ch17 z5e electrochem

1

Electrochemistry pp

Applications of Redox

2

17.1 Galvanic CellsOxidation reduction reactions involve a

transfer of electrons.OIL- RIGOxidation Involves LossReduction Involves GainLEO-GER Lose Electrons OxidationGain Electrons Reduction

3

ApplicationsMoving electrons = electric current. 8H++MnO4

-+ 5Fe+2 Mn+2 + 5Fe+3 +4H2O

It helps to break the reactions into half reactions.

8H++MnO4-+5e- Mn+2 +4H2O

5(Fe+2 Fe+3 + e-) In the same mixture this happens without

doing useful work, but if separate . . .

4

H+

MnO4-

Fe+2

Connected this way the reaction starts.Stops immediately because charge builds

up.

5

H+

MnO4-

Fe+2

Galvanic Cell

Salt Bridge allows current to flow

6

H+

MnO4-

Fe+2

Galvanic Cell

Electrons flow in the wire, ions flow through the salt bridge

7

H+

MnO4-

Fe+2

e-

Electricity travels in a complete circuit Instead of a salt bridge . . .

8

H+

MnO4-

Fe+2

Porous Disk

9

Reducing Agent

Oxidizing Agent

e-

e-

e- e-

e-

e-

Oxidation

at AnodeReduction at Cathode

10

Cell PotentialOxidizing agent pulls the electron.Reducing agent pushes the electron. The push or pull (“driving force”) is called

the cell potential EcellAlso called the electromotive force (emf). Unit is the volt (V). = 1 joule of work per coulomb of charge

transferred (1 V = 1 J/C).Measured with a voltmeter.

11

17.2 Standard Reduction PotentialsThe reaction in a galvanic cell is a redox

reaction.So, break it down into two half-reactions.Assign a potential to each.Sum the half-cell potentials to get the

overall cell potential. “Active anodes” - the more active metal is

always the anode (good for multiple choice questions).

12

Zn+2 SO4-

2

1 M HCl

Anode

0.76

1 M ZnSO4

H+

Cl-

H2 in

Cathode

13

1 M HCl

H+

Cl-

H2 in

Standard Hydrogen ElectrodeThis is the reference

all other oxidations are compared to.

Eº = 0 º indicates standard

states of 25ºC, 1 atm, 1 M solutions.

14

Z5e 841 Figure 17.5: Zn/H Galvanic Cell. Notice the electron flow also.

15

Cell Potential Zn(s) + Cu+2 (aq) Zn+2(aq) + Cu(s)

The total cell potential is the sum of the potential at each electrode.

Eº cell = EºZn Zn+2 + Eº Cu+2 Cu

We can look up reduction potentials in a table (see, p. 796).

For Br, s/b 1.07 (not 1.09) need for online HW!!) Since one of the 1/2 reactions is oxidation its

table value must be reversed, so change its sign.

16

Z5e 842 Fig 17.6 Zn/Cu Galvanic Cell

17

Cell Potential pp

Determine the cell potential for a galvanic cell based on the redox reaction . . . Cu(s) + Fe+3(aq) Cu+2(aq) + Fe+2(aq)

steps followFe+3(aq) + e- Fe+2(aq) Eº = 0.77 V

Cu+2(aq)+2e- Cu(s) Eº = 0.34 VSince one of these must be oxidation,

one of them needs to be reversed.Which one?

18

Cell Potential pp

Cu(s) + Fe+3(aq) Cu+2(aq) + Fe+2(aq)

This is spontaneous only if Eºcell = (+), so

reverse the copper half-reaction.

Fe+3(aq) + e- Fe+2(aq) Eº = 0.77 V

Cu(s) Cu+2(aq)+2e- Eº = -0.34 V

Must balance the e-s, so multiply the Fe 1/2-

reaction by 2, BUT do not multiply Eº. Why?

Cell potential is an intensive property (doesn’t depend on number of times the reaction occurs).

19

Cell Potential pp

Cu(s) + Fe+3(aq) Cu+2(aq) + Fe+2(aq) 2Fe+3(aq) + 2e- 2Fe+2(aq) Eº = 0.77 V Cu(s) Cu+2(aq)+2e- Eº = -0.34 V Cu(s) + 2Fe+3(aq) Cu+2(aq) + 2Fe+2(aq)

Eº = 0.43 V

20

Cell Potential

Be sure to use the correct 1/2-reactions!

For example, table 17.1, p. 796 lists three 1/2-reactions for MnO4

1- Find them.

Answers next slide.

21

Cell Potential Three 1/2-reactions for MnO4

1- : MnO4

1- + 4H1+ 3e- MnO2 + 2H2O Eº = 1.68 MnO4

1- + 8H1+ 5e- Mn2+ + 4H2O Eº = 1.51 MnO4

1- + e- MnO42- Eº = 0.56

Pick the reaction that “works” with your overall reaction (look at reactants and products).

22

Line Notation pp

solidAqueousAqueoussolidAnode on the leftCathode on the rightSingle line to show different phases.Double line porous disk or salt bridge. If all the substances on one side are

aqueous, a platinum electrode is used.For: Cu(s) + Fe+3(aq) Cu+2(aq) + Fe+2(aq)

Cu(s)Cu+2(aq)Fe+3(aq),Fe+2(aq)Pt(s) Remember to show the electrodes!!

23

Complete Galvanic Cell Description (AP Test) pp

The reaction always runs spontaneously in the direction that produced a positive cell potential.

Four things for a complete description:1. Cell Potential and balanced reaction2. Direction of flow3. Designation of anode and cathode4. Nature of all components -- electrodes

& ions (plus inert conductor like Pt if needed). Use line notation.

24

Practice pp

Completely describe the galvanic cell based on the following half-reactions under standard conditions.

MnO4- + 8 H+ +5e- Mn+2 + 4H2O

Eº = 1.51 V Fe+2

+2e- Fe(s) Eº = -0.44 V

25

Practice - Item 1 pp

Determine cell potential & balanced reaction MnO4

- + 8 H+ +5e- Mn+2 + 4H2O Eº = 1.51 V Fe+2 +2e- Fe(s) Eº = -0.44 V

Since (+) E required, reverse the 2nd reaction Eºcell =1.51 + (+0.44) = 1.95 V Complete, balanced reaction is

2MnO4- + 5Fe(s) + 16 H+ 2Mn+2 + 5Fe2+(aq) + 8H2O(l)

Note: multiplying the half reactions to balance the reaction does NOT multiply the Eº values!!! (intensive property).

26


Determine e- flow by inspecting 1/2 rxns & using the direction that gives a (+) Eºcell

MnO4- + 8 H+ +5e- Mn+2 + 4H2O Eº = 1.51 V

Fe(s) Fe+2 +2e- Eº =+0.44 V Eºcell =1.51 + (+0.44) = 1.95 V

So, electrons flow from Fe(s) to MnO4- (aq)

27


Designate the anode and cathode MnO4

- + 8 H+ +5e- Mn+2 + 4H2O Eº = 1.51 V Fe(s) Fe+2 +2e- Eº =+0.44 V Eºcell =1.51 - (0.44) = 1.95 VElections flow from Fe(s) to MnO4

- So, oxidation occurs in the compartment

containing Fe(s) -- the anode Reduction occurs in the compartment

containing MnO4- -- Use Pt for the cathode

Note: e- always flow from anode to cathode ”Red cat ate an ox". Red/cat = reduct/cathode

28


Describe nature of each electrode & ions present (use line notation)

Complete, balanced reaction is 2MnO4

- + 5Fe(s) + 16 H+ 2Mn+2 + 5Fe2+(aq) + 8H2O(l)

Electrode in Fe/Fe2+ compartment is iron metal

An inert conductor like Pt must be used in MnO4

-1/ Mn+2 compartment (don’t forget).

Line notation is: Fe(s)Fe+2(aq)MnO4-

1(aq),Mn+2(aq)Pt(s)

29

pp Figure 17.7: A Schematic of the previous Galvanic Cell

Eº = 1.95 v

Be able to draw this as well as write the line notation for the AP exam.

30

17.3 Cell Potential, Work & Gemf = potential (V) = work (J) / Charge(C)E = work done by system / chargeE = -w/q (emf & work have opposite signs)

Use (-)w because it is flowing out from system.Charge is measured in coulombs. -w = qE (where q = the charge)Faraday = 96 485 C/mol e-

q = nF = moles of e- x charge per mole e-

w = -qE = -nFE = G

31

Potential, Work and G pp

Gº = -nFE º (at standard conditions) if E º > 0, then Gº < 0 spontaneous if E º < 0, then Gº > 0 nonspontaneous In fact, reverse is spontaneous.

32

Potential, Work and GCalculate Gº for the following reaction:

Cu+2(aq)+ Fe(s) Cu(s)+ Fe+2(aq) Fe+2(aq) + 2e-Fe(s) Eº = -0.44 V Cu+2(aq)+2e- Cu(s) Eº = 0.34 V

Gº = -nFE º Answer? -1.5 x 105 J Calculation with units is . . . -(2 mol e-)(96 485 C/mol e-)(0.78 J/C)

33

Putting It Together pp

Using Table 17.1, predict if 1 M HNO3 will dissolve gold to form a 1 M Au3+ solution?

What are the half reactions? . . .Gold needs to be oxidized so HNO3 must

be reduced. Look for a half-reaction with HNO3 where NO3

1- is being reduced . . . NO3

1- + 4H1+ + 3e- NO + 2H2O Eº = 0.96 v Au Au3+ + 3e- Eº = -1.50 v

34

Putting It Together pp

Using Table 17.1, predict if 1 M HNO3 will dissolve gold to form a 1 M Au3+ solution?

NO31- + 4H1+ + 3e- NO + 2H2O Eº = 0.96 v

Au Au3+ + 3e- Eº = -1.50 v Au + NO3

1- + 4H1+ Au3+ + NO + 2H2O Eºcell = -0.54 v

Since Eºcell = negative, this cannot be spontaneous because . . .

Gº = -nFE º = (-)(3)(96 485)(-0.54) = +156kJ

Since Gº is (+), not spontaneous.

35

17.4 Cell Potential and Concentration pp

Notes for online HW1 lb = 453.6 g1 Faraday = 96 485 c/s (not 96 500 c/s)Use 0.0592 in Nernst equation (not

0.0591)

36

17.4 Cell Potential and Concentration17.4 Cell Potential and Concentration pppp

Qualitatively - Can predict direction of

change in E from LeChâtelier.

2Al(s) + 3Mn+2(aq) 2Al+3(aq) + 3Mn(s)

Predict if Ecell will be greater or less than

Eºcell of 0.48 v if

[Al+3] = 1.5 M and [Mn+2] = 1.0 M if [Al+3] = 1.0 M and [Mn+2] = 1.5 M if [Al+3] = 1.5 M and [Mn+2] = 1.5 M

Steps . . .

37

Cell Potential pp

2Al(s) + 3Mn2Al(s) + 3Mn+2+2(aq)(aq) 2Al2Al+3+3(aq)(aq) + 3Mn(s) + 3Mn(s)

Predict if Predict if EEcellcell will be greater or less than will be greater or less than

EEººcellcell of 0.48 v of 0.48 v if if [Al[Al+3+3] = 1.5 ] = 1.5 MM and and [Mn[Mn+2+2] = 1.0 ] = 1.0 MM

Answer . . .Answer . . . Since a Since a productproduct [ ] [ ] has been raised above has been raised above

1.0 1.0 MM, Le Chatelier predicts a shift , Le Chatelier predicts a shift leftleft and Eand Ecellcell < Eº < Eºcellcell

38

Cell Potential pp

2Al(s) + 3Mn2Al(s) + 3Mn+2+2(aq)(aq) 2Al2Al+3+3(aq)(aq) + 3Mn(s) + 3Mn(s)

Predict if Predict if EEcellcell will be greater or less than will be greater or less than

EEººcellcell of 0.48 v of 0.48 v if if [Al[Al+3+3] = 1.0 ] = 1.0 MM and and [Mn[Mn+2+2] = 1.5 ] = 1.5 MM

Answer . . .Answer . . . Since a Since a reactantreactant [ ] [ ] has been raised has been raised

above 1.0 above 1.0 MM, Le Chatelier predicts a shift , Le Chatelier predicts a shift rightright and E and Ecellcell > Eº > Eºcellcell

39

Cell Potential pp

2Al(s) + 3Mn+2(aq) 2Al+3(aq) + 3Mn(s)

Predict if Ecell will be greater or less than

Eºcell of 0.48 v if [Al+3] = 1.5 M and [Mn+2] = 1.5 M

Answer . . .Since both [ ]s have been raised above

1.0 M, cannot use Le Chatelier!Must use the Nernst Equation (coming to

your community soon!)

40

Le Chatelier, ∆G, & Concentration Cells Figure 17.9

A Concentration Cell That Contains a Sliver Electrode and Aqueous Silver Nitrate in Both Compartments

Since the right compartment has higher [Ag1+] there is a shift right of e-

So, Ag metal is deposited on the right side while [Ag1+] decreases on right side and increases on the left.

41

Le Chatelier, ∆G, & Concentration Cells pp

So, Ag metal is deposited on the right side while [Ag1+] decreases on right side and increases on the left.

You will need to recognize this concept to solve for ∆G on the AP exam.

Hint: the electrode with the largest [ ] will always be the cathode (where reduction occurs).

42

The Nernst EquationG = Gº +RTln(Q), since G = -nFE . . . -nFE = -nFEº + RTln(Q)

E = Eº - RTln(Q) nF

What is n in Al(s) + Mn2+ Al3+ + Mn(s)? Always have to figure out “n” by balancing.

2Al(s) + 3Mn+2(aq) 2Al+3(aq) + 3Mn(s) Eº = 0.48 V n = 6. Why? . . . n = mole of e-, not mole of compound.

43

The Nernst Equation continued G = Gº +RTln(Q)

-nFE = -nFEº + RTln(Q)

E = Eº - RTln(Q) = Eº - 2.303RT log (Q)

nF nF

Since we know R and F, at 25o C, above is aka

E = Eº - 0.0592log(Q)

n Textbook has typo: 0.0591 should be 0.0592 Use 0.0592 for all your calculations!

44

The Nernst Equation pp

E = Eº - 0.0592log(Q)

n For concentration cells (i.e., not at 1 M) this

equation must be done separately for each 1/2 cell, then subtract the results (or flip one and add) to get the cell potential.

See the following problem . . .

45

We’ll do “a,” “b,” & “c”.We’ll do “a,” “b,” & “c”.

pppp

46

Ecell when [Ag1+] on the right = 1.0 M pp

Since [Ag1+] is same on both sides Ecell = Eºcell which is 0 because . . .

Ag1+ + e- Ag Eº = .80v Ag Ag1+ + e- Eº = -.80v

Eºcell = 0

E = Eº - 0.0592log(Q) n

Since log(1) = 0, Ecell = Eºcell

and Eº = 0 from above, so Ecell also = 0.

47


Cathode always has the higher [ ] and e- always flow from the anode to the cathode.

Q = [ ]anode ÷ [ ]cathode

Here, [cathode] is on the right (2.0 M), & in the denominator for Q

E = Eº - 0.0592log(Q)

n

E = 0 - 0.0592log 1.0 = .018 v

1 2.0

48


Cathode always has the higher [ ] and e- always flow from the anode to the cathode.

So, [cathode] is on the left & in the denominator for Q

E = Eº - 0.0592log(Q)

n

E = 0 - 0.0592log 0.1 = .059 v

1 1.0

You will have a test question on this and it is NOT on the pre-test, so . . .

Do p. 832 #53!!

49


Notes for all of these: Eº is always 0 for

concentration cells because the 1/2 reactions cancel.

E = Eº - 0.0592log(Q)

n

E = 0 - 0.0592log anode .

n cathode Memorize this equation!

You will have a test question on this and it is NOT on the pre-test, so . . .

Do p. 832 #53!!

50

The Nernst Equation As reactions proceed concentrations of

products increase and reactants decrease. Reaches equilibrium where Q = K and Ecell = 0

Since at equilibrium Ecell = 0 = Eº - RTln(K)

nF

Eº = RTln(K)

nF

nFEº = ln(K) at 25º C aka log(K) = nEº RT 0.0592

Use ln(K) version in online HW even if at 25ºC!!

51

Nernst Equation & K pp

Calculate KCalculate Kspsp of silver iodide at 298 K of silver iodide at 298 K AgI(s) AgI(s) Ag Ag++ + I + I-- where Ewhere Eoo

AgI(s) + eAgI(s) + e-- Ag(s) + I Ag(s) + I-- -0.15 v-0.15 vII22(s) + 2e(s) + 2e- - 2I 2I-- +0.54 v+0.54 vAgAg++ + e + e-- Ag(s) Ag(s) +0.80 v+0.80 v

logK = nEº/0.0592logK = nEº/0.0592 1st, get Eº1st, get Eºcellcell for overall reaction. Steps. for overall reaction. Steps. Find overall reaction, then EºFind overall reaction, then Eºcellcell

Only need 1st & 3rd equations to get . . .Only need 1st & 3rd equations to get . . . AgI(s) AgI(s) Ag Ag++ + I + I-- where Ewhere Eoo

cellcell = -0.95 v = -0.95 v

52

Nernst Equation & K pp

Calculate Ksp of silver iodide at 298 K AgI(s) Ag+ + I-

where Eocell = -0.95 v

at 25ºC log(K) = nEº = (1)(-0.95) = -16.05 0.0592 0.0592

Ksp = 10-16.05 = 9.0 x 10-17

53

17.5 Batteries are Galvanic CellsCar batteries are lead storage batteries.Pb +PbO2 +H2SO4 PbSO4(s) +H2OBe able to recognize the anode &

cathode from the half reactions.

54

Figure 17.13One of the Six Cells in Storage Battery a 12-V Lead Storage Battery

55

Batteries are Galvanic Cells Dry CellDry Cell - - acidacid battery battery

Zn + 2NHZn + 2NH44++ + 2MnO + 2MnO2 2 Zn Zn+2 +2 + 2NH+ 2NH3 3 + Mn+ Mn22OO33 + H + H22OO

Dry Cell - Dry Cell - alkalinealkaline batterybattery

Zn + 2MnOZn + 2MnO22 ZnO + Mn ZnO + Mn22OO3 3 (in base)(in base)

NiCadNiCad - can be re-charged indefinitely - can be re-charged indefinitelyNiONiO22 + Cd + 2H + Cd + 2H22O O Cd(OH) Cd(OH)2 2 +Ni(OH)+Ni(OH)22

Fuel CellFuel Cell - - reactants are continuously supplied reactants are continuously supplied

CHCH44 + 2O + 2O22 CO CO22 + 2H + 2H22O + energy O + energy

56

Figure 17.14A Common Dry Cell Battery

57

17.6 Corrosion RustingRusting - spontaneous - spontaneous oxidationoxidation.. Most structural metals have Most structural metals have reductionreduction

potentials that are potentials that are lessless positive than O positive than O22 . .

So, they So, they oxidizeoxidize while O while O22 is is reduced.reduced.

FeFe+2+2 +2e +2e-- Fe Fe EEº= º= --0.44 V O0.44 V O22

+ 2H+ 2H22O + 4eO + 4e- - 4OH4OH-- EEº= º= ++0.40 V0.40 V Reverse top half reaction, then add bothReverse top half reaction, then add both Fe + OFe + O2 2 + H+ H22O O FeFe2 2 OO33(rust)(rust) + H+ H++ EºEºcellcell = 0.84 v = 0.84 v

Reaction happens in two places . . . Reaction happens in two places . . .

58

Water

Rust

Iron Dissolves - Fe Fe+2

e-

Salt speeds up process by increasing conductivity

59

Figure 17.17The Electrochemical

Corrosion of Iron

60

Preventing CorrosionCoating - to keep out air and water.Galvanizing - Putting on a zinc coatZinc has a lower reduction potential than

iron, so it is more easily oxidized. So, zinc is a more active metal than iron.Alloying with metals that form oxide coats.Cathodic Protection - Attaching large

pieces of a more active metal like magnesium that get oxidized instead of iron (iron stays reduced).

61

Preventing Corrosion Cathodic Protection - Attaching large Cathodic Protection - Attaching large

pieces of an active metal like magnesium pieces of an active metal like magnesium that get oxidized instead of iron.that get oxidized instead of iron.Attach Mg wireAttach Mg wire to iron pipe (and replace to iron pipe (and replace periodically).periodically).

Attach titanium barsAttach titanium bars to ships’ hulls. In to ships’ hulls. In salt water the Ti acts as the anode and is salt water the Ti acts as the anode and is oxidized instead of the steel hull, which oxidized instead of the steel hull, which now acts as the cathode.now acts as the cathode.

62

Figure 17.18Cathodic

Protection

63

Running a galvanic cell backwards. Put a voltage whose magnitude is bigger than

the potential which reverses the direction of the redox reaction.

Produces a chemical change which would not normally happen because the potential is negative.

Remember: 1 A = 1 C/s and 1 F = 96 485 C Used for electroplating -- depositing the neutral

metal onto the electrode by reducing the metal ions in solution.

17.7 Electrolysis

64

1.0 M

Zn+2

e- e-

Anode Cathode

1.10

Zn Cu1.0 M

Cu+2

Galvanic Cell - spontaneous

65

1.0 M

Zn+2

e- e-

AnodeCathode

A battery >1.10V

Zn Cu1.0 M

Cu+2

Electrolytic Cell Forces the opposite reaction.

66

Figure 17.19(a) A Standard Galvanic Cell

(b) A Standard Electrolytic Cell

67

Calculating plating Have to include the charge.Have to include the charge. Measure current Measure current II (in amperes) (in amperes) 1 amp = 1 coulomb of charge per second1 amp = 1 coulomb of charge per second

1 A = 1 C/s1 A = 1 C/s q = q = II x t = the charge x t = the charge q/nF = moles of metalq/nF = moles of metal Mass of plated metalMass of plated metal

68

Calculating plating ppHow many minutes must a 5.00 amp

current be applied to produce 10.5 g of Ag from Ag+ Steps follow . . .

Set up a picket fence that includes – Current & time– Quantity of charge (in coulombs)– Moles of electrons– Moles of metal (may be different)– Grams of metal

Arrange the picket fence so the units give you what you’re looking for.

69

Calculating plating pp How many minutes must a 5.00 amp

current be applied to produce 10.5 g of Ag from Ag+ Steps follow . . .

The pieces of the picket fence are:– 5.00 amp (rewrite as 5.00

C/s)– 10.5 g Ag– 107.868 g Ag/1mol Ag– 1 mol e-/mol Ag (Ag Ag1+ + 1e-)– 96 485 C/mol e-

– 60 sec/min Your answer? . . .

70

Calculating platingCalculating plating pppp

How many minutes must a 5.00 amp current be applied to produce 10.5 g of Ag from Ag+ (amp = C/s)

31.3 minutes31.3 minutes (10.5 g Ag)(1 mol Ag/107.868 g Ag)(1 mol e-/1 mol

Ag)(96 485 C/1 mol e-)(1 s/5.00C)(1 min1 min/60 s)

71

Calculating plating pp

An antique automobile bumper is to be chrome plated by dipping into an acidic Cr2O7

2- solution serving as the cathode of an electrolytic cell. MCr = 51.996

Using 10.0 amperes, how long to deposit 1.00 x 102 grams of Cr(s)? Steps . . .

Find overall reaction from 1/2-reactions (to get moles of e-). Steps follow.

The only substances you start with are Cr2O7

2- and H2O. Which is oxidized? Reduced? . . .

72


Using 10.0 amperes, how long to deposit Using 10.0 amperes, how long to deposit 1.00 x 101.00 x 1022 grams of Cr(s) by dipping into grams of Cr(s) by dipping into acidic Cracidic Cr22OO77

2-2-(aq) , (aq) , MMCrCr = 51.996? = 51.996?

The only substances you start with are The only substances you start with are CrCr22OO77

2-2- and H and H22O. O. Which is oxidized? Which is oxidized? Which is reduced?Which is reduced? . . . . . .

CrCr22OO772-2- must be reduced must be reduced to get to Cr to get to Cr(s)(s) so so

HH22O must be oxidizedO must be oxidized. Use Table 17.1 p. . Use Table 17.1 p. 796 to find the 1/2-reaction of H796 to find the 1/2-reaction of H22O . . . O . . .

73


Use Table 17.1 p. 843 to find the 1/2-Use Table 17.1 p. 843 to find the 1/2-reaction of Hreaction of H22O. Which is it?O. Which is it?HH22OO22 + 2H + 2H1+ 1+ + 2e+ 2e-- 2H 2H220 0 2H2H220 0 O O22 + 4H + 4H1+1+ + 4e + 4e--

OO22 + 2H + 2H220 + 4e0 + 4e- - 4OH4OH1-1- 2H 2H220 0 + 2e+ 2e-- H H22 + 2OH + 2OH1-1-

*****

#2 above is the only one that starts #2 above is the only one that starts with water and is oxidized.with water and is oxidized.

74


Using 10.0 amperes, how long to deposit 1.00 x 102 grams of Cr(s) by dipping into acidic Cr2O7

2-(aq) ? , MCr = 51.996

So, the 1/2-reactions needed are . . .6e- + 14H1+ + Cr2O7

2- 2Cr3+ + 7H2O 2H20 O2 + 4H1+ + 4e-

and one more! (Why?)You only got to Cr3+, but you need Cr(s)

Also need Cr3+ + 3e- Cr(s)Now, write the balanced equation . . .

75



2-(aq) ? , MCr = 51.996

Now, write the balanced equation . . .2H1+ + Cr2O7

2- H2O + 3O2 + 2Cr(s)How many mol e- changed in this reaction

(comprised of three 1/2-reactions)?12 mol e- Now, you can start to do the problem.Your answer in days? . . .

76



2-(aq)? , MCr = 51.996 with 2H1+

+ Cr2O72- H2O + 3O2 + 2Cr(s) and 12

mol e- moving.Your answer is . . .1.29 days . . . 100.g Cr(s) • 1mol Cr/51.996g Cr(s) • 12 mol e-/2 mol

Cr(s) • 96 486 C/1 mol e- • 1s/10.0C • 1 h/3600 s • 1 d/24 h = 1.29 days

77

Calculating plating Electrolysis of a molten salt , MCl, using Electrolysis of a molten salt , MCl, using

3.86 amps for 16.2 min deposits 1.52 g of 3.86 amps for 16.2 min deposits 1.52 g of metal. What is the metal?metal. What is the metal?

Use picket fence to get moles of metal. Use picket fence to get moles of metal. Since g/mol = M, use calculated moles of Since g/mol = M, use calculated moles of metal and 1.52 g to get M. Answer . . . metal and 1.52 g to get M. Answer . . .

Potassium.Potassium. The solution is . . . The solution is . . .

78

Calculating plating Electrolysis of a molten salt , MCl, using Electrolysis of a molten salt , MCl, using

3.86 amps for 16.2 min deposits 1.52 g of 3.86 amps for 16.2 min deposits 1.52 g of metal. What is the metal?metal. What is the metal? 3.86 C/s • 16.2 min • 60s/1m • 1mol e-/96 485 C • 1 mol 3.86 C/s • 16.2 min • 60s/1m • 1mol e-/96 485 C • 1 mol MM1+1+/1mol e- = 0.039 = moles of metal./1mol e- = 0.039 = moles of metal.

1.52g/0.039 mol = 39.1 g/mol = potassium1.52g/0.039 mol = 39.1 g/mol = potassium

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Other uses pp

Electrolysis of water.Electrolysis of water. Separating mixtures of ions Separating mixtures of ions A more positive reduction potential A more positive reduction potential

means that reaction proceeds forward. means that reaction proceeds forward. The metal with the most The metal with the most positivepositive

reductionreduction potential is easiest to plate out potential is easiest to plate out of solution (and is the best oxidizer).of solution (and is the best oxidizer).

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Relative Oxidizing Abilities pp An acidic solution has CeAn acidic solution has Ce4+4+, VO, VO22

1+1+, & Fe, & Fe3+3+. Use . Use Table 17.1 p. 796 to predict the order of Table 17.1 p. 796 to predict the order of oxidizingoxidizing ability. Answer . . . ability. Answer . . .

Order of oxidizing ability is the same as the Order of oxidizing ability is the same as the order for order for beingbeing reduced, So. . . reduced, So. . . CeCe4+4+ + e + e-- CeCe3+3+ E = 1.70 vE = 1.70 v VOVO22

1+1+ + 2H + 2H1+1+ + e + e-- VOVO2+2+ + H + H22OO E = 1.00 vE = 1.00 v FeFe3+3+ + e + e-- FeFe2+2+ E = 0.77 vE = 0.77 v

Also, predict which one will be Also, predict which one will be reducedreduced at the at the cathode of an cathode of an electrolyticelectrolytic cell at the cell at the lowestlowest voltage. Answer . . .voltage. Answer . . .

Since Since CeCe4+4+ is the greatest oxidizer it is the most is the greatest oxidizer it is the most easily reduced (needs least voltage).easily reduced (needs least voltage).

Do section 17.8 on your own. Do section 17.8 on your own.

ch17 z5e electrochem

Education

e mno

e fes e

e cell mno

reaction e cell

e flowby

e cuse

galvanic cell h

agent oxidizing agent