Electrochemical Cells – Practice Problems KEY
1. Write half-cell reactions and give standard reduction potentials for these reduction
reactions. (use table & go to 2 decimal places)
+
+
a) Cu
Cu Cu + 1e Cu E°= +0.52 V
b) Cl2 2Cl Cl2 + 2e 2Cl E°= +1.36 V
2+
2+
Cu Cu + 2e Cu E°= +0.34 V
c) Cu
2+
2+
Ba Ba + 2e Ba E°= -2.91 V
d) Ba
+
+
e) Ag
Ag Ag + 1e Ag E°= +0.80 V
2. Determine which metal will be the anode (more negative number), and E°cell
(= cathode - anode) for the following sets of half-reactions:
3+
a)
Al + 3e ===> Al E° =-1.66 V (anode)
3+
Au + 3e ===> Au E° =+1.50 V
E° = (1.50 V) – (-1.66 V) = 3.16 V
+
b)
-
Li + e ===> Li E° = -3.05 V (anode)
+
Ag + e ===> Ag E° = +0.80 V
E° = (0.80 V) – (-3.05 V) = 3.85 V
2+
c)
-
Mg +2e ===> Mg E° = -2.37 V (anode)
3+
Fe + 3e ===> Fe E° = -0.036 V
E° = (-0.036 V) – (-2.37 V) = 2.33 V
2+
+
3. Two half-cells, one containing Fe 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. E° = 1.2466 V = SPONTANEOUS
4. Use a Standard Reduction Potential table to determine the E°cell value for the
spontaneous reaction (which means these E° must be POSITIVE for the rxn to be
spontaneous) of each pair of half-cells listed below:
LOOK AT THE HALF-CELL RXNS… THEY MAY BE SWITCHED AROUND BASED ON
WHAT YOU SEE IN THE CHART, BUT DO NOT SWITCH THE NUMBERS FOR E°.
LEAVE THOSE NUMBERS AS THEY ARE IN THE CHART TO DETERMINE THE
OVERALL E°
+
-
2+
-
a) Ag ===> Ag + 1e ; 0.80 V Cathode Fe + 2e ===> Fe; -0.45 V Anode
2+
2+
b) Mg ===> Mg + 2e ; -2.37 V Anode Sn + 2e ===> Sn; -0.14 V Cathode
+
2+
c) Li ===> Li + 1e ; -3.04 V Anode Mn + 2e ===> Mn; -1.19 V Cathode
3+
2+
d) Cr ===> Cr + 3e ; -0.74 V Anode Hg + 2e ===> Hg; +0.85 V Cathode
2+
2+
e) Ni ===> Ni + 2e ; -0.26V Anode Cu + 2e ===> Cu; +0.34 V Cathode
E° = +1.25 V
E° = 2.23 V
E° = 1.85 V
E° = 1.59 V
E° = 0.60 V
2+
+
5. Two half cells, one containing Ca 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 (spontaneous means the E° must be positive) and the
value for E° cell. Diagram the cell and label its parts. Give equations for the half reactions.
Ca2+ + 2 eCa = -2.87 V = ANODE
Ag+ + eAg = 0.80 V = CATHODE
E° = 0.80 V – (-2.87 V) = 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:
2+
Ca + 2e Ca E° = -2.87 V (anode)
3+
2+
Fe + e Fe 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).
2+
Anode: Ca
-
Ca + 2e = -2.87 V
in the Table 21-1 even though the equation is written backwards here.
USE THE NUMBER IN THE TABLE AS IT IS GIVEN!
3+
-
Cathode: Fe + e
8.
E°= (0.77 V) – (-2.87 V) = +3.64 V
2+
Fe = 0.77 V in the table
What is the standard cell potential for a voltaic cell composed of the following half-cells:
2+
Cu + 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).
2+
-
Anode: Cu
Cu + 2e
+
Ag
Cathode: Ag + e
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!)
+
2Ag + Ni
2+
2+
2Ag + Ni
-
E°= -0.26 V
Anode: Ni
Ni + 2e
+
Cathode: Ag + e Ag E°= 0.80 V = 0.80 V
E°= (0.80 V) – (-0.26 V) = +1.06 V YES, spon. as written
10. Decide if the following redox reaction is spontaneous as written: (see hint in #9)
3+
Cr + Al
3+
-
Anode: Al
Al + 3e
3+
Cathode: Cr + 3e Cr
Cr + Al
3+
E°= -1.66 V
E°= -0.74 V
E°= (-0.74 V) – (-1.66 V) = +0.92 V YES, spon. as written