 ¡x2

4Dt

Then,

dC

Co=

1p

4¼Dt

µx2

4Dt¡ 1

¶

dx

1

Co

Z C

o

dC =1

p4¼Dt

Z x

o

µx2

4Dt¡ 1

¶

dx

C

Co=

1p

4¼Dt

µx3

12Dt¡ x

¶


6.24 Derive eq. (5.122) by converting eq. (3.29) into chemical potential(¹) if only one ion participates in a reaction. The reaction constant is de…nedby Kr = °C, where ° is the activity coe¢cient and C is the concentration. Letthe molar di¤usion and migration be equal and Jc = 0 in eq. (6.10).

Solution:

¹ = ¹o + kT ln (Kr) = ¹o + kT ln (°C)

¹ = ¹o + kT ln (°) + ln (C)

d¹

dC=

kT

C

Letd¹

dx=

d¹

dC

dC

dx=

kT

C

dC

dx

and

Jd = Jm

¡DdC

dx= ¡CB

d¹

dx= ¡BkT

dC

dxD = BkT

6.25 Follow the statement given in Example 6.3 for determining the averageconcentration as the starting point in this problem. Let Co;1 and Co;2 be theoriginal concentrations of columns 1 and 2, respectively.

Solution:

Start withCx ¡ C

Co;1 ¡ C= erf

µx

p¼Dt

¶

Average concentration:

C =Co;1 + Co;2

2=

Co

2

C =Co

2

6.1. PROBLEMS 89

Thus,

Cx

Co=

1

2

·

erf

µx

p¼Dt

¶¸

Cx

Co=

1

2@ x = 0

The rest of the problem is worked out in Example 6.3.

6.26 This problem requires the determination of the total molar ‡ux andcurrent density pro…les due to di¤usion, migration and convection for the reduc-tion of nickel from an electrolyte containing 8x10¡ 4 mol=cm3 of Ni+2 cations.The electrochemical cell has a continuos and direct ‡uid ‡ow system of 20 cm=sat 35oC. The di¤usivity of Ni+2 cations is D = 1:44 £ 10¡ 5 cm2=s. a) Use0 · t · 100 seconds for Jx = f(t; x = 0) at the electrode surface and b)0 · x · 10 centimeters for Jx = f(x; t = 10), and c) analyze the resultantpro…les very succinctly and determine which molar ‡ux dominates in this hy-pothetical reduction process for Ni+2 cations from solution. Can this processachieve a steady-state?

Solution:

Ni+2 + 2e = Ni z = 2 F = 96; 500 A:s=molCb = 8 £ x10¡ 4 mol= cm3

vb = 20 cm= sD = 1:44 £ 10¡ 5 cm2= s

Electrode Surface: Using eqs. (6.5), (6.64a), and (6.54b) yields the molar‡uxes on the electrode surface (x = 0)

Jc = Cbvb = 0:002 mol=cm2:s

Jm = (DCb) =p

¼Dt =¡1:712 8 £ 10¡ 3

¢=p

t

Jd = (DCb) =p

¼Dt =¡1:712 8 £ 10¡ 3

¢=p

t

Jx = Jd + Jm + Jc

Jx =(2)

¡1:712 8 £ 10¡ 3

¢

pt

+ 0:002

Jx =3:4256 £ 10¡ 3

pt

+ 0:002


where t is in seconds and Jx in mol:cm¡ 2s¡ 1. Thus, the Jx vs: t pro…le isshown below.

1009080706050403020100

0.80.70.60.50.40.30.20.1

0-0.1-0.2-0.3-0.4-0.5-0.6-0.7

Jx

Problem 4.26a

This pro…le indicates that Jx ! ¡1 as t ! 0 and this is a transient process.However Jx ! 0 as t ! 1 which is the condition for a steady-state process.These theoretical analysis can further be con…rmed by letting t = 10 seconds andvary the distance x from the electrode surface. Hence,

Fixed Time: Using eqs. (6.5), (6.64), (6.4) and (6.3) yieldsJc = Cbvb = 0:002 mol=cm2:s

t = 10 sÁ = Áo exp (¡x=±) = RT (zF )

¡ 1exp

³¡x=

p¼Dt

´

dÁ=dx = RT³zF

p¼Dt

´¡ 1

exp³¡x=

p¼Dt

´

Jm = (DCb) (RT )¡ 1 dÁ=dx

Jm = (DCb)³p

¼Dt´¡ 1

exp³¡x=

p¼Dt

´

Jm = 5:416 2 £ 10¡ 7 exp (¡47:016x)

Jd = ¡D@Cx=@x = DCb

³p31:416D

´¡ 1

exp³¡x2 (40D)¡ 1

´

Jd =¡5:416 2 £ 10¡ 7

¢exp

¡¡1736:1x2

¢

Jx = Jd + Jm + Jc

6.1. PROBLEMS 91

Jx =¡5:416 2 £ 10¡ 7

¢exp

¡¡1736:1x2

¢+

¡5:416 2 £ 10¡ 7

¢exp (¡47:016x) +

0:002

109876543210

0.1

0

Jx

This con…rms that a steady-state is achieved at t ¸ 10 seconds. It can alsobe concluded that the convective molar ‡ux controls the electrodeposition sinceJx ¼ Jc ¼ 0:002 mol=cm2:s. Therefore, an e¤ective electrolyte ‡ow system isimportant in electrodeposition. Similarly pro…les result for the current density.

Jx =(2)

¡1:712 8 £ 10¡ 3

¢

pt

+ 0:002 =3:425 6 £ 10¡ 3

pt

+ 0:002

Using eq. (6.9) yield i in A=cm2 and time t in seconds

i = zFJx = (2) (96500)³

3: 425 6£ 10¡3p

t+ 0:002

´


0.050.03750.0250.01250

0.175

0.15

0.125

0.1

0.075

0.05

0.025

0

t (sec)

i

This pro…le indicates that the current density in A=cm2 is su¢ciently high,but it rapidly decreases as the electrochemical process progresses. For instance,at t = 0:01 sec the current density is i = 3:63x10¡ 2 A=cm2. In addition,

i = zFJx

i = (2) (96500)

µ ¡5:416 2 £ 10¡ 7

¢exp

¡¡1736:1x2

¢

+¡5:416 2 £ 10¡ 7

¢exp (¡47:016x) + 0:002

¶

i = 0:104 53 exp¡¡1736:1x2

¢+ 0:104 53 exp (¡47:016x) + 386:0

i = 386 A=cm2

107.552.50

390

387.5

385

382.5

380

i

6.27 (a) Derive an expression for the concentration in g=l as a functionof temperature T and distance x from an oxidizing pure copper electrode inan electrolyte under an electrostatic …eld. Use the chemical potential gradient,d¹=dx = ¡ (zF ) dÁ=dx with an electric potential decay, Áx = Áo exp (¡¸x) Let

6.1. PROBLEMS 93

the free energy change be de…ned as ¢G = ¢Gx @ x and ¢G = ¢G1 @ x = 1.(b) Plot the concentration Cx pro…le at two di¤erent temperatures. Let

Á1 = 0 ¸ = 0:2 cm¡ 1

Áo = 8:314x10¡ 3 V 0 · x · 3:8 cmT = 298oK

¡curve

¢C1 = 1:57x10¡ 3 mol=l = 0:1 g=l

T = 318oK (¢ ¢ ¢ curve) F = 96; 500 J=mol:V

Solution:

a) The oxidation reaction is Cu = Cu+2 + 2e. Combining eqs. (a) and (b)yields

Z ¹x

¹1

d¹ = ¡zF

Z Á1

Áx

dÁ (c)

¹x ¡ ¹1 = ¡zFÁx (d)

¹x ¡ ¹1 = ¡zFÁo exp (¡¸x) (e)

From eq. (2.27),

¢Gx = ¹x;prod ¡ ¹x;react = ¹x;prod (f)

¢G1 = ¹1 ;prod ¡ ¹1 ;react = ¹1 ;prod (g)

since ¹x;prod = ¹1 ;react = 0 for the oxidation reaction. From eq. (2.31b),the change

¢Gx = ¡RT ln (Kx) = ¡RT ln (Cx) (h)

¢G1 = ¡RT ln (K1 ) = ¡RT ln (C1 ) (i)

Combining eqs. (f) through (i) gives

¹x = ¡RT ln (Cx) (j)

¹1 = ¡RT ln (C1 ) (k)

¹x ¡ ¹1 = RT ln

µC1Cx

¶

(l)

Finally, equating eqs. (e) and (l) yields

Cx = f(x; T ) = C1 exp

·zFÁo

RTexp (¡¸x)

¸

(m)

Cx = (0:1 g=l) exp

·193

Texp (¡0:2x)

¸

(n)


43210

0.2

0.175

0.15

0.125

0.1

Cx

(g/liter)

x (cm)

Cx 0.1 g/l exp 193T

exp0.2x

T 298oK

T 318o K

6.28 Derive an expression for iL = f (Cx; T; vx) and explain its physical sig-ni…cance with respect to concentration polarization and metal reduction. Plot´c = f (i=iL) when iL = 5 £ 10¡ 3 A and iL = 10¡ 2 A to support your explana-tion. When is the concentration polarization process discernible (apparent)?

Solution:

Combining eqs. (6.92), (6.97), (6.72), and (6.76) yields the sought expressionalong with the overpotential equation

iL =zF·CxvxT

Fx

p¼Dt

= f(Cx; vx; T ) (a)

´c =2:303RT

zFlog

µ

1 ¡i

iL

¶

(b)

The schematic polarization diagram is shown below for two di¤erent cases.

6.1. PROBLEMS 95

Thus, iL = f(Cx; vx; T ) represents the maximum rate of reduction, and it isthe limiting di¤usion current density. In fact, iL = f(Cx; vx; T ) is signi…cantduring reduction processes and negligible during metal dissolution. With respectto iL = f(Cx; vx; T ), it increases with increasing concentration, species velocityand temperature as predicted by equation (a). Finally, concentration polarizationbecomes apparent when i ! iL and ć ! 1 and ć ! 0 when iL ! 1. Thelatter can be de…ned as

ć =2:303RT

zFlog

Ã

1 ¡iFx

p¼Dt

zF·CxvxT

!

(c)

It is also apparent that concentration polarization is possible when jćj > 0due to presence of Fx for ionic mobility at an ionic velocity vx.

Chapter 7

CORROSIVITY ANDPASSIVITY

7.1 PROBLEMS

7.1 Determine (a) the electric potential E in millivolts, (b) the time in seconds,and (c) the growth rate in ¹m=s for electroplating a 3-¹m thick Cr …lm on aNi-undercoated steel part, provided that the electrolyte contains 10¡ 4 mol=lof Cr+3 cations at 25oC, the Ni-steel part has a 10-cm2 surface area, and thecell operates at 50% current e¢ciency and at a passive current of 0:8 amperes.Assume that the oxide passive …lm has a density of 7:19 g=cm3.

Solution:Cr+3 + 3e = Cr Eo

Cr = ¡0:744 V z = 3 x = 3x10¡ 4 cm½ = 7:19 g=cm3 Aw;Cr = 52 g=mol i = I=As = 0:08 A=cm2

As = 10 cm2 ² = 0:50 [Cr+3] = aCr+3 = 10¡ 4 mol=l(a) Using Nernst equation yields

E > Eo = Eo +RT

zFln (aCr+3)

E > Eo = ¡0:744V +(8:314 J=mol:K) (298 K)

(3) (96; 500 J=mol:V )ln

¡10¡ 4

¢

E > Eo = ¡0:823 V

(b) Integrating equation eq. (7.21) yields the time

t =

Z x

0

µzF½

²iAw

¶

dx =zF½x

²iAw

t =(3) (96; 500 A:s=mol)

¡7:19 g=cm3

¢ ¡3x10¡ 4 cm

¢

(0:50) (0:08 A=cm2) (52 g=mol)

t = 300:22 sec ' 5 min

97

98 CHAPTER 7. CORROSIVITY AND PASSIVITY

(c) The …lm growth rate is

dx

dt=

¢x

¢t=

3 ¹m

300:22 secdx

dt= 0:01

¹m

sec

7.2 (a) If chromiun oxide …lm frow rate is 0:03 ¹m= sec, how long will ittake to grow a 5 ¹m …lm on a Ni-substrate. Assume that the cell operates at50% current e¢ciency. Calculate (b) the current density.

Solution:(a) The time is

dx

dt=

¢x

t

t =¢x

dx=dt=

5¹m

0:05 ¹m= sec= 100 sec

(b) From eq.(7.51), the current density is

i =zF½x

²tAw=

(3) (96500) (7:19)¡5 £ 10¡ 4

¢

(0:8) (100) (52)

i = 0:25 A=cm2

7.3 Brie‡y, explain why the corrosion rate and the corrosion potential in-crease on the surface of some metallic materials in contact with an oxygen-containing acid solution. In this situation oxygen acts as an oxidizer. Use aschematic polarization diagram to support your explanation. Let the exchangecurrent densities and the open-circuit potentials be io;O2 < io;H2 < io;M andEo;O2

> io;H2> io;M . What e¤ect would have oxygen on the metal dissolution

if io;O2<< io;H2

< io;M and Eo;O2> io;H2

> io;M .

Solution:

NO OXYGEN: The polarization diagram below implicitly explain the rea-son why the corrosion rate increases in the presence of oxygen in an acid so-lution. According to the …gure below, the corrosion rate in terms of currentdensity and the corrosion potential are

i0corr = i0H No oxygen

i0net = i0corr ¡ i0H = 0 at equilibrium

E0corr = E0H No oxygen

7.1. PROBLEMS 99

WITH OXYGEN: However, when oxygen is present as an oxidizer the cor-rosion rate and corrosion potential become

icorr = (iH + iO2) > i0corr

Ecorr > E0corr

Consequently, hydrogen evolution decreases because iH < i0H and oxygenevolution evolves consuming additional electrons from the metal. In certaincases when io;O2 << io;H2 < io;M the presence of oxygen as an oxidizer wouldhave insigni…cant e¤ect on the rate of metal dissolution.

iL

EO2

Eo,H

Eo,M

io,M

io,H

i O2

icorr

Ecorr

Evans DiagramStern Diagram

Corrosion Point

ia = ir – if

ic = if – ir

i’corr

E’corr

iH iO2

E(V

)

Log (iL)

7.4 Assume that a stainless steel pipe is used to transport an aerated acidsolution containing 1:2x10¡ 6 mol=cm3 of dissolved oxygen at room temperatureand that the electric-double layer is 0:7 mm and 0:8 mm under static and ‡owingvelocity, respectively. If the critical current density for passivation and thedi¤usion coe¢cient of dissolved oxygen are 300 ¹A=cm2 and 10¡ 5 cm2=s, thendetermine (a) whether or not corrosion will occur under the given conditionsand (b) the passive …lm thickness at 5-minute exposure time if ip = 50 ¹a=cm2.Also assume that the density of the …lm and the molecular weight of the stainlesssteel are 7:8 g=cm3 and 55:85 g=mol, respectively and that the predominantoxidation state is 3. Given data:

Solution:Given data:


F = 96; 500 A:s=mol Cx = 1:2x10¡ 6 mol=cm3

z = 3 ±(static) = 0:07 cmD = 10¡ 5 cm2=s ±(flowing) = 0:008 cmicrit = 300 ¹A=cm2 ½ = 7:8 g=cm3

The reduction reaction is

O2 + 2H2O + 4e = 4OH¡ with EoO2

= 0:401 VHSE

(a) From eq. (6.99), the limiting current density for both conditions is

iL =zFDCx

±=

3 (96500 A:s=mol)¡10¡ 5 cm2=s

¢ ¡1:2x10¡ 6 mol=cm3

¢

0:07 cm

iL ' 49:63 ¹A=cm2 < icrit for corrosion under static solution

iL ' 43:43 ¹A=cm2 > icrit for passivation under ‡owing solution

(b) The limiting current density for passivation under ‡owing solution be-comes the passive current density, then the passive …lm thickness can be esti-mated using eq. (7.51) along with a density of 7:8 g=cm3 and a molecular weightof 55:85 g=mol. Hence,

x =ipAwt

zF½

x =

¡50x10¡ 6 A=cm2

¢(55:85 g=mol) (300 s)

(3) (96; 500 A:s=mol) (7:8 g=cm3)

x = 0:371 ¹m = 371 nm

This is a reasonable result despite the crude approximation technique beingused in this example problem. The passive oxide …lm may be CrOx(OH)3¡ 2x:nH2Oas per Figure 7.20.

7.5 For the reinforced concrete specimen (concrete slab) shown in Example7.2, calculate (a) the chloride ion penetration depth (x) when the potentialgradient due to di¤usion and migration are equal and (b) the potential gradients.[Solution: (a) x = 1:03 mm].

Solution:(a) From eqs. (6.12) and (6.13),

µdÁ

dx

¶

diffusion

=RT

xFµ

dÁ

dx

¶

electric

=E

L

7.1. PROBLEMS 101

Equating these equation yields

x =RTL

EF=

(8:314) (298) (8)

(2) (96500)= 1:03 mm

Then the potential gradients are

µdÁ

dx

¶

electric

=

µdÁ

dx

¶

diffusion

=RT

xF=

(8:314) (298)

(0103) (96500)µ

dÁ

dx

¶

electric

=

µdÁ

dx

¶

diffusion

= 2:49 £ 10¡ 4 V=cm

7.6 For a reinforced concrete slab having a concrete cover 50-mm deep,the threshold and the surface concentrations of chloride anions are Cth = 0:60Kg=m3 and Cs = 19 Kg=m3. Also, D = 32 mm2=yr is the di¤usivity of chlorideanions. Determine (a) the chloride molar ‡ux, (b) the time (Ti) to initiatecorrosion on an uncoated reinforcing steel surface (Figure 7.21c). Explain. (c)Plot the concentration pro…le of chloride anions at the corrosion initiation timeTi. Calculate (d) the rate of iron hydroxide Fe(OH)2 production (molar ‡uxand mass ‡ux) at the anodic regions for an anodic current density of 1:5 ¹A=cm2,(e) the corrosion rate (CR) of steel in mm=yr, (f) the time required for crackingand spalling due to the formation of a critical rust (Fe(OH)3) volume if the bardiameter is reduced only 25 ¹m and (g) the time for repair and the apparentmass of rust per area.

Solution:

(a) The chloride molar ‡ux at ti = 11 years can be calculated using eq.(6.8). Thus,

JCl¡ =i

zF=

1:5 ¹A=cm2

(1) (96; 500 A:s=mol)' 1:5510¡ 11 mol

cm2: sec

(b) The time to initiate corrosion, eq. (7.63), is

ti =x2

4D

·

erf ¡ 1

µ

1 ¡Cth

Cs

¶¸¡ 2

ti =(50 mm)

2

4 (31:61 mm2=yr)

·

erf ¡ 1

µ

1 ¡0:6 Kg=m3

18 Kg=m3

¶¸¡ 2

ti ' 13:43 years

(c) The concentration pro…le of chloride ions, C (x; ti) at the approximatedinitiation time of 13:43 years can be determined using the solution of Fick’ssecond law of di¤usion, eq. (7.59).


C (x; t) = Cs

·

1 ¡ erf

µx

2p

Dt

¶¸

(7.59)

C (x; t) =¡19 Kg=m3

¢"

1 ¡ erf

Ã10x

2p

(32 mm2=yr) (13:43 yr)

!#

C (x; t) = y = 19:0 erfc (0:241 19x)

53.752.51.250

20

15

10

5

0

x

y

(d) From eq. (7.69) along with ia = icorr,

JFe(OH)2 =icorr

zF=

1x10¡ 6 A=cm2

(2) (96500 A: sec =mol)

JFe(OH)2 = 5:18 £ 10¡ 12 mol=¡cm2: sec

¢

Multiplying this result by the iron molar mass, 55:85 g=mol, yields

JFe(OH)2 = 2:893 £ 10¡ 10 g=¡cm2: sec

¢= 9:123 £ 10¡ 3 g=

¡cm2:yr

¢

JFe(OH)2 ' 91:23 g=¡m2:yr

¢' 0:091 Kg=

¡m2:yr

¢

Thus, the rate of rust [Fe(OH)3] production, eq. (7.68), is

7.1. PROBLEMS 103

JFe(OH)3 =

·MFe(OH)3

MFe(OH)2

¸

JFe(OH)2 (Rust)

JFe(OH)3 =

µ106:85 g=ml

89:85 g=mol

¶£91:23 g=

¡m2:yr

¢¤

JFe(OH)3 ' 108:49 g=¡m2:yr

¢(Rust)

(e) The uniform corrosion rate (CR) of ferrous ions (Fe2+), eq. (7.74) atthe end of the 11th year

CR =iaAw

zF½=

¡1:5 £ 10¡ 6 A=cm2

¢(55:85 g=mol)

(2) (96500 A:sec=mol) (7:86 g=cm3)

CR = 5:5225 £ 10¡ 11 cm= sec ' 0:017 mm=yr

CR = 17 ¹m=yr

Therefore, the corrosion rate, 127 ¹m=yr, is rather low, but su¢cient to benoticed after 13:43 years. This corrosion rate corresponds to pitting formationat the 13th year; however, if the chloride ion supply continues, then the corrosioneventually propagates.

(f) If the reduction in diameter of the uncoated steel bar is 25 ¹m, then thetime for some rust formation that causes cracking and spalling, eq. (7.64), is

tc =25 ¹m

17 ¹m=yr= 1:47 yr

(g) The repair time, eq. (7.65), is

tr = ti + tc = 13:43 yr + 1:47 yr

tr = 14:9 yr

Therefore, the …rst concrete repair is required at 14:9 years. On the otherhand, The apparent mass of rust per area is

mrust = Jrust:tc =£108:49 g=

¡m2:yr

¢¤(1:47 yr)

mrust = 159:48 g=m2 = 0:159 g=cm2

7.7 Assume that all electrochemical reactions are governed by the Nernstequation, which speci…es the relationship between the potential of an electrodeand the concentrations of the two species Fe3+ and Fe+2 involved in the re-versible redox reaction at the working electrode. Assume a hydrogen-rich elec-trolyte. (a) Write down the electrochemical reaction for Fe3+ and Fe2+ and


the potential equation. (b) What is the driving force when the concentrations ofthe species at the working electrode surface are equal? (c) What is the drivingforce at the working electrode surface when CFe3+ (1; t) > 2CFe3+ (0; t) andCFe2+ (0; t) = 0:5CFe3+ (0; t), where t > 0 at 25oC? (d) Derive the currentequation for the case described in part (c). Recall that the current is simply the‡ow rate of electrons. (e) Draw schematic concentration pro…les for CO = CFe3+

when I = 0, I > 0 for reduction (Fe3+ cation moves to the electrode surface)and I < 0 for oxidation so that the Fe3+ cation moves away from the electrodesurface.

Solution:

(a) The electrochemical reaction for O and R and the potential equation are

O + ze¡ = R

Fe3+ + e¡ = Fe2+ for z = 1

E = Eo ¡RT

zFln

µCFe2+

CFe3+

¶

(b) There is no driving force because CFe3+ = CFe2+ and E = Eo sinceln (CFe2+=CFe3+) = ln (1) = 0. Therefore, there is no driving force for transportof analyte to or from the working electrode surface.

(c) For CFe3+ (1; t) > 2CFe3+ (0; t),

Fe3+ + e¡ = Fe2+ for z = 1

E = Eo ¡RT

zFln

µCFe2+ (0; t)

CFe3+

¶

E = Eo ¡RT

zFln

µ0:5CFe3+ (0; t)

CFe3+

¶

E = Eo ¡RT

zFln (0:5)

From Table 2.2, Eo = 0:77 VSHE and the driving force is

E = 0:44 VSHE ¡

¡8:314510 J:mol¡ 1K¡ 1

¢(298 K)

(1) [96; 500 J= (mol:VSHE)]ln (0:5)

E = 0:46 VSHE

(d) The current equation at the working electrode surface (x = 0) is

J = ¡D

µ@C

@x

¶

x=0

Q = zFN

7.1. PROBLEMS 105

Figure 7.1:

where N is the mole number. Then, the fundamental relationship for thecurrent in any amperometric experiment derived as

I =dQ

dt= ¡zFAsJ = zFAsD

µ@C

@x

¶

x=0

(e) The schematic concentration pro…les for CO = CFe3+ is

7.8 Consider the reversible reaction O+ze¡ = R (as in Fe3+ +e¡ = Fe2+)for deriving the half-wave potential (E1=2) using I = 0:5IL, where IL is thelimiting current for a voltammogram. Assume that the initial solution containsonly the species O and that di¤usion is the form of mass transport so that thecurrent in a voltammetric cell is

I = KO ([O]b ¡ [O]x=0) (a)

I = KR ([R]x=0 ¡ [R]b) = KR [R]x=0 (b)

where KO = zFAsDO=± and KR = zFAsDR=±, [R]b = 0 in the bulk solutioninitially. Use the Nernst equation for de…ning the voltammetric cell potential.Recall that the applied potential reduces O to R and that the current dependson the rate at which O di¤uses through the di¤usion layer (±).

Solution:

The Nernst equation gives the potential at the electrode surface. Thus,

E = Eo ¡RT

zFln

[R]x=0

[O]x=0

(c)

Initially, [O]x=0 = 0 in eq. (a) so that the current becomes the limitingcurrent (IL)

IL = KO [O]b (d)


From eqs. (b) and (d),

[R]x=0 =I

KR(e)

Substitute eq. (d) into (a) yields

I = IL ¡ KO [O]x=0

[O]x=0 =IL ¡ I

KO(f)

Substitute eqs. (e) and (f) into (c) gives

E = Eo ¡RT

zFln

[R]x=0

[O]x=0

E = Eo ¡RT

zFln

µI

KR

KO

IL ¡ I

¶

E = Eo ¡RT

zFln

·µKO

KR

¶µI

IL ¡ I

¶¸

and

E = Eo ¡RT

zFln

µKO

KR

¶

¡RT

zFln

µI

IL ¡ I

¶

Use I = 0:5IL to get E = E1=2 and

E1=2 = Eo ¡RT

zFln

µKO

KR

¶

since ln [I= (IL ¡ I)] = ln [0:5IL= (IL ¡ 0:5IL)] ln [1] = 0. Hence,

Chapter 8

DESIGN AGAINSTCORROSION


CATHODIC PROTECTION8.1 Why cathodic protection is not generally recommended for stress cor-

rosion problems on high-strength ferritic steels?

Answer: Hydrogen embrittlement may be the main cause of corrosion onferritic steel cathodes when the hydrogen evolution occurs. In this case, cathodi-cally discharged atomic hydrogen readily di¤uses into the iron matrix, speci…callyin voids, making the steel susceptible to hydrogen embrittlement due to a highpressure (P ) build-up by molecular hydrogen within voids. This phenomenonmay be represented by the reaction H + H ! H2. If P > ¾ys (yield strength),then the steel loses its strength and ductility.

8.2 A steel structure exposed to seawater is to be cathodically protectedusing the sacri…cial anode technique. calculate a) the number of Zn anodes (N)that will be consumed in a year if the Zn anode capacity is 770 A:h=Kg (averagevalue as per Table 8.2) and the current is 0:70 amps; and b) the individual weight(M) of anodes so that NM ¸ W , where W is the total theoretical weight of theanodes.

Solution:

Zn = Zn+2 + 2e

Ca = 770 A:h=Kg = 2:772x106 A:s=Kgt = 1 year = 31:536x106 secAw = 65:37 g=mol I = 0:7 Az = 2 F = 96; 500 A:s=mol

107

108 CHAPTER 8. DESIGN AGAINST CORROSION

Using Faraday’s law, Eq. (3.12c) or (8.60), the total theoretical weight ofanodes is

W =ItAw

zF

W =(0:70 A)

¡31:536x106 s

¢(65:37 g=mol)

(2) (96; 500 A:s=mol)

W = 7:48 Kg

Using the anode life (Ld = t) equation given in Table 8.4 yields the theoreticalnumber of anodes. Hence,

N =7CaAw

3¼zF

N =(7)

¡2:772x106 A:s=Kg

¢(65:37 g=mol)

(3¼) (2) (96; 500 A:s=mol)

N ' 1

Then,

M >W

N= 7:48 Kg

8.3 Cathodically protect a vertical pressurized steel tank (¾ys = 414 MPaand P = 12 MPa) using one ‡ush mounted Mg-Al-Zn bracelet (Bi = 0:9525cm). Assume 2% tank coating damage per year and neglect any installationdamage if the tank is coated after installation. Use a bracelet with dimensionsequal to (0:50 cm)x (3 cm)x (6 cm). Calculate (a) the maintenance currentand current density if the initial current density and the design lifetime are0:43 ¹A=cm2 and 25 years. Will it be convenient to protect the tank having athickness of 1:8 cm? The tank height and the internal diameter are 65 cm and15:24 cm, respectively. (b) Will the tank explode at the end of the anode lifetime? Assume that the anode Mg-alloy and ½x = 5; 000 ohm:cm for the soil.Given data for the Mg-alloy anode: ½ = 1:74 g=cm3, Aw = 24:31 g=mol andCa = 1; 230 A:h=Kg for the Mg-Al-Zn anode (Table 8.2).

Solution:The tank and the anode surface areas are, respectively

At = ¼diht = (¼) (15:24 cm) (25:40 cm)

At = 1; 216 cm2

Aa = NwL = (1) (3 cm) (6 cm) = 18 cm2

The initial current for polarizing the structure is

Ii = Aai =¡18 cm2

¢ ¡0:43 ¹A=cm2

¢= 7:74 ¹A


From Table 8.4, the average initial weight of the anode is

Wi =3¼IiLd

7Ca

Wi =(3¼)

¡7:74x10¡ 6 A

¢(219; 000 h)

(7) (1; 230 A:h=Kg)

Wi = 1:86 grams

From Table, 8.5, the anode electrical resistance is

Rf =¼½x

10p

Aa

Rf =(¼)

¡5x103ohm; cm

¢

10p

18 cm2= 370:24 ohm

Using Ohm’s law yields the initial potential (Ei) using a recti…er having acapacity greater than Ei

Ei = IiRf =¡7:74x10¡ 6 A

¢(370:24 ohm)

Ei = 2:87 mV

Erect > Ei

From Table 8.3, the polarized current density is chosen as 0:5 ¹A=cm2 sothat

Ip = Aaip =¡18 cm2

¢ ¡0:5 ¹A=cm2

¢= 9 ¹A

The …nal current for maintaining the structure polarized and cathodicallyprotected is

If = (1 ¡ ¸i ¡ ¸cLd) (Aai) (8.61)

If = 1

·

1 ¡ 0 ¡

µ2%

100%

¶

(25)

¸

(Aai)

If = 0:5Aai = 0:5Ii

If = (0:5)¡18 cm2

¢ ¡0:43 ¹A=cm2

¢= 3:87 ¹A

For maintenance, the current and the …nal current density are, respectively

I =Ip + If

2=

9 ¹A + 3:87 ¹A

2= 6:44 ¹A

if =I

Aa=

6:44 ¹A

18 cm2= 0:36 ¹A=cm2

Finally, the maintenance or …nal potential is

Ef = IRf =¡6:44x10¡ 6 A

¢(370:24 V=A)

Ef = 2:38 mV


a) There will be a potential drop and a current change of

¢E = Ei ¡ Ef = Ei = 2:87mV ¡ 2:38 mV = 0:49 mV

¢I = Ii ¡ I = 7:74 ¹A ¡ 6:44 ¹A = 1:30 ¹A

Both ¢E and ¢I represent a 17% reduction. Therefore, it is convenient tocathodically protect the steel tank. The initial and …nal corrosion rates of the an-ode in terms of current density are 0:43 ¹A=cm2 and 0:36 ¹A=cm2, respectively.The initial corrosion rate of the anode in mm=y is

CR;i =iiAw

zF½=

¡0:43x10¡ 6 ¹A=cm2

¢(24:31 g=mol)

(2) (96; 500 A:s=mol) (1:74 g=cm3)

CR;i = 3:11x10¡ 11 cm=s = 0:0010 mm=y

and the …nal corrosion rate becomes

CR;;f =ifAw

zF½=

¡0:36x10¡ 6 ¹A=cm2

¢(24:31 g=mol)

(2) (96; 500 A:s=mol) (1:74 g=cm3)

CR;;f = 2:61x10¡ 11 cm=s = 0:0008 mm=y

b) The …nal thickness of the Mg-alloy anode is Bf = (0:0008 mm=y) (25 y) =0:02 mm, which is a thin foil and it must be replaced by a new anode.

The hoop stress in the steel tank is

¾h =Pd

2B=

(12 MPa) (15:24 cm)

2 (1:8 cm)= 50:8 MPa

¾h << ¾ys

As a result, he steel tank will not deform neither explode because ¾h << ¾ys

and there are no cracks; therefore, it will not corrode at all the end of the designlife of 25 years.

8.4 A 2-m diameter steel tank containing water is pressurized at 200 kPa.The hoop stress, thickness, and height are 420MPa, 2 cm, and 8 cm, respec-tively. If the measured corrosion rate is 0:051 mm=y, determine a) the tank lifeand b) the developed current. Data: ½ = 7:86 g=cm3 and Aw = 55:86 g=mol.

Solutions:

Fe ! Fe+2 + 2eP = 200 kPa d = 2 m¾ = 420 MPa B = 2 cmCR = 0:051 mm=y h = 8 m


a) From eq. (8.64),

CR =iAw

zF½=

Pd

2¾t(8.64)

t =Pd

2¾CR=

(200 kPa)¡2x103 mm

¢

(2) (420; 000 kPa) (0:051 mm=y)

t = 9:34 years = 2:94x108 sec

b) From eq. (8.61),

i =zF½B

tAw(8.63)

i =(2) (96; 500 A:s=mol)

¡7:86 g=cm3

¢(2 cm)

(2:94x108 s) (55:85 g=mol)

i = 1:84x10¡ 4 A=cm2 = 184 ¹A=cm2

For a lateral area As = ¼dh, the current becomes

I = Asi = ¼dhi

I = (¼)¡2x102 cm

¢ ¡8x102 cm

¢ ¡1:84x10¡ 4 A=cm2

¢

I = 92:49 A

8.5 (a) Derive eq. (8.58) and (b) calculate the theoretical anode capacity(Ca) for magnesium (Mg) and zinc (Zn).

Solutions:

(a) From eqs. (5.30) with ² = 1 and Table 8.4,

m =ItAw

zFeq. (5.30)

W =3¼It

7CaTable 8.4

Let m = W since both represent mass. Thus, the anode capacity equationsis

ItAw

zF=

3¼It

7Ca

Ca =3¼zF

7Aw


(b) Anode capacity for Mg and Zn:

Mg = Mg+2 + 2e

Ca;Mg =3¼zF

7Aw=

(3¼) (2) (96; 500 A:s=mol)

(7) (24:31 g=mol)

Ca;Mg = 2; 969A:h

Kg

and

Zn = Zn+2 + 2e

Ca;Zn =3¼zF

7Aw=

(3¼) (2) (96; 500 A:s=mol)

(7) (65:37 g=mol)

Ca;Zn = 1; 104A:h

Kg

8.6 Show that ® =p

Rs=RL.

Solution:

From eq. (8.20),

´x = ó exp (¡®x) (8.20)

d´x

dx= ¡®ó exp (¡®x) (a)

d2´x

dx2= ®2ó exp (¡®x) (b)

Substituting eqs. (a) and (b) into (8.18) yields

d2´x

dx2=

µRs

RL

¶

´x (8.18)

®2ó exp (¡®x) =

µRs

RL

¶

ó exp (¡®x) (c)

® =

rRs

RL(d)

8.7 Use an annealed (@ 792oC) 1040 steel plate of B = 0:635 cm thicknessto design a cylindrical pressurized vessel using a safety factor (SF ) of 2. Theannealed steel has a yield strength of 355 MPa. Calculate (a) the averageallowable pressure according to the theory for thin-walled vessels. Will thevessel fail when tested at 2 MPa for 10 minutes? If it does not fail, thenproceed to cathodically protect the cylindrical pressure vessel as a domesticor home water heater having a cylindrical magnesium (Mg) anode rod. All


dimensions shown below. Calculate (b) the applied hoop stress, the current,the potential for polarizing the heater and the resistance of the system, (c) thetheoretical anode capacity and the anode lifetime of the anode in years and (d)how much electric energy will delivery the Mg anode rod? The steel vessel has a98% internal glass coating (vitreous porcelain enamel lining) for protecting thesteel inner wall against the corrosive action of portable water. Assume a watercolumn of 76:20 cm and an internal pressure of 1 MPa. Assume that the exteriorsurface of the steel vessel is coated with an appropriate paint for protecting itagainst atmospheric corrosion and that the anode delivers 1:5 VCu=CuSO4

: TheNACE recommended potential for steels is ¡85 VCu=CuSO4

.

emf: (+) (-)

Anode

ColdH2O

Hot H2O

H

Di

L

Steel TankH = 85 cmDi = 40.64 cm

Ρw = 900 ohm.cmAw, s = 55.85 g/molPS = 7. 86 g/cm3

Mg Anode RodL = 0.75H cmd = 5.08 cmAw,Mg = 24.31 g/molΡMg = 1.74 g/cm3

Drain Valve

Electric HeatingElements

+-

-+

Solution:

(a) The design hoop stress (tangential stress) is given

¾ =¾ys

SF=

355 MPa

2= 177:50 MPa

The hoop stress (average tangential stress) equation is used for determiningthe average allowable pressure

¾ =PDi

2B

P =2B¾

Di=

(2) (0:635cm) (177:50 MPa)

40:64 cm= 5:55 MPa

Assuming that the pressure vessel, as well as the glass coating, is crack-freeone can deduce that the pressure vessel will not fail because P > Ptest = 2 MPa.


(b) Since the steel major element is iron, the reaction Fe+2 + 2e = Fe isassumed in this problem. Using eqs. (8.66) and (8.65) yield the applied hoopstress and the protective current density, respectively

¾ =PDi

2B=

(1 MPa) (40:64 cm)

(2) (0:635 cm)= 32 MPa

i =zF½B

tAw;Fe=

(2) (96; 500 A: sec =mol)¡7:86 g=cm3

¢(0:635 cm)

(6:312 £ 108 sec) (55:85 g=mol)

i = 27:33 ¹A=cm2

The current and the potential for protecting the steel tank are based on thelateral area of the tank in contact with water. Letting the water column beh = 76:20 cm yields

As = Ao (1 ¡ ¸s)

As = ¼Dih (1 ¡ ¸s) = (¼) (40:64 cm) (76:20 cm) (1 ¡ 0:98)

As = 194:58 cm2

I = iAs =¡27:33 ¹A=cm2

¢ ¡194:58 cm2

¢= 5:32 mA

The cell potential along with the NACE recommended potential of ¡0:85VCu=C SO4

is

E = Ec + Ea = 0:85 V + 1:5 V = 0:65 V

Thus, the system resistance is

Rs =E

I=

0:65 V

5:32x10¡ 3 A= 122:18 ohm

(c) The anode lifetime is based on the oxidation reaction Mg ! Mg+2 +2e.From eq. (8.4), the anode capacity is

Ca =²zF

Aw=

(0:50) (2) (96; 500 A:s=mol)

24:31 g=mol= 1; 103 A:h=Kg

Thus, anode e¤ective length is La = 0:75H = (0:75) (85 cm) = 63:75 cm sothat the anode mass becomes

Va =¼d2La

4=

(¼) (5:08 cm)2 (63:75 cm)

4= 1; 292:10 cm3

ma = ²½aVa = (0:50)¡1:74 g=cm3

¢ ¡1; 292:10 cm3

¢

ma = 1:12 Kg

From Table 8.4, the average anode lifetime is


Ld =7Cama

3¼Ia=

(7) (1; 103 A:h=Kg) (1:12 Kg)

(3¼) (5:32x10¡ 3 A)

Ld = 172:468:11 h = 19:7 years

This result seems reasonable.

(d) A 1:12 Kg of anode mass will deliver an electric energy of

EE = Cama

EE = (1; 103 A:h=Kg) (1:12 Kg) = 1; 235:36 A:h

EE = 4:45 MJ

Recall that 1 J = 1 A:s.

ANODIC PROTECTION

8.8 For anodic protection, the interface potential of the structure is in-creased to passive domain, provided that the metal exhibits active-passive be-havior. Thus, prevention of corrosion is through impressed anodic current.Explain including the mathematical de…nition of the applied anodic currentdensity.

Explanation: The impressed anodic current imparts an initial corrosionrate (icorr) is shifted to a low value at ipass (passive current density). Otherwise,passivation will not occur. The applied anodic currents density

i = iox ¡ ired

where iox is the oxidation and ired is the reduction current densities.

8.9 What are the implications of cathodic protection of steels in acid solu-tions?

Answer: The cathodic protection currents in acid solution can lead to hy-drogen liberation (H2) and embrittlement of steels. This makes anodic protectionthe most preferred choice for protection of chemical process equipment.

8.10 Why is passivation important for anodic protection? What is themechanisms that explains the electron motion for reactions to take place?

Answer: First of all, passivation is the process for oxide …lm formation onthe metal exposed to an acid solution. Once passivation is completed, an anodiccurrent-potential pair has to be kept within a reasonable range to keep the metal


in its passivated stated. The mechanism is quantum mechanical electron tunnel-ing through a thin oxide …lm, which may be an insulator or a semiconductor.However, ionic tunneling seems to be restricted by the …lm width acting as abarrier.

8.11 What are the passive …lm preferred break-down sites?

Answer: Grain boundaries (if any), nonmetallic inclusions and ‡aws onthe metal oxide surface.

8.12 Calculate (a) the potential for the two titanium reactions given be-low. These reactions represent the passivation process for titanium-oxide …lmformation. Thus, anodic protection is natural because the T iO2 …lm acts asa semiconductor barrier to protect the underlying metallic Ti from corrosion,Ti ! Ti2+ + 2e¡ at Eo = 1:63 VSHE . (b) Determine the Ph.

Ti + O2 = TiO2 ¢Go = ¡852:70 kJ=mol

T i + 2H2O = TiO2 + 4H+ + 4e¡ ¢Go = ¡346:94 kJ=mol

Solutions:(a) The potential for Ti + O2 = TiO2,

E = ¡¢G

zF= ¡

(¡852:70 kJ=mol)

2 (96500 J=mol:V )= 4:42 mV

Using the Nernst equation,

E = Eo ¡RT

zFlnKsp = Eo ¡

RT

zFln

[TiO2]

[Ti] [O2]

E = Eo since [TiO2] = [Ti] = [O2] = 1

For Ti + 2H2O = T iO2 + 4H+ + 4e¡ at Eo = ¡0:860 V (Table 2.3 ),

E = ¡¢G

zF= ¡

(¡346:94 kJ=mol)

2 (96500 J=mol:V )= 1:80 mV

(b) Using the Nernst equation for Ti + 2H2O = TiO2 + 4H+ + 4e¡ givesthe pH

E = Eo ¡RT

zFlnKsp = Eo ¡

RT

zFln

[TiO2] [H+]

4

[Ti] [H2O]2

E = Eo ¡4RT

zFln

£H+

¤since [TiO2] = [Ti] = [H2O] = 1


Thus,

£H+

¤= exp

·

¡zF (E ¡ Eo)

4RT

¸

£H+

¤= exp

·

¡2 (96500) (0:0018 + 0:86)

4 (8:314) (298)

¸

£H+

¤= 5:1417 £ 10¡ 8

Then,

pH = ¡ log£H+

¤= ¡ log

¡5:1417 £ 10¡ 8

¢

pH = 16:78

8.13 The e¤ect of sulfuric acid (H2SO4) concentration on polarization be-havior of a metal at 25oC can be assessed obtaining experimental potentiody-namic polarization curves. Below is a data set for the critical current density(ic) of a hypothetical metal.

C (%) 0 20 30 40 50 60 70ic (A=cm2) 46 30 23 17 12 8 5

Plot ic = f (C) and explain whether or not passivation is a¤ected by theconcentration of sulfuric acid.

Solution:Curve …tting yields a polynomial …t : ic = 46:096 ¡ 0:907 23C + 4:552 3 £

10¡ 3C2 with R2 = 0:98. The plot is


For anodic protection, ic must exist for passivation to take place by the for-mation of an oxide …lm. The plotted data indicates that ic nonlinearly decreasewith increasing percent concentration. This trend means that passivation is read-ily as ic decreases because at the applied current density i = ic ! ipass untilE ! Epass > Ecorr. Therefore, passivation is enhanced by increasing H2SO4

concentration.

8.14 Below is a data set taken from Table 12.7 in reference [10] for de-termining the e¤ect of concentration of sulfuric acid (H2SO4) at 24± C on thecritical current density of 316 (UNS S31600) stainless steel at 24oC.

%CH2SO40 40 45 55 65 75

ic¡mA=cm2

¢4:7 1:6 1:4 1:0 0:7 0:4

CR:icorr(mm=y) 0 2:2 5:6 8:9 7:8 6:7

(a) Do non-linear curve …tting on ic versus %CH2SO4and plot ic = f (%CH2SO4

).(b) Assume that Fe, Cr, Mo and Mn oxides simulteneously go into solution.Now calculate CR for these four elements using the given experimental ic values,do curve …tting on the new CR data set and the experimental CR:icorr

valuesversus %CH2SO4

. Plot the CR functions for Fe, Cr, Mo and Mn, and plotthe experimental CR:icorr

data ant its linear curve …t equation in order to com-pare the corrosion behavior of the stainless steel at icorr and these elements ic.Explain.

Solution:

(a) Non-linear curve …tting. Let y = ic and x = %CH2SO4

ic = 4:6924 ¡ 9:8094 £ 10¡ 2C + 5:5224 £ 10¡ 4C2

with a correlation coe¢cient R2 = ¡0:97. The plot is


806040200

5

3.75

2.5

1.25

0

x

y

Actually, Aw has been used as the atomic weight of an element, but now Aw

becomes the molecular weight of the alloy.

Element Fe C Cr Mn Mo Ni Si P S% 62 0.08 18 2 3 14 1 0.045 0.03

Aw =1

100

XfiAw;i =

1

100(62 ¤ 55:845 + 0:08 ¤ 12 + 18 ¤ 51:996 + ::::)

Aw = 75:684g=cm3

Oxidation of iron (Fe):Aw;Fe = 55:85 g=mol ½Fe = 7:857 g=cm3 Fe ! Fe2+ + 2e¡ with

z = 2

CR;Fe;ic =icAwF e

zF½Fe

=(55:85 g=mol) (ic)

(2) (96500 A:s=mol) (7:857 g=cm3)

CRF e =icAwFe

zF½Fe=

(55:85)(0:4)(10¡3)(2)(96500 )(7:857) (10) (60 ¤ 60 ¤ 24 ¤ 365) = 4:6460

%CH2SO40 40 45 55 65 75

ic¡mA=cm2

¢4:7 1:6 1:4 1:0 0:7 0:4

CR;Fe:ic(mm=y) 54.59 18.58 16.26 11.62 8.13 4.65

Polynomial …t: yFe = 6:4166 £ 10¡ 3x2 ¡ 1:1395x + 54:501 with R2 = ¡0:97

Oxidation of chromium (Cr):


Aw;Cr = 52 g=mol ½Cr = 7:19 g=cm3 Cr ! Cr3+ + 3e¡ with z = 3

CR;Cr;ic =icAwCr

zF½Cr

=(52 g=mol) (ic)

(3) (96500 A:s=mol) (7:19 g=cm3)

CRFe=

(52)(0:4)(10¡3)(3)(96500 )(7:19) (10) (60 ¤ 60 ¤ 24 ¤ 365) = 3:1513

%CH2SO40 40 45 55 65 75

ic¡mA=cm2

¢4:7 1:6 1:4 1:0 0:7 0:4

CR;Cr:ic(mm=y) 37.03 12.61 11.03 7.88 5.51 3.15

Polynomial …t: yCr = 4:5731 £ 10¡ 3x2 ¡ 0:772 92x + 36:862 with R2 = ¡96

Oxidation of molybdenum (Mo):Aw;Mo = 95:96 g=mol ½Mo = 10:22 g=cm3 Mo ! Mo3+ + 2e¡ with

z = 3

CR;Mo;ic=

icAwMo

zF½Mo

=(95:96 g=mol) (ic)

(3) (96500 A:s=mol) (10:22 g=cm3)

CRFe=

(95:96)(0:4)(10¡3)(3)(96500 )(10:22) (10) (60 ¤ 60 ¤ 24 ¤ 365) = 4:0913

%CH2SO40 40 45 55 65 75

ic¡mA=cm2

¢4:7 1:6 1:4 1:0 0:7 0:4

CR;Mo:ic (mm=y) 48.07 16.37 14.32 10.23 7.16 4.09

Polynomial …t: yMo = 5:3368£10¡ 3x2 ¡0:967 2x+47:000 with R2 = ¡0:98

Oxidation of manganese (Mn):Aw;Mn = 54:94 ½Mn = 7:44 g=cm3 Mn ! Mn2+ + 2e¡ with z = 2

CR;Mn;ic=

icAwMn

zF½Mn

=(54:94 g=mol) (ic)

(2) (96500 A:s=mol) (10:22 g=cm3)

CRFe=

(54:94)(0:4)(10¡3)(2)(96500 )(7:44) (10) (60 ¤ 60 ¤ 24 ¤ 365) = 4:8264

%CH2SO40 40 45 55 65 75

ic¡mA=cm2

¢4:7 1:6 1:4 1:0 0:7 0:4

CR;Mn:ic (mm=y) 56.71 19.31 16.89 12.07 8.45 4.83

Plots: y = CR;icand x = %CH2SO4

. Red dot are experimental data fromthe icorr point of the polarization curves. So ic > icorr at CH2SO4 < 55%.

Red (2): yFe = 6:4166 £ 10¡ 3x2 ¡ 1:1395x + 54:501Green (4): yCr = 4:5731 £ 10¡ 3x2 ¡ 0:772 92x + 36:862Purple (3): yMo = 5:3368 £ 10¡ 3x2 ¡ 0:967 2x + 47:000Black (1 Top Curve): yMn = 6: 6637 £ 10¡ 3x2 ¡ 1:1836x + 56:618Experimental: yCR;exp = 0:148 87x ¡ 5:1658 £ 10¡ 4x2 ¡ 0:32656

with R2 = 0:85


806040200

50

37.5

25

12.5

0

x

y

The above …gure shows that the AISI 316 stainless steel is a corrosion resis-tant alloy, but the curves for Fe, Cr, Mo and Mn indicate that these elementsoxide very rapidly at CH2SO4

< 55%. Beyond this concentration the experi-mental data and the calculated CR values are similar. In other words, whenCH2SO4

¸ 55% the icorr and ic position on the polarization curves are not farapart on the normal log scale (x-axis).

8.15 This problem deals with the nature of the anodic oxide …lm that formson titanium in 0:9% NaCl. This media is isotonic with human blood for surgicalimplants made out of titanium and its alloys. Normally, the passive oxide …lmon titanium mainly consists of TiO2. The table given below contains data forthe passive current density (ip) as a function of sweep rate (de/dt in mV/s) [52]on titanium in 0:9% NaCl containing electrolyte. (a) Convert the ip data tocorrosion rate CR in mm=y. Then plot both ip and CR as functions of sweeprate v = dE=dt. (b) Explain if this titanium material would be suitable for ahuman transplant for a prolong time.

v = dE=dt (mV=s) 10 50 100 300 500ip

¡mA=cm2

¢0:13 0:25 0:50 1:58 2:10

Solution:The oxidation reaction for titanium is Ti ! Ti2++2e¡ @ Eo = 1:630 VSHE.

The corrosion rate is

CR =ipAw

zF½


For Aw = 47:867 g=mol and ½ = 4:54 g=cm3,

Aw

zF½=

47:867 g/mol

(2) (96500 A.s/mol)³4:54 g/cm3

´ = 5:4629 £ 10¡ 5 cm3

A:s

Thus,

CR =

µ

5:4629 £ 10¡ 5 cm3

A:s

¶

ip

v = dE=dt (mV=s) 10 50 100 300 500ip

¡mA=cm2

¢0:13 0:25 0:50 1:58 2:10

CR

¡£10¡ 8 mm=y

¢7:10 13:66 27:32 86:31 114:72

Let y = ip and x = v. The plot for ip = f (v) is

6255003752501250

3

2.5

2

1.5

1

0.5

0

x

y

Curve …tting gives

ip =

µ

0:09932mA

cm2

¶

+

µ

4:2327 £ 10¡ 3 mA

cm2

s

mV

¶

v

with a correlation coe¢cient R2 = 0:99.

Let y = CR and x = v. The plot for CR = f (v) is


v = dE=dt (mV=s) 10 50 100 300 500ip

¡mA=cm2

¢0:13 0:25 0:50 1:58 2:10

CR

¡£10¡ 8 mm=y

¢7:10 13:66 27:32 86:31 114:72

Polynomial …t: y = (0:23122x + 5:4276)¡10¡ 8 mm=y

¢with a correlation

coe¢cient R2 = 0:99

6255003752501250

150

125

100

75

50

25

0

x

y

x

y

These results indicate that a titanium transplant is suitable for a humanbeing. For instance, the lowest and highest corrosion rates are

CR;min = (0:23122 (10) + 5:4276)¡10¡ 8

¢= 7:74 £ 10¡ 8mm=y

CR;max = (0:23122 (500) + 5:4276)¡10¡ 8

¢= 1:21 £ 10¡ 6mm=y

The corrosion rates are extremely low and therefore, the titanium transplantwould not cause detrimental health e¤ects in humans.

Chapter 9

ELECTRODEPOSITION

9.1 PROBLEMS

9.1 Copper cations are reduced on a cathode for 8 hours at a current of 10amps. and 25oC: Calculate a) the theoretical weight of copper deposited on thecathode and b) the number of coulombs of electricity in this electrodepositionprocess. Data: Aw;Cu = 63:55 g=mol and F = 96; 500 C:

Solution:Cu+2 + 2e = Cu

a) From eq. (9.8g) with F = 96; 500 C = 96; 500 A:s=mol

W =ItAw

zF=

(10 A) (8x60x60 s) (63:55 g=mol)

(2) (96; 500 A:s=mol)

W = 94:83 grams @ 100% efficiency

b) From eq. (9.8b),

Q = It = (10 A) (8x60x60 s)

Q = 288; 000 C (= A:s)

9.2 An electrochemical cell operates at 6 A, 25oC and 85% current e¢ciencyin order to electrolytically deposit copper ions (Cu+2) from a leaching solutionin a 8 ¡ hour shift. Calculate a) the amount of electrolytically deposited Cuand b) the thickness of the deposit layer of copper. Data: As = 100 cm2 (totalcathode surface area), Aw;Cu = 63:55 g=mol, ½ = 8:96 g=cm3 and F = 96; 500C=mol:

Solutions:

Cu+2 + 2e = Cu

125

126 CHAPTER 9. ELECTRODEPOSITION

a) From eq. (9.9) with F = 96; 500 C = 96; 500 A:s=mol

W =²ItAw

zF=

(0:85) (6 A) (8x60x60 s) (63:55 g=mol)

(2) (96; 500 A:s=mol)

W = 48:36 grams

b) Letting m = W = 48:36 grams yields

V = xAT =m

½

x =m

½AT=

48:36 g

(8:96 g=cm3) (100 cm2)

x = 0:54 mm

9.3 It is desired to electroplate chromium (Cr) onto a ferritic-martensiticcarbon steel automobile bumper for an attractive appearance and corrosionresistance. An electroplating cell is operated at I = 6:68 A and 25oC. TheCr+3 cations are reduced on the bumper (cathode) to form a thin …lm of 1:5 ¹mthick. How long will it take to produce a 1:5 ¹m thick electroplated chromium…lm on the bumper if the total surface area is 100 cm2 and ² = 60%?

Solution:

Cr+3 + 3e = Cr

a) Using eq. (9.9) with F = 96; 500 C = 96; 500 A:s=mol; ½ = 7:20 g=cm3

and Aw = 52 g=mol yields

W =²ItAw

zF=

(0:85) (6 A) (8x60x60 s) (63:55 g=mol)

(2) (96; 500 A:s=mol)

W = V ½ = Asx½ =¡100 cm2

¢ ¡1:5x10¡ 4 cm

¢ ¡7:20 g=cm3

¢

W = 0:1080 g

t =zFW

²IAw=

(2) (96; 500 A:s=mol) (0:1080 g)

(0:60) (6:68 A) (52 g=mol)

t = 100 sec

9.4 An electrowinning cell contains 2x10¡ 4 mol=cm3 of Cu+2 ions andoperates for 10 minutes at 40oC and 1 atm. Assume an ionic copper di¤usivityequals to 2:34x10¡ 5 cm2=s. Calculate the current density due to di¤usion masstransfer. Compare your result with a typical industrial current density value of200 A=m2. Explain any discrepancy.

Solution:

Cu+2 + 2e = Cu

9.1. PROBLEMS 127

From eq. (9.68),

Jd =DCop¼Dt

= Co

rD

¼t

Jd =¡2x10¡ 4mol=cm3

¢s

(2:34x10¡ 5cm2=s)

(¼) (600 s)

Jd = 2:23x10¡ 8 mol

cm2:s

From eq. (9.71),

id = zFJd = (2) (96; 500 A:s=mol)

µ

2:23x10¡ 8 mol

cm2:s

¶

id = 4:30x10¡ 3 A=cm2 ' 43 A=m2

Obviously, the assume di¤usion mechanism for reducing copper ions is justpart of the total mass transfer. In fact, id = 43 A=m2 represents approximately22% of the total current density. If the electrolyte is continuously replenishedwith fresh electrolyte, then the remaining 78% current density has to come frommigration and convection mass transfer.

The total molar ‡ux using 200 A=m2 current density is

Jx =i

zF=

200x10¡ 4 A=cm2

(2) (96; 500 A:s=mol)

Jx = 1:04x10¡ 7 mol

cm2:s

Thus, the migration and convection molar ‡ux becomes

Jm + Jc = Jx ¡ Jd = 8:13x10¡ 8 mol

cm2:s

9.5 Calculate the amount of silver that can be electroplated in an electro-chemical cell containing Ag+ ions and operates at 7 A, 25oC and 101 kPa for10 minutes. Assume a current e¢ciency of 95%.

Solution:

Ag+ + e = Ag

From eq. (9.9),

W =²ItAw

zF=

(0:95) (7 A) (10x60 s) (107:87 g=mol)

(1) (96; 500 A:s=mol)

W = 4:46 grams


9.6 What are the di¤erences between galvanic and electrolytic cells? Recallthat both are electrochemical cells.

Answer: Figure 2.2 shows the di¤erences between these cells. However,in a galvanic cell, a chemical reaction generates an electric current ‡ow andconsequently,chemical energy is converted into electrical energy and the cathodeis the positive electrode. On the other hand, in an electrolytic cell, electricalenergy is converted into chemical energy and the cathode is the negative electrodesince the supply of electrons comes from an external source, such as a galvaniccell or dc generator.

9.7 Consider the bipolar electrode for electrore…ning metal M . Thus, therate of formation (reduction) and dissolution (oxidation) are treated as steady-state quantities. Despite that the initial formation of a thin …lm at the cathodeface and dissolution rates are controlled by reactions at the electrode-electrolyteinterfaces, assume a steady-state di¤usion mechanism. Derive expressions forthe weight gain at the cathode side and the weight loss at the anode side.

Solution:

Ele

ctro

de

JM JM+2

xx'

Reduction OxidationJM = ¡D@C=@x0 JM+2 = ¡D@C=@x

JM = ¡D0³C

0

o ¡ Cx0

´=x0 JM+2 = ¡D

³C

0

x ¡ Co

´=x

Co > Cx0 Cx > Co

W0= V 0½0= x0A0½0 W = V ½ = xA½

Eliminating x0and x yields

W0= A0½0D0³C

0

o ¡ Cx0

´=JM W = A½D

³C

0

x ¡ Co

´=JM+2

According to the given conditions, W0= W since A0= A, ½0= ½ andJM = JM+2 .

9.8 A steel plate is Ni-plated for corrosion protection, but its appearance isnot so appealing. Therefore, a chromium electroplating process is carried out fordecorative purposes. Eventually, 0:25 g of Cr+6 cations are electroplated on the

9.1. PROBLEMS 129

Ni-plated surface for a period of 15 minutes at 8 amps. If the density and atomicweight of chromium are 7:19 g=cm3 and 52 g=mol, respectively, calculate (a)the thickness of the electroplated …lm of chromium from its highest oxidationstate and (b) the deposition rate. Data: As = 100 cm2 (cathode area).

Solution:

H+

Cr+6 + 6e¡ = Cr

(a) Thickness:

x =W

As½=

0:25 g

(100 cm2) (7:19 g=cm3)

x = 3:48 ¹m

(b) Production rate:

PR =W

t=

0:25 g

900 s= 2:78x10¡ 4 g=s

PR = 1 g=h

(c) From eq. (9.16),

² =zFW

ItAw

² =(6)

¡96; 500 A:s

mol

¢(0:25 g)

(8 A) (900 s) (52 g=mol)= 0:39

² = 39%

This is a low, but typical current e¢ciency in Cr electroplating.

9.9 It is desired to produce a 20 ¹m thick …lm of chromium on a Ni-platedsteel part, which has a surface area of 65 cm2. This can be accomplished bysetting up an electroplating cell to operate at 7 amps and current e¢ciencyof 70%. The density and atomic weight of chromium are 7:19 g=cm3 and 52g=mol, respectively. Calculate a) the amount of Cr being plated and b) the timeit takes to plate the 20 ¹m thick …lm of chromium from a solution containingCr+6 cations at 30oC.

Solution:

Cr+6 + 6e¡ = Cr

a) Amount of Cr:

W = V ½ = ½xAs

W =¡7:19 g=cm3

¢ ¡20x10¡ 4 cm

¢ ¡65 cm2

¢

W = 0:94 g


b) From eq. (9.16),

t =zFW

²IAw

t =(6)

¡96; 500 A:s

mol

¢(0:94 g)

(0:70) (7 A) (52 g=mol)= 2; 124:67 sec

t = 35:40 min

9.10 Predict (a) the electric potential E, (b) the time in minutes, and (c)the …lm growth rate in ¹m=min for electroplating a 10 ¹m-thick Cr …lm ona Ni-undercoated steel part when the electrolyte contains 10¡ 4 mol=l of Cr+3

cations at 25oC. The Ni-plated steel part has a 10 cm2 surface area. The celloperates at 50% current e¢ciency and at a passive current density of 5:12x10¡ 3

A=cm2.

Solution:

Cr+3 + 3e = Cr EoCr = ¡0:744 V z = 3

½ = 7:19 g=cm3 Aw;Cr = 52 g=molAs = 10 cm2 ² = 0:50 aCr+3 = 10¡ 4 mol=l

(a) Using Nernst equation yields

E > Eo = Eo +RT

zFln (aCr+3)

E > Eo = ¡:0744V +(8:314 J=mol:K) (298 K)

(3) (96; 500 J=mol:V )ln

¡10¡ 4

¢

E > Eo = ¡0:82 V

(b) Integrating eq. (7.21) yields the time

t =

Z x

0

µzF½

²iAw

¶

dx =zF½x

²iAw

t =(3) (96; 500 A:s=mol)

¡7:19 g=cm3

¢ ¡10¡ 4 cm

¢

(0:50) (5:12x10¡ 3 A=cm2) (52 g=mol)

t = 26 min

(c) The …lm growth rate is

dx

dt=

¢x

¢t=

10 ¹m

26 mindx

dt= 0:38

¹m

min

9.1. PROBLEMS 131

9.11 The dissociation of silver hydroxide, AgOH, is 1:10x10¡ 4 at 25 ± C inan aqueous solution. (a) Derive an expression for the degree of dissociation as afunction of total activity [CT ] and plot the resultant expression for 0 < [CT ] <0:50 mol=l, (b) calculate the degree of dissociation constant ®1 when the molarconcentration is [CT ] = 0:144 mol=l, and (c) determine the Gibbs free energychange ¢Go.

Solution:

a) The expression and the plot are

AgOH = Ag+ + OH¡

Ke =[Ag+] [OH ¡ ]

[AgOH¡ ]= 1:10x10¡ 4

From eq. (2.67) with x = y = 1;

®1 =¡Kex

¡ xy¡ y¢1=(x+y)

[CT ]¡ 1+1=(x+y)

= K1=2e [CT ]

¡ 1=2

®1 =¡1:05x10¡ 2

¢[CT ]

¡ 1=2

0.50.3750.250.1250

0.25

0.2

0.15

0.1

0.05

0

)(mol/literCT

1

b) The degree of dissociation:

®1 =¡1:05x10¡ 2

¢[0:144]

¡ 1=2= 0:028

c) The free energy change is


¢Go = ¡RT ln (Ke) = ¡

µ

8:314J

mol:oK

¶

(298oK) ln¡1:1x10¡ 4

¢

¢Go ¼ ¡22:58 kJ=mol

Therefore, the reaction as written will proceed since ¢Go < 0.

9.12 Calculate (a) the concentration of AgOH, Ag+ and OH ¡ at 25 ± Cand (b) the pH if [CT ] = 0:144 mol=l and ®1 = 0:028.

Solution:

a) AgOH = Ag+ + OH ¡ and K = 1:10x10¡ 4 (From Example 2.11). Let[Ag+] = [OH ¡ ] for charge balance so that

Ke =[Ag+] [OH ¡ ]

[AgOH]=

[OH ¡ ]2

[AgOH](a)

£OH ¡

¤=

pK [AgOH] (b)

The dissociation of water ant its rate constant are

H2O = H+ + OH ¡ (c)

Kw =[H+] [OH¡ ]

[H2O]= 10¡ 14 (d)

From eq. (2.64a) along with [H2O] = 1 mol=l,

£OH ¡

¤= x®1 [CT ] = (1) (0:028) [0:144 mol=l] = 4:03x10¡ 3 mol=l (e)

COH¡ =£OH ¡

¤Aw;OH =

µ

4:03x10¡ 3 mol

l

¶³17

g

mol

´= 6:85x10¡ 2 g

l(f)

£Ag+

¤=

£OH ¡

¤= 4:03x10¡ 3 mol=l (g)

CAg+ =£Ag+

¤Aw;Ag =

µ

4:03x10¡ 3 mol

l

¶³107:87

g

l

´= 0:43

g

l(h)

From eq. (2.64c),

[AgOH] = (1 ¡ ®1) [CT ] = 0:14 mol=l

CAgOH = [AgOH]Aw;[AgOH] = (0:14 mol=l) (124:87 g=mol)

CAgOH = 17:48 g=l

From eq. (d),

9.1. PROBLEMS 133

£H+

¤=

Kw

[OH ¡ ]=

10¡ 14

4:03x10¡ 3 mol=l= 2:48x10¡ 12 mol=l

pH = ¡ log£H+

¤= ¡ log

¡2:48x10¡ 12

¢= 11:61

9.13 An electrochemical cell contains 75 g=l of Cu+2 ions at 35oC. Calcu-late (a) the Reynold and Sherwood numbers,(b) the di¤usion ‡ux (Jx), (c) thecurrent density (i) and (d) the the applied potential (E). Data:

d Characteristic distance = 12 cm .L = x Cathode height = 100 cm .

w Cathode width = 80 cm .N Number of cathodes = 50 .Kv Kinematic viscosity = 0:80 cm2=s .D di¤usivity = 1:24x10¡ 5 cm2=svx Flow velocity = 46:90 cm=s .P Electrical Power = 32 kW .

Solution:

Cu+2 + 2e = Cu z = 2CCu+2 = (75g=l) = (63:55 g=mol) = 1:18 mol=CCu+2 = 1:18x10¡ 3 mol=cm3

² = 0:85 F = 96; 500 Aa:s=mol

(a) From eq. (9.4), (7.5) and (7.25), the Reynold and Sherwood numbers are

Re =vxd

Kv=

(46:90 cm=s) (12 cm)

(0:80 cm2=s)= 703:50

Sc =Kv

D=

0:80 cm2=s

1:24x10¡ 5 cm2=s= 64; 516:13

Sh =2

3(Sc)

1=3(Re)

1=2= 709:19

(b) The di¤usion ‡ux Jx:

Jx = ¡D@C

@x= ¡

DCCu+2

x=

Jx =

¡1:24x10¡ 5 cm2=s

¢ ¡1:18x10¡ 3 mol=cm3

¢

100 cm

Jx = 1:46x10¡ 10 mol

cm2:s

(c) From (7.13), the current density is


i = zFJxSh

i = (2)

µ

96; 500A:s

mol

¶µ

1:46x10¡ 10 mol

cm2:s

¶

(709:19)

i = 2x10¡ 2 A=cm2 = 200 A=m2

(d) From elementary physics, the electrical power divided by the current isthe applied potential. Thus,

Ac = wL = (80 cm) (100 cm) = 8; 000 cm2

As = 2NAc = (2) (50)¡8; 000 cm2

¢

As = 800; 000 cm2 = 80 m2

I = iAs =¡200 A=m2

¢ ¡80 m2

¢= 16 kA

E =P

I=

32 kW

16 kA=

32 kV=A

16 kAE = 2 volts

9.14 An electrolyte containing Pb+2 ions at 35oC is used in an electrowin-ning cell. The electrolyte is under steady laminar force convection. Use thedata given below to determine (a) the the Sherwood number, (b) the di¤usion‡ux and (c) the current.

Cb Bulk concentration = 75 g=l .D di¤usivity = 1:24x10¡ 5 cm2=s

L = x Cathode length = 100 cmSc Schmidt Number = 64; 516:13 .Re Reynolds Number = 703:50 .As Total cathode area = 800; 000 cm2

² Current e¢ciency = 85%

Solution:

Pb+2 + 2e = Pb

Cb = (75 g=l) = (207:19 g=mol)

Cb = 0:362 mol=l = 3:62x10¡ 4 mol=cm3

(a) From eq. (9.25), the Sherwood number is

Sh =2

3(Sc)

1=3(Re)

1=2

Sh =2

3(64; 516:13)

1=3(703:50)

1=2= 709:19

9.1. PROBLEMS 135

(b) The di¤usion ‡ux is

Jx = ¡D@C

@x= ¡

DCb

x=

Jx =

¡1:24x10¡ 5 cm2=s

¢ ¡3:62x10¡ 4 mol=cm3

¢

100 cm

Jx = 4:49x10¡ 11 mol

cm2:s

(c) The current is

i = zFJxSh

i = (2)

µ

96; 500A:s

mol

¶µ

4:49x10¡ 11 mol

cm2:s

¶

(709:19)

i = 6:14x10¡ 3 A=cm2

I = iAs =¡6:14x10¡ 3 A=cm2

¢ ¡800; 000 cm2

¢

I = 4:91 kA

9.15 A hypothetical rotating-disk cell is shown below for electrowinningcopper cations from a solution containing Co = 65 g=l at 40oC and 101 kPa.Assume that the di¤usivity and the electrolyte kinematic viscosity are 10¡ 5

cm2=s and 0:60 cm2=s, respectively. Each disk has a radius of 50 cm; a widthof 6 cm, and only a 160o segment is immersed in the electrolyte. Assume a cellcurrent e¢ciency range of 0:50 · ² · 1:

Theoretically, analyze the e¤ect of a) the angular velocity ! on the currentdensity and b) the angular velocity (!) and the current density (²) on theproduction rate of metal powder, and c) the potential (E) and the currentdensity (²) on the energy consumption °.

Solution:

Data: Cu+2 + 2e = Cu @ T = 40oC


Co = 65 g=l =¡65x10¡ 3 g=l

¢= (63:55 g=mol) Aw = 63:55 g=mol

Co = 1:02x10¡ 3 mol=cm3 r = 50 cmD = 10¡ 5 cm2=s Kv = 0:60 cm2=sµ = 160o w = 6 cm

Total e¤ective surface area:

h1 = r cos (µ=2) = 8:68 cm L = r sin (µ=2) = 49:24 cmh2 = r = h1 = 41:32 cm S = ¼rµo=180o = 139:63 cm

Aseg = r2

2

¡¼µo

180o

¢

Aseg =r2

2

µ¼µo

180o¡ sin (µ)

¶

= 3; 063:13 cm2

Aarc = wL = 295:44 cm2

As = 4 (2Aseg + Aarc) = 25; 686:80 cm2 ' 2:57 m2

a) From eq. (9.107),

i =0:62zFCoD

2=3p

!

K1=6

i =(0:62) (2) (96; 500 A:s=mol)

¡1:02x10¡ 3 mol=cm3

¢ ¡10¡ 5 cm2=s

¢2=3 p!

(0:60 cm2=s)1=6

i =

µ

616:87A:s1=2

m2

¶p

!

i(A/m2)

ω (Rad/sec)

0

1000

2000

3000

4000

5000

6000

7000

8000

20 40 60 80 100

9.1. PROBLEMS 137

0100200300400

0.5 0.6 0.7 0.8 0.9

0

2

4

6

8

10

12

ε ω (Hz)

PR

(g/s)

Figure 9.1:

b) From 7.16,

PR =²iAsAw

zF=

³616:87 A:s1=2

m2

´AsAw²

p!

zF

PR =

³616:87 A:s1=2

m2

´ ¡2:57 m2

¢(63:55 g=mol) ²

p!

(2) (96; 500 A:s=mol)

PR =³0:52 g:s1=2=s

´²p

!

0

2

4

6

8

10

12

100 200 300 400 500

ω (Hz)

PR

(g/s)

ε

0.7

0.8

0.91


c) From eq. (9.18),

° =zF

Aw

E

²

° =(2) (96; 500 A:s=mol)

(63:55 g=mol)

E

²

° =

µ

3; 036:98A:s

g

¶E

²

° =

µ

0:84kW:h

Kg

¶E

²where E is in volts

0

1

2

3

4

5

6

7

8

0.5 0.6 0.7 0.8 0.9 1

γ (kW.h/Kg)

ε

E (V)

2

3

4

The energy consumption does not depend directly on the angular velocity,but on the applied potential and current density, which in turn, depends on theapplied current density.

9.16 Use the hypothetical rotating-disk voltammetry data given in the tablebelow for determining the di¤usivity D of a metal cation M+2 in solution atroom temperature.

! (1= sec) 100 225 400 625 900 1; 225

!1=2¡s1=2

¢10 15 20 25 30 35

i¡mA=cm2

¢4:81 7:22 9:58 12:10 14:38 16:80

Co = 2:21x10¡ 6 mol=cm3

Kv = 0:056 cm2=sz = 2

9.1. PROBLEMS 139

Solution:

Linear least squares on the given data, i vs: !1=2, yields the Levich plot

i = ¡0:488 57mA:cm¡ 2 +³0:49771 mA:s1=2:cm¡ 2

´!1=2

0

2

4

6

8

10

12

14

16

18

20

10 20 30 40

ω1/2 (Hz)

i (mA/cm2)

From eq. (9.107),

i =³0:62zFCoD

2=3K¡ 1=6v

´p!

Slope =³0:62zFCoD

2=3K¡ 1=6v

´

D =

ÃSlope

0:62zFCoK¡ 1=6v

!3=2

D =

"0:49771x10¡ 3 A:s1=2:cm¡ 2

(0:62) (2)¡96; 500 A:s

mol

¢ ¡2:21x10¡ 6 mol

cm3

¢(0:056 cm2=s)

¡ 1=6

#3=2

D = 3:97 £ 10¡ 5cm2=s

9.17 It is desired to produce copper according to the electrolytic cell diagram

(¡)SteeljCu+2=Cu; H+=H2

¯¯ jH2O=O2 jPbO2 (+)

for 24 hours at which time the cathodes are mechanically stripped (removed).Copper is reduced from an aqueous copper sulfate (CuSO4) and sulfuric acid


(H2SO4) solution at 35oC. The anodes operate at 100% current e¢ciency withrespect to oxygen evolution and that for the cathodes is ² = 90% due to hydrogenevolution. These type of electrodes are connected in parallel within 100 tanksthat operate at 2:4 volts each and the electrodes have a speci…c submergedarea Ae. Use the following relevant data to perform some calculations givenbelow. Cost of electricity per kW.h: Ce = 0:30 $=kW:h, Market price of copper:Cmarket = $0:70=Kg, and

P = 72 kW (Power) Aw = 63:54 g=molE = 2:4 V DCu+2 = 1:24x10¡ 5 cm2=sNT = 100 (Tanks) Kv = 0:80 cm2=sNa = 30 (Anodes per tank) vx = 0:312 cm=s (Velocity of Cu+2)Nc = 29 (Cathodes per tank) d = 6 cm (electrode distance)Ae = 0:80 m2 (Electrode area) Ce = 0:30 $=kW:h (Electricity)

(a) Write down all the reactions and the standard half-cell potential involvedin this electrowinning process. What should the applied potential be? Calculate(b) the total current, the anodic and cathodic current densities, and (c) Sc, Re,Sh and Cb in g=l per tank. Let the anode characteristic length be L =

pAe.

Use the Sherwood number (Sh < 100) and the Reynolds number (Re < 2; 300)to assure force laminar ‡ow. Determine (d) the production rate (PR) and thetotal weight produced (WT ) in 24 hours, (e) the energy consumption (°). Useyour engineering economics skills to estimate (f) the cost of energy consumed(Cte), the total cost of production (Cprod) and the gross income (GI) in 24hours if the industrial cost of electric energy and the market price of copperare Ce = $0:30 kW:h and Cmarket = $3:193=lb = $7:041=Kg, respectively.[Solutions: (a) Eo

Half ¡ cell ¼ ¡1:57 V , (b) ic = 0:001293 A=cm2, (c) Sh ¼ 39and Cb = 78:80 g=l].

Solution:(a) The reactions, the standard half-cell potential and the applied potential

are

Anode: 2H2O ) 4H+ + O2 + 4e

Cathode: Cu+2 + 2e ) Cu

Cathode: 2H+ + 2e ) H2

Overall : 2H2O + Cu+2 + 2H+ ) Cu + 4H+ + O2 " +H2 "

EoHalf ¡ cell = Eo

H2O + EoCu + Eo

H2= ¡1:23 V ¡ 0:337 V + 0

EoHalf ¡ cell ¼ ¡1:57 V

E ¸¯¯Eo

Half ¡ cell

¯¯

(b) The total current

9.1. PROBLEMS 141

I =P

E=

72 £ 103 AV

2:4 V= 3:0 £ 104 A

Electrode areas per tank :

Aa = NT NaAe = (100) (30)¡8; 000 cm2

¢= 24:0x106 cm2

Ac = NT NcAe = (100) (29) (8; 000 cm)2

= 23:2x106 cm2

Current densities per tank :

ia =I

Aa=

3:0 £ 104 A

24:0x106 cm2= 0:001250 A=cm2 = 1:250

A

m2

ic =I

Ac=

3:0 £ 106 A

23:2x106 cm2= 0:001293 A=cm2 = 1:293

A

m2

(c) Assume force laminar ‡ow. The Schmidt (Sc), Reynolds (Re) numberand the anode length are

Sc = Kv=D =¡0:80 cm2=s

¢=¡1:24x10¡ 5 cm2=s

¢= 64:52x103

Re = vxd=Kv = (0:312 cm=s) (6 cm) =¡0:80 cm2=s

¢= 2:34

From eq. (9.101) along with the anode characteristic length (L), the Sher-wood number is

L =p

Ae =p

0:8 m2 = 0:8944 m2 = 89:44 cm2

Sh =

µ9

5

¶µvxd2

LD

¶1=3

=

µ9

5

¶"(0:312 cm=s) (6 cm)

2

(89:44 cm) (1:24x10¡ 5 cm2=s)

#1=3

Sh ¼ 39

or from eq. (9.97)

Sh =

µ9

5

¶·d

L(ScRe)

¸1=3

Sh =

µ9

5

¶·µ6

89:44

¶¡64:52 ¤ 103

¢(2:34)

¸1=3

¼ 39

Using eq. (9.97) for one tank yields the cathodic current density as thelimiting current density ( ic = iL) and the bulk concentration (Cb) in g=l pertank.


ic =9zFCb

5

µvxd2D2

L4

¶1=3

Cb =5ic9zF

µL4

vxd2D2

¶1=3

Cb =(5)

¡0:001293 A=cm2

¢

(9) (2)¡96; 500 A:s

mol

¢

"(89:44 cm)

4

¡0:312 cm

s

¢(6 cm)

2 ¡1:24x10¡ 5 cm2

s

¢2

#1=3

Cb = 1:24 £ 10¡ 3 mol=cm3 = 1:24 mol=l

Cb = (1:24 mol=l) (63:55 g=mol) ' 78:80 g=l

Alternatively, using eqs. (9.94) and (7.87b) with ¢C = Cb gives per tank

Sh =

µ2

3

¶

(Sc)1=3

(Re)1=2

=

µ2

3

¶¡64:52x103

¢1=3(2:34)

1=2¼ 41

Cb =icL

zFDSh=

¡0:001293 A

cm2

¢(89:44 cm)

(2)¡96; 500 A:s

mol

¢ ¡1:24 £ 10¡ 5 cm2

s

¢(41)

Cb = 1:18x10¡ 3 mol=cm3 = 1:18 mol=l

Cb = (1:18 mol=l) (63:55 g=mol) = 75 g=l

Both equations give fairly similar results. Let’s use the results given by eq.(4.94) in eq. (9.87b).

(d) Production rate (PR) and total weight produced (WT ) in 24 hours:

PR =²icAcAw

zF=

(0:90)¡0:001293 A

cm2

¢ ¡23:2x106 cm2

¢ ¡63:55 g

mol

¢

(2) (96; 500 A:s=mol)

PR = 8:8897 g= sec = 32:003 Kg=h = 768:07 Kg=day (Total)

WT = (32:003 Kg=h) (24 h) = 768:07 Kg (Total)

wt = WT =NT = (768:07 Kg) =100 = 7:68 Kg (Per tank)

wc = (7:68 Kg) =29 = 0:26483 Kg = 264:83 grams (Per cathode)

The energy consumption is

° =P

PR=

72 kW

32:003 Kg=h= 2:25 kW:h=Kg (Total)

° = 0:0225 kW:h=Kg (Per tank)

° = 7:76x10¡ 4 kW:h=Kg (Per cathode)

9.1. PROBLEMS 143

If the cost of electricity per kW.h is Ce = 0:30 $=kW:h, then the cost ofenergy consumed is

Cte = Ce°WT = (0:30 $=kW:h) (2:25 kW:h=Kg) (768:07 Kg)

Cte = $518:45 (Total)

Ct = Cte=NT = ($518:45) =100 = $5:1845 (Per tank)

Cc = Cte= (Nc) = ($518:45) =29 = $17:88 (Per Cathode)

Thus, the cost of production is

Cprod =Cte

WT=

$518:45

768:07 Kg

Cprod = $0:68=Kg

If the market price is Cmarket = $3:193=lb = $7:041=Kg, then the grossincome in 24 hours is

GI = (Cmarket ¡ Cprod) (WT ) = ($7:041=Kg ¡ $0:68=Kg) (768:07 Kg)

GI = $4; 885:70 (For a 24-hour run)

This result is rather low. What can be improved for a higher GI?

9.18 A Hall-Heroult (HH) cell is used to produce molten aluminum con-taining hydrogen. The hydrogen content can be removed by allowing a thin …lmof molten aluminum ‡ow down on an inclined plane during pouring in vacuum.The incline plane is shown below and it is in the L-w-± space, where ± is the…lm thickness and w is the width.

The weight-induced and gravity-induced free ‡uid ‡ow occurs in the x-direction. Using Newton’s law of viscosity yields the force acting on the ‡uidin the direction of the ‡ow, together with the moment balance. Thus, thisphenomenon is described by an ordinary di¤erential equation.


´v

d2vx

dy2= ¡½g cos (µ) with BC’s

½vx = 0 @ y = ±vx > 0 @ y > 0

¾

where ´v = Fluid viscosity, vx = Fluid velocity and ½ = Density of the‡uid: Solve this di¤erential equation and calculate the average a) value of theboundary layer, b) velocity vx, c) molar ‡ux of hydrogen due to both di¤usionand convection e¤ects, d) time it takes to remove the dissolved hydrogen alongthe channel length L and the amount of hydrogen being removed, and e) theReynolds number. Given data:

½ = ½Al = 2:5 g=cm3 w = 50 cm´v = 40 g:cm¡ 1s¡ 1 L = 100 cmDH = 5x10¡ 3 cm2=s µ = 2o

CH;b = 30 mol=cm3 ¢m = 100 g=s

CH = 0 @ y = 0 g = 981 cm=s2

Solution:

The solution of the above di¤erential equation is

dvx

dy= ¡

½g cos (µ)

´v

y

Z vx

0

dvx = ¡½g cos (µ)

´v

Z y

±

ydy

vx =½g cos (µ)

2´v

¡±2 ¡ y2

¢

vx;max =½g±2 cos (µ)

2´v

The average ‡uid velocity is

vx =1

±

Z ±

0

vxdy =1

±

Z ±

0

½g cos (µ)

2´v

¡±2 ¡ y2

¢dy

vx =1

±

·½g cos (µ)

2´v

¸ ·

±2y ¡y3

3

¸±

o

vx =½g±2 cos (µ)

3´v

vx =2

3vx;max

The cross-sectional of the ‡owing ‡uid is A = ±w and the average mass ‡owrate is

9.1. PROBLEMS 145

¢m = vx½A

¢m =

½g±3w cos (µ)

3´v

a) Thus, the average boundary layer becomes

± =

"3´v

¢m

½gw cos (µ)

#1=3

= 0:46 cm

b) The average ‡uid velocity is

vx =½g±2 cos (µ)

3´v

= 4:34 cm=s

c) The average molar ‡ux mathematical model is de…ned in Chapter 4. Usingeq. (4.9a) along with dÁ=dx = 0 gives

Jx = Jd + Jc = ¡DH@C

@x+ vxCH;b

Jx = ¡DHCs ¡ Cx

±+ vxCH;b

Using the boundary conditions Cs = 0 at y = 0 and Cx = CH;b at y = ±yields

Jx =DHCH;b

±+ vxCH;b

Jx = CH;b

µDH

±+ vx

¶

= 130:53mol

cm2s

d) The average time along the channel length is

t =L

vx= 23:04 sec

Then, the amount of hydrogen removed is

m = t¢m = 2; 304 g ¼ 2:30 Kg

e) Using eq. (9.78) yields the Reynolds number along with d = ±

Re =½vx±

´v

= 0:13

9.19 An electrowinning cell produces 10 Kg=h of nickel (Ni) at a currente¢ciency (²) of 80%: The cell contains 20 cathodes and operates at a current of


215 A=m2 and a potential of 2 V . Determine (a) the energy consumption, (b)the ‡ow rate of the electrolyte and (c) the cathode length for a width of 1 m.The concentration of nickel cations is 35 g=l. (d) Plot eq. (9.25) and explainthe result.

Solution:

Ni2+ + 2e = Ni @ E = 2 V and i = 215 A=m2 with z = 2

PR = 10 Kg=h = 104 g=h; N = 20; Aw;Ni = 58:71 g=mol

Co = 35 g=l = 5:96x10¡ 4 mol=cm3

(a) From, eq. (9.23),

As = zFPr= (²iAw) = 53:10 m2

I = iAs = 11; 416:20 A

P = EI = 22:83 kW

° = P=PR = 2:28 kW:h=Kg

(b) The ‡ow rate of the electrolyte:

Fr = Pr=Co = 4:76 l=min

(c) The cathode length:

Ac = As= (2N) = 1:3275 m2

L = Ac=w = 1:3275 m

(d) From eq. (9.25), the simpli…ed energy consumption equation becomes

° = 1:83=² (in kW:h=Kg)

which is plotted in …gure given below indicating that the energy consumptionis strongly dependent on the current e¢ciency since ° is inversely proportionalto ²: However, this strong dependency begins to have a lesser e¤ect when ² ¸ 0:25and when ² > 0:80 the energy consumption is signi…cantly reduced.

Let y = ° and x = ² so that y = 1:83=x.

9.1. PROBLEMS 147

0.80.60.40.20

50

37.5

25

12.5

0

x

y

9.20 Below is an electrolytic cell diagram used in an electrowinning factoryfor producing solid copper on cathode electrodes.

(¡)SteeljCu+2=Cu; H+=H2

¯¯ jH2O=O2 jPbO2 (+)

Recall that the cathodes are mechanically stripped (removed) after the elec-trowinning run is accomplished. The electrolyte used in this electrowinningoperation contains aqueous copper sulfate (CuSO4) and sulfuric acid (H2SO4)solution, and it is maintained at 35oC. Assume that the anodes operate at 100%current e¢ciency with respect to oxygen evolution and that for the cathodes is² = 88% due to hydrogen evolution. These type of electrodes are connected inparallel within 80 tanks that operate at 2:6 volts each and the electrodes havea speci…c submerged area Ae. Use the following relevant data to perform somecalculations given below. The average retail price of electricity (cost of electric-ity) per kWh: Ce = 0:14 $=kW:h, Market price of copper: Cmarket = $0:17=Kg,and

P = 72 kW (Power) Aw = 63:54 g=molE = 2:6 V DCu+2 = 1:24x10¡ 5 cm2=sNT = 100 (Tanks) Kv = 0:80 cm2=sNa = 30 (Anodes per tank) vx = 0:312 cm=s (Velocity of Cu+2)Nc = 29 (Cathodes per tank) d = 6 cm (electrode distance)Ae = 0:80 m2 (Electrode area) Ce = 0:14 $=kW:h (Electricity)

(a) Write down all the reactions and the standard half-cell potential in-volved in this electrowinning process. What should the applied potential be


compared to the half-cell potential? Is the given operating potential (2:6 volts)suitable for this electrowinning? Calculate (b) the total current, the anodicand cathodic current densities, and (c) Sc, Re, Sh and Cb in g=l per tank.Let the anode characteristic length be L =

pAe. Use the Sherwood num-

ber (Sh < 100) and the Reynolds number (Re < 2; 300) to assure force laminar‡ow. Determine (d) the production rate (PR) and the total weight produced(WT ) in 24 hours, (e) the energy consumption (°). Use your engineering eco-nomics skills to estimate (f) the cost of energy consumed (Cte), the total costof production (Cprod) and the gross income (GI) in 24 hours if the indus-trial cost of electric energy and the market price of copper are Ce = $0:30kW:h and Cmarket = $3:193=lb = $7:041=Kg, respectively. [Solutions: (a)Eo

Half ¡ cell ¼ ¡1:57 V , (b) ic = 0:001194 A=cm2, (c) Sh ¼ 39 and Cb = 73:08g=l].

Solution:(a) The reactions, the standard half-cell potential and the applied potential

are

Anode: 2H2O ) 4H+ + O2 + 4e

Cathode: Cu+2 + 2e ) Cu

Cathode: 2H+ + 2e ) H2

Overall : 2H2O + Cu+2 + 2H+ ) Cu + 4H+ + O2 " +H2 "

EoHalf ¡ cell = Eo

H2O + EoCu + Eo

H2= ¡1:23 V ¡ 0:337 V + 0

EoHalf ¡ cell ¼ ¡1:57 V

E ¸¯¯Eo

Half ¡ cell

¯¯

The applied potential should be E ¸¯¯¯Eo

Half ¡ cell

¯¯¯. So the given operating

potential (2:6 volts) is suitable for this electrowinning cell because 2:6 volts >¯¯¯Eo

Half ¡ cell

¯¯¯.

(b) The total current

I =P

E=

72 £ 103 AV

2:6V= 27; 692 A

Electrode areas per tank :

Aa = NT NaAe = (100) (30)¡8; 000 cm2

¢= 24:0x106 cm2

Ac = NT NcAe = (100) (29) (8; 000 cm)2

= 23:2x106 cm2

Current densities per tank :

9.1. PROBLEMS 149

ia =I

Aa=

27; 692A

24:0x106 cm2= 0:001154 A=cm2 = 1:250

A

m2

ic =I

Ac=

27; 692A

23:2x106 cm2= 0:001194 A=cm2 = 1:293

A

m2

(c) Assume force laminar ‡ow. The Schmidt (Sc), Reynolds (Re) numberand the anode length (L) are

Sc = Kv=D =¡0:80 cm2=s

¢=¡1:24x10¡ 5 cm2=s

¢= 64:52x103

Re = vxd=Kv = (0:312 cm=s) (6 cm) =¡0:80 cm2=s

¢= 2:34

From eq. (9.101) along with the anode characteristic length (L), the Sher-wood number is

L =p

Ae =p

0:8 m2 = 0:8944 m2 = 89:44 cm2

Sh =

µ9

5

¶µvxd2

LD

¶1=3

=

µ9

5

¶"(0:312 cm=s) (6 cm)

2

(89:44 cm) (1:24x10¡ 5 cm2=s)

#1=3

Sh ¼ 39

or from eq. (9.97)

Sh =

µ9

5

¶·d

L(ScRe)

¸1=3

Sh =

µ9

5

¶·µ6

89:44

¶¡64:52 ¤ 103

¢(2:34)

¸1=3

¼ 39

Using eq. (9.97) for one tank yields the cathodic current density as thelimiting current density ( ic = iL) and the bulk concentration (Cb) in g=l pertank. Thus,

ic =9zFCb

5

µvxd2D2

L4

¶1=3

Cb =5ic9zF

µL4

vxd2D2

¶1=3

Cb =(5)

¡0:001194 A=cm2

¢

(9) (2)¡96; 500 A:s

mol

¢

"(89:44 cm)

4

¡0:312 cm

s

¢(6 cm)

2 ¡1:24x10¡ 5 cm2

s

¢2

#1=3

Cb = 1:15 £ 10¡ 3 mol=cm3 = 1:24 mol=l

Cb = (1:15 mol=l) (63:55 g=mol) = 73:08 g=l


Alternatively, using eqs. (9.94) and (7.87b) with ¢C = Cb gives per tank

Sh =

µ2

3

¶

(Sc)1=3

(Re)1=2

=

µ2

3

¶¡64:52x103

¢1=3(2:34)

1=2¼ 41

Cb =icL

zFDSh=

¡0:001194 A

cm2

¢(89:44 cm)

(2)¡96; 500 A:s

mol

¢ ¡1:24 £ 10¡ 5 cm2

s

¢(41)

Cb = 1:189x10¡ 3 mol=cm3 = 1:19 mol=l

Cb = (1:19 mol=l) (63:55 g=mol) = 75:63 g=l

Both equations give fairly similar results. Let’s use the results given by eq.(4.94) in eq. (9.87b).

(d) Production rate (PR) and total weight produced (WT ) in 24 hours:

PR =²icAcAw

zF=

(0:88)¡0:001194 A

cm2

¢ ¡23:2x106 cm2

¢ ¡63:55 g

mol

¢

(2) (96; 500 A:s=mol)

PR = 8:0266 g= sec = 28:90 Kg=h ' 694 Kg=day (Total)

WT = (28:90 Kg=h) (24 h) = 694 Kg (Total)

wt = WT =NT = (694 Kg) =100 = 6:94 Kg (Per tank)

wc = (6:94 Kg) =29 = 0:23931 Kg = 239:31 grams (Per cathode)

The energy consumption is

° =P

PR=

72 kW

28:90 Kg=h= 2:49 kW:h=Kg (Total)

° = (2:49 kW:h=Kg) =100 = 0:0249 kW:h=Kg (Per tank)

° = 8:59x10¡ 2 kW:h=Kg (Per cathode)

If the cost of electricity per kW.h is Ce = 0:14 $=kW:h, then the cost ofenergy consumed is

Cte = Ce°WT = (0:14 $=kW:h) (2:49 kW:h=Kg) (694 Kg)

Cte = $241:93 (Total)

Ct = Cte=NT = ($241:93) =100 = $2:4193 (Per tank)

Cc = Cte= (Nc) = ($518:45) =29 = $8:34 (Per Cathode)

Thus, the cost of production is

Cprod =Cte

WT=

$241:93

694 Kg

Cprod = $0:35=Kg

9.1. PROBLEMS 151

If the market price is Cmarket = $3:193=lb = $7:041=Kg, then the grossincome in 24 hours is

GI = (Cmarket ¡ Cprod) (WT ) = ($7:041=Kg ¡ $0:35=Kg) (694 Kg)

GI = $4; 643:60 (For a 24-hour run)

This result is rather low. What can be improved for a higher GI?

Chapter 10

HIGH-TEMPERATURECORROSION

10.1 A silicon oxide crucible is used to melt pure aluminum in the presenceof oxygen at 1300oC. a) Will the silicon oxide (SiO2) corrode? b) What’s theoxygen pressure? [Hint: Use Figure 10.1].

Solution:From Figure 10.1 @ 1300oC,

43Al + O2 = 2

3Al2O3 ¢GoAl2O3

= ¡774 kJ=molSiO2 = Si + O2 ¢Go

SiO2= +586 kJ=mol

43Al + SiO2 = 2

3Al2O3 + Si ¢Go = ¡188 kJ=mol

¢Go = ¢GoAl2O3

+ ¢GoSiO2

or

4Al + 3O2 = 2Al2O3 ¢GoAl2O3

= ¡774 kJ=mol3SiO2 = 3Si + 3O2 ¢Go

SiO2= +586 kJ=mol

4Al + 3SiO2 = 2Al2O3 + 3Si ¢Go = ¡188 kJ=mol

The SiO2 crucible will corrode at 1300oC since molten aluminum is reducedAl2O3 (alumina) because ¢Go < 0.

b) From eq. (10.5b) and Po = 1 atm (standard pressure),

¢Go = ¡RT ln [K]

¢Go = ¡RT ln[Al2O3]

2=3

[Al]4=3

[Si] [O2]= ¡RT ln

µ1

[O2]

¶

¢Go = RT ln [O2] = RT ln

·PO2

Po

¸

153

154 CHAPTER 10. HIGH-TEMPERATURE CORROSION

Thus,

PO2= Po exp

µ¢Go

RT

¶

= (1 atm) exp

·¡188x103 J=mol

(8:314 J=mol:oK) (1573oK)

¸

PO2= 5:71x10¡ 7 atm

10.2 An Alumina (Al2O3) crucible contains molten copper and oxygen (O2)at 1300oC. Determine a) if the crucible will corrode, if so, calculate the oxygenpressure.

Solution:From Figure 10.1 @ 1300oC,

4Cu + O2 = 2Cu2O ¢GoCu2O = ¡130 kJ=mol

2Al2O3 = 43Al + O2 ¢Go

Al2O3= +774 kJ=mol

or

4Cu + O2 = 2Cu2O ¢GoCu2O = ¡130 kJ=mol

2Al2O3 = 43Al + O2 ¢Go

Al2O3= +774 kJ=mol

4Cu + 2Al2O3 = 2Cu2O + 43Al ¢Go = +644 kJ=mol

The Al2O3 crucible will not corrode at 1300oC because ¢Go > 0. Therefore,there is no need to calculate the oxygen pressure.

10.3 In the case of thick oxide formation at high temperature, the Pilling-Bedworth theory may have a limited applicability. Explain why this may be thecase.

Solution: Basically, the Pilling-Bedworth theory fails when strains developduring the formation and growth of the oxide scale. Consequently, the scale mayrupture or become nonprotective.

10.4 If the protective nature of an oxide …lm at room temperature is lostat relatively high temperatures, explain the sequence of the oxide thickeningprocess.

Explanation: This is the case of oxide …lm growth in which the mechanicalproperties and the thermal coe¢cient of expansion are di¤erent from the basemetal. Therefore, the scale cracks and spalls. If the nature of the thick oxideis volatile, such as tungsten oxide (WO3) at T > 550oC, the oxide vaporizes athigh pressure.

10.5 Determine the values of Kw from Figure 10.10 and plot Kw vs. T .What can you conclude from this plot?

Solution:From Figure 10.10,

155

T (oC) Kw

¡mg:cm¡ 4h¡ 1

¢

900 10:40950 25:001050 56:001100 83:001150 125:01200 250:01350 750:0

According to these data, the trend is described by the curve …tting equation¡R2 = 0:99

¢

Kw =¡7 £ 10¡ 30 mg:cm¡ 4h¡ 1

¢T 10:234

1375125011251000875

800

600

400

200

0

T

K

10.6 Use the data given below to plot Kp vs. Aw;oxide (oxide molecularweight) at T = 1000oC and PO2

= 1 atm. What can you conclude from theplot?

Element Atomic Weight Molecular Weight Kp

(g=mol) (g=mol)£¡

g2O2

¢=cm4:s

¤

Co 58:93 CoO ! 74:93 2:10x10¡ 10

Cu 63:54 Cu2O ! 143 6:30x10¡ 9

Fe 55:85 FeO ! 71:85 4:80x10¡ 10

Si 28:10 SiO2 ! 60:10 1:20x10¡ 12

Solution:Polynomial …t with R2 = 0:98

Kp

¡x10¡ 10

¢= ¡52:69 6 +

µ

0:8:022 7mol

g

¶

Aw;oxide


Figure 10.1:

One can conclude that Kp increases linearly with molecular weight of theoxide, but more data is needed to be certain of this linear relationship.

10.7 Calculate a) the equilibrium constant K and b) the dissociation oxygenpressure PO2

for the oxidation of aluminum at 1100OC:

Solution:

a) From Figure 10.1 @ 1100oC,

43Al + O2 = 2

3Al2O3 ¢GoAl2O3

= ¡820 kJ=mol of O2

¢Go = ¡RT ln [K]

K = exp

µ

¡¢Go

RT

¶

= exp

µ820x103 J=mol

(8:314 J=mol:oK) (1373oK)

¶

K = 1:58x1031

b) Dissociation oxygen pressure and Po = 1 atm (standard pressure):

K =[Al2O3]

2=3

[Al]4=3

[O2]=

Po

PO2

PO2=

Po

K=

1 atm

1:58x1031= 6:35x10¡ 32 atm

157

10.8 If 100 grams of pure aluminum (Al) oxidizes according to the reaction4Al + 3O2 = 2Al2O3, calculate the PB ratio de…ned by

PB = VAl2O3=VAl

whereV 0s are volumes. Data: ½Al = 2:70 g=cm3 and ½Al2O3= 2:70 g=cm3 .

Solution:

Aw;Al = 26:98 g=mol ½Al = 2:70 g=cm3

Aw;Al2O3= 101:96 g=mol ½Al2O3

= 3:80 g=cm3

If WAl = 100 g, then

WAl2O3= WAl

·2Aw;Al2O3

4Aw;Al

¸

WAl2O3= (100 g)

·(2) (101:96 g=mol)

(4) (26:98 g=mol)

¸

WAl2O3= 188:95 g

Thus,

VAl =WAl

½Al

=100 g

2:70 g=cm3= 37:04 cm3

VAl2O3=

WAl2O3

½Al2O3

=188:95g

3:80 g=cm3= 49:72 cm3

and

PB =VAl2O3

VAl=

49:72 cm3

37:04 cm3

PB = 1:34

10.9 A chromium bar is exposed to oxygen gas at 900oC. Calculate a) theoxygen partial pressure and b) the oxygen activity.

Solution:

a) Using Figure 10.1 yields

4Cr + 3O2 = 2Cr2O3

¢G = ¡550 kJ=mol @ T = 900oC


and

¢G = ¡RT ln (K)

K = exp

µ

¡¢G

RT

¶

= exp

·550x103 J=mol

(8:314 J=mol.oK) (1173oK)

¸

K = 3:11x1024

The equilibrium constant with the standard pressure Po = 1 atm = 101 kPais

K =a2 (Cr2O3)

a4 (Cr) a3 (O2)=

1

a3 (O2)=

·1

PO2=Po

¸3

PO2= PoK

¡ 1=3 = (1atm)¡3:11x1024

¢

PO2= 6:85x10¡ 9 atm = 6:92x10¡ 7 kPa

b) The activity of oxygen is

aO2= K¡ 1=3 =

¡3:11x1024

¢¡ 1=3

aO2= 6:85x10¡ 9 mol=l = 6:85x10¡ 12 mol=cm3

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