Problem Set 23 Answers
1.
(,) Use your table of standard reduction potentials to choose the electron that has greater potential
energy in each case.
For each of these, I would ask myself which electron is more easily removed. An electron that is more easily
removed is at higher potential energy. In order to tell if an electron is more easily removed, I would look at the
standard reduction potential table. However, since the table lists reductions (adding electrons), I would think of
each reaction in reverse (removing electrons). Therefore, I would flip the sign on each reduction potential.
For example, in part (a), I would look at these two reactions:
Al  Al3+ + 3 eE˚ = + 1.66 V
+
Cu  Cu + e
E˚ = -0.52 V
The first reaction, where aluminum is oxidized, has a higher potential, so it is more likely to happen. Therefore,
electrons on Al are more easily removed. So an electron on Al is at a higher potential energy.
a.
b.
c.
d.
An electron on Al or an electron on Cu?
An electron on Fe or an electron on Fe2+?
An electron on Co2+ or an electron on Fe2+?
An electron on H2 or an electron on O2?
2. (,) Using your table of reduction potentials, explain why gold and silver are good materials for jewelry
and coins.
Gold and silver both have relatively high positive reduction potentials:
Au3+ + 3 e-  Au
E˚ = +1.50 V
Ag+ + e-  Ag
E˚ = +0.80 V
A positive reduction potential means that these metal ions are likely to be reduced. The reverse is also true – the
metals themselves are unlikely to be oxidized. Since we are surrounded by an atmosphere of oxygen (a good oxidizing
agent), metal oxidation can be a problem. We want jewelry and coins to last through time. If we made these things
out of iron, they would rust and eventually get worn away. Silver and gold, however, do not oxidize easily, so they
last much longer.
Note: In this cell, because neither Fe nor Co is solid, we would probably use a non-reactive conductor as the
electrode. I chose platinum in the drawings above, but others (such as graphite) would also work. Platinum
is a common choice because it is very hard to oxidize.
17.52. (,) Consider only the species (at standard conditions)
Ce4+, Ce3+, Fe2+, Fe3+, Fe, Mg2+, Mg, Ni2+, Sn
17.56. (,) Consider the concentration cell below:
Predict the direction of electron flow, and designate the anode and cathode compartments.
The “goal” of a concentration cell is to make the two concentrations equal (since neither side is favored at
equilibrium). In order for that to happen, [Fe2+] must go down in the left cell, and [Fe2+] must go up in the
right cell.
So on the left, we want this half-reaction to occur: Fe2+(aq) + 2 e−  Fe(s)
On the right, we want this half-reaction to occur: Fe(s)  Fe2+(aq) + 2 e−
Therefore, electrons will flow from right to left. The right side is the anode, and the left side is the cathode.
17.58. (,) Calculate the Ecell for the concentration cell described in Exercise 56. (Assume T = 25oC.)
The overall reaction occurring here is Feright(s) + Fe2+left(aq)  Fe2+right(aq) + Feleft(s)
So for this reaction, 𝑸 =
[𝑭𝒆𝟐+ ]𝒓𝒊𝒈𝒉𝒕
[𝑭𝒆𝟐+ ]𝒍𝒆𝒇𝒕
The number of electrons (n) is 2.
We’ll use the Nernst equation:
𝑬𝒄𝒆𝒍𝒍
𝑬𝒄𝒆𝒍𝒍
[𝑭𝒆𝟐+ ]𝒓𝒊𝒈𝒉𝒕
𝑹𝑻
=𝑬 −
𝒍𝒏 �
�
𝒏𝑭
[𝑭𝒆𝟐+ ]𝒍𝒆𝒇𝒕
𝟎
𝑱
� (𝟐𝟗𝟖 𝑲)
𝟓 × 𝟏𝟎−𝟔
𝒎𝒐𝒍
∙
𝑲
=𝟎−
𝒍𝒏 �
�
𝑪
𝟐. 𝟓
(𝟐 𝒎𝒐𝒍 𝒆− ) �𝟗𝟔𝟓𝟎𝟎
�
𝒎𝒐𝒍 𝒆−
�𝟖. 𝟑𝟏
𝑱
𝑬𝒄𝒆𝒍𝒍 = − �𝟎. 𝟎𝟏𝟐𝟖 � 𝒍𝒏(𝟐 × 𝟏𝟎−𝟔 ) = 𝟎. 𝟏𝟔𝟖 𝑽
𝑪
Note: 1 Joule per Coulomb is defined as 1 Volt.
17.72. (,) A disproportion reaction involved a substance that acts as both an oxidizing and reducing
agent, producing the higher and lower oxidation states of the same element in the products. Which of the
following disproportionation reactions are spontaneous under standard conditions? (Circle the ones that are.)
Calculate ΔGo and K at 25o C for each reaction.