1. The Molecular Nature of Matter and Change
CHEMISTRY
The Molecular Nature of Matter and Change
Third Edition
Chapter 21
Lecture Outlines*
*See PowerPoint Image Slides for all figures and tables pre-inserted into PowerPoint without notes.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Chapter 21
Electrochemistry:
Chemical Change and Electrical Work

3. Electrochemistry: Chemical Change and Electrical Work
21.1 Half-Reactions and Electrochemical Cells
21.2 Voltaic Cells: Using Spontaneous Reactions to Generate
Electrical Energy
21.3 Cell Potential: Output of a Voltaic Cell
21.4 Free Energy and Electrical Work
21.5 Electrochemical Processes in Batteries
21.6 Corrosion: A Case of Environmental Electrochemistry
21.7 Electrolytic Cells: Using Electrical Energy to Drive a
Nonspontaneous Reaction

4. 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.

5. A summary of redox terminology
Figure 21.1
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 x to +2.
REDUCTION
Other reactant gains electrons.
Hydrogen ion gains electrons.
Oxidizing agent is reduced.
Hydrogen ion is the oxidizing agent and becomes reduced.
Oxidation number decreases.
The oxidation number of H decreases from +1 to 0.

6. Half-Reaction Method for Balancing Redox Reactions
Summary: 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 some integer to make the number of electrons gained and lost equal.
The half-reactions are then recombined to give the balanced redox equation.
Advantages:
The separation of half-reactions reflects actual physical separations in electrochemical cells.
The half-reactions are easier to balance especially if they involve acid or base.
It is usually not necessary to assign oxidation numbers to those species not undergoing change.

7. Balancing Redox Equations in Acidic Conditions
Write the skeletons of the oxidation and reduction half-reactions.
Balance all elements other than O and H.
Balance the oxygen atoms by adding H2O molecules on the side of the arrow where O atoms are needed.
Balance the hydrogen atoms by adding H+ ions on the side of the arrow where H atoms are needed.
Balance the charge by adding electrons, e-.
If the number of electrons lost in the oxidation half-reaction is not equal to the number of electrons gained in the reduction half-reaction, multiply one or both of the half- reactions by a number that will make the number of electrons gained equal to the number lost.
Add the 2 half-reactions as if they were mathematical equations.
Check to make sure that the atoms and the charge are balanced.

8. Balancing Redox Equations in Basic Conditions
Steps #1-8 
Begin by balancing the equation as if it were in acid solution. If you have H+ ions in your equation at the end of these steps, proceed to Step #9. Otherwise, skip to Steps 9-12
Add enough OH- ions to each side to cancel the H+ ions.
Be sure to add the OH- ions to both sides to keep the charge and atoms balanced.
Combine the H+ ions and OH- ions that are on the same side of the equation to form water.
Cancel or combine the H2O molecules.
Check to make sure that the atoms and the charge balance. If they do balance, you are done. If they do not balance, re-check your work in Steps 1-11.

9. Balancing Redox Reactions in Acidic Solution
Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq)
1. Divide the reaction into half-reactions -
Determine the O.N.s 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: +6
Add 6e- to left.
net: +12 2
Cr2O Cr3+
+ 7H2O(l)
14H+(aq) +
6e- +

10. Balancing Redox Reactions in Acidic Solution
continued
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(s)
+ 7H2O(l)
14H+(aq) +
Do a final check on atoms and charges.

11. Balancing Redox Reactions in Basic Solution
Balance the reaction in acid and then add OH- so as 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.

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

13. 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 ion to form carbonate ion and solid mangaese dioxide. Balance the skeleton ionic reaction that occurs between NaMnO4 and Na2C2O4 in basic solution:
MnO4-(aq) + C2O42-(aq) MnO2(s) + CO32-(aq)
PLAN:
Proceed in acidic solution and then neutralize with base.
SOLUTION:
MnO MnO2
C2O CO32- +7 +4 +3 +4
MnO MnO2
C2O CO32- 2
4H+ +
MnO MnO2
+ 2H2O
C2O CO32-
+ 2H2O
+ 4H+
+3e-
+2e-

14. Sample Problem 21.1:
Balancing Redox Reactions by the Half-Reaction
Method
continued:
4H+ + MnO4- +3e MnO2+ 2H2O
C2O H2O CO H+ + 2e-
X 2
4H+ + MnO4- +3e MnO2+ 2H2O
X 3
C2O H2O CO H+ + 2e-
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)

15. General characteristics of voltaic and electrolytic cells
Figure 21.3
General characteristics of voltaic and electrolytic cells
VOLTAIC CELL
ELECTROLYTIC CELL
System does work on its surroundings
Energy is released from spontaneous redox reaction
Energy is absorbed to drive a nonspontaneous redox reaction
Surroundings(power supply)
do work on system(cell)
X(s)
Y(s)
A(s)
B(s)
Oxidation half-reaction
X X+ + e-
Oxidation half-reaction
A A + e-
Reduction half-reaction
Y++ e- Y
Reduction half-reaction
B++ e B
Overall (cell) reaction
X + Y X+ + Y DG < 0
Overall (cell) reaction
A- + B A + B DG > 0

16. The spontaneous reaction between zinc and copper(II) ion
Figure 21.4
The spontaneous reaction between zinc and copper(II) ion
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)

17. A voltaic cell based on the zinc-copper reaction
Figure 21.5
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)

18. 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

19. A voltaic cell using inactive electrodes
Figure 21.6
A voltaic cell using inactive electrodes
Oxidation half-reaction
2I-(aq) I2(s) + 2e-
Reduction half-reaction
MnO4-(aq) + 8H+(aq) + 5e-
Mn2+(aq) + 4H2O(l)
Overall (cell) reaction
2MnO4-(aq) + 16H+(aq) + 10I-(aq) Mn2+(aq) + 5I2(s) + 8H2O(l)

20. 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.
PLAN:
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). e-
SOLUTION:
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)

21. 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)

22. 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

23. Figure 21.7
Determining an unknown E0half-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)

24. Sample Problem 21.3:
Calculating an Unknown E0half-cell from E0cell
PROBLEM:
A voltaic cell houses the reaction between aqueous bromine and zinc metal:
Br2(aq) + Zn(s) Zn2+(aq) + 2Br-(aq) E0cell = 1.83V
Calculate E0bromine given E0zinc = -0.76V
PLAN:
The reaction is spontaneous as written since the E0cell is (+). Zinc is being oxidized and is the anode. Therefore the E0bromine can be found using E0cell = E0cathode - E0anode.
SOLUTION:
anode: Zn(s) Zn2+(aq) + 2e- E = +0.76
E0Zn as Zn2+(aq) + 2e Zn(s) is -0.76V
E0cell = E0cathode - E0anode = = E0bromine - (-0.76)
E0bromine = = 1.07V

25. Table 21.2 Selected Standard Electrode Potentials (298K)
Half-Reaction
E0(V)
F2(g) + 2e F-(aq)
+2.87
Cl2(g) + 2e Cl-(aq)
+1.36
MnO2(g) + 4H+(aq) + 2e Mn2+(aq) + 2H2O(l)
+1.23
NO3-(aq) + 4H+(aq) + 3e NO(g) + 2H2O(l)
+0.96
Ag+(aq) + e Ag(s)
+0.80
Fe3+(g) + e Fe2+(aq)
+0.77
O2(g) + 2H2O(l) + 4e OH-(aq)
+0.40
Cu2+(aq) + 2e Cu(s)
+0.34
2H+(aq) + 2e H2(g)
0.00
N2(g) + 5H+(aq) + 4e N2H5+(aq)
-0.23
Fe2+(aq) + 2e Fe(s)
-0.44
2H2O(l) + 2e H2(g) + 2OH-(aq)
-0.83
Na+(aq) + e Na(s)
-2.71
Li+(aq) + e Li(s)
-3.05

26. 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 E0cell.
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

27. 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 E0cell for each.
(b) Rank the relative strengths of the oxidizing and reducing agents:
E0 = 0.96V
(1) NO3-(aq) + 4H+(aq) + 3e NO(g) + 2H2O(l)
E0 = -0.23V
(2) N2(g) + 5H+(aq) + 4e N2H5+(aq)
E0 = 1.23V
(3) MnO2(s) +4H+(aq) + 2e Mn2+(aq) + 2H2O(l)
PLAN:
Put the equations together in varying combinations so as to produce (+) E0cell for the combination. Since the reactions are written as reductions, remember that as you reverse one reaction for an oxidation, reverse the sign of E0. Balance the number of electrons gained and lost without changing the E0.
In ranking the strengths, compare the combinations in terms of E0cell.

28. Sample Problem 21.4:
Writing Spontaneous Redox Reactions and Ranking
Oxidizing and Reducing Agents by Strength
continued (2 of 4)
SOLUTION:
(1) NO3-(aq) + 4H+(aq) + 3e NO(g) + 2H2O(l)
E0 = 0.96V
(a)
(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e-
E0 = +0.23V
Rev
E0cell = 1.19V
(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)
E0 = -0.96V
(1) NO(g) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e-
Rev
E0 = 1.23V
(3) MnO2(s) +4H+(aq) + 2e Mn2+(aq) + 2H2O(l)
E0cell = 0.27V
(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) + 3Mn3+(aq) + 2H2O(l)
(B)

29. Sample Problem 21.4:
Writing Spontaneous Redox Reactions and Ranking
Oxidizing and Reducing Agents by Strength
continued (3 of 4)
E0 = +0.23V
(2) N2H5+(aq) N2(g) + 5H+(aq) + 4e-
Rev
E0 = 1.23V
(3) MnO2(s) +4H+(aq) + 2e Mn2+(aq) + 2H2O(l)
E0cell = 1.46V
(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+

30. Sample Problem 21.4:
Writing Spontaneous Redox Reactions and Ranking
Oxidizing and Reducing Agents by Strength
continued (4 of 4)
A comparison of the relative strengths of oxidizing and reducing agents produces the overall ranking of
Oxidizing agents: MnO2 > NO3- > N2
Reducing agents: N2H5+ > NO > Mn2+

31. 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

32. The reaction of calcium in water
Figure 21.8
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)

33. Free Energy and Electrical Work
DG a -Ecell
DG = wmax = charge x (-Ecell)
-Ecell =
-wmax
charge
DG = -n F Ecell
In the standard state -
DG0 = -n F E0cell
DG0 = - RT ln K
E0cell = - (RT/n F) ln K
charge = n F
n = # mols e-
F = Faraday constant
F = 96,485 C/mol e-
1V = 1J/C
F = 9.65x104J/V*mol e-

34. The interrelationship of DG0, E0, and K
Figure 21.9
The interrelationship of DG0, E0, and K
DG0 K
Reaction at standard-state conditions
E0cell
DG0
< 0
> 1
> 0
spontaneous
at equilibrium
nonspontaneous 1
> 0
< 1
< 0
DG0 = -nFEocell
DG0 = -RT lnK
By substituting standard state values into E0cell, we get
E0cell = (V/n) log K (at 250C)
E0cell K
E0cell = -RT lnK nF

35. Sample Problem 21.5:
Calculating K and DG0 from E0cell
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 DG0 at 250C for this reaction.
Pb(s) + 2Ag+(aq) Pb2+(aq) + 2Ag(s)
PLAN:
Break the reaction into half-reactions, find the E0 for each half-reaction and then the E0cell. Substitute into the equations found on slide
SOLUTION:
Pb2+(aq) + 2e Pb(s)
E0 = -0.13V
E0 = 0.13V
E0 = 0.80V
E0cell = 0.93V
Ag+(aq) + e Ag(s)
E0 = 0.80V
Ag+(aq) + e Ag(s)
Pb(s) Pb2+(aq) + 2e- 2X
E0cell =
log K
0.592V n
DG0 = -nFE0cell
= -(2)(96.5kJ/mol*V)(0.93V)
n x E0cell
0.592V
(2)(0.93V)
0.592V =
DG0 = -1.8x102kJ
log K =
K = 2.6x1031

36. 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+] = 0.010M [H+] = 2.5M P = 0.30atm H2
Calculate Ecell at 250C.
PLAN:
Find E0cell and Q in order to use the Nernst equation.
SOLUTION:
Determining E0cell :
Q =
P x [Zn2+] H2
[H+]2
E0 = 0.00V
2H+(aq) + 2e H2(g)
E0 = -0.76V
Zn2+(aq) + 2e Zn(s)
Q =
(0.30)(0.010)
(2.5)2
Zn(s) Zn2+(aq) + 2e-
E0 = +0.76V
Ecell = E0cell -
0.0592V n
log Q
Q = 4.8x10-4
Ecell = (0.0592/2)log(4.8x10-4) =
0.86V

37. The Effect of Concentration on Cell Potential
DG = DG0 + RT ln Q
-nF Ecell = -nF Ecell + RT ln Q
Ecell = E0cell -
ln Q RT nF
When Q < 1 and thus [reactant] > [product], lnQ < 0, so Ecell > E0cell
When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell
When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell
Ecell = E0cell -
log Q
0.0592 n

38. The relation between Ecell and log Q for the zinc-copper cell
Figure 21.10
Overall (cell) reaction
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

39. Overall (cell) reaction
Figure 21.11
A concentration cell based on the Cu/Cu2+ half-reaction
Oxidation half-reaction
Cu(s) Cu2+(aq, 0.1M) + 2e-
Reduction half-reaction
Cu2+(aq, 1.0M) + 2e Cu(s)
Overall (cell) reaction
Cu2+(aq,1.0M) Cu2+(aq, 0.1M)

40. 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 4.0x10-4M AgNO3. What is the cell potential at 298K? Which electrode has a positive charge?
PLAN:
E0cell 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, 0.010M) half-cell A Ag+(aq, 4.0x10-4M) half-cell B
Ecell = E0cell -
0.0592V 1
log
[Ag+]dilute
[Ag+]concentrated
Ecell = 0 V log 4.0x10-2
= V
Half-cell A is the cathode and has the positive electrode.

41. The laboratory measurement of pH
Figure 21.12
The laboratory measurement of pH Pt
Reference (calomel) electrode
Glass electrode Hg
Paste of Hg2Cl2 in Hg
AgCl on Ag on Pt
KCl solution
1M HCl
Porous ceramic plugs
Thin glass membrane

42. Table 21.3 Some Ions Measured with Ion-Specific Electrodes
Species Detected
Typical Sample
NH3/NH4+
Industrial wastewater, seawater
CO2/HCO3-
Blood, groundwater F-
Drinking water, urine, soil, industrial stack gases
Br-
Grain, plant tissue I-
Milk, pharmaceuticals
NO3-
Soil, fertilizer, drinking water K+
Blood serum, soil, wine H+
Laboratory solutions, soil, natural waters

43. Figure 21.13
The corrosion of iron

44. Enhanced corrosion at sea
Figure 21.14
Enhanced corrosion at sea

45. The effect of metal-metal contact on the corrosion of iron
Figure 21.15
The effect of metal-metal contact on the corrosion of iron
cathodic protection
faster corrosion

46. The use of sacrificial anodes to prevent iron corrosion
Figure 21.16
The use of sacrificial anodes to prevent iron corrosion

47. Figure 21.17
The tin-copper reaction as the basis of a voltaic and an electrolytic cell
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)

48. ELECTROLYTIC(recharge)
Figure 21.18
The processes occurring during the discharge and recharge of a lead-acid battery
VOLTAIC(discharge)
ELECTROLYTIC(recharge)

49. Table 21.4 Comparison of Voltaic and Electrolytic Cells
Electrode
Cell Type DG
Ecell
Name
Process
Sign
Voltaic
< 0
> 0
Anode
Oxidation -
Voltaic
< 0
> 0
Cathode
Reduction +
Electrolytic
> 0
< 0
Anode
Oxidation +
Electrolytic
> 0
< 0
Cathode
Reduction -

50. 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

51. Overall (cell) reaction
Figure 21.19
The electrolysis of water
Overall (cell) reaction
2H2O(l) H2(g) + O2(g)
Oxidation half-reaction
2H2O(l) 4H+(aq) + O2(g) + 4e-
Reduction half-reaction
2H2O(l) + 4e H2(g) + 2OH-(aq)

52. Sample Problem 21.9:
Predicting the Electrolysis Products of Aqueous
Ionic Solutions
PROBLEM:
What products form during electrolysis of aqueous solution 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 0.6V 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:
E0 = -2.93V
(a) K+(aq) + e K(s)
E0 = -0.42V
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.
E0 = 1.07V
2Br-(aq) Br2(g) + 2e-
2H2O(l) O2(g) + 4H+(aq) + 4e-
E0 = 0.82V
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.

53. Sample Problem 21.9:
Predicting the Electrolysis Products of Aqueous
Ionic Solutions
continued
E0 = -0.80V
(b) Ag+(aq) + e Ag(s)
E0 = -0.42V
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-
E0 = -2.37V
(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.

54. A summary diagram for the stoichiometry of electrolysis
Figure 21.20
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)

55. Sample Problem 21.10:
Applying the Relationship Among Current, Time,
and Amount of Substance
PROBLEM:
A technician is plating a faucet with 0.86g 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
0.86g (mol Cr) (3mol e-)
(52.00gCr) (mol Cr)
= 0.050mol e-
3mol e-/mol Cr
mol of e- transferred
0.050mol e- (9.65x104C/mol e-) = 4.8x103C
charge (C)
9.65x104C/mol e-
4.8x103C
12.5min
(min)
(60s)
= 6.4C/s = 6.4 A
current (A)
divide by time

56. Dry Cell

57. Alkaline Battery

58. Mercury and Silver (Button) Batteries

59. Lead-Acid Battery

60. Nickel-Metal Hydride (Ni-MH) Battery

61. Lithium-Ion Battery