Fullerton College Office of Special Programs
SI Review 111B
Handout #13 Solutions
1.
a – c:
e- flow
v
Pb, anode
Pb  Pb2+ + 2e-
d.
Ag, cathode
Ag+ + e-  Ag
cations 
 anions
To obtain the standard cell potential, we first obtain the potentials for the half-reactions. The
standard reduction and oxidation potentials are added together to obtain the standard cell
potential. Note that for spontaneous electrochemical cells (galvanic cells), the standard cell
potential must be positive.
Ag+ + ePb


Eored = +0.80 V
Eoox = +0.13 V
Ag
Pb2+ + 2e-
Eoredox = Eored + Eoox = 0.80 V + 0.13 V = 0.93 V
e.
The line notation is written according to the following format: anode || cathode. Where the
components on the anode side of the cell are broken by commas (if in same phase) or a single
bar (if in different phases). If known, the concentrations and pressures are included in the
line notation. Lastly, an inert, conductive material must be used for the electrode when the
neither the reactant nor product can serve as an electrode. Commonly, platinum metal is
used for this purpose.
Pb(s) | Pb2+ (aq, 1.0 M) || Ag+ (aq, 1.0 M) | Ag(s)
2. To begin, we determine the standard reduction potentials for all applicable species:
Cl2 + 2eAg+ + ePb2+ + 2eZn2+ + 2eNa+ + e-





2ClAg
Pb
Zn
Na
Eored
Eored
Eored
Eored
Eored
=
=
=
=
=
1.36 V
0.80 V
-0.13 V
-0.76 V
-2.71 V
Possible Choices: Na+, Cl-, Ag+, Ag, Zn2+, Zn, Pb
a.
The strongest oxidizing agent will be the most easily reduced and, therefore, the chemical
species with the greatest reduction potential: Ag+
(Note: If listed as a possible choice, Cl2 would be the strongest oxidizing agent.)
b.
The strongest reducing agent will be most easily oxidized and, therefore, the chemical species
with the lowest reduction potential: Zn
(Note: If listed as a possible choice, Na would be the strongest reducing agent.)
c.
To be oxidized by sulfate implies that sulfate undergoes a reduction reaction:
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SO42- + 4H+ + 2e-

Eored = 0.20 V
H2SO3 + H2O
To participate in a redox reaction would require that the oxidation potential be greater (more
positive) than -0.20 V. The only choices with oxidation potentials greater than -0.20 V are:
Zn and Pb.
(Note: An oxidation potential greater than -0.20 V ensures that the redox potential (Eoredox) is
positive.)
d.
To be reduced by aluminum implies that aluminum undergoes an oxidation reaction:
Al

Al3+ + 3e-
Eoox = 1.66 V
To participate in a redox reaction would require that the reduction potential be greater (more
positive) than -1.66 V. The only choices with reduction potentials greater than -1.66 V are:
Ag+ and Zn2+.
(Note: A reduction potential greater than -1.66 V ensures that the redox potential (Eoredox) is
positive.)
3. We begin this problem by writing the chemical equation:
Pb + Cu2+  Pb2+ + Cu
Note that this is an educated guess (lead is more active than copper). If it were incorrect, then
the cell potential would be negative (nonspontaneous in direction written) and I would reverse the
direction of the reaction.
Given the above chemical equation, we might then determine the standard cell potential:
Cu2+ + 2ePb
Pb(s) + Cu2+(aq)
a.
Cu
Pb2+ + 2ePb2+(aq) + Cu
Eored = +0.34 V
Eoox = +0.13 V
o
E redox = +0.47 V
To obtain the initial cell potential, we make use of the Nernst equation:
Eredox  Eoredox -
b.



 [P b2 ] 
0 .0 2 5 7V
  0 .4 7V - 0 .0 2 5 7V Ln  0 .0 5 0 0  0 .5 1V
Ln 
 [C u2 ] 
n
2
 1 .5 0 


We can determine the cell potential after at certain amount of chemical change by first
determining the concentrations of all species. This is best accomplished by using an
amounts table:
Pb(s)
+ Cu2+(aq)
1.50
-x
1.50 - x

I
C
F
Pb2+(aq)
0.0500
+x
0.0500 + x
+ Cu
We know that the concentration of copper has changed to 0.200 M. Therefore,
[Cu2+] = 0.200 = 1.50 – x
and x = 1.30
such that,
[Pb2+] = 0.0500 + x = 1.35 M
Lastly, we solve for the nonstandard potential using the Nernst equation:
Eredox  Eoredox -
c.
 [P b2 ] 
0 .0 2 5 7V
  0 .4 7V - 0 .0 2 5 7V Ln  1 .3 5   0 .4 5V
Ln 
 [C u2 ] 
n
2
 0 .2 0 0


This is problem is almost the exact opposite of (b). We begin by calculating the reaction
quotient, and solve for the change. (Note that the change is represented by x, just as
above in (b).)
 [Pb2  ] 
0.0257 V

Ln 
 [Cu2  ] 
n


0.0257 V
 0.0500  x 

0.35 V  0.47 V Ln 
2
 1.50 - x 
Eredox  Eoredox -
11367 
0.0500  x
1.50 - x
x  1.499
So,
[Cu2  ]  1.50 - 1.499  0.0 M
[Pb2  ]  0.0500  1.499  1.5 M
4. The line notation describes a Cu/Cu+ concentration cell with solution containing iodide (1.0 M) on
the anode side of the cell. The iodide limits the concentration of the copper(I) according the K sp
that is given for copper(I) iodide:
Ksp  [C u ][I-]

[C u ]  1 .1 x 1 0-12 / 1 .0  1 .1 x 1 0-12 M
The nonstandard cell potential can then be determined from the expression given in #3 above:
Eredox 
 [C u ]

- 0 .0 2 5 7V
Low 
Ln



n
 [C u ]High 
Eredox 
 1 .1 x 1 0-12 
- 0 .0 2 5 7V

Ln


1
1 .0


Eredox  0 .7 1V
5. In predicting the products in the electrolysis of aqueous (salt) solutions, we will use the following
nonstandard potentials for the oxidation / reduction of water:
red
4H2O + 4e-

2H2 + 4OH-
Ered = -0.41 V
ox
2H2O

O2 + 4H+ + 4e-
Eox = -1.4 V
Note: The oxidation potential is the “overvoltage” for production of oxygen!
In solving this problem, we are interested in identifying the half-reactions with the largest, most
positive reduction (or oxidation) potentials. These will be the reactions likely to occur upon
electrolysis…
a.
red
Ni2+ + 2e-

Ni
Eored = -0.23 V
ox
2Br-

Br2 + 2e-
Eoox = -1.09 V
It will be easier to reduce Ni2+ than H2O, and easier to oxidize Br- than H2O. Therefore,
we should expect to see Ni form at the cathode and Br2 form at the anode if we pass
electricity through a solution that is 1.0 M in Ni2+ and Br-. The half-reactions at the anode
and cathode will be
b.
anode
2Br-

Br2 + 2e-
cathode
Ni2+ + 2e-

Ni
red
Al3+ + 3e-

Al
Eored = -1.66 V
ox
2F-

F2 + 2e-
Eoox = -2.87 V
It will be easier to reduce H2O than Al3+, and easier to oxidize H2O than F-. Therefore, we
should expect to see H2 form at the cathode and O2 form at the anode if we pass
electricity through a solution that is 1.0 M in Al3+ and F-. The half-reactions at the anode
and cathode will be
anode
2H2O

O2 + 4H+ + 4e-
cathode
4H2O + 4e-

2H2 + 4OH-
6. The reduction of sodium ion to form sodium metal requires one mole of electrons per mole of
sodium metal reduced:
Na+ + e-  Na
To determine the amount of current required to produce 1.0 kg of Na, we will calculate moles of
electrons involved in the reaction:
1 .0 kg N a x
n
e-

It
F
1 0 0 0g
1 mol N a
1 mol ex
x
 43.4 9 mol e1 kg
2 2 .9 9g N a 1 mol N a

I
n
e-
t
F

(43.4 9 mol e- )(9 .6 5x 1 0 4C /mole- )
 116 5 .7C /s  1 2 0 0A
(1 .0 hr)(6 0 min/hr)(6 0s /min)