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Electrochemistry Review Questions
1. The net equation for a given voltaic cell is:
Sn (s) + 2 Ag+  Sn2+ + 2 Ag (s)
a. Write the two half-reactions involved, and identify each in terms of (1) site of oxidation or
reduction and (2) anode or cathode.
b. Calculate the net potential of the cell (the voltage).
c. Draw a fully labeled diagram of the voltaic cell. Be sure to indicate the flow of electrons in the
external circuit (through the wire) and the flow of ions
2. The following questions pertain to a voltaic cell
Cl2 (g) + 2e-  2Cl - (aq)
Ca2+ (aq) + 2e-  Ca (s)
a) Which half-reaction is a reduction and which is an oxidation? How do you know?
b) Rewrite the half-reactions in the correct directions.
c) Write a balanced equation for the overall cell reaction.
d) Calculate the cell potential.
3. A voltaic cell is constructed based on the following half reactions:
Pb2+ (aq) + 2e- -> Pb(s)
Au3+(aq) +3e- -> Au(s)
a) Which is the anode and which is the cathode in this cell?
b) What is the cell potential?
4. A voltaic cell is constructed based on the following half reactions:
Mn2+ (aq) + 2e- -> Mn(s)
Cu2+ (aq) +2e- -> Cu(s)
a) Which is the anode and which is the cathode in this cell? The overall voltage is 1.52 V
And since Mn is not on the table you must figure out where it would be placed on the table and
then whether it was an anode or cathode.
b) What is the standard cell potential for Mn?
5. Indicate whether the following reactions are spontaneous or not and what the reaction potential
between the substances are.
a) Zn + HCl
b) Mg + FeCl2
c) Pb + SnCl2
d) Mg + Fe SO4
e) Ag + Cu(NO3)2
f) Al + CuSO4
6. Write the appropriate half reactions, which metal is the cathode/anode, and calculate the cell potential
for each of the following electrochemical cells.
a) Copper and aluminum
b) Silver and zinc
c) Copper and nickel
7. Given the following experimental data, arrange in an activity series.
Co2+ + In(s) → Co(s) + In2+
Cu2+ + Co(s) → Cu(s) + Co2+
Cu2+ + Pd(s) → no reaction
Electrochemistry Review ANSWERS
1.
Sn(s)  Sn2+ + 2eAg+ + e-  Ag(s)
b. Eo = 0.94 V
 Oxidation, anode
 Reduction, cathode
c. Sorry, in this drawing, replace copper with tin
Sorry, in this drawing, replace copper with tin
2.a) Chlorine will be reduced, higher on chart
b) Ca(s)

Ca2+
+ 2eCl2 (g) + 2e 
2Cl -(aq)
d) Eo = 4.23 V
3. a) Anode: lead
Cathode: gold
b) Eo = 1.63 V
4.
a) Anode: Mn
Cathode: Cu
b) Eo = for Mn if placed on the table would be 1.86 as Cu would now be flipped to -0.34 V
a)
b)
c)
d)
e)
f)
Zn + HCl Spontaneous  Eo = 0.76 V
Mg + FeCl2 Spontaneous  Eo = 1.93 V
Pb + SnCl2 Non-spontaneous  Eo = -0.01 V
Mg + Fe SO4 Spontaneous  Eo = 1.93 V
Ag + Cu(NO3)2 Non-spontaneous  Eo = -0.46 V
Al + CuSO4 Spontaneous  Eo = 2.00
6. a)Cathode: Cu2+ + 2e-  Cu(s) Anode: Al(s)  Al3+ + 3e-Eo = 2.18
b) Cathode: Ag+ + 1e-  Ag(s)
c) Cathode: Cu2+ + 2e-  Cu(s)
7. In2+ + 2e-  In(s)
Co2+ + 2e-  Co(s)
Cu2+ + 2e-  Cu(s)
Pd2+ + 2e-  Pd(s)
Anode: Zn(s)  Zn2+ + 2eAnode: Ni(s)  Ni2+ + 2e-
Eo = 1.56 V
Eo = 0.59 V