CHEM 163

Chapter 21

Spring 2009

1

3-minute review

• What is a redox reaction?

2

Half-Reactions

Split overall reaction into two reactions

3

Oxidation Reduction

Step 1. Divide reaction into half reactions.

Step 2. Balance atoms in each half reaction.Do O and H last!

Step 3. Balance charges in each half reaction.Add e-

Step 4. Make # e- gained equal # e- lost.

Multiply by integer!

Step 5. Add reactions together.

Step 6. Check that atoms and charges are balanced.

Need O?

Add H2ONeed

H? Add H+

Electrochemical Cells

4

Voltaic (Galvanic) Cell

Electrolytic Cell

∆G < 0 ∆G > 0

Sys does work on surr Surr do work on sys

Erxt > Eprod Elost electricity Erxt < Eprod Electricity rxn

• Electrodes: • Conduct electricity between cell and

surroundings• Anode (oxidation)• Cathode (reduction)

• Electrolyte: contains ions

5Fig. 21.3

Voltaic CellsHalf-cells: to complete the circuit, electrons must flow externally

• Oxidation half-cell: • Anode (Zn)

“reactant”• Electrolyte

• Reduction half-cell: • Cathode (Cu)

“product”• Electrolyte Fig. 21.5 6

Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)

Voltaic Cells

• Electrode charges: • e- flow left to right• e- created at anode, used up at

cathode• Anode has excess e-

• Salt bridge:• Completes circuit• Keeps each cell

neutral• Direction of ions

7

Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)

Anode

- Cathode

+

)()()()( 22 sCuaqCuaqZnsZn

Anode Cathode

Electrodes• Conduct electricity between cell and

surroundings

Active Electrodes:

electrodes are components of half-reactions

Inactive Electrodes:conduct electrons but are not reactants or productsEx. Graphite, Pt

8

2I- (aq) I2 (s) + 2e- MnO4

- (aq) + 8H+ (aq) + 5e- Mn2+ (aq) + 4H2O (l)

graphite(aq)Mn(aq),MnO(aq),H(s)I(aq)Igraphite 242

Anode Cathode

How much electricity?

• Zn gives up electrons more easily• Zn is a stronger reducing agent• Potential difference between two

electrodes– Cell potential (Ecell)

– Cell voltage– Electromotive force (emf)

9

Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)

Zn (s) Zn2+ (aq) + 2 e- Cu (s) Cu2+ (aq) + 2 e-

Ecell > 0

(spontaneous process)

Standard Cell Potentials• Ecell at standard conditions

– Specific T (usually 298 K)– All components in standard states

• 1 M (aq)

• 1 atm (g)

• Pure solid

• Standard Electrode Potential– Half-cell potential – Always shown as a reduction

10

ocellE

ocellhalfE

ocellE o

anodeocathode EE

reduction

oxidation

How can you measure a half-cell?

• Half-cell potentials are relative to a standard

Standard Hydrogen Electrode (SHE)

11

2H+ (aq; 1 M) + 2e- H2 (g; 1 atm)

V 00.0oreferenceE

Stronger oxidizing agents…• are easily reduced themselves

Reduction reaction occurs more easily• have more positive Eo

• are weaker reduction agents

M+(aq) + e- M (s)

Writing spontaneous redox reactions

1. Which is the oxidizing agent?

2. Write reduction rxn for oxidizing agent (incl. Eo)

3. Flip oxidation rxn for reducing agent (incl. -Eo)

4. Multiply to make # e- lost = # e- gained

5. Add together12

2 Ag+ (aq) + Sn (s) 2 Ag (s) + Sn2+

(aq)?

Ag+ (aq) + e- Ag (s)

Sn2+ (aq) + 2e- Sn (s)

V 08.0oEV 14.0oE

Ag

Eo value does not change!

Activity Series of Metals

1. Metals that can displace H2 from acid– Ecell is positive for reaction with H+

– Any negative Ehalf-cell (reduction potential)

2. Metals that cannot displace H2 from acid– Ecell is negative for reaction with H+

– Any positive Ehalf-cell (reduction potential)

3. Metals that can displace H2 from water– Ecell is positive for reaction with water

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How much Work?

14

CJ 1V 1

Voltelectrical

potential

Jouleenerg

y

Coulombelectrical

charge

G

Max work: maxw charge cellE G

How much charge flows? FnFaraday

constantCharge of 1 mol of e-= 96,485 C / mol e-

# mols of e- transferred

= 96,485 J/V mol e-nFEcell

oG nFE ocell (standard state)

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Spontaneous

At equilibrium

Nonspontaneous

oG nFE ocell oG KRT ln

KRT ln nFE ocell

ocellE K

nF

RTln

0 oG 1K 0ocellE

0 oG 1K 0ocellE

0 oG 1K 0ocellE

Effect of Concentration on Ecell

16

G QRTGo ln

cellnFE ocellnFE

QRTnFE ocell ln cellnFE

nF

QRTE ocell

lncellE Nernst

Equation

Qn

E ocell logV 0592.0

cellE (at 298 K)

Concentration CellsCells with different concentrations of same half-reaction

17

0ocellENot standard

conditions!0cellE ?

Primary BatteriesNonrechargable• Alkaline

Zn (s) + MnO2 (s) + H2O (l) ZnO (s) + Mn(OH)2 (s)

• Mercury and Silver– Zn anode; Hg or Ag cathode– Steady output

• Primary Lithium Batteries– High energy/mass ratio– Lithium metal anode– Implanted medical devices, watches

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E = 1.5 V

Secondary Batteries

RechargeableReverse reaction using electricity

• Lead-Acid PbO2 (s) + Pb (s) + 2H2SO4 (aq) 2 PbSO4 (s) + 2 H2O

(l)Ecell = 2.1 V

• Nickel-Metal Hydride (Ni-MH)

• Lithium-Ion– Anode: Li atoms between graphite sheets– Cathode: Lithium metal oxide

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Corrosion

Natural redox:metal metal oxides and metal sulfides

Anodic regions: – Dents, ridges– Iron loss

Cathodic regions:– Surface– Forms water

Fe2+ reacts with O2:– Rust deposits

20

Electrolytic Cells electrical energy nonspontaneous reaction

21

Ecell < 0

• oxidation at anode

• reduction at cathode• anode is positive • cathode is negative

Electrolysis• Splitting a substance using electrical energy• Way to harvest elements (for industrial use)

from substances

What types of substances?• Pure molten salts

– Isolate metal or nonmetal

• Mixed molten salts– Isolate more easily reduced metal (based on

EA)22

(l)(l) 2Cl Ca 2 (g)(s) 2Cl Ca

Electrolysis of Water

AnodeCathode

Net

23

(l)O2H2

e(l) 2O2H2

(g)(g)(l) 222 2H OO2H (l)O6H2

Not at standard state:

Ecell determined using Nernst equation:

[H+] = [OH-] = 10-7 M

Qn

EE ocellcell log

V0592.0

)(H4O2 aq(g) )(OH42H 2 aq(g)

e(aq)(g) 44H O2

(aq)(g) 2OH H2

e(l) 4O4H2 (aq)(g) 4OH 2H2

2

Electrolysis of Aqueous Salts

Which is going to react: water or salt?– Reduction with less negative Eelectrode occurs

– Oxidation with less positive Eelectrode occurs

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Example: KI (aq)

Reduction:

)(K)(K seaq

el 2)(O2H2 )(OH2)(H2 aqg V93.2oEV42.0E

Oxidation:

esaq 2)(I)(2I 2

)(O2H2 l eaqg 4)(4H)(O2

V53.0oE

V82.0EH2 forms at cathodeI2 forms at anode

Electrolysis of Aqueous Salts (con’t)

• Overvoltage: Additional voltage used to produce gases (including H2 and O2) at electrodes– Usually 0.4 – 0.6 V

So what forms?1.Cations of less active metals are reduced 2.Cations of more active metals are not

reduced; Water is reduced instead3.Anions that are oxidized are typically halides4.F-, common oxoanions are not oxidized;

water is oxidized instead25

How much product forms?The amount of product is directly proportional

to quantity of charge that flows

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How long does it take to produce 0.0423 mol of Cl2 (g) by electrolysis of NaCl (aq) with power supply current of 12 A?

Cl 2Cl2 e2 2Cl 0423.0 mol2Cl

2

mol

emol -

-emol 0846.0

ee-

mol

C1065.9 mol 0846.0

4

C102.8 3 tA 12

C 102.8 3s 680

Homework due TUESDAY, May 19th

Chap 21:

#16, 21, 30, 33, 38, 42, 56, 60, 70, 89, 94, 105

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