1. Electrochemistry: Chemical Change and Electrical Work
Chapter 21
Electrochemistry: Chemical Change and Electrical Work

2. Electrochemistry: Chemical Change and Electrical Work
Half-reactions and electrochemical cells
Voltaic cells: Using spontaneous reactions to generate electrical energy
Cell potential: Output of a voltaic cell
Free energy and electrical work
Electrochemical processes in batteries
Corrosion: A case of environmental electrochemistry
Electrolytic cells: Using electrical energy to drive a non-spontaneous reaction

3. Key points about redox reactions
Oxidation (electron loss) always accompanies reduction (electron gain).
The oxidizing agent is reduced and the reducing agent is oxidized.
The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent.

4. A summary of redox terminology
Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)
OXIDATION
One reactant loses electrons.
Zn loses electrons.
Reducing agent is oxidized.
Zn is the reducing agent and becomes oxidized.
Oxidation number increases.
The oxidation number of Zn increases from 0 to +2 (each Zn loses 2 electrons).
REDUCTION
Other reactant gains electrons.
Oxidizing agent is reduced.
Hydrogen ion gains electrons.
Oxidation number decreases.
Hydrogen ion is the oxidizing agent and becomes reduced.
The oxidation number of H decreases from +1 to 0 (each H+ ion gains one electron).
Figure 21.1

5. Half-reaction method for balancing redox reactions
This method divides the overall redox reaction into oxidation and reduction half-reactions.
Each reaction is balanced for mass (atoms) and charge.
One or both are multiplied by an integer to make the number of electrons gained and lost equal.
The half-reactions are recombined to give the balanced redox equation.
Advantages
The separation of half-reactions reflects the actual physical separations in electrochemical cells.
The half-reactions are easier to balance, especially if they involve acid or base.
It is usually unnecessary to assign oxidation numbers to those species not undergoing change.

6. Balancing redox reactions in acidic solution
Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq)
1. Divide the reaction into half-reactions
Determine ONs for the species undergoing redox. +6 -1 +3
Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq)
Cr2O Cr3+
Cr is going from +6 to +3
I I2
I is going from -1 to 0
2. Balance atoms and charges in each half-reaction
14H+(aq) +
Cr2O Cr3+ 2
+ 7H2O(l)
net: +12
net: +6
Add 6 e- to left. 2
Cr2O Cr3+
+ 7H2O(l)
14H+(aq) +
6e- +

7. 3. Multiply each half-reaction by an integer, if necessary
Cr2O Cr3+
+ 7H2O(l)
14H+(aq) +
6e- + 2 2
I I2
+ 2e-
Cr(+6) is the oxidizing agent and I(-1) is the reducing agent.
3. Multiply each half-reaction by an integer, if necessary
x 3
+ 2e-
I I2 2
4. Add the half-reactions together
Cr2O Cr3+
+ 7H2O(l)
14H+ +
6e- + 2 3
I I2 6
+ 6e-
Cr2O72-(aq) + 6 I-(aq) Cr3+(aq) + 3I2(aq)
+ 7H2O(l)
14H+(aq) +
Do a final check on atoms and charges.

8. Balancing redox reactions in basic solution
Balance the reaction in acid and then add OH- to neutralize the H+ ions.
Cr2O72-(aq) + 6 I-(aq) Cr3+(aq) + 3I2(s)
+ 7H2O(l)
14H+(aq) +
+ 14OH-(aq)
+ 14OH-(aq)
Cr2O I Cr3+ + 3I2
+ 7H2O + 14OH-
14H2O +
Reconcile the number of water molecules.
+ 14OH-
Cr2O I Cr3+ + 3I2
7H2O +
Do a final check on atoms and charges.

9. The redox reaction between dichromate and iodide anions
Cr2O72- I-
Cr3+ + I2
Figure 21.2

10. Sample Problem 21.1
Balancing redox reactions by the half-reaction method
PROBLEM:
Permanganate ion is a strong oxidizing agent and its deep purple color makes it useful as an indicator in redox titrations. It reacts in basic solution with the oxalate anion to form carbonate anion and solid manganese dioxide. Balance the skeleton ionic reaction that occurs between KMnO4 and Na2C2O4 in basic solution.
MnO4-(aq) + C2O42-(aq) MnO2(s) + CO32-(aq)
PLAN:
Proceed in acidic solution and then neutralize with base.
SOLUTION: R O
MnO MnO2
C2O CO32- +7 +4 +3 +4
MnO MnO2
C2O CO32- 2
4H+ +
MnO MnO2
+ 2H2O
C2O CO32-
+ 2H2O
+ 4H+
+ 3e-
+ 2e-

11. x3 x2 Sample Problem 21.1 (continued) C2O42- + 2H2O 2CO32- + 4H+ + 2e-
4H+ + MnO4- +3e MnO2+ 2H2O
8H+ + 2MnO4- + 6e MnO2+ 4H2O
3C2O H2O CO H+ + 6e-
8H+ + 2MnO4- + 6e MnO2 + 4H2O
3C2O H2O CO H+ + 6e-
2MnO2-(aq) + 3C2O42-(aq) + 2H2O(l) MnO2(s) + 6CO32-(aq) + 4H+(aq)
+ 4OH-
+ 4OH-
2MnO2-(aq) + 3C2O42-(aq) + 4OH-(aq) MnO2(s) + 6CO32-(aq) + 2H2O(l)

12. Types of Electrochemical Cells
Voltaic (or galvanic) cell: uses a spontaneous reaction (∆G < 0)
to generate electrical energy.
Electrolytic cell: uses electrical energy to drive a non-spontaneous
reaction (∆G > 0).
Contain two electrodes (anode and cathode) dipped into an
aqueous electrolyte solution.
The oxidation half-reaction occurs at the anode; the reduction
half-reaction occurs at the cathode.

13. General characteristics of voltaic and electrolytic cells
VOLTAIC CELL
ELECTROLYTIC CELL
Energy is released from spontaneous redox reaction
system does work on its surroundings
Energy is absorbed to drive a non-spontaneous redox reaction
surroundings (power supply)
do work on the system (cell)
oxidation half-reaction:
X X+ + e-
oxidation half-reaction:
A A + e-
reduction half-reaction:
reduction half-reaction:
B+ + e B
Y+ + e Y
overall (cell) reaction:
X + Y X+ + Y (DG < 0)
overall (cell) reaction:
A- + B A + B (DG > 0)
Figure 21.3

14. The spontaneous reaction between zinc and copper(II) ion
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
Figure 21.4
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15. A voltaic cell based on the zinc-copper reaction
oxidation half-reaction:
Zn(s) Zn2+(aq) + 2e-
reduction half-reaction:
Cu2+(aq) + 2e Cu(s)
overall (cell) reaction:
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Figure 21.5

16. Notation for a voltaic cell
components of anode compartment
(oxidation half-cell)
components of cathode compartment
(reduction half-cell)
phase of lower oxidation state
phase of higher oxidation state
phase of higher oxidation state
phase of lower oxidation state
phase boundary between half-cells
examples:
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s)
Zn(s) Zn2+(aq) + 2e-
Cu2+(aq) + 2e Cu(s)
graphite | I-(aq) | I2(s) || H+(aq), MnO4-(aq) | Mn2+(aq) | graphite
inert electrode

17. A voltaic cell using inactive electrodes
reduction half-reaction:
MnO4-(aq) + 8H+(aq) + 5e-
Mn2+(aq) + 4H2O(l)
oxidation half-reaction:
2I-(aq) I2(s) + 2e-
overall (cell) reaction:
2MnO4-(aq) + 16H+(aq) + 10I-(aq) Mn2+(aq) + 5I2(s) + 8H2O(l)
Figure 21.6
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18. Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)
Sample Problem 21.2
Diagramming voltaic cells
PROBLEM:
Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO3)3 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode.
Identify the oxidation and reduction reactions and write each half-reaction. Associate the (-)(Cr) pole with the anode (oxidation) and the (+) pole with the cathode (reduction).
PLAN:
SOLUTION: e-
voltmeter
salt bridge
oxidation half-reaction:
Cr(s) Cr3+(aq) + 3e- Cr
Cr3+ Ag
Ag+ K+
NO3-
reduction half-reaction:
Ag+(aq) + e Ag(s)
overall (cell) reaction:
Cr(s) + Ag+(aq) Cr3+(aq) + Ag(s)
Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)

19. Why does a voltaic cell work?
The spontaneous reaction occurs as a result of the different abilities of materials (such as metals) to give up their electrons and the ability of the electrons to flow through the circuit.
Ecell > 0 for a spontaneous reaction
1 volt (V) = 1 joule (J)/ coulomb (C)

20. Table 21.1 Voltages of Some Voltaic Cells
voltage (V)
common alkaline battery
1.5
lead-acid car battery (6 cells = 12V)
2.0
calculator battery (mercury)
1.3
electric eel (~5000 cells in 6-ft eel = 750V)
0.15
nerve of giant squid (across cell membrane)
0.070

21. Determining an unknown Eohalf-cell with the standard reference (hydrogen) electrode
oxidation half-reaction:
Zn(s) Zn2+(aq) + 2e-
reduction half-reaction:
2H3O+(aq) + 2e H2(g) + 2H2O(l)
overall (cell) reaction:
Zn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l)
Figure 21.7
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22. Sample Problem 21.3
Calculating an unknown Eohalf-cell from Eocell
PROBLEM:
A voltaic cell houses the reaction between aqueous bromine and zinc metal:
Br2(aq) + Zn(s) Zn2+(aq) + 2Br-(aq) Eocell = 1.83 V
Calculate Eobromine given Eozinc = V
PLAN:
The reaction is spontaneous as written since the Eocell is (+). Zinc is being oxidized and is the anode. Therefore the Eobromine can be found using Eocell = Eocathode - Eoanode.
SOLUTION:
anode: Zn(s) Zn2+(aq) + 2e- E = +0.76
EoZn as Zn2+(aq) + 2e Zn(s) is V
Eocell = Eocathode - Eoanode = = Eobromine - (-0.76)
Eobromine = = 1.07 V

24. By convention, electrode potentials are written as reductions.
When pairing two half-cells, you must reverse one reduction half-cell to produce an oxidation half-cell. Reverse the sign of the potential.
The reduction half-cell potential and the oxidation half-cell potential are added to obtain the Eocell.
When writing a spontaneous redox reaction, the left side (reactants) must contain the stronger oxidizing and reducing agents.
example:
Zn(s) Cu2+(aq) Zn2+(aq) Cu(s)
stronger reducing agent
stronger oxidizing agent
weaker oxidizing agent
weaker reducing agent

25. Sample Problem 21.4
Writing spontaneous redox reactions and ranking oxidizing and reducing agents by strength
PROBLEM:
(a) Combine the following three half-reactions into three spontaneous, balanced equations (A, B and C), and calculate Eocell for each.
(b) Rank the relative strengths of the oxidizing and reducing agents:
Eo = 0.96 V
(1) NO3-(aq) + 4H+(aq) + 3e NO(g) + 2H2O(l)
Eo = V
(2) N2(g) + 5H+(aq) + 4e N2H5+(aq)
Eo = 1.23 V
(3) MnO2(s) + 4H+(aq) + 2e Mn2+(aq) + 2H2O(l)
PLAN:
Put the equations together in varying combinations so as to produce (+) Eocell for the combination. Since the reactions are written as reductions, remember that as you reverse one reaction for an oxidation, reverse the sign of Eo. Balance the number of electrons gained and lost without changing the Eo.
In ranking the strengths, compare the combinations in terms of Eocell.

26. Sample Problem 21.4
(continued)
SOLUTION:
(1) NO3-(aq) + 4H+(aq) + 3e NO(g) + 2H2O(l)
Eo = 0.96 V
(a)
(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e-
Eo = V
Rev
Eocell = 1.19 V
(1) NO3-(aq) + 4H+(aq) + 3e NO(g) + 2H2O(l) x4
(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- x3
4NO3-(aq) + 3N2H5+(aq) + H+(aq) NO(g) + 3N2(g) + 8H2O(l)
(A)
Eo = V
(1) NO(g) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e-
Rev
Eo = 1.23 V
(3) MnO2(s) + 4H+(aq) + 2e Mn2+(aq) + 2H2O(l)
Eocell = 0.27 V
(1) NO(g) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e- x2
(3) MnO2(s) + 4H+(aq) + 2e Mn2+(aq) + 2H2O(l) x3
2NO(g) + 3MnO2(s) + 4H+(aq) NO3-(aq) + 3Mn2+(aq) + 2H2O(l)
(B)

27. Sample Problem 21.4
(continued)
Eo = V
(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e-
Rev
Eo = 1.23 V
(3) MnO2(s) + 4H+(aq) + 2e Mn2+(aq) + 2H2O(l)
Eocell = 1.46 V
(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e-
(3) MnO2(s) + 4H+(aq) + 2e Mn2+(aq) + 2H2O(l) x2
N2H5+(aq) + 2MnO2(s) + 3H+(aq) N2(g) + 2Mn2+(aq) + 4H2O(l)
(C)
(b) Ranking oxidizing and reducing agents within each equation:
(A): oxidizing agents: NO3- > N2
reducing agents: N2H5+ > NO
(B): oxidizing agents: MnO2 > NO3-
reducing agents: NO > Mn2+
(C): oxidizing agents: MnO2 > N2
reducing agents: N2H5+ > Mn2+

28. Sample Problem 21.4
(continued)
A comparison of the relative strengths of oxidizing and reducing agents produces the following overall ranking.
oxidizing agents: MnO2 > NO3- > N2
reducing agents: N2H5+ > NO > Mn2+

29. Relative reactivities (activities) of metals
1. Metals that can displace H2 from acid
2. Metals that cannot displace H2 from acid
3. Metals that can displace H2 from water
4. Metals that can displace other metals from solution

30. The reaction of calcium in water
oxidation half-reaction:
Ca(s) Ca2+(aq) + 2e-
reduction half-reaction:
2H2O(l) + 2e H2(g) + 2OH-(aq)
overall (cell) reaction:
Ca(s) + 2H2O(l) Ca2+(aq) + H2(g) + 2OH-(aq)
Figure 21.8
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31. Free energy and electrical work
DG a -Ecell
DG = wmax = charge x (-Ecell)
-Ecell =
-wmax
charge
DG = -n F Ecell
In the standard state:
DGo = -n F Eocell
DGo = - RT ln K
Eocell = - (RT/n F) ln K
charge = n F
n = # mols e-
F = Faraday constant
F = 96,485 C/mol e-
1 V = 1 J/C
F = x 104 J/V.mol e-

32. reaction at standard-state conditions
The interrelationship of DGo, Eo, and K
DGo K
reaction at standard-state conditions
Eocell
DGo
< 0
> 1
> 0
spontaneous
at equilibrium
non-spontaneous 1
> 0
< 1
< 0
DGo = -nFEocell
DGo = -RT ln K
By substituting standard state values into Eocell, we get:
Eocell = ( V/n) log K (at 25 oC)
Eocell K
Eocell = -RT ln K nF
Figure 21.9
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33. Sample Problem 21.5
Calculating K and DGo from Eocell
PROBLEM:
Lead can displace silver from solution:
As a consequence, silver is a valuable by-product in the industrial extraction of lead from its ore. Calculate K and DGo at 25 oC for this reaction.
Pb(s) + 2Ag+(aq) Pb2+(aq) + 2Ag(s)
PLAN:
Break the reaction into half-reactions, find the Eo for each half-reaction and then the Eocell.
SOLUTION:
Pb2+(aq) + 2e Pb(s)
Eo = V
Eo = V
Eo = V
Eocell = 0.93 V
Ag+(aq) + e Ag(s)
Eo = V
Ag+(aq) + e Ag(s)
Pb(s) Pb2+(aq) + 2e- 2x
Eocell =
log K
0.592V n
= -(2)(96.5 kJ/mol.V)(0.93 V)
DGo = -nFEocell
n x Eocell
0.592 V
(2)(0.93 V)
0.592 V =
log K =
K = 2.6 x 1031
DGo = x 102 kJ

34. Using the Nernst equation to calculate Ecell
Sample Problem 21.6
Using the Nernst equation to calculate Ecell
PROBLEM:
In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the following conditions:
[Zn2+] = M [H+] = 2.5 M P = 0.30 atm H2
Calculate Ecell at 25 oC.
PLAN:
Find Eocell and Q in order to use the Nernst equation.
SOLUTION:
Determining Eocell :
Q =
P x [Zn2+] H2
[H+]2
Eo = 0.00 V
2H+(aq) + 2e H2(g)
Eo = V
Zn2+(aq) + 2e Zn(s)
Q =
(0.30)(0.010)
(2.5)2
Zn(s) Zn2+(aq) + 2e-
Eo = V
Ecell = Eocell - V n
log Q
Q = 4.8 x 10-4
Ecell = (0.0592/2) log (4.8 x 10-4) =
0.86 V

35. The effect of concentration on cell potential
DG = DGo + RT ln Q
-nF Ecell = -nF Ecell + RT ln Q
Ecell = Eocell -
ln Q RT nF
When Q < 1 and thus [reactant] > [product], ln Q < 0, so Ecell > Eocell
When Q = 1 and thus [reactant] = [product], ln Q = 0, so Ecell = Eocell
When Q >1 and thus [reactant] < [product], ln Q > 0, so Ecell < Eocell
Ecell = Eocell -
log Q
0.0592 n

36. The relation between Ecell and log Q for the zinc-copper cell
Figure 21.10
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37. overall (cell) reaction:
A concentration cell based on the Cu/Cu2+ half-reaction
oxidation half-reaction:
Cu(s) Cu2+(aq, 0.1 M) + 2e-
reduction half-reaction:
Cu2+(aq, 1.0 M) + 2e Cu(s)
overall (cell) reaction:
Cu2+(aq,1.0 M) Cu2+(aq, 0.1 M)
Figure 21.11
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38. Sample Problem 21.7
Calculating the potential of a concentration cell
PROBLEM:
A concentration cell consists of two Ag/Ag+ half-cells. In half-cell A, electrode A dips into M AgNO3; in half-cell B, electrode B dips into 4.0 x 10-4 M AgNO3. What is the cell potential at 298 K? Which electrode has a positive charge?
PLAN:
Eocell will be zero since the half-cell potentials are equal. Ecell is calculated from the Nernst equation with half-cell A (higher [Ag+]) having Ag+ being reduced and plating out, and in half-cell B Ag(s) will be oxidized to Ag+.
SOLUTION:
Ag+(aq, M) half-cell A Ag+(aq, 4.0 x 10-4 M) half-cell B
Ecell = Eocell - V 1
log
[Ag+]dilute
[Ag+]concentrated
Ecell = 0 V log 4.0 x 10-2
= V
Half-cell A is the cathode and has the positive electrode.

39. The laboratory measurement of pHPt
reference (calomel) electrode
glass electrode Hg
paste of Hg2Cl2 in Hg
AgCl on Ag on Pt
KCl solution
1 M HCl
porous ceramic plugs
thin glass membrane
Figure 21.12
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41. The corrosion of iron Figure 21.13
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42. Enhanced corrosion at sea
Figure 21.14

43. The effect of metal-metal contact on the corrosion of iron
cathodic protection
faster corrosion
Figure 21.15
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44. The use of sacrificial anodes to prevent iron corrosion
Figure 21.16

45. The tin-copper reaction as the basis of a voltaic and an electrolytic cell
Figure 21.17
voltaic cell
electrolytic cell
oxidation half-reaction:
Sn(s) Sn2+(aq) + 2e-
oxidation half-reaction:
Cu(s) Cu2+(aq) + 2e-
reduction half-reaction:
Cu2+(aq) + 2e Cu(s)
reduction half-reaction:
Sn2+(aq) + 2e Sn(s)
overall (cell) reaction
Sn(s) + Cu2+(aq) Sn2+(aq) + Cu(s)
overall (cell) reaction:
Sn(s) + Cu2+(aq) Sn2+(aq) + Cu(s)
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46. ELECTROLYTIC (recharge)
The processes occurring during the discharge and recharge of a lead-acid battery
VOLTAIC (discharge)
Figure 21.18
ELECTROLYTIC (recharge) .

48. Sample Problem 21.8
Predicting the electrolysis products of a molten salt mixture
PROBLEM:
A chemical engineer melts a naturally occurring mixture of NaBr and MgCl2 and decomposes it in an electrolytic cell. Predict the substance formed at each electrode, and write balanced half-reactions and the overall cell reaction.
PLAN: Consider the metal and nonmetal components of each compound and then determine which will recover electrons(be reduced; strength as an oxidizing agent) better. This is the converse to which of the elements will lose electrons more easily (lower ionization energy).
SOLUTION:
Possible oxidizing agents: Na+, Mg2+
Possible reducing agents: Br-, Cl-
Na, the element, is to the left of Mg in the periodic table, therefore the IE of Mg is higher than that of Na. So Mg2+ will more easily gain electrons and is the stronger oxidizing agent.
Br, as an element, has a lower IE than does Cl, and therefore will give up electrons as Br- more easily than will Cl-.
Mg2+(l) + 2Br-(l) Mg(s) + Br2(g)
cathode
anode

49. overall (cell) reaction:
The electrolysis of water
oxidation half-reaction:
2H2O(l) 4H+(aq) + O2(g) + 4e-
reduction half-reaction:
2H2O(l) + 4e H2(g) + 2OH-(aq)
overall (cell) reaction:
2H2O(l) H2(g) + O2(g)
Figure 21.19
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50. Sample Problem 21.9
Predicting the electrolysis products of aqueous ionic solutions
PROBLEM:
What products form during electrolysis of aqueous solutions of the following salts: (a) KBr (b) AgNO3 (c) MgSO4?
PLAN:
Compare the potentials of the reacting ions with those of water, remembering to consider the 0.4 to 0.6 V overvoltage.
The reduction half-reaction with the less negative potential, and the oxidation half-reaction with the less positive potential will occur at their respective electrodes.
SOLUTION:
Eo = V
(a) K+(aq) + e K(s)
Eo = V
2H2O(l) + 2e H2(g) + 2OH-(aq)
The overvoltage would make the water reduction to but the reduction of K+ is still a higher potential so H2(g) is produced at the cathode.
Eo = 1.07 V
2Br-(aq) Br2(g) + 2e-
2H2O(l) O2(g) + 4H+(aq) + 4e-
Eo = 0.82 V
The overvoltage would give the water half-cell more potential than the Br-, so the Br- will be oxidized. Br2(g) forms at the anode.

51. Sample Problem 21.9
(continued)
Eo = V
(b) Ag+(aq) + e Ag(s)
Eo = V
2H2O(l) + 2e H2(g) + 2OH-(aq)
Ag+ is the cation of an inactive metal and therefore will be reduced to Ag at the cathode.
Ag+(aq) + e Ag(s)
The N in NO3- is already in its most oxidized form so water will have to be oxidized to produce O2 at the anode.
2H2O(l) O2(g) + 4H+(aq) + 4e-
Eo = V
(c) Mg2+(aq) + 2e Mg(s)
Mg is an active metal and its cation cannot be reduced in the presence of water. So as in (a) water is reduced and H2(g) is produced at the cathode.
The S in SO42- is in its highest oxidation state; therefore water must be oxidized and O2(g) will be produced at the anode.

52. A summary diagram for the stoichiometry of electrolysis
MASS (g)
of substance oxidized or reduced
M (g/mol)
AMOUNT (MOL)
of substance oxidized or reduced
AMOUNT (MOL)
of electrons transferred
Faraday constant (C/mol e-)
balanced half-reaction
CHARGE (C)
time(s)
CURRENT (A)
Figure 21.20
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53. Sample Problem 21.10
Applying the relationship among current, time, and amount of substance
PROBLEM:
A technician is plating a faucet with 0.86 g of Cr from an electrolytic bath containing aqueous Cr2(SO4)3. If 12.5 min is allowed for the plating, what current is needed?
PLAN:
mass of Cr needed
SOLUTION:
Cr3+(aq) + 3e Cr(s)
divide by M
mol of Cr needed
0.86 g (mol Cr) (3 mol e-)
(52.00 g Cr) (mol Cr)
= mol e-
3 mol e-/mol Cr
mol of e- transferred
0.050 mol e- (9.65 x 104 C/mol e-) = 4.8 x 103 C
charge (C)
9.65 x 104 C/mol e-
4.8 x 103 C
12.5 min
(min)
(60 s) x
= 6.4 C/s = 6.4 A x
current (A)
divide by time

54. dry cell

55. alkaline battery

56. mercury and silver (button) batteries

57. lead-acid battery

58. nickel-metal hydride (Ni-MH) battery

59. lithium-ion battery