Chemistry 163
Study Guide Answer Key
Chapter 19
1. The anode in a galvanic cell and in an electrolytic (electrolysis) cell is
a. positive in both cells.
b. the site of oxidation and reduction respectively.
c. the site of reduction and oxidation respectively. d. the site of oxidation.
e. the site of reduction.
Oxidation always occurs at the anode in either type of cell. In a galvanic cell the reaction is spontaneous,
i.e., the oxidation or giving up of electrons is spontaneous as is the removal of electrons from the cathode
by the spontaneously reducing species. The result is that the anode has more electrons than the cathode,
i.e., is negative with respect to the cathode. The opposite is true in an electrolytic cell. Here the species are
being forced to oxidize by having their electrons stripped from them by a strongly positive (with respect
to the cathode) electrode potential. Thus a. is not true and only d. is the correct definition of the anode.
2. The overall reaction for the lead storage battery is
Pb(s) + PbO2(s) + 4H+(aq) + 2SO42- (aq) → 2PbSO4(s) + 2H2O
When the battery discharges
b. PbO2 is the oxidizing agent.
c. the pH increases.
a. PbSO4 is formed at both electrodes.
d. the density of the solution decreases.
e. All of these statements are true.
a. Is true because PbSO4 is the product of both the reduction Pb4+ → Pb2+ , and the oxidation Pb(s) →
Pb2+
b. is true since Pb4+ is being reduced to 2+. c is true since the reaction is spontaneous left-to-right, H+ is
being consumed in the reaction. d is true primarily viewed from the standpoint that ions are disappearing
in favor of a solid and water - ionic solutions are more dense than pure water.
3. The following reaction takes place in a lead storage battery.
discharge
PbO2 + Pb + 2H2 SO4
2PbSO4 + 2H2O
charging
Which statement is true ?
a. The concentration of H2SO4 increases as the battery discharges. Acid is being consumed during
discharge
b. Pb is formed at the anode during discharge.Formation of Pb is a reduction/cathode process
c. PbO2 is formed at the anode during charging.Formation of lead oxide is an oxidation process
d. The mass of Pb decreases during charging. No, it increases as we go right-to-left
e. The mass of PbSO4 remains constant during charging and discharging. It is formed during discharge
and consumed during charging.
4. The voltaic cell is represented as
Zn(s) | Zn2+ (1.0 M) || Cu2+(1.0 M) | Cu(s)
Which of the following statements is false ?
a. The mass of the zinc electrode increases during discharge. The Zn electrode is the place of
oxidation, or loss of electrons. Thus the Zn electrode is decreasing in mass.
b. The copper electrode is the cathode. Reduction occurs at the cathode and is written on the right in
electrochemical cell notation.
c. Electrons flow through the external circuit from the zinc electrode to the copper electrode.Electron flow
is from the oxidation to the reduction sides of the reaction for a cell.
d. Reduction occurs at the copper electrode during discharge. By the notation given above this must be
true by definition of the notation.
e. The concentration of Cu2+ decreases during discharge. Again this is what reduction means for this cell.
5. When Al is obtained by electrolysis from Al2O3 , the minimum number of coulombs required to produce 1.00
mol of aluminum is? There are 96,500 C per mole of electrons. Since Al has an oxidation number of +3,
each mole of Al(s) requires 3 moles of electrons. 3 x 96,500 C = 289,500 C or 290,000 using 3 SF
6. How many Faradays are required to convert a mole of MnO4- to Mn2+ ? The oxidation state of Mn in the
permanganate ion is +7. Thus to convert 1 mole of permanganate to Mn2+ requires 5 moles of electrons.
One mole of electrons has a charge of 1 Faraday (96,500 C), thus we need 5 Faradays for the conversion.
7. A cell with a potential of 0.15 V has the cell reaction
2Cu2+ (aq) + H2
→ 2Cu+(aq) + 2H+ (aq)
Chemistry 163
Study Guide Answer Key
Chapter 19
If the concentrations of the ions were 1.0 molar and the
pressure of H2 was 1.0 atmosphere, then Eo for the half-reaction
Cu2+ (aq) + e- →
o
o
would be: E cell = E = 0.15 V, because the hydrogen electrode has a potential of 0 V.
a. -0.075 V.
b. -0.15 V.
c. 0.075 V.
d. 0.15 V.
e. 0.30 V.
Cu+(aq)
8. Consider the following standard reduction potentials: 2H+(aq) + 2e- →
H2 (g)
Eo = 0.00 V
Ni2+(aq) + 2e- → Ni(s)
Eo = -0.25 V
2+
Cd (aq) + 2e →
Cd(s)
Eo = -0.40 V
Which pair of substances will react spontaneously?
a. Ni 2+ with Cd2+
b. Ni with Cd2+
c. Ni2+ with H+
d. Cd with Ni
e. Cd with Ni2+
For a spontaneous reaction, a positive voltage must come from the cell, thus Ereduction - Eoxidation > 0 V. For
this to occur (using the choices given) Cd must be oxidized and Ni2+ reduced.
Ag +(aq) + e- →
Ag
Eo = 0.80 V
2+
Mn + 2e →
Mn
Eo = -1.18 V
2+
+
All the following statements are correct except
for the standard cell Mn | Mn (aq) || Ag (aq) | Ag
a. The cell voltage will be 1.98 V.
b. The one cell reaction is Mn →
Mn2+ + 2e. Oxidation is shown on the left in cell notation.
c. The overall cell potential will decrease with time. True of all spontaneous reactions therefore true of
all galvanic cells.
d. Anions will be attracted to the Ag electrode compartment. This is false. Ag+ is being reduced, thus
cations are being removed from the Ag electrode compartment. Thus, cations must be attracted to the
compartment to replace the Ag+.
e. Ions migrate in the solution, and electrons flow through the external circuit. This a brief but accurate
decription of a voltaic cell.
9. Consider the following standard electrode potential.
10. What is the value of the reaction quotient, Q, for the cell that is constructed from the two half-reactions
Zn2+ + 2e- → Zn -0.76 V
Ag+ + e→ Ag
0.80 V
when the Zn2+ concentration is 0.0100 M and the Ag+ concentration is 1.25 M? This cell must be assumed
to be a galvanic cell. Then, the reaction is spontaneous left-right when Ag+ is reduced and on the left and
Zn is oxidized and Zn2+ is on the right, thus Q = [Zn2+]/[Ag+]2 = 0.0100/(1.25)2 = 6.40 x 10-3
11. What is the equilibrium constant, K, of the net cell reaction involving the two half-reactions below?
Cd2+ + 2e- →
Cd
-0.40 V
+
Ag + e
→ Ag
0.80 V
nFEcell/RT
We worked through this relationship in class and came up with K = e
= e2x96500x1.20/8.31x298
n = 2, F and R are constants, T is assumed to be 298 K, Eocell = 1.20 V
K = 4.1 x 1040
12. Calculate the equilibrium constant K for the reaction
Sn + Pb2+(aq) → Sn2+(aq) + Pb(s)
where Ecell = 0.014 V Same process as the problem above
K = 3.0 to 2 SF
13. What is the emf at 25oC for the following cell?
Cr | Cr3+(0.010 M) || Ag+ (0.00010 M) | Ag
Cr3+ + 3e- → Cr
Eo = -0.74 V
+
Ag + e → Ag
Eo = 0.80 V
o
3+
+ 3
3
10
E cell = 1.54 V, T = 298 K, n = 3, Q = [Cr ]/[Ag ] = .010/(.00010) = 1.0 x 10
This requires the Nernst equation Ecell = Eocell - RT/nF LnQ = 1.54 - (8.31x298)/(3x 96500)x Ln 1.0 x 1010
Ecell = 1.54 - .197 = 1.34 V
14. What half-reaction occurs at the anode during the electrolysis of molten lithium bromide? Oxidation
always occurs at the anode. Since oxidation is the loss of electrons we want the species that is losing
electrons , i.e., the anion here. so...
2Br- → Br2 + 2e-
Chemistry 163
Study Guide Answer Key
Chapter 19
15. For a certain oxidation-reduction reaction, Ecell is negative. This means that
a. ΔG is negative and K is less than 1.
b. ΔG is negative and K is greater than 1.
c. ΔG is zero and K is greater than 1.
d. ΔG is positive and K is greater than 1.
e. ΔG is positive and K is less than 1. If the cell voltage is negative the reaction is non-spontaneous.
Therefore, ΔG > 0. It also indicates that the reaction proceeds towards reactants in a spontaneous
fashion, in other words reactants are favored, meaning K < 1.
16. Zn2+ + 2e→ Zn
2+
Cu + 2e
→ Cu
Cr2O72- + 14H+ + 6e→
2Cr3+ + 7H2O
Eo = -0.76 V
Eo = +0.34 V
Eo = +1.33 V
On the basis of the foregoing standard electrode potentials, determine which of the following is the strongest
reducing agent. The best reducing agent is the material which is oxidized easiest, this is the same as having
the lowest reduction potential and this is Zn.
b. Zn
c. Cu
d. Cr2O72e. Cr 3+
a. Zn2+
17. A substance that will reduce Hg2+ to Hg but will not reduce Cd2+ to Cd is
a. Fe.
b. Sn.
c. Au.
d. Zn.
e. Cr. To be able to reduce one of these ions but not the other
requires that the material be a better reducing agent or oxidizer than Hg and less so than Cd, of those
listed Sn is higher on the table (page 791) Than Hg and lower than Cd. Fe3+ doesn’t work even though it
is between also, because it will preferentially reduce to Fe2+ which is not between Hg and Cd.
18. Which of the following species is the best reducing agent? The question is really asking the same
question as 17 above. Which of these is the best at oxidizing? This translates to the one having the lowest
reduction potential, which of the group listed is: Cd, it is highest on the table page 791 for the direction
the reaction must go.
c. I
d. Sn
e. F
a. Cd
b. Mg2+
19. Which of the following reactions is or are spontaneous under standard electrochemical conditions?
1. Ni(s) + 2Fe3+
2. 2Fe2+ + Cu2+
3. Mg(s) + 2H+
→ 2Fe2+ + Ni2+
→ 2Fe3+ + Cu
→ Mg2+ + H2(g)
Eocell = 0.77 - (-0.23) = 1.00
Eocell = 0.34 - 0.77 = -0.43
Eocell = 0.00 - (-2.38) = 2.38
a. 1
b. 3
c. 1 and 2
d. 1 and 3
e. 1, 2, and 3
Cells are spontaneous as written when the cell voltage is positive. ½ cell reduction potentials determine
from the textbook.