dsj294 – hw10 – Holcombe – (53570)
1
This
printout
should
have
25
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
You
must
use
the
Table
of
Stan
dard
Reduction
Potentials
on
Dr.
Mc
Cord’s
course
website
for
all
prob
lems
in
which
values
are
not
given.
http://courses.cm.utexas.edu/pmccord/
ch302/help/redtable.html
Note: emf = electromotive force, which is
analogous to potential.
001
10.0 points
A tin electrode in 0
.
601 M Sn(NO
3
)
2
(aq) is
connected to a hydrogen electrode in which
the pressure of H
2
is 1 bar.
Sn

Sn
2+

H
+

H
2

Pt
If the cell potential is 0
.
061 V at 25
◦
, what
is the pH of the electrolyte at the hydrogen
electrode?
Correct answer: 1
.
44592.
Explanation:
The cell reaction is
Sn(s) + 2 H
+
→
Sn
2+
+ H
2
(g)
.
Using the Nernst equation,
E
=
E
◦

0
.
025693
n
ln
[Sn
2+
]
P
H
2
[H
+
]
2
0
.
061 = 0
.
14

0
.
025693
2
ln
(0
.
601)(1)
[H
+
]
2

0
.
079 =

(0
.
0128465) ln
0
.
601
[H
+
]
2
6
.
14953 = ln
0
.
601
[H
+
]
2
e
6
.
14953
=
0
.
601
[H
+
]
2
[H
+
]
2
=
0
.
601
e
6
.
14953
= 0
.
00128282
[H
+
] =
√
0
.
00128282 = 0
.
0358165
pH =

log(0
.
0358165) = 1
.
44592
002
10.0 points
How much Au will be plated out (from a so
lution containing Au
3+
) by the same amount
of current that plates 10.5 g of Cu (from a
solution of Cu
2+
)?
1.
32.5 g
2.
0.110 g
3.
10.5 g
4.
21.6 g
correct
5.
48.8 g
Explanation:
003
10.0 points
A concentration cell consists of the same
redox couples at the anode and the cathode,
with di
ff
erent concentrations of the ions in the
respective compartments. Find the unknown
concentration for the following cell.
Pt(s)

Fe
3+
(aq
,
0
.
1 M)
,
Fe
2+
(aq
,
1 M)

Fe
3+
(aq
,
?)
,
Fe
2+
(aq
,
0
.
001 M)

Pt(s)
E
= 0
.
1 V
Correct answer: 0
.
00490142 M.
Explanation:
E
cell
= 0
.
1 V
M
1
= 0
.
1 M
M
2
= 1 M
M
3
= 0
.
001 M
F
= 96485 C/mol
R T
F
= 0
.
025693 V
At the cathode,
Fe
3+
(aq
, x
) +
e

→
Fe
2+
(aq
,
0
.
001 M)
E
◦
= 0
.
77 V
At the anode,
Fe
2+
(aq
,
1 M)
→
Fe
3+
(aq
,
0
.
1 M) +
e


E
◦
=

0
.
77 V
Equate the
e

:
Fe
3+
(aq
, x
) +
e

→
Fe
2+
(aq
,
0
.
001 M)
At the anode,
Fe
2+
(aq
,
1 M)
→
Fe
3+
(aq
,
0
.
1 M) +
e

The overall reaction is
Fe
3+
(aq
, x
) + Fe
2+
(aq
,
1 M)
→
Fe
2+
(aq
,
0
.
001 M) + Fe
3+
(aq
,
0
.
1 M)
E
cell
= 0
.
77 V

0
.
77 V = 0 V
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dsj294 – hw10 – Holcombe – (53570)
2
Using the Nernst equation,
E
cell
=
E
◦
cell

R T
n F
ln
Q
R T
n F
ln
M
1
M
3
x M
2
=
E
◦
cell

E
cell
ln
M
1
M
3
x M
2
=
n F
(
E
◦
cell

E
cell
)
R T
M
1
M
3
x M
2
= exp
n F
(
E
◦
cell

E
cell
)
R T
1
x
=
M
2
M
1
M
3
×
exp
n F
R T
(
E
◦
cell

E
cell
)
=
1 M
(0
.
1 M)(0
.
001 M)
×
exp
1(0 V

0
.
1 V)
0
.
025693 V
= 204
.
022
Thus
x
= [Fe
3+
] =
1
204
.
022
= 0
.
00490142 M
.
004
10.0 points
Consider the voltaic cell
In

In
3+
(1 M)

Ru
3+
(1.0 M), Ru
2+
(0.010 M)

C
In
3+
+ 3
e

→
In
E
0
=

0
.
34 V
Ru
3+
+ 1
e

→
Ru
2+
E
0
= +0
.
25 V
The experimental cell potential for the cell
is approximately
1.
+0.71 V.
correct
2.
+0.30 V.
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 Spring '07
 Holcombe
 Electrochemistry, Correct Answer, Ecell, Ecell − Ecell

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