KEY
Practice Problems: Electrochemical Cells
CHEM 1B
1) Given the galvanic cell below and the table of Standard Reduction Potentials at 25°C in Appendix D
(pages A-19 and A-20) of your textbook, complete the diagram. A check list is provided to help you
include all of the required elements.



use standard reduction potentials to determine the spontaneous direction of flow for this cell
fill in the cell notation, the redox ½ reactions and potentials, and determine the standard cell
potential
label the diagram with the following:
 direction of species flow
o the species plating, eroding, or otherwise reacting on the electrode surfaces
o the electrons in the wire
o the ions in the salt bridge
 anode and cathode (and whether they are plating, eroding, or neither)
 source and sink of electrons
 positive and negative terminals
Cell Notation:
Al | Al3+ || Ni2+ | Ni
Al  Al3+ + 3 e–
Oxidation ½ reaction:
Ni2+ + 2 e–  Ni
Reduction ½ reaction:
–
Eºred =
– 0.26 V
Eºcell =
+ 1.40 V
–
e
e–
Eºox = – (– 1.66 V)
e
e–
+
K
NO3–
Cathode
(plating)
Anode
(eroding)
sink of
electrons
source of
electrons
Ni
Al
+
terminal
Ni
2+
Al
3+
–
terminal
2) Given the galvanic cell below and the table of Standard Reduction Potentials at 25°C in Appendix D
(pages A-19 and A-20) of your textbook, complete the diagram. A check list is provided to help you
include all of the required elements.
 use standard reduction potentials to determine the spontaneous direction of flow for this cell
 determine an appropriate metal to be used as the electrode in the left ½ cell
 fill in the cell notation, the redox ½ reactions and potentials, and determine the standard cell
potential
 label the diagram with the following:
 direction of species flow
o the species plating, eroding, or otherwise reacting on the electrode surfaces
o the electrons in the wire
o the ions in the salt bridge
 metal to be used as the electrode in the left ½ cell
 anode and cathode (and whether they are plating, eroding, or neither)
 source and sink of electrons
 positive and negative terminals
Pt | Sn2+ , Sn4+ || Ag+ | Ag
Cell Notation:
Oxidation ½ reaction:
Reduction ½ reaction:
Sn2+  Sn4+ + 2 e–
Ag+ + e–  Ag
e–
e–
Eºox = – (+ 0.15 V)
Eºred =
+ 0.80 V
Eºcell =
+ 0.65 V
e–
e–
+
K
NO3–
Anode
(neither)
Cathode
(plating)
source of
electrons
sink of
electrons
–
Pt
___
Ag
Sn
terminal
Sn4+
+
2+
terminal
Ag
+
3) Given the galvanic concentration cell below, the table of Standard Reduction Potentials at 25°C in
Appendix D and the constants in the back cover of your textbook, and the Nernst equation, complete
the diagram. A check list is provided to help you include all of the required elements.
 use the solution concentrations to determine the spontaneous direction of flow for this cell
 fill in the cell notation, the redox ½ reactions and potentials, and determine the standard cell
potential
 use the Nernst equation to determine the actual cell potential at 20.0°C
 label the diagram with the following:
 direction of species flow
o the species plating, eroding, or otherwise reacting on the electrode surfaces
o the electrons in the wire
o the ions in the salt bridge
 anode and cathode (and whether they are plating, eroding, or neither)
 source and sink of electrons
Nernst notes:
 positive and negative terminals
 the full Nernst eqn.
must be used because
the temp. is not 25°C
 solid Ni does not
appear in the reaction
quotient
 0.00 V and 2 mol e–
have infinite sig. fig.s
 J/C = V, potential is
an intensive property,
therefore the extra per
mol can be ignored
Ecell
Cell Notation:
Ni | Ni2+ || Ni2+ | Ni
Oxidation ½ reaction:
Reduction ½ reaction:
Nernst Equation:
Ni  Ni2+ + 2 e–
Ni2+ + 2 e–  Ni
Ecell = E°cell –
Eºred =
– 0.26 V
Eºcell =
0.00 V
RT
RT
ln Q = E°cell –
ln
nF
nF
(0.50 M Ni2+)
(3.00 M Ni2+)
(8.314462 J/K·mol) (293.15 K)
= 0.00 V –
ln
(2 mol e–) (96,485.34 C/mol e–)
e–
e–
Eºox = – (– 0.26 V)
[Ni2+]products
[Ni2+]reactants
= 0.0226314527 V = 0.023 V
e–
e–
+
K
NO3–
Cathode
(plating)
Anode
(eroding)
sink of
electrons
source of
electrons
Ni
Ni
+
terminal
Ni
2+
3.0 M Ni(NO3)2
Ni
20.0°C
2+
0.50 M Ni(NO3)2
–
terminal
4) Given the electrolytic cell below and the table of Standard Reduction Potentials at 25°C in
Appendix D, complete the diagram. A check list is provided to help you include all of the required
elements.
 use the external power supply symbol and/or standard reduction potentials to determine the
direction of flow for this cell (Hint: the non-spontaneous direction)
 fill in the redox ½ reactions and potentials, and determine the standard cell potential
 label the diagram with the following:
 direction of species flow
o the species plating, eroding, or otherwise reacting on the electrode surfaces
o the electrons in the wire
o the ions in the salt bridge
 anode and cathode (and whether they are plating, eroding, or neither)
 source and sink of electrons
 positive and negative terminals
Fe  Fe2++ 2 e–
Oxidation ½ reaction:
Reduction ½ reaction:
Cr3+ + 3 e–  Cr
source of
External
electrons Power Supply
e–
–
Eºred =
– 0.74 V
Eºcell =
– 0.29 V
sink of
electrons
e–
+
terminal
e–
Eºox = – (– 0.45 V)
terminal
e–
+
K
NO3–
Cathode
(plating)
Anode
(eroding)
Cr
Fe
Cr3+
Fe2+
Why is Fe2+ shown instead of Fe3+?
Fe2+ forms because it has a reduction potential that
is closer to that of Cr3+, therefore less potential is
required and Fe2+ will form preferentially over Fe3+.
5) Aluminum metal is formed from a molten mixture containing Al3+ ions in an electrolytic cell. A
current of 250,000 A is applied to this cell at a potential of 6.0 V for 1.0 hour. Given the
relationships below, calculate the indicated quantities.
Charge (Coulombs): 1 mole e– = 96,485.34 C, Faraday’s constant: F = 96,485.34 C / mol e–
Current (amperes, amps): A = C/s (s = seconds)
Energy (Joules): J = kg·m2/s2
Potential (volts): V = J/C
Power (watts): W = J/s, W = A·V
a) What mass (in kg) of aluminum metal is formed?
Hint: use dimensional analysis, starting with the given 1.0 hour
1.0 hour
60 min
60 s
250,000 C
1 mol e–
1 mol Al
26.98 g Al
1 kg
1h
1 min
1s
96,485.34 C
3 mol e–
1 mol Al
103 g
current
Faraday’s
constant
molar ratio
molar mass
83.88839175 kg Al =
84 kg Al
Answer: _______________________
b) If the market price of electricity is $ 0.1323 / kWh (kilowatt hour), what would the cost be for
the amount of electricity consumed?
Hint: find the watts using the provided equation, then use dimensional analysis, starting with
the given 1.0 hour multiplied by this number of watts
W = A·V = (250,000 C/s) (6.0 J/C) = 1,500,000 J/s = 1,500,000 W
1.0 hour (1,500,000 W)
1 kW
$ 0.1323
103 W
1 kWh
price
kWh
$ 2.0 x 102
Answer: _______________________
= $ 198.45