Samples: Multiple-Choice Questions
- What mass of copper would be produced by the reduction of the Cu2+(aq) ion by passing
96.487 amperes of current through a solution of copper(II) chloride for 100.00 minutes?
(1 Faraday = 96,487 coulombs)
A. 95.325 g
B. 190.65 g
C. 285.975 g
D. 381.30 g
E. cannot be determined from the information provided
Answer: B
Step 1: Write the reaction that would occur at the cathode.
Cu2+(aq) +2e–→Cu(s)
Step 2: This problem can be solved by using the factor-label method:
(Note all the conversion factors that you should be comfortable with.)
..
,
.
.
min
min
sec
sec
amperes utes
utes
onds
ampere ond
coulomb
coulombs
Faraday
Faraday
mole e
moles e
mole Cu
mole Cu
gCu
gCu
1
96 487
1
100 0
1
60
1
1
96 487
1
1
1
2
1
1
63 55
190 65
##
# : ##
##=
- Find E°for a cell composed of silver and gold electrodes in 1 molar solutions of their
respective ions: E°redAg = +0.7991 volts; E°redAu = +1.68 volts.
A. −0.44 volt
B. 0 volt
C. 0.44 volt
D. 0.88 volt
E. 2.48 volts
Answer: D
Notice that E°redfor silver is lower than E°redfor gold. This means that because silver is higher in
the activity series, silver metal will reduce the gold ion.
Reduction and Oxidation