Chapter 18 Chem1010 Msj

1

John W. MooreConrad L. StanitskiPeter C. Jurs

Stephen C. Foster • Mississippi State University

http://academic.cengage.com/chemistry/moore

Chapter 18Electrochemistry and Its

Applications

Electrochemistry

ElectrochemistryElectrochemistry is the study and use of e- flow inchemical reactions.

Redox reactions generate (and use) e-

• Those e- can be harnessed (batteries).

• Corrosion is an electrochemical reaction.

Applied e- flow:

• Can drive reactant-favored reactions toward products.

• Rechargeable batteries, electrolysis, andelectroplating…

Oxidation Number RefresherOxidation Number Refresher

• Pure element = 0.

• Monatomic ion = charge of ion.

• (ox. numbers in a species) = overall charge.

Element ox. no. Exceptions?

F −1 None

Cl, Br, I −1 Interhalogens

H +1 Metal hydrides = -1

O −2 Metal peroxides = -1

Halogen oxides

Redox Reactions

Oxidation & reduction (Redox) always occur together.

2 HCl(aq) + Mg(s) H2(g) + MgCl2(aq)

•• ReductionReduction = gain of e- = decrease in ox. no.

•• OxidationOxidation = loss of e- = increase in ox. no.

+2 e-

-2 e-+1 -1 0 0 +2 -1

Redox Reactions

H+ is reduced, Mg is oxidized.

Give oxidation numbers for each atom. Identify the oxidizingand reducing agents:

6 Fe2+ + Cr2O72- + 14 H3O

+ 6 Fe3+ + 2 Cr3+ + 21 H2O

Species Ox. number Explanation

Fe2+ +2 charge on ion

Cr2O72- O = -2; Cr = +6 O is usually -2; 2(Cr) + 7(-2) = -2

H3O+ O = -2; H = +1 O is usually -2; H is usually

+1

Fe3+ +3 charge on ion

Cr3+ +3 charge on ion

H2O O = -2; H = +1 O is usually -2, H is usually +1

Fe2+ → Fe3+ oxidationCr(+6) → Cr3+ reduction

Fe2+ = reducing agentCr2O7

2- = oxidizing agent

Redox Reactions

Redox reactions split into half reactions:half reactions:

Using Half-Reactions to Understand Redox

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Half-reactions may include different numbers of e-:

Al(s) Al3+(aq) + 3 e-

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

e- must balance in the full reaction.

2[ Al(s) Al3+(aq) + 3 e- ]

3[ Zn2+(aq) + 2 e- Zn(s) ]

2 Al(s) + 3 Zn2+(aq) 2 Al3+(aq) + 3 Zn(s)

Using Half-Reactions to Understand Redox

Redox in acidic or basic solutions are harder

(H2O, H3O+ or OH- are often omitted…).

Balance:

H3AsO4 + I2 → HAsO2 + IO3-

which occurs in aqueous acidic solution.

Balancing Redox Equations

H3AsO4 + I2 HAsO2 + IO3- (acidic solution)

(i) What is oxidized? Reduced?

I2 (I = 0) → IO3- (I = +5) oxidationoxidation

H3AsO4 (As = +5) → HAsO2 (As = +3) reductionreduction

(iii) Balance atoms (except H and O).

H3AsO4 → HAsO2 I2 → 22 IO3-

(ii) Write unbalanced half-reactions:

H3AsO4 → HAsO2 I2 → IO3-

Balancing Redox Equations in Acidic Solution

(iv) Balance O (add H2O as needed).

H3AsO4 → HAsO2 + 2 H2 H22OO I2 + 6 H6 H22OO → 2 IO3-

(vi) Balance charges (add e- ).

H3AsO4 + 2 H+ + 2 e2 e-- → HAsO2 +2 H2O

I2 + 6 H2O → 2 IO3- + 12 H+ + 10 e10 e--

zero charge 2(-1) + 12(+1)} 10(-1)0 = 10 + -10

(v) Balance H (add H+ as needed).

H3AsO4 + 2 H2 H++ → HAsO2 +2 H2O

I2 + 6 H2O → 2 IO3- + 12 H12 H++


(viii) Make H3O+ (H2O + H+). Add H2O if needed.

5 H5 H33AsOAsO44 + I+ I22 →→ 5 HAsO5 HAsO22 + 2 H+ 2 H22O + 2 IOO + 2 IO33-- + 2 H+ 2 H33OO

++

(vii) Equalize e- and add.

5 [ H3AsO4 + 2 H+ + 2 e- → HAsO2 +2 H2O ]

1 [ I2 + 6 H2O → 2 IO3- + 12 H+ + 10 e- ]

5 H3AsO4 + 10 H+ + 10 e- + I2 + 6 H2O

→ 5 HAsO2 + 10 H2O + 2 IO3- + 12 H+ + 10 e-

5 H3AsO4 + I2 → 5 HAsO2 + 4 H2O + 2 IO3- + 2 H+

2H+

4H2O


Balance the following (basic conditions):N2 + S2- S + N2H4

(i) Oxidized? Reduced?

N2 (N = 0) → N2H4 (N = -2) reductionreduction

S2- (S = -2) → S (S = 0) oxidationoxidation

(iii) Balance (except H and O).

N2 → N2H4 S2- → S

(ii) Unbalanced half-reactions:

N2 → N2H4 S2- → S

Balancing Redox Equations in Basic Solution

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(iv) Balance O (add H2O as needed).

N2 → N2H4 S2- → S

(v) Balance H (add H+ as needed).

N2 + 4 H4 H++ → N2H4 S2- → S

(vi) Balance charges (add e- ).

N2 + 4 H+ + 4 e4 e-- → N2H4

S2- → S + 2 e2 e--


(vi) Equalize e- and add.1 [ N2 + 4 H+ + 4 e- → N2H4 ]

2 [ S2- → S + 2 e- ]

N2 + 4 H+ + 4 e- + 2 S2- → N2H4 + 2 S + 4 e-

N2 + 4 H+ + 2 S2- → N2H4 + 2 S

(vii) Make H2O (H+ + OH-). Add OH-.

N2 + 4 H+ + 4 OH4 OH-- + 2 S2- → N2H4 + 2 S + 4 OH4 OH--

NN22 + 4 H+ 4 H22O + 2 SO + 2 S22-- →→ NN22HH44 + 2 S + 4 OH+ 2 S + 4 OH--


Linked oxidation and reductionreactions.

e- move across an externalconductor.

Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)

Also called a voltaic cellvoltaic cell or a batterybattery.

• A battery is strictly a series of linked voltaic cells.

e- e-

Electrochemical CellsCu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)

e- e-

Electrochemical Cells

Anode(oxidation)

Cathode(reduction)

Saltbridge

ElectrodesElectrodes (anode & cathode)

Allow e- to pass in and out ofsolution.

A salt bridge (or porous barrier)is required...

Salt bridgeSalt bridge

Contains a salt solution (e.g. K2SO4 ).

Ions pass into the cells (restricts bulk flow).

Stops charge buildup.

SO42- released

as Zn → Zn2+

2 K+ releasedas Cu2+ → Cu

K2SO4

SO42-

K+

Zn Cu

Zn2+Cu2+

porousplug

Electrochemical Cells Electrochemical Cells

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Zn Zn2+ + 2 e-

Zinc is removed:

•• OOxidation at the aanode (both vowels).

• Zn supplies e-.

• Anode has “-” charge.

Copper is deposited:

•• RReduction at the ccathode (both consonants).

• Cu2+ accepts e-.

• Cathode has “+” charge.

Cu2+ + 2 e- Cu


A compact notation:

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


Current flows from anode to cathode.

| = phase boundary.

|| = salt bridge.

Details (e.g. concentration) listed after each species.

anode cell cathode cell

Electrochemical Cells & Voltage

Electrical work = charge x ΔEp

= (number of e-) ΔEp

SI UnitsSI Units

Charge: 1 coulomb (C) = 1 ampere x second = 1 As

Potential: 1 volt (V) = 1 J C-1

Voltage depends on cellchemistry ≠ size.

Charge depends on nreactants ≈size.

Cell voltage varies if conditions vary.

A standard voltagestandard voltage ( E° ) occurs if:occurs if:

• All [solute] = 1 M.

or saturated if the solubility < 1 M.

• All gases have P = 1 bar.

• All solids are pure.


E°cell is positivepositive = product favoredproduct favored reaction

(E°cell < 0 is reactant favoredreactant favored)


Absolute voltages cannot be measured.

• They are measured relative to a standard electrode.

Standard hydrogen electrode (SHE)Standard hydrogen electrode (SHE)

Pt | H2(1bar), 1M H3O+ ||

E° = 0 V (oxidation & reduction).


2 H3O+(aq, 1M) + 2 e- H2(g, 1 bar) + 2 H2O (ℓ)

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Reduction Half Reaction E° (V)

F2(g) + 2 e- → 2 F-(aq) +2.87

H2O2(aq) + 2 H3O + 2 e- → 4 H2O(ℓ) +1.77

MnO4-(aq)+8 H3O

+ + 5 e- → Mn2+(aq) + 12 H2O(ℓ) +1.51

Cl2(g) + 2 e- → 2 Cl-(aq) +1.358

Br2(g) + 2 e- → 2 Br-(aq) +1.066

Ag+(aq) + e- → Ag(s) +0.799

Cu2+(aq) + 2 e- → Cu(s) +0.337

2 H3O+(aq) + 2 e- → H2(g) + 2 H2O(ℓ) 0.00

Ni2+(aq) + 2 e- → Ni(s) -0.25

Fe2+(aq) + 2 e- → Fe(s) -0.44

Zn2+(aq) + 2 e- → Zn(s) -0.763

Al3+(aq) + 3 e- → Al(s) -1.66

Li+(aq) + e- → Li(s) -3.045

Using Standard Cell Potentials

Tabulated asreductions.

If an equation is reversed, E° → -1 x E°

E°oxidation = - E°reduction.

E° tables are reduction values, so:

The overall voltage:

E°cell = E°Zn2+, reduction + E°Cu, oxidation


EE°°cellcell == EE°°reductionreduction ++ EE°°oxidationoxidation

EE°°cellcell == EE°°cathodecathode -- EE°°anodeanode

Cu2+(aq) + 2 e- Cu(s) reduction

Zn(s) Zn2+(aq) + 2 e- oxidation


E°cell = E°cathode - E°anode

E°cell = 0.34 – (-0.76) V

= 1.10 V

Cu2+ + 2 e- → Cu reduction +0.34 V +0.34 V

Zn → Zn2+ + 2 e- oxidation -0.76 V +0.76 V

Reaction Process E°red (Table) E°

Or E°cell = E°oxid + E°red

= 0.34 + 0.76 V = 1.10 V


Ag+(aq) + e- → Ag(s) +0.799. .. .. .

Ni2+(aq) + 2 e- → Ni(s) -0.25

What is E° for a Ni(s) | Ni2+|| Ag+ | Ag(s) cell?

Anode written on the left; cathode the right.

E° = E°cathode - E°anode = 0.799 – (-0.25) V = 1.05 V


Ni loses mass:

Ni(s) → Ni2+(aq) + 2 e-

Ni is ooxidized (aanode).

If Ni(s)| Ni2+(aq, 1M) is connected to SHE, the Nielectrode loses mass over time. E°cell = 0.25 V. Is Nioxidized or reduced? What is E° for the Ni half cell?

Since: E°cell = E°cathode - E°anode

0.25 V = E°SHE - E°anode = 0 - E°anode

E°anode = -0.25 V

(Note: this is the tabulated reduction value)


1. Tabulated half-cell E° are reductions.

2. Reactions can be reversed.

3. More positive E° = easier reduction.

4. Less positive E° = easier oxidation (for the reversereaction).

5. A “left” species, will oxidize any “right” below it.

6. E° depends on [reactant] and [product], but not onnreactant or nproduct (i.e notnot stoichiometric coefficient).


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Will Zn(s) react with a 1 M iron(III) solution? If so whatis E° for the reaction?

Yes!Yes! Fe3+ (“left”) is higher than Zn (“right”).



F2(g) + 2 e- → 2 F-(aq) +2.87

Fe3+(aq) + e- → Fe2+(aq) +0.771

2 H3O+(aq) + 2 e- → H2(g) + 2 H2O(ℓ) +0.0

Zn2+(aq) + 2 e- → Zn(s) -0.763

2 Fe3+ + Zn → 2 Fe2+ + Zn2+ E°cell = +1.534 V

Will Zn react with a 1 M iron(III) solution? If so what is E° for the reaction.

2 Fe3+ + 2 e- → Fe2+ E° = +0.771 VZn → Zn2+ + 2 e- E° = -1(-0.763 V)


Note2 x (Fe3+ reaction) to balance e-.

E° (Fe3+) is notnot doubled.

Sn4+ + 2 e- → Sn2+ (s) +0.15 V

Sn2+ + 2 e- → Sn (s) -0.14 V

Fe2+ + 2 e- → Fe (s) -0.44 V

Al3+ + 3 e- → Al (s) -1.66 V

a) Will Al(s) react with a 1 M tin(IV) solution?

b) Will 1 M Fe2+ react with Sn(s)?

(a) Yes – “left” (Sn4+) above “right” (Al).

(b) No – “left” (Fe2+) below “right” (Sn).


What’s the voltage of: Al(s)|Al3+,1M || Sn2+,1M|Sn(s) ?

Balance e-:

3(Sn2+ + 2 e- → Sn) E° = -0.14 V

2(Al → Al3+ + 3 e- ) E° = +1.66 V

3 Sn2+ + 2 Al → 3 Sn + 2 Al3+ E°cell = +1.52 V


Sn2+ + 2 e- → Sn E° = -0.14 V

Al → Al3+ + 3 e- E° = -1(-1.66 V)

E° and Gibbs Free Energy

Product-favored reactions: ΔG° < 0.

Spontaneous cell reactions: E°cell > 0.

ΔΔGG°° == ––n F En F E°°cellcell

with n = moles of e- transferred,

F = Faraday constantFaraday constant = charge/(mol of e-).

= (e- charge) x (Avogadro’s number).

= (1.60218 x 10-19 C)(6.02214 x 1023 mol-1).

F = 96,485 C/mol = 96,500 C/mol (3 sig. fig.)

Cu2+ + Zn(s) → Cu(s) + Zn2+ E°cell = 1.10 V

Spontaneous.

ΔG° = − nFE°cell = −2 mol (96500 C/mol)(1.10 V)

= −2 .12 x105 J (1 J = 1 C V)

= −212 kJ

E° and Gibbs Free Energy

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Since: ΔG° = − RT ln K° = −nFE°cell

At 298 K: E°cell = ln K° or,

E°cell = log K°

0.0257 Vn

0.0592 Vn

R Tn F

E°cell = ln K° = (2.303) log K°

ΔG°, E°cell, and K°

R Tn F

Determine K° for:

Cu2+ + Zn(s) → Cu(s) + Zn2+ E°cell = 1.10 V

0.0592 V2

1.10 V = log K°

log K° = 37.16

K° = 1037.16 = 1.5 x 1037

E°cell = log K°0.0592 Vn

ΔG°, E°cell, and K°

2 mol e-

E° values apply if [solute] = 1 M (or saturated).

Other conditions:

Ecell = E°cell − ln QRTnF

Nernst equationNernst equation

At 298 K:

Ecell = E°cell − log Q

Ecell = E°cell − ln Q

0.0592 Vn

0.0257 Vn

Effect of Concentration on Cell Potential

What is the voltage for:

Cu2+ + Zn(s) → Cu(s) + Zn2+

if [Cu2+] = 0.1 M and [Zn2+] = 5.0 M. E°cell = 1.10 V.

E = E° − log0.0592

2

[Zn2+]

[Cu2+]

E = 1.10 − log0.0592

2

5.0

0.1

= 1.05 V

Effect of Concentration on Cell Potential

Concentration Cells

Concentration dependence leads to:

Zn | Zn2+ (dilute) || Zn2+ (conc.) | Zn Ecell ≠ 0 V

ExampleExample Zn | Zn2+ (0.01M) || Zn2+ (1M) | Znanode = oxidation cathode = reduction

Zn(s) → Zn2+ (0.01M) + 2 e-

Zn2+ (1M) + 2 e- → Zn(s)

Zn2+(1M) → Zn2+(0.01M) (net reaction)

Concentration Cells

E = E° − log0.0592

2

[Zn2+]dilute

[Zn2+]conc

E = 0.0 − log0.0592

2

0.010

1

E = 0.0592 V

E = 0.0 − log 10-20.0592

2

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Common Batteries

Primary batteryPrimary battery

One time use. Not easily rechargeable

Secondary batterySecondary battery

Rechargeable battery.

Primary BatteriesAlkaline Battery:

Zn(s) + 2 OH-(aq) → ZnO(aq) + H2O(ℓ) + 2 e-

MnO2(s) + H2O(ℓ) + e- → MnO(OH)(s) + OH-(aq)

OverallOverall

Zn(s) + H2O(ℓ) + 2 MnO2(s) → ZnO(aq) + 2 MnO(OH)(s)

Ecell = 1.54 V when new.

Primary Batteries

Mercury battery

Zn(s) + 2 OH- → ZnO(aq) + H2O(ℓ) + 2 e-

HgO(s) + H2O(ℓ) + 2 e- → Hg(ℓ) + 2 OH-(aq)

Zn(s) + HgO(s) → Hg(ℓ) + ZnO(aq) Ecell = 1.35 V

Secondary Batteries

Pb(s) + HSO4-(aq) + H2O(ℓ) → PbSO4(s) + H3O

+(aq) + 2 e-

PbO2(s) + 3 H3O+(aq) + HSO4

-(aq) + 2 e- → PbSO4(s) + 5 H2O

Pb + PbO2(s) + 2 H3O+ + 2 HSO4

-(aq) → 2 PbSO4(s) + 4 H2O

Net E° = +2.041 V

Lead-Acid Battery (high capacity, high current).

• Insoluble PbSO4 stays on the electrodes• The reaction is reversed by recharging.

Lead-Acid Storage Battery

6 cells in series (12 V).

Secondary Batteries

Cd(s) + 2 OH-(aq) → Cd(OH)2(s) + 2 e-

2[NiO(OH)(s) + H2O(ℓ) + e- → Ni(OH)2(s)+ OH-(aq)]

Cd(s) + 2 NiO(OH)(s) + 2 H2O → Cd(OH)2(s) + 2 Ni(OH)2(s)

net: E° = +1.299 V

Nickel-Cadmium (Nicad).

InsolubleInsoluble(Rechargable…)

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Secondary Batteries

MH(s) + OH-(aq) → M(s) + H2O(ℓ) + e-

NiO(OH)(s) + H2O(ℓ) + e- → Ni(OH)2(s) + OH-(aq)

MH(s) + NiO(OH)(s) → M(s) + Ni(OH)2(s) E°cell =+1.4 V

Nickel-metal hydride (NiMH). Doesn’t use toxic Cd.

M is a metal alloy in KOH

Secondary Batteries

Li(s) (in polymer) → Li+ (in polymer) + e-

Li+ ( in CoO2 ) + e- + CoO2 → LiCoO2

Li(s) + CoO2(s) → LiCoO2(s)

Ecell = 3.4 V

Lithium Ion. Low mass, high energy density.

V

e-e-

O2 in

H2O out

H2 in

H2 out

Anode Cathode

H+ exchangemembrane

Pt catalyst

gases flowthrough

channels

H+

H+

Fuel CellsConvert bond energy into electricity.

ProtonProton--ExchangeExchangeMembraneMembrane (PEM) fuel cell.

H2 → 2 H+ + 2 e-

½ O2 + 2 H+ + 2 e- → H2O

Graphite electrodes.

Pt catalyst coated on bothsides of the membrane.

Electrolytic cellElectrolytic cell: Applied voltage forces a reaction tooccur. e.g. electrolysis of molten NaCl:

2 Na+ + 2 e- → 2Na (ℓ)

2 Cl- → Cl2 (g) + 2 e-

2 Na+ + 2 Cl- → Cl2 (g) + 2 Na(ℓ)

Na(ℓ) and Cl2(g) produced if > 4.1 V is applied.

However, melting NaCl takes lots of energy…

Electrolysis

E° = -2.714 V

E° = -1.358 V

E° = -4.072 V

Aqueous solutions?Aqueous solutions? Other reactions can occur.Consider KI(aq):

2 I-(aq) → I2(aq) + 2 e-

6 H2O(ℓ) → O2(g) + 4 H3O+(aq) + 4 e-

Electrolysis

K+(aq) + e- → K(s)

2 H2O(ℓ) + 2 e- → H2(g) + 2 OH-(aq)

The most positive E half reactions occur…

2 H2O + 2 I- → H2 + I2 + 2 OH- Ecell = -1.363 V

E° = -2.925 V

E° = -0.828 V

E° = -0.535 V

E° = -1.229 V

E°ox → -1 x E°red

red

uctio

ns

oxid

atio

ns

Electrolysis

Brown = I2 ; Purple = OH- (phenolphthalein)

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F2(g) +2 e- → 2 F-(aq) +2.87

H2O2(aq) + 2 H3O+2 e- → 4 H2O(ℓ) +1.77

Cl2(g) +2 e- → 2 Cl-(aq) +1.358

O2(g) + 4 H3O+(aq) + 4 e- → 6 H2O(ℓ) +1.229

Br2(g) +2 e- → 2 Br-(aq) +1.066

Ag+(aq) + e- → Ag(s) +0.799

2 H3O+(aq) + 2 e- → H2(g) + 2 H2O(ℓ) 0.00

Ni2+(aq) + 2 e- → Ni(s) -0.25

Fe2+(aq) + 2 e- → Fe(s) -0.44

Zn2+(aq) + 2 e- → Zn(s) -0.763

2 H2O(ℓ) + 2 e- → H2(g) + 2 OH-(aq) -0.8277

Al3+(aq) + 3 e- → Al(s) -1.66

Na+(aq) + e- → Na(s) -2.714

K+(aq) + e- → K(s) -2.925

Electrolysis

Aqueousoxidation willnot occur.

Aqueousreduction willnot occur.

SummarySummary

Metal ions are reduced if E°red > −0.8 V

Aqueous Na+, K+, Mg2+, Al3+ … cannot be reduced.

Anions can be oxidized if E°red < +1.2 V

Aqueous F- … cannot be oxidized.

In practice Erequired > E calculated.

Overvoltage is needed

• Cl-(aq) can be oxidized to Cl2(g).

Electrolysis

ChlorineChlorine is produced from NaCl(aq) by the chlor-alkaliprocess.

2 Cl-(aq) → Cl2(g) + 2 e-

2 H2O(ℓ) + 2 e- → 2 OH-(aq) + H2(g)

NaOH(aq) is 21- 30% NaOH by weight.

Electrolysis of Brine Counting Electrons

2 mol e- ≡ 1 mol Cu.

Also charge = current x time

1 coulomb = 1 ampere x 1 second

1 C = 1 A s

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

Counting ElectronsDetermine the mass of Cu plated onto an electrode froma Cu2+ solution by the application of a 10. A current for10. minutes.

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

Charge = 10. A x 10. min x (60 s/min) = 6.0 x 103 C

1 mol e-96500 C

63.55 g1 mol Cu

1 mol Cu2 mol e-

6.0 x103 C = 2.0 g

Counting ElectronsWhat is the cost to produce 14. g of Al (mass of a sodacan) by the reduction of Al3+. Assume V = 4.0 V and 1kWh of electricity costs 10 cents.

Al3+ + 3 e- Al

96500 C1 mol e-

1 mol Al26.98 g

3 mol e-

1 mol Al14. g = 1.5 x 105 C

Energy used = (1.5 x105 C)(4.0V)1 J

1 C x 1V1 kWh

3.60 x 106 J

= 0.17 kWh (or 1.7 cents)

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Anode:Anode: M(s) Mn+ + n e-

Cathode:Cathode: often involve water and/or O2

O2(g) + 2 H2O(ℓ) + 4 e- → 4 OH-(aq)

2 H2O(ℓ) + 2 e- → 2 OH-(aq) + H2(g)

Anode and cathode must be electrically connected.(The metal itself acts as the conductor).

Corrosion: Product-Favored Reactions

Oxidation of metals by the environment.

Anode: 2[Fe(s) Fe2+ + 2 e-]

Cathode: O2(g) + 2 H2O(ℓ) + 4 e- 4 OH-(aq)

2 Fe(s) + O2(g) + 2 H2O(ℓ) 2 Fe(OH)2(s)

Iron(II) hydroxide is converted to rust by O2 and H2O:


Rust(Fe2O3·xH2O(s); x = 2 - 4)

Iron “rusts”

Iron nails corroding:

• Blue = Fe2+ indicator

• Purple = OH- indicator


Stress points corrodequickly.

Corrosion Protection

Anodic inhibitionAnodic inhibition

• Paint or coat the surface.

• Form thin metal oxide coat:

2 Fe(s) + 2 Na2CrO4(aq) + 2 H2O(ℓ)

→ Fe2O3(s) + Cr2O3(s) + 4 NaOH(aq)

Cathodic protectionCathodic protection – Attach a more reactive metalwhich will corrode first.

Insoluble coating;impervious to O2 and H2O

Insoluble coating;impervious to O2 and H2O

Corrosion PreventionIron can also be galvanized (coated with Zn):

Zn(OH)2 (insoluble film) forms on thesurface.

Zn → Zn2++ 2 e- E° = 0.763 V

Fe → Fe2+ + 2 e- E° = 0.44 V

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Chapter 18 Chem1010 Msj