Chapter 17 Review Question Answers
Choose the best answer for the following questions.
1.
What value of ℰ do you expect for Pb(s) → Pb+2 + 2e-1 in 0.015m Pb+2 solution? (ℰ = +0.13 V)
a. ℰ = +0.18v
ℰ = ℰ  0.0592/n log Q
b. ℰ = +0.27v
ℰ = 0.13  (0.0592/2)* log [0.015]
c. ℰ = +0.13v
ℰ = 0.13 + 0.05
ℰ = 0.18 V
d. ℰ = 0.00v
e. ℰ = +0.83v
2.
For the galvanic cell described below, the correct line notation is:
Cl2 + 2e-1 → 2Cl-1
+1
-1
Cu + e → Cu
a.
b.
c.
d.
e.
+1
(ℰ = 1.36 V)
(ℰ = 0.52 V)
-1
Pt(s)|Cu(s)|Cu (aq)||Cl2(g)|2Cl (aq)|Pt(s)
Pt(s)|Cl2(g)|2Cl-1(aq)||Cu(s)|Cu+1(aq)
Cu(s)|Cu+1(aq)||Cl2(g)|2Cl-1(aq)
Pt(s)|Cl2(g)|2Cl-1(aq)||Cu(s)|Cu+1(aq)|Pt(s)
Cu(s)|Cu+1(aq)||Cl2(g),2Cl-1(aq)|Pt(s)
Cl2 + 2e-1 → 2Cl-1
(ℰ = 1.36 V)
+1
2 Cu → 2Cu + 2e
-1
+1
Cl2 + 2Cu → 2Cl + 2Cu
In the electrolytic decomposition of water: 2H2O → 2H2 + O2
a. hydrogen is formed at the anode and oxygen is formed at the cathode
b. hydrogen is formed at both the anode and the cathode
c. oxygen is formed at both the anode and the cathode
d. hydrogen is formed at the cathode and oxygen is formed at the anode
4.
Reduction potential is:
a. an extensive property
b. measured in amperes
c. an intensive property
d. always positive
+1 2
5.
0
0
2H2O → 2H2 + O2
change
reduced H +1  0
1
oxidized O 2  0
+2
Hydrogen is reduced so H2 is formed at the cathode.
Oxygen is oxidized so O2 is formed at the anode.
This picture of an electrochemical cell can best be described as:
a. a complete electrolytic cell
b. a complete galvanic cell
c. a galvanic cell, but missing at least one
essential component
d. an electrolytic cell, but missing at least one
essential component
This is a galvanic cell because the current is flowing from
anode to cathode. However, the salt bridge or porous disk
is missing so that there isn’t a complete circuit.
(ℰ = +0.84 V)
The reversed reaction is at the anode. It comes first
and then the cathode comes last. The solids are on the
outside and the aqueous are on the inside. Cl2 gas is
included with the aqueous ions because it is a gas.
Therefore, Pt must be used as a physical electrode.
3.
Reduction potential is an
intensive property because
it does NOT depend on
how much is there. (You
can multiply the reaction
by any coefficient and not
affect the cell potential.)
(ℰ = 0.52 V)
-1
6.
With a current of 20 amps, how long would it take to generate 1.00 kilograms of
aluminum by the reaction: Al+3 + 3e-1 → Al(s)
a. 1.78 x 105 sec
1s
96,485 C 3 mole e-1 1 mol Al 103 g 1.00 kg = 5.36 x 105 s
b. 1.68 x 102 sec
20 C 1 mole e-1 1 mol Al 27.0 g
1 kg
c. 5.36 x 105 sec
3
d. 3.90 x 10 sec
e. 2.19 x 104 sec
7.
The equation that represents a reaction that is not a redox reaction is:
1. 2H2 + O2 → 2H2O
0
0
+1 2
2. Zn + CuSO4 → ZnSO4 + Cu
1. 2H2 + O2 → 2H2O
0
+2 +6 2
+2 +6 2
0
3. 2H2O2 → 2H2O + O2
2.
Zn
+
CuSO
→
ZnSO
+
Cu
4
4
4. H2O + CO2 → H2CO3
+1 1
+12
0
a. Reaction 2
3. 2H2O2 → 2H2O + O2
b. Reaction 4
+1 2
+4 2
+1 +4 2
c. Reaction 1
4. H2O + CO2 → H2CO3
d. Reaction 3
The first three reactions change oxidation
e. All of these are redox reactions
numbers, but the fourth does not.
8.
How many electrons are transferred in the following reaction?
2NaCl + 2H2SO4 + MnO2 → Na2SO4 + MnSO4 + 2H2O + Cl2
a. 8
+1 1
+1 +6 2
+4 2
+1 +6 2
+2 +62
+12
0
b. 2
2NaCl + 2H2SO4 + MnO2 → Na2SO4 + MnSO4 + 2H2O + Cl2
c. 6
change
d. 4
reduced Mn +4  +2 2  1 = 2
Two electrons are
e. 1
oxidized Cl 1  0
+1  2 = +2
transferred.
0
9.
The equilibrium constant (K) for the reaction below is approximately:
a.
b.
c.
d.
e.
10.
Cl2 + 2e-1 → 2Cl-1
(ℰ = 1.36 V)
Cu+1 + e-1 → Cu
(ℰ = 0.52 V)
-34
10
1011
1017
1028
102
Calculate ΔG for the reaction:
Cl2 + 2e-1 → 2Cl-1
+1
-1
Cu + e → Cu
-3.1 x 104 J
-8.1 x 104 J
8.1 x 105 J
-1.6 x 105 J
-6.2 x 104 J
(ℰ = 1.36 V)
(ℰ = 1.36 V)
2 Cu → 2Cu + 2e
(ℰ = 0.52 V)
Cl2 + 2Cu → 2Cl-1 + 2Cu+1
(ℰ = +0.84 V)
+1
0
a.
b.
c.
d.
e.
Cl2 + 2e-1 → 2Cl-1
-1
log K = n ℰ/0.0592
log K = (2 0.84/0.0592
log K = 28
K = 1028
(ℰ = 0.52 V)
ΔG = nFℰ
ΔG =  (2 mole e-1)(96,485 C/s)(0.84 V)
ΔG =  1.6 x 105 J
11.
A mole of electrons has a charge of 96,485 coulombs per mole of electrons. This quantity
is known to chemists as:
a. 1 faraday
b. 1 ampere
A faraday is 96,485 C/mole e1.
c. 1 watt
d. 1 joule
e. 1 volt
12.
The standard hydrogen electrode is assigned a potential of:
a. 1.00 volts
b. -1.00 volts
The standard hydrogen electrode has a potential of 0.00 V.
c. zero volts
d. 0.76 volts
13.
In all electrochemical cells, the process that takes place at the anode is ______________
and the process that takes place at the cathode is _________________.
a. reduction, oxidation
b. oxidation, oxidation
Oxidation occurs at the anode and reduction occurs at the cathode.
c. oxidation, reduction
d. reduction, reduction
14.
Two cells, one containing aqueous AgNO3 and the other containing CuSO4 are set up in
series. In a given electrolysis that results in depositing 1.25 g of silver in the first cell,
how much copper should deposit simultaneously in the second cell?
a. 0.368 g
b. 0.736 g
63.5 g Cu 1 mol Cu 1 mole e-1 1 mol Ag 1.25 g Ag = 0.367 g Cu
c. 1.25 g
d. 2.65 g
1 mol Cu 2 mole e-1 1 mol Ag 108 g Ag
e. 1.47 g
15.
The current in a given wire is 1.80 amp. How many coulombs will pass a given point on
the wire in 1.36 minutes?
a. 1.32 C
b. 2.45 C
1.80 C 60 s
1.36 min = 147 C
c. 147 C
1s
1 min
d. 45.3 C
e. 253 C
16.
What is the oxidation state of S in H2SO3?
a. +4
+1 +4 2
b. 0
c. -3
H2SO3
d. +2
e. +6
17.
If a constant current of 8.00 amperes is passed through a cell containing Zn2+ for 2.00
hours, how many grams of zinc will plate out onto the cathode?
a. 39.0 g
65.4 g Zn 1 mol Zn 1 mole e-1 8.00 C 60 s
60 min 2.00 hr = 19.5 g Zn
b. 126 g
c. 0.985 g
1 mol Zn 2 mole e-1 96,485 C 1 s
1 min 1 hr
d. 19.5 g
e. 1.43 x 103 g
18.
Which of the following would be the best reducing agent?
a. O-2
change
b. Na+1
oxidized O
2  0
+2
c. Na
reduced
Na
+1

0
1
d. F-1
oxidized Na
0  +1 +1
e. Cl2
oxidized F
1  0
+1
oxidized Cl
0  1 +1
The best reducing agent would be the most easily oxidized. This eliminates Na +1.
The remaining choices are all elements from the right side of the periodic table which
have high numbers of valence electrons except for Na which has a small number of
valence electrons. Na would be the most likely.
19.
Calculate the value of E0cell for a galvanic cell that contains the following half cells:
Cl2 + 2e-1 → 2Cl-1
+1
a.
b.
c.
d.
e.
20.
-1
Cu + e → Cu
E cell = 0.32 volts
E0cell = 0.84 volts
E0cell = 1.88 volts
E0cell = -0.32 volts
E0cell = -0.84 volts
(ℰ = 1.36 V)
(ℰ = 0.52 V)
0
Cl2 + 2e-1 → 2Cl-1
+1
(ℰ = 1.36 V)
(ℰ = 0.52 V)
-1
2 Cu → 2Cu + 2e
-1
+1
Cl2 + 2Cu → 2Cl + 2Cu
(ℰ = +0.84 V)
In the reaction below, the oxidizing agent is:
2NaCl + 2H2SO4 + MnO2 → Na2SO4 + MnSO4 + 2H2O + Cl2
+1
a. H
b. Mn+4
+1 1
+1 +6 2
+4 2
+1 +6 2
+2 +62
+12
0
c. Cl-1
-2
2NaCl
+
2H
SO
+
MnO
→
Na
SO
+
MnSO
+
2H
O
+
Cl
2
4
2
2
4
4
2
2
d. O
+1
r.a.
o.a.
e. Na
change
reduced
oxidized
Mn +4  +2 2
Cl 1  0 +1