Chemistry Reference
In-Depth Information
0.0592
n
1
[Ag
]
MnO
4
(aq)
Cl
(aq) £ Mn
2
(aq)
ClO
4
PP16.13
Step 1:
Step 2:
(aq)
?
PP17.3
ε
ε°
log
MnO
4
(aq) £ Mn
2
(aq)
1
(0.100)
0.74 V
No change.
0.80
0.0592 log
MnO
4
(aq) £ Mn
2
(aq)
4 H
2
O(
/
)
Step 3:
Step 4:
Step 5:
Step 1:
Step 2:
PP17.4
ε
°
0.34 V
(
0.13 V)
0.47 V
8 H
(aq)
MnO
4
(aq) £ Mn
2
(aq)
4 H
2
O(
/
)
log
[Pb
2
]
[Cu
2
]
0.0592
n
ε
ε°
5 e
8 H
(aq)
MnO
4
(aq) £ Mn
2
(aq)
4 H
2
O(
/
)
Cl
(aq) £ ClO
4
(aq)
log
(0.100)
0.0592
2
0.47
(0.200)
0.48 V
No change.
Scince
ε
is positive, Cu is reduced.
Cl
(aq) £ ClO
4
Step 3:
Step 4:
Step 5:
Step 7:
4 H
2
O(
/
)
(aq)
Electricity
Heat
Cl
(aq) £ ClO
4
(aq)
8 H
(aq)
4 H
2
O(
/
)
PP17.5
3 C(s)
Al
2
O
3
(special solution) -------------£
Cl
(aq) £ ClO
4
(aq)
8 H
(aq)
8 e
4 H
2
O(
/
)
2 Al(
/
)
3 CO(g)
40 e
64 H
(aq)
8 MnO
4
(aq)
20 H
2
O(
)
5 Cl
(aq)
5 ClO
4
(aq)
40 H
(aq)
8 Mn
2
(aq)
32 H
2
O(
)
40 e
Simplifying:
£
a
1 mol e
96,500 C
a
3600 s
1 h
a
10.0 C
1 s
b
b
b
PP 17.6
3.00 h
a
1 mol Ag
1 mol e
a
108 g Ag
1 mol Ag
b
b
121 g Ag
24 H
(aq)
8 MnO
4
(aq)
5 Cl
(aq) £
5 ClO
4
(aq)
8 Mn
2
(aq)
12 H
2
O(
/
)
a
3 mol e
1 mol Al
a
1 mol Al
27.0 g Al
a
96,500 C
1 mol e
1 s
500 C
1.00
10
6
g Al
PP 17.7
b
b
b
a
b
Step 8:
24 H, 8 Mn, 32 O, 5 Cl, 11 net positive charges on each side
2
(aq) £ HCHO
2
(aq)
Cr
3
(aq)
PP16.14
Step 1:
Step 2:
CH
3
OH(aq)
Cr
2
O
10
7
s
2.14
248 days
7
CH
3
OH £ HCHO
2
No change.
Step 3:
Step 4:
Step 5:
Step 6:
H
2
O
CH
3
OH £ HCHO
2
18 Chemical Equilibrium
PP18.1
H
2
O
CH
3
OH £ HCHO
2
4 H
The equilibrium will shift to the right to use up some of the
added
H
2
O
CH
3
OH £ HCHO
2
4 H
4 e
H
2
O.
1 C, 6 H, 2 O, zero net charge, change in oxidation number 4
(C from
PP18.3
The equilibrium will shift to the left to use up some nitrogen.
2
to
2).
[SO
3
]
2
[SO
2
]
2
[O
2
]
PP18.10
K
2
(aq) £ 2 Cr
3
(aq)
Step 1:
Step 2:
Cr
2
O
7
No change.
[N
2
O]
2
[N
2
]
2
[O
2
]
[N
2
O]
[N
2
][O
2
]
1
2
(aq) £ 2 Cr
3
(aq)
7 H
2
O(
/
)
PP18.11 (a)
K
(b)
K
Step 3:
Step 4:
Step 5:
Cr
2
O
7
2
14 H
(aq)
Cr
2
O
2
(aq) £ 2 Cr
3
(aq)
7 H
2
O(
/
)
7
The constant in part (b) is the square root of the one in part (a).
6 e
14 H
(aq)
Cr
2
O
2
(aq) £
7
PP18.14
2 Cr
3
(aq)
7 H
2
O(
/
)
2 SO
2
(g)
O
2
(g)
E
2 SO
3
(g)
Step 6:
14 H, 2 Cr, 7 O, net charge
6
,
change in oxidation number
6.
Initial:
0.0150
0.00750
0.00000
3 CH
3
OH(aq)
16 H
(aq)
2 Cr
2
O
2
Step 7:
(aq) £
Change:
Equilibrium:
0.00760
0.00380
0.00760
7
4 Cr
3
(aq)
11 H
2
O(
/
)
3 HCHO
2
(aq)
0.0074
0.00370
0.00760
Step 8:
3 C, 28 H, 4 Cr, 17 O, 12 positive net charge on each side.
PP16.15
Both the oxidizing agent and the reducing agent are
The
[SO
3
]
2
[SO
2
]
2
[O
2
]
(0.00760)
2
(0.0074)
2
(0.0037)
2.8
5
10
2
K
H
2
O
2
.
H
2
O
2
disproportionates according to the equation
The equilibrium concentrations are the same as in Example
18.14, and the value of
K
is the reciprocal of that in the exam-
ple because the equation is written in the opposite direction.
2 H
2
O
2
(aq) £ 2 H
2
O(
/
)
O
2
(g)
17 Electrochemistry
PP17.1
PP18.17
Wire
2 H
2
S(g)
E
2 H
2
(g)
S
2
(g)
Initial:
0.200
0.200
0.000
K
NO
3
e
e
Change:
Equilibrium:
2
x
2
x
x
Ag
Salt bridge
0.200
2
x
0.200
2
x
x
Cu
[H
2
]
2
[S
2
]
[H
2
S]
2
9.3
10
8
K
Cu
2
Ag
NO
3
SO
4
2
(0.200)
2
(
x
)
(0.200)
2
AgNO
3
CuSO
4
9.3
10
8
x
Ag
e
£ Ag
Cu £ Cu
2
2 e
The initial concentration of
H
2
has affected the equilibrium
concentration of
S
2
markedly.
Cu
2 Ag
£ 2 Ag
Cu
2
