chapter 20 electrochemistry • electrochemistry deals with the

Chapter 20

20-1

Chapter 20Electrochemistry

• Electrochemistry deals with the relationships between electricity and chemical reactions.• Oxidation-reduction (redox) reactions were introduced in Chapter 4• Can be simple displacement reactions:

Zn(s) + Cu2+(aq) � Zn2+(aq) + Cu(s)Cu(s) + 2Ag+(aq) � Cu2+(aq) + 2Ag(s)

20.1 Oxidation-Reduction Reactions• Redox reactions can also be more complex, with structural and composition changes as well

as an exchange of electrons.• 5VO2+(aq) + MnO4

-(aq) + H2O(l) � 5VO2+(aq) + Mn2+(aq) + 2H+(aq)

• 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) � 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

Oxidation and ReductionYou can�t have one without the other!

OxidationDecrease in number of electrons

(loss of electrons)Increase in oxidation number

Oxidation number: -3 -2 -1 0 +1 +2 +3

Increase in number of electrons(gain of electrons)

Decrease in oxidation numberReduction

Half-Reactions• Oxidation-reduction reactions can be written as separate oxidation and reduction reactions,

called half-reactions.• Oxidation:

VO2+(aq) + H2O(l) � VO2+(aq) + 2H+(aq) + e-

• VO2+ is called the reducing agent (or reductant), because it causes the reduction of anothersubstance; the reducing agent is oxidized in the process

• Reduction:MnO4

-(aq) + 8H+(aq) + 5e- � Mn2+(aq) + 4H2O(l)• MnO4

- is called the oxidizing agent (or oxidant), because it causes the oxidation of anothersubstance; the oxidizing agent is reduced in the process

• Oxidation-reduction can be considered to be the competition between two substances forelectrons. The one with the greater attraction for additional electrons becomes the oxidizingagent; the one with the least is the reducing agent.

Chapter 20

20-2

Spontaneous Oxidation-Reduction Reactions• If we place a metal in a solution of another metal ion, sometimes we get metal deposition,

sometimes not. How do we decide?Zn + Sn2+ � Zn2+ + SnSn + Zn2+ � no rxn

• How do we know which substances will act as oxidizing agents or as reducing agents?• An activity series gives the relative activity of substances as oxidizing or reducing agents.

Review from Chapter 4.• Determine an activity series in several ways:• activity in displacing H2 from water• activity in displacing metals from metal ion solutions (more active metal displaces a less

active metal from solution)• generation of an electrochemical potential or voltage

Spontaneous Redox• Na displaces H2 from H2O• Zn displaces Pb from a solution of Pb2+

• Pb + 2Ag+ � Pb2+ + 2Ag• Ag + Pb2+ � no rxn• The reverse reactions can be made to occur with electric current, but they are not

spontaneous.• General rule, using an activity series:

stronger stronger weaker weakeroxidizing + reducing � reducing + oxidizingagent agent agent agent

• Depending on the relative strengths, the reaction can go to completion, or reach a state ofequilibrium.

Activity Series

Chapter 20

20-3

• Predict which of the following combinations will undergo an oxidation-reduction reaction:Mg + K+

Mg + Zn2+

H2 + Ni2+

H2 + Cu2+

20.2 Balancing Oxidation-Reduction Equations• Some redox equations can be balanced by inspection, just like other types of reactions.

Zn + CuCl2 � ZnCl2 + Cu• Net ionic equation:

Zn + Cu2+ � Zn2+ + Cu• Need to balance atoms and charge• What is wrong with the following?

Al + Cu2+ � Al3+ + Cu

Balancing Equations• 5Cr3+ + 3MnO4

- + 8H2O � 5CrO42- + 3Mn2+ + 16H+

• How do we balance an equation as complex as this?• Two methods:

• half-reaction method• oxidation number change method

• Will focus on the half-reaction method since it is useful in more circumstances.

Half-Reaction Method• Write an unbalanced half-reaction for either oxidation or reduction.• Balance the half-reaction:

a. Balance all atoms other than H and O. b. Balance O by adding H2O to the equation. c. Balance H by adding H+ to the equation. d. Balance ionic charges by adding electrons to the equation. e. If in basic solution, add OH- to each side of the equation to neutralize all the H+. f. Cancel any H2O occurring on both sides.

• Write an unbalanced half-reaction for the other process.• Balance by the same procedure.• Equalize the number of electrons lost and gained by multiplying each coefficient in each

half-reaction by the appropriate constant.• Add the two half-reactions and cancel equal amounts of anything occurring on both sides.• Make a final check of atom and charge balances.

Balancing Equations• Cr3+ + MnO4

- � CrO42- + Mn2+

• How do we balance this equation in acidic solution?1. Cr3+ � CrO4

2-

2a. Cr already balanced2b. Cr3+ + 4H2O � CrO4

2-

2c. Cr3+ + 4H2O � CrO42- + 8H+

Chapter 20

20-4

2d. Cr3+ + 4H2O � CrO42- + 8H+ + 3e-

3. MnO4- � Mn2+

4a. Mn already balanced4b. MnO4

- � Mn2+ + 4H2O4c. MnO4

- + 8H+ � Mn2+ + 4H2O4d. MnO4

- + 8H+ + 5e- � Mn2+ + 4H2O5. 3 e- lost, but 5 e- gained. Equalize by multiplying half-reactions:

5(Cr3+ + 4H2O � CrO42- + 8H+ + 3e-)

3(MnO4- + 8H+ + 5e- � Mn2+ + 4H2O)

5Cr3+ + 20H2O � 5CrO42- + 40H+ + 15e-

3MnO4- + 24H+ + 15e- � 3Mn2+ + 12H2O

6. Add the equations:5Cr3+ + 3MnO4

- + 20H2O + 24H+ + 15e- � 5CrO42- + 3Mn2+ + 12H2O + 40H+ + 15e-

Cancel any substances on both sides of the equation:5Cr3+ + 3MnO4

- + 8H2O � 5CrO42- + 3Mn2+ + 16H+

7. Check atom and charge balance:5 Cr, 3 Mn, 20 O, 16 H, 12 + charges on each side

Balancing Equations in Basic Solution• Neutralize any H+ with OH- to form H2O and add the same number of OH- to the other side

of the equation.Cr3+ + 4H2O � CrO4

2- + 8H+ + 3e-

• Add 8OH- to each side of the equation:Cr3+ + 4H2O + 8OH- � CrO4

2- + 8H+ + 8OH- + 3e-

• Form water by neutralization:Cr3+ + 4H2O + 8OH- � CrO4

2- + 8H2O + 3e-

• Cancel any water occurring on both sides:Cr3+ + 8OH- � CrO4

2- + 4H2O + 3e-

• Balance the following equation in acidic solution:MnO4

- + Cl- � Mn2+ + ClO3-

20.3 Voltaic Cells• Voltaic (or galvanic) cells: spontaneous redox � electricity (or electrical work)• Electrolysis cells: electricity � redox• To produce electricity, we must direct the electron flow through an external circuit. We

cannot have direct redox.• Daniell cell: Zn - Cu

Voltaic Cells• To produce electricity, we need:

• Isolated half-reactions, using half-cells• Conductive solids (electrodes) connected by external circuits

• May consist of a reactant/product or be an inert substance such as platinum orgraphite

• Anode: oxidation half-reaction

Chapter 20

20-5

• Cathode: reduction half-reaction• Externally, the anode has the negative charge; internally, it has a positive charge)• Anions flow towards the anode; cations move away from it and towards the cathode.• Two half-cells must be connected to pass ions• Salt bridge or porous glass

20.4 Cell EMF• If the half-reactions are carried out separately (but coupled), we find they generate an

electrical current characterized by a voltage (or electromotive force or electrical potential).This is the force pushing electrons through the circuit.

• The voltage produced by a voltaic cell is called the cell potential, Eocell (also the reaction

potential, Eorxn, when the half-reactions are not separated)

• The voltage is called the standard electromotive force (emf), Eo, under standard conditions.

Electromotive Force• Eo

cell = Eoox + Eo

red• Values are determined relative to the standard hydrogen electrode, with Eo = 0 V• Eo

red = emf for the reduction half-rxn• Eo

ox = emf for the oxidation half-rxn• Eo

cell = Eored + Eo

ox• We can only measure Eo

cell, so we must define a standard reference to get Eoox and Eo

red.

Eo Values• Reference for Eo is the standard hydrogen electrode, using the reaction:• 2H+(aq) + 2e- � H2(g) Eo = 0 1 M 1 atm (std conditions)

Cell Potential• Other values are then measured from Eo

cell with the standard hydrogen electrode or withother known half-cells.• AgCl + e- � Ag + Cl- Eo = 0.22 V• Hg2Cl2 + 2e- � 2Hg + 2Cl- Eo = 0.2802 V

• Half-cells such as these are used as reference electrodes. The Ag/AgCl electrode, alongwith a glass electrode, is used in a pH meter.

Eo Values• Then get other half-reaction potentials from measured Eo

cell values.• Zn + 2H+ � Zn2+ + H2 Eo

cell = 0.76 V• Zn � Zn2+ + 2e- Eo

ox = 0.76 V since Eored = 0.00 V

Reference to Hydrogen Electrode• The Eo of the hydrogen electrode is defined as 0.00 V. What would be the values of Eo for

other half-reactions if this were defined as 1.00 V?• The Eo of the hydrogen electrode is defined as 0.00 V. What would be the values of Eo for

other half-reactions if this were defined as 1.00 V?

Chapter 20

20-6

• What would be the Eo values of overall redox reactions if the reference were defined as 1.00V?

• What would be the Eo values of overall redox reactions if the reference were defined as 1.00V?

Eo Values• If we reverse a half-reaction, we change the sign of Eo.

Zn2+ + 2e- � Zn Eored = -0.76 V

• Measure a value of Eocell = 0.63 V for the following reaction.

Zn + Pb2+ � Zn2+ + Pb• What is the Eo

red ofPb2+ + 2e- � Pb ?

• Zn � Zn2+ + 2e- Eoox = 0.76 V

Pb2+ + 2e- � Pb Eored = ?

Zn + Pb2+ � Zn2+ + Pb Eocell = 0.63 V

• Eocell = Eo

red + Eoox

0.63 V = Eored + 0.76 V

Eored = 0.63 V - 0.76 V = -0.13 V

• Values determined in this way are listed in Table 20.1 and Appendix E.

Reduction Potentials

20.5 Spontaneity of Redox Reactions• A redox reaction is spontaneous if Eo

rxn > 0.• To determine spontaneity, add the two half-reactions and their Eo values to see if Eo

rxn has apositive value.

• What will happen if we place a piece of Zn and a piece of Cu in a solution that contains amixture of Zn2+ and Cu2+?

Chapter 20

20-7

• Two possibilities:• Zn + Cu2+ � Zn2+ + Cu• Cu + Zn2+ � Cu2+ + Zn

Spontaneous Redox• Two possible reduction half-reactions: Zn2+ + 2e- � Zn Eo

red = -0.76 VCu2+ + 2e- � Cu Eo

red = 0.34 V• Two possible oxidation half-reactions:

Zn � Zn2+ + 2e- Eoox = 0.76 V

Cu � Cu2+ + 2e- Eoox = -0.34 V

• Two ways to combine them:Zn2+ + 2e- � Zn Eo

red = -0.76 VCu � Cu2+ + 2e- Eo

ox = -0.34 V�� Cu + Zn2+ � Cu2+ + Zn Eo

rxn = -1.10 V• Two ways to combine them:

Cu2+ + 2e- � Cu Eored = 0.34 V

Zn � Zn2+ + 2e- Eoox = 0.76 V

�� Zn + Cu2+ � Zn2+ + Cu Eo

rxn = 1.10 V• The combination with Eo

rxn > 0 is spontaneous.

Predicting Reactions• Vanadium Reduction Potentials:

VO2+ + 2H+ + e- � VO2+ + H2O E°red = +1.00 V

VO2+ + 2H+ + e- � V3+ + H2O E°red = +0.36 VV3+ + e- � V2+ E°red = -0.26 VV2+ + 2e- � V E°red = -1.20 V

• Chromium Reduction Potentials:Cr2O7

2- + 14H+(aq) + 6e- � 2Cr3+(aq) + 7H2O(l) E°red = +1.33 VCr3+(aq) + e- � Cr2+(aq) E°red = -0.41 VCr2+(aq) + 2e- � Cr(s) E°red = -0.91 V

• Manganese Reduction Potentials:MnO4

-(aq) + e- � MnO42-(aq) E°red = +0.56 V

MnO42-(aq) + 4H+(aq) + 2e- � MnO2(s) + 2H2O(l) E°red = +2.26 V

MnO2(s) + 4H+(aq) + e- � Mn3+(aq) + 2H2O(l) E°red = +0.95 VMn3+(aq) + e- � Mn2+(aq) E°red = +1.51 VMn2+(aq) + 2e- � Mn(s) E°red = -1.18 V

Predicting Reactions• Add excess Cr2+ to VO2

+. What is the product?Cr3+(aq) + e- � Cr2+(aq) E°red = -0.41 VCr2+(aq) � Cr3+(aq) + e- E°ox = 0.41 VVO2

+ + 2H+ + e- � VO2+ + H2O E°red = +1.00 VVO2+ + 2H+ + e- � V3+ + H2O E°red = +0.36 V

Chapter 20

20-8

V3+ + e- � V2+ E°red = -0.26 VV2+ + 2e- � V E°red = -1.20 V

• Product is V2+, because successive Eorxn = 1.41, 0.77, 0.15, -0.79 V

Predicting Reactions• Add excess Cr to Cr2O7

2-. What is the product?Cr2O7


• Product is Cr2+; the successive values of Eorxn are 2.24 and 0.50 V.

Predicting Reactions• Add excess Cr2+ to Cr2O7

2-. What is the product?Cr2O7


Predicting Reactions• Add excess V2+ to MnO4

-. What is the product?V2+ � V3+ + e- E°ox = 0.26 VMnO4

-(aq) + e- � MnO42-(aq) E°red = +0.56 V



Predicting Reactions• Add Mn2+ to MnO4

-. What is the product?MnO4

-(aq) + e- � MnO42-(aq) E°red = +0.56 V



• Predict no reaction, but reaction actually occurs to form Mn3+ or MnO2. We will deal withthis discrepancy in the next section. It arises from the fact that there are more half-reactionsthat could be considered, which arise from combinations of these half-reactions.

Stability in Aqueous Systems• Disproportionation Reactions

A substance reacts with itself to form new substances with higher and lower oxidationnumbers.

• No examples with V or Cr.• MnO4

2- will disproportionate:MnO4

-(aq) + e- � MnO42-(aq) E°red = +0.56 V


Chapter 20

20-9

��3MnO4

2- + 4H+ � 2MnO4- + MnO2 + 2H2O Eo

rxn = 1.70 VStability in Aqueous Systems

• Mn3+ will disproportionate:MnO2(s) + 4H+(aq) + e- � Mn3+(aq) + 2H2O(l) E°red = +0.95 VMn3+(aq) + e- � Mn2+(aq) E°red = +1.51 V��2Mn3+ + 2H2O � Mn2+ + MnO2 + 4H+ Eo

rxn = 0.56 V

Stability in Aqueous Systems• Reaction with Water Reduce hydronium ion to release hydrogen gas:

2H+(aq) + 2e- � H2(g) E°red = 0.000 V• Any substance with Eo

ox > 0 will reduce H+ to H2• Examples are V, V2+, Cr, Cr2+, Mn• The ions will react, but tend to react only very slowly. There seems to be a kinetic factor

that results in a fast reaction only if Eorxn > 0.4-0.5 V (called an overvoltage).

• Reaction with Water Oxidize water to release oxygen gas:

2H2O(l) � O2(g) + 4H+(aq) + 4e- E°ox = -1.23 V• Any Eo

red > 1.23 V will result in production of O2. Generally need Eorxn > 0.4-0.5 V for fast

reaction.• Examples are Cr2O7

2- (very slow), MnO42- (disproportionates faster), Mn3+

(disproportionates faster)

Stability in Aqueous Systems• Oxidation by O2 in Air

O2(g) + 4H+(aq) + 4e- � 2H2O(l) E°red = 1.23 V• Any Eo

ox > -1.23 V will result in oxidation by air. Many substances fall into this category(Eo

rxn > 0.4-0.5 V for fast reaction).V Cr MnV2+ Cr2+ not Mn2+

V3+ not Cr3+ Mn3+

VO2+ (very slow) not MnO2 MnO4

2- (disproportionates faster)

EMF and Free Energy Change• We have seen three criteria for spontaneity:

• Eo > 0• ∆Go < 0 1 J = 1 coul x 1 V• K >> 1

• These criteria are related:• ∆Go = -RT lnK• ∆Go = - nFEo, ∆G = - nFE

where n = number of e- transferred and F = Faraday constant (charge on 1 mole e-)• 1 F = 96,500 coul/mol e- = 96,500 J/V mol e-

Chapter 20

20-10

Thermodynamics• These relationships work for half-reactions or complete redox reactions.

Zn + Cu2+ � Zn2+ + Cu Eo = 1.10n = 2∆Go = -2mol e- x 96500 J/V mol e- x 1.10 V∆Go = -212,300 J = -212.3 kJ

• ∆Go depends on the number of moles, but Eo does not

Voltage and Moles• Note that different size alkaline cells all deliver the same voltage, in spite of different

number of moles of reactants.

Thermodynamics• We can add Eo

ox to Eored to give Eo

cell or Eorxn in the same way that we can add half-reactions

to give an overall reaction.• Fe � Fe2+ + 2e- Eo

ox = +0.44 V∆Go

ox = - 2 x 96500 x 0.44 = -84900 J• Cl2 + 2e- � 2Cl- Eo

red = 1.36 V∆Go

red = - 2 x 96500 x 1.36 = -262500 J• Fe + Cl2 � Fe2+ + 2Cl- Eo

rxn = 1.80 V∆Go = - 2 x 96500 x 1.80 = -347400 J

• ∆Gorxn = ∆Go

ox + ∆Gored

= -84.9 + -262.5 = -347.4 kJ• From Chapter 19, we know that ∆Go values are additive when we add reactions.• Eos are additive when we add half-reactions to give a complete reaction because the value of

n is the same for the half-reactions and the complete reaction.• Eos are not additive when adding two half-reactions to give a third half-reaction because the

value of n is not constant.• We can add ∆Go under all circumstances:

∆Go3 = ∆Go

1 + ∆Go2

-n3FEo3 = -n1FEo

1 - n2FEo2

n3Eo3 = n1Eo

1 + n2Eo2

Eo3 = (n1Eo

1 + n2Eo2)/n3

• V � V2+ + 2e- Eo1 = 1.20 V

V2+ � V3+ + e- Eo2 = 0.26 V

V � V3+ + 3e- Eo3 < 1.20 + 0.26

Eo3 = (2 x 1.20 + 1 x 0.26)/3 = 0.887 V

• Given the first two half-reactions, what is the value of E°red for the third?MnO4

-(aq) + e- � MnO42-(aq) E°1,red = +0.56 V

MnO42-(aq) + 4H+(aq) + 2e- � MnO2(s) + 2H2O(l) E°2.red = +2.26 V

MnO4-(aq) + 4H+(aq) + 3e- � MnO2(s) + 2H2O(l) E°3,red = ?

• E°3,red = 1.69 V• This is why we could not make correct predictions for this system earlier.

Chapter 20

20-11

Thermodynamics• We can calculate Keq from Eo:

∆Go = - nFEo = - RT ln KEo = (RT/nF) ln K = 2.303 (RT/nF) log K

• At 25oC, 2.303 RT/F = 0.05916Eo = (0.05916/n) log K at 25oC

• Thus, we can measure Eo for a redox reaction and then calculate the equilibrium constant forthat reaction.

20.6 Effect of Concentraiton on Cell EMF• So far, we have been using standard state conditions, but we don�t always have 1 M

solutions. We can correct Eo to E by using the Nernst equation.∆G = ∆Go + RT ln QBut, ∆G = - nFE and ∆Go = -nFEo, so- nFE = -nFEo + 2.303 RT log QE = Eo - (2.303 RT/nF) log Q

• At 25oC, E = Eo - (0.05916/n) log Q

Nernst Equation• At 25oC, E = Eo - (0.05916/n) log Q• E = Eo if Q = 1• When the system reaches equilibrium, Q = K, and E = 0, because Eo = (0.05916/n) log K,

and the cell has �run down�.• Consider the Zn/Cu2+ reaction if more Cu2+ is added to the cell. The voltage becomes

greater than 1.10 V.• What is E of the Zn/Cu2+ reaction if [Cu2+] = 0.010 M and [Zn2+] =1.99 M? Note that this

corresponds to starting with standard conditions and changing to 99% completion ofreaction. Eo = 1.10 V (with [Cu2+] = [Zn2+] = 1.00 M)

• Zn + Cu2+ � Zn2+ + CuFor this reaction, n = 2.Q = [Zn2+]/[Cu2+]E = Eo - (0.05916/n) log QE = 1.10 V - (0.05916/2) log (1.99/0.010)E = 1.10 V - (0.05916/2) log 199E = 1.10 - 0.068 = 1.03 V

• For 99.9% reaction (1.999 M Zn2+, 0.001 M Cu2+), E = 1.10 - 0.098 = 1.00 V• For 99.99% reaction, 1.9999 M Zn2+, 0.0001 M Cu2+), E = 1.10 - 0.127 = 0.97 V

Concentration Cells• We can generate a voltage with a cell that contains the same materials in the cathode and

anode compartments, but at different concentrations.

20.7 Batteries (Voltaic Cells)• Voltaic cells are used as portable electricity sources.• one unit = cell

Chapter 20

20-12

• several cells = battery• Commercial cells:

• Anode: usually a metal that can be oxidized• Cathode: usually an inert conductor surrounded by a substance with a high oxidation

number that can be reduced• Electrolyte: aqueous solution or moist paste for ion movement• Electrolytes: acidic, alkaline, organic liquid, molten salt, solid state

• Primary cells: one use, then discard• Secondary cells: rechargeable, reusable• Consider some commercial cells and batteries

• What are the anode and cathode?• What is used as an electrolyte?• What half-reactions are used?• What voltage is generated?

LeClanche Dry Cell• Used in flashlight batteries.• Electrolyte is NH4Cl• Zn � Zn2+ + 2e- Eo = 0.763 V

2MnO2 + 2NH4+ + 2H2O + 2e- � 2NH4OH + 2MnO(OH) Eo = 0.5 V

Zn + 2MnO2 + 2NH4+ + 2H2O � Zn2+ + 2NH4OH + 2MnO(OH) Eo = 1.26 V

• The measured voltage is actually 1.5 V, because concentrations are higher than standardstate concentrations.

Alkaline Dry Cell• Electrolyte is KOH• Zn + 2OH- � ZnO + H2O + 2e-

2MnO2 + H2O + 2e- � 2OH- + Mn2O3Zn + 2MnO2 � ZnO + Mn2O3 E = 1.5 V

• Reactions are similar to the acidic dry cell• Can be recharged a few times

Lead Storage Battery• Used in automobiles• PbO2 + 4H+ + SO4

2- + 2e- � PbSO4 + 2H2O Eo = 1.69 VPb + SO4

2- � PbSO4 + 2e- Eo = 0.35 vPb + PbO2 + 2SO4

2- + 4H+ � 2PbSO4 + 2H2O Eo = 2.04 V• With 6 M H2SO4, the cell produces 2.0 V; the voltage decreases as the cell reactions occur.• Six cells connected in series makes up the common 12 V car battery.• Can be recharged:

2PbSO4 + 2H2O � Pb + PbO2 + 2SO42- + 4H+

• Must pass 2 V per cell back through the battery to recharge.• Rapid recharging can electrolyze the acid, producing hydrogen gas, which might explode, or

which can dislodge the lead oxide or sulfate, which shortens the battery life.

Chapter 20

20-13

NiCad Cell• Used in rechargeable power tools.• Rechargeable because the reaction products stick to the electrodes.• Cd + 2OH- � Cd(OH)2 + 2e- Eo = 0.81 V

NiO2 (on Ni) + 2H2O + 2e- � Ni(OH)2 + 2OH- Eo = 0.49 VCd + NiO2 + 2H2O � Cd(OH)2 + Ni(OH)2 Eo = 1.30 V

Silver Cell• Used in cameras, watches, hearing aids• Electrolyte is 20-40% KOH• Zn + 2OH- � ZnO + H2O + 2e- Eo = 1.25 V

2AgO + H2O + 2e- � Ag2O + 2OH- Eo = 0.61 VZn + 2AgO + H2O � Zn(OH)2 + Ag2O Eo = 1.86 V

• Rechargeable if don�t pass this point.• After the AgO is depleted:

Zn + Ag2O + H2O � Zn(OH)2 + 2Ag Eo = 1.58 V

Fuel Cell• Used for space travel, and in new hydrogen fuel automobiles• Reactants are stored external to the cell and introduced to the electrodes as they are needed.• The reactants are usually gaseous, such as O2 and H2, or O2 and CH4, or O2 and NH3.• O2 + 2H2O + 4e- � 4OH- Eo = 0.40 V

2H2 + 4OH- � 4H2O + 4e- Eo = 0.83 V2H2 + O2 � 2H2O Eo = 1.23 V

• Recent development uses butane

20.8 Corrosion• Corrosion of Iron• Since E°red(Fe2+) < E°red(O2) iron can be oxidized by oxygen.• Cathode: O2(g) + 4H+(aq) + 4e- → 2H2O(l).• Anode: Fe(s) → Fe2+(aq) + 2e-.• Dissolved oxygen in water usually causes the oxidation of iron.• Fe2+ initially formed can be further oxidized to Fe3+ which forms rust, Fe2O3.xH2O(s).• Oxidation occurs at the site with the greatest concentration of O2.

Preventing the Corrosion of Iron• Corrosion can be prevented by coating the iron with paint or another metal.• Galvanized iron is coated with a thin layer of zinc.• Zinc protects the iron since Zn is the anode and Fe the cathode:

Zn2+(aq) +2e- → Zn(s), E°red = -0.76 VFe2+(aq) + 2e- → Fe(s), E°red = -0.44 V

• With the above standard reduction potentials, Zn is easier to oxidize than Fe.• To protect underground pipelines, a sacrificial anode is added.• The water pipe is turned into the cathode and an active metal is used as the anode.• Often, Mg is used as the sacrificial anode:

Chapter 20

20-14

Mg2+(aq) +2e- → Mg(s), E°red = -2.37 VFe2+(aq) + 2e- → Fe(s), E°red = -0.44 V

• Also used in water heaters, on gasoline storage tanks, and on ship hulls

20.9 Electrolysis• Since chemical oxidation-reduction involves the transfer of electrons from one substance to

another, it should be possible to harness the flow of electrons to produce electricity. We dothis with voltaic cells.

• Electricity can also be used to cause non-spontaneous chemical reactions. This process iscalled electrolysis.

Electrolytic Processes with Metals• A variety of metals can be prepared by electrolysis, if a cheap source of electricity is

available. In addition, some metals* are purified by electrolysis.aluminum cadmiumcalcium copper*gold* lead*magnesium sodiumzinc

Purification of Copper• Recovered from its ores by chemical reduction.• Purified by electrolysis.• Recover impurities:

Mo (25%)Se (93%)Te (96%)Au (32%)Ag (28%)

Production of Aluminum• Recovered as Al2O3

.nH2O from clays and bauxite.• The oxide is difficult to reduce.• The metal forms a protective coating of the oxide.• Al is very active but can be used in air because of the oxide coating.• The oxide coating can be thickened and colored by anodizing.• Electrolysis of Al2O3 (melts at 2045oC) dissolved in cryolite, Na3AlF6 (melts at 1000oC).• Uses 5% of U.S. electricity production

Electrolysis• Electrolysis is used for isolating active elements, purifying metals, and electroplating.• Pure compounds: H2O, molten salts• Use inert electrodes in the liquid and pass electricity through the system• The negative electrode (cathode) attracts cations; reduction occurs.• The positive electrode (anode) attracts anions; oxidation occurs.

Chapter 20

20-15

Electrolysis of NaCl• Cathode:

Na+(l) + e- � Na(l) Eo = -2.71 V• Anode:

2Cl-(l) � Cl2(g)+ 2e- Eo = -1.36 V• 2Na+(l) + 2Cl-(l) � 2Na(l) + Cl2(g) Eo = -4.08 V• Must supply at least 4.08 V to electrolyze molten sodium chloride.• NaCl melts at 804oC, where Na vaporizes and burns.• Lower the temperature by adding CaCl2. (Why does this work?)• Na reacts with Cl2, even at room temperature.• Commercial operations use a Downs Cell.• How does the Downs Cell solve the problem of reaction between Na and Cl2?

Electrolysis of H2O• Cathode:

2H2O + 2e- � H2(g) + 2OH-

Eo = -0.82 V at pH 14, -0.41 V at pH 7, 0.00 V at pH 0• Anode:

2H2O � O2(g) + 4H+ + 4e-

Eo = -0.41 V at pH 14, -0.82 V at pH 7, -1.23 V at pH 0• 2H2O � 2H2(g) + O2(g)

Eo = -1.23 V at any pH• Must supply at least 1.23 V to electrolyze water. Add an electrolyte to increase electrical

conductivity

Electrolysis of Aqueous Solutions• Products depend on whether it is easier to oxidize or reduce the dissolved ions or water.• Consider a solution of VCl3 under standard conditions at pH 7.• Cathode:

V3+ + e- � V2+ Eo = -0.26 V2H2O + 2e- � H2 + 2OH- Eo = -0.41 V at pH 7

• Because of lower voltage, will reduce V3+, not H2O.• Anode:

2Cl- � Cl2 + 2e- Eo = -1.36 V2H2O � O2 + 4H+ + 4e- Eo = -0.82 V at pH 7

• Because of lower voltage, will oxidize H2O, not Cl-.• Products are V2+ and O2.

• What are the products of electrolysis of VBr3 at pH 7?• Cathode:


Chapter 20

20-16

• Anode:2Br- � Br2 + 2e- Eo = -1.07 V2H2O � O2 + 4H+ + 4e- Eo = -0.82 V at pH 7

• What are the products of electrolysis of VI3 at pH 7?• Cathode:


• Anode:2I- � I2 + 2e- Eo = -0.54 V2H2O � O2 + 4H+ + 4e- Eo = -0.82 V at pH 7

• What are the products of electrolysis of a mixture of CuCl2 and VCl3 at pH 7?• Cathode:

V3+ + e- � V2+ Eo = -0.26 VCu2+ + 2e- � Cu Eo = 0.34 V2H2O + 2e- � H2 + 2OH- Eo = -0.41 V at pH 7

• Anode:2Cl- � Cl2 + 2e- Eo = -1.36 V2H2O � O2 + 4H+ + 4e- Eo = -0.82 V at pH 7

Faraday�s Law• Faraday�s Law: the mass of product produced by a given amount of current is proportional

to the equivalent weight• Equivalent weight: molar mass/no. e- transferred• Equivalent weight is the mass of substance oxidized or reduced by 1 mole of electrons• We can pass electrons through a series of cells and compare the amount of substance

deposited.

• What is the equivalent weight of � ?AgNO3 CuSO4AuCl3 HCl

• Moles deposited:• Ag+, 1 mole e- � 1 mol Ag• Cu2+, 1 mole e- � 1/2 mol Cu• Au3+, 1 mole e- � 1/3 mol Au• H+, 1 mole e- � 1/2 mol H2

• F = charge on 1 mol e- = 96500 coul/mol• charge = current x time• 1 coul = 1 A s• moles e- = charge (coul) x 1 mol/96500 coul• moles e- = current (A) x time (s) x 1 coul/1 A s x 1 mol/96500 coul• If we electrolyze molten NaCl with a current of 5000 A for 30 min (or 1800 s), what mass of

Na is produced?

Chapter 20

20-17

• Na+ + e- � Na• moles e- = 5000 A x 1800 s x 1 mol/96500 coul = 93.26 mol• moles Na = 93.26 mol e- x 1 mol Na/1 mol e- = 93.26 mol• mass Na = 93.26 mol x 22.99 g/mol = 2144 g• We can also calculate how much electrical energy it will take for an electrolysis. We will

not pursue these calculations.

• moles e- = current (A) x time (s) x 1 coul/1 A s x 1 mol/96500 coul• How long would we have to electrolyze molten NaCl with a current of 3000 A to produce

150 g of Na?

chapter 20 electrochemistry • electrochemistry deals with the

Documents