KEY
Electrochemical Cells – Practice Problems
1. Write half-cell reactions and give standard reduction potentials for these reduction
reactions. (use table)
a) Cu+  Cu
Cu+ + 1e-  Cu
E= +0.52 V
b) Cl2  2Cl
Cl2 + 2e  2Cl
E= +1.36 V
c) Cu2+  Cu
Cu2+ + 2e-  Cu
E= +0.34 V
2+
2+
d) Ba  Ba
Ba + 2e  Ba
E= -2.91 V
e) Ag+  Ag
Ag+ + 1e-  Ag
E= +0.80 V
2. Determine which metal will be the anode, and E°cell for the following sets of halfreactions:
a)
Al3+ + 3e- ===> Al
E° =-1.66 V (anode)
Au3+ + 3e- ===> Au
E° =+1.50 V
E = (1.50 V) – (-1.66 V) = 3.16 V
b)
Li+ + e- ===> Li
Ag+ + e- ===> Ag
E° = -3.05 V (anode)
E° = +0.80 V
E = (0.80 V) – (-3.05 V) = 3.85 V
c)
Mg2+ +2e- ===> Mg
Fe3+ + 3e- ===> Fe
E° = -2.37 V (anode)
E° = -0.036 V
E = (-0.036 V) – (-2.37 V) = 2.33 V
3. Two half-cells, one containing Fe2+ and Fe and the other containing Ag+ and Ag, are
connnected to form a voltaic cell. Use a Standard Reduction Potential table to determine
the direction of spontaneous reaction and the value for E°cell. Diagram the cell and label its
parts. Give equations for the half reactions.
The drawing should have two half cells, one with a solid iron electrode in a solution of Fe2+
ions. The solid Fe is the anode, where the following half reaction takes place:
Fe (s)  Fe2+ (aq) + 2eThis half cell is separated from the other half cell by either a semi-permeable membrane,
or it is connected only by a salt bridge (allows ions to pass through).
The second half cell has a solid silver electrode (the cathode) in a solution of Ag+ ions. At
the source of the cathode in the second half cell, the following half reaction takes place:
Ag+ (aq) + 1 e- Ag (s)
The electrons will flow through the circuit from the anode (Fe) to the cathode (Ag)
Eocell = Eocathode - Eoanode
Eocell = 0.80 V – (-0.45 V)
Eocell = +1.25 V
4. Use a Standard Reduction Potential table to determine the E°cell value for the
spontaneous reaction of each pair of half-cells listed below:
a)
Ag ===> Ag+ + 1e-; +0.80 V
Fe2+ + 2e- ===> Fe; -0.45 V
E° = 1.25 V
2+
2+
b)
Mg ===> Mg + 2e ; -2.37 V
Zn + 2e ===> Zn; -0.76 V
E° = 1.61 V
c)
Li ===> Li+ + 1e-; -3.04 V
Mn2+ + 2e- ===> Mn; -1.19 V
E° = 1.85 V
3+
2+
d)
Cr ===> Cr + 3e ; -0.74 V
Hg + 2e ===> Hg; +0.85 V
E° = 1.54 V
e)
Ni ===> Ni2+ + 2e-; -0.26V
Cu2+ + 2e- ===> Cu; +0.34 V
E° = 0.60 V
5. Two half cells, one containing Ca2+ and Ca and the other containing Ag+ and Ag, are
connected to form a voltaic cell. Use a Standard Reduction Potential table to determine the
direction of spontaneous reaction and the value for E° cell. Diagram the cell and label its
parts. Give equations for the half reactions.
The drawing should have two half cells, one with a solid calcium electrode in a solution of
Ca2+ ions. The solid Ca is the anode, where the following half reaction takes place:
Ca (s)  Ca2+ (aq) + 2eThis half cell is separated from the other half cell by either a semi-permeable membrane,
or it is connected only by a salt bridge (allows ions to pass through).
The second half cell has a solid silver electrode (the cathode) in a solution of Ag + ions. At
the source of the cathode in the second half cell, the following half reaction takes place:
Ag+ (aq) + 1 e- Ag (s)
The electrons will flow through the circuit from the anode (Ca) to the cathode (Ag)
Eocell = Eocathode - Eoanode
Eocell = 0.80 V – (-2.87 V)
Eocell = +3.67 V
6. Explain why a rechargeable battery can be considered a combination voltaic-electrolytic
cell.
When operating in a device (calculator, phone, etc.) it is a voltaic cell (chem. energy 
electric energy); when being recharged, it is operating as an electrolytic cell (electric
energy  chem. energy).
7. A voltaic cell is composed of the following half-cells:
Ca2+ + 2e-  Ca E° = -2.87 V (anode)
Fe3+ + e-  Fe2+ E° = +0.77 V
Write the reaction that takes place at the anode (oxidation) and the reaction that takes
place at the cathode (reduction). Calculate the standard cell potential (E° cell).
Anode:
Ca  Ca2+ + 2eCathode:
Fe3+ + e-  Fe2+
E= (0.77 V) – (-2.87 V) = +3.64 V
8. What is the standard cell potential for a voltaic cell composed of the following half-cells:
Cu2+ + 2e-  Cu E° = +0.34 V (anode)
Ag+ + e-  Ag E° = +0.80 V
Write the reaction that occurs at the anode (oxidation) and the cell reaction that occurs at
the cathode (reduction).
Anode:
Cu  Cu2+ + 2eCathode:
Ag+ + e-  Ag
E= (0.80 V) – (0.34 V) = +0.46 V
9. Is the following redox reaction spontaneous as written? (Hint: write oxidation and
reduction half-reactions; look up the standard reduction potentials, and find E° cell; if E° is
positive, the reaction is spontaneous!)
Anode:
Cathode:
2Ag+ + Ni  2Ag + Ni2+
Ni  Ni2+ + 2eE= -0.26 V
+
2Ag + 2e  2Ag E= 2(0.80 V) = 1.60 V
E= (1.60 V) – (-0.26 V) = +1.86 V
YES, spon. as written
10. Decide if the following redox reaction is spontaneous as written: (see hint in #9)
Anode:
Cathode:
Cr3+ + Al  Cr + Al3+
Al  Al3+ + 3eE= -1.66 V
Cr3+ + 3e-  Cr
E= -0.74 V
E= (-0.74 V) – (-1.66 V) = +0.92 V
YES, spon. as written (positive value)