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Chemistry 132, Winter 2008
Solutions to Homework No. 3
Problem 1.
See Levine problem 14.12.
(a) True. Taking the derivative of
E
(Levine eq. 14.41) with
respect to the activity
a
j
of substance
j
gives
∂
E
/∂a
j
=

ν
j
RT/nFa
j
, which is negative when
ν
j
>
0, i.e., when
substance
j
is a product. Therefore, the emf decreases when
the activity of a product is increased.
(b) True. The cell’s charge number
n
is the stoichiometric
coeﬃcient of the electrons on the left terminal (Levine eq.
14.37). These electrons are considered to be a product of
the electrochemical reaction. Stoichiometric coeﬃcients have
no units (Levine p. 132) and the stoichiometric coeﬃcients
of products are positive (Levine p. 132). Therefore,
n
is
positive and has no units.
(c) True. Doubling all the stoichiometric coeﬃcients
ν
i
in a
cell’s reaction causes twice as many electrons to be trans
ferred, thus causing
n
to also double. For a reversible cell,
E
=

∑
0
i
ν
i
μ
i
/nF
(Levine eq. 14.39), thus doubling both
the
ν
i
’s and
n
leaves
E
unchanged.
(d) False. The standard emf of the cell
E
◦
is the limiting
value of the lefthand side of Levine eq. 14.49 as all molalities
approach zero. The lefthand side of that equation is not the
same as
E
.
Problem 4.
See Levine problem 14.26. The halfreactions for this cell
are shown in the text (Levine eq. 14.33). From these half
reactions and from Levine table 14.1, the standard emf for
the cell is found to be
E
◦
=
E
◦
R

E
◦
L
= (0
.
2222 V

0 V) =
0
.
2222 V. The activity quotient is given by
Q
= exp[
nF
(
E
◦

E
)
/RT
] (Levine eq. 14.43) and for this cell
n
= 2.
(a)
E
=

1
.
00 V
⇒
E
◦

E
= (0
.
2222 V)

(

1
.
00 V) = 1
.
2222 V
,
Q
= exp
±
2(96485 C
/
mol)(1
.
2222 V)
(8
.
3145 J
/
K
·
mol)(298
.
15 K)
²
=
2
.
08
×
10
41
.
(b)
E
= 1
.
00 V. The calculations are the same as in part (a),
but in this case
E
◦

E
= (0
.
2222 V)

(1
.
00 V) = 0
.
7778 V,
so that
Q
= 5
.
07
×
10

27
.
Problem 7.
See Levine problem 14.31. The given half reactions and their
combination are as follows, where
E
◦
3
is to be determined:
Cr
3+
(
aq
) +
e

→
Cr
2+
(
aq
)
E
◦
1
=

0
.
424 V
n
1
= 1
Cr
2+
(
aq
) + 2
e

→
Cr(
s
)
E
◦
2
=

0
.
90 V
n
2
= 2
Cr
3+
(
aq
) + 3
e

→
Cr(
s
)
E
◦
3
=?
n
3
= 3
When each of the above halfreactions takes place on the
right electrode of a cell that has a hydrogen electrode as the
left electrode, the standard molar Gibbs energy change of
each overall reaction is Δ
G
◦
j
=

n
j
F
E
◦
j
(Levine eq. 14.42),
where
j
= 1
,
2
,
3. Since the third halfreaction is the sum of
the ﬁrst and second halfreactions
Δ
G
◦
3
= Δ
G
◦
1
+ Δ
G
◦
2

n
3
F
E
◦
3
=

n
1
F
E
◦
1

n
2
F
E
◦
2
E
◦
3
=
n
1
E
◦
1
+
n
2
E
◦
2
n
3
=
(

0
.
424 V) + 2(

0
.
90 V)
3
=

0
.
7413 V
.
Problem 10.
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 Winter '08
 Lindenberg
 Chemistry

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