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(b) 0.77 V
(c)
0.14 V
26 H (aq) 3 Cr 2 O 7 2 (aq) 2 CN (aq) £
(b)
6 Cr 3 (aq) 2 CO 2 (g) 2 NO 2 (g) 13 H 2 O( / )
[Mn 2 ]
[MnO 4 ][H ] 8
a 0.0592
n
0
(d)
ε ε
b
log
a
b
38 H (aq) 5 Cr 2 O 7 2 (aq) 2 CNS (aq) £
(c)
10 Cr 3 (aq) 2 CO 2 (g) 2 NO 2 (g)
a 0.0592
5
0.500
(2.00) (1.00) 8
2 SO 4 2 (aq) 19 H 2 O( / )
1.51
b
log
a
b
1.52 V
16.82
Titanium(IV) chloride and titanium tetrachloride
Note that water is not included in the argument of the log,
since in dilute aqueous solutions it is regarded as pure.
17 Electrochemistry
17.1
[Mn 2 ][Fe 3 ] 5
[MnO 4 ][H ] 8 [Fe 2 ] 5
a 0.0592
n
0
17.28
(a)
ε ε
b
log
a
b
In the lead storage cell, all the reactants and products are
solids and cannot migrate in the sulfuric acid, so both elec-
trodes can be placed in the same solution.
(0.500) (0.900) 5
(0.125) (1.00) 8 (0.600) 5
a 0.0592
5
0.74
b
log
a
b
17.2
Ye s
0.72 V
17.3
Since electrons appear in the equation, there must be at least
one ionic species to balance the charge.
The reaction goes as written. Note that water is not included
in the argument of the log, since in dilute aqueous solutions
it is regarded as pure.
(b)
17.4
(a) The coefficients in the equation for any systems do not
matter; standard state is 1 M for any solute and 1 atm for
any gas.
Sn 2
Pb 2 £ Pb
Sn 4
0
ε
0.26 V
[Sn 4 ]
[Sn 2 ][Pb 2 ]
17.5
(a) 0.90 V (Changing the coefficients does not alter the
potential.)
a 0.0592
2
ε ε
0
b
a
b
log
(b)
0.90 V
(Reversing the equation requires changing the
sign of the potential.)
(0.500)
(3.00) (2.00)
0.26 0.0296 log
a
b
17.6
No. No matter what its potential, a half-reaction cannot pro-
ceed without another half-reaction.
0.23 V
Opposite direction
(c)
1.58 V
As written
17.7
(a) Coulomb
(b) Ampere
a 1 mol e
96,500 C
17.30
Hydrogen and oxygen. (Water is more easily reduced than
aluminum ion and more easily oxidized than sulfate ion.)
10 5 mol e
17.9
(a)
5.00 C
b
5.18
17.31
Iron(II) sulfate
e £ 2 H 2
OH
(b)
H 2 O
17.35
(a) iii
(b) i
(c) ii
e
1 mol of
17.36
(a) Voltaic (The reaction goes spontaneously.)
(b) Electrolysis (The reaction does not go spontaneously.)
(c) Voltaic
1 mol e
6.02
2.00 10 11 e
(c)
a
b
10 23 e
(d) Voltaic
(e) Electrolysis
3.32 10 13 mol e
17.37
The charge that passes through the first cell is given by
2 mol e
1 mol e
(d)
(e)
a 1 mol Ag
107.9 g Ag
a 1 mol e
1 mol Ag
0.01390 mol e
1.500 g Ag
b
b
Heat
17.10
Al 2 O 3 (in solution)
3 C(s) ¡
2 Al( / )
3 CO(g)
a 3 mol e
1 mol Al
2 mol Al
1 mol Al 2 O 3
The mass of the other elements is given by
15.0 mol e
2.50 mol Al 2 O 3
a
b
b
a 63.55 g Cu
1 mol Cu
a 1 mol Cu
2 mol e
0.01390 mol e
b
b
0.4417 g Cu
17.12
The mass of copper deposited is
25.782 g 20.171 g
5.611 g.
a 69.72 g Ga
1 mol Ga
a 1 mol Ga
3 mol e
0.01390 mol e
b
b
0.3230 g Ga
5.611 g
10.69 g
100%
52.49% Cu
a 1 mol Hg 2 2
2 mol e
a 1 mol e
96,500 C
17.38
19,200 C
b
b
17.13
Merely change the zinc and zinc ion in the Daniell cell (Fig-
ure 17.1) to lead and lead(II) ion. Lead will be oxidized.
2 mol Hg
1 mol Hg 2 2
a 200.6 g Hg
1 mol Hg
a
b
b
39.9 g Hg
MnO 4
Fe 3
17.15
(a)
(b) Equal
(c)
2 H
2 e £ H 2
17.17
(a)
(b)
17.40
The oxidation of the chloride ion does not affect the hydro-
gen ion concentration:
2 H 2 O 2 e £ H 2
2 OH
17.18
It turns to heat.
2 Cl £ Cl 2
2 e
Cu £ Cu 2
2 e
0
17.19
(a)
ε
0.34 V
The reduction reaction produces hydroxide ion:
Fe 2
2 e £ Fe
0
ε
0.44 V
H 2 O e £ 2 H 2 OH
Fe 2
Cu £ Fe Cu 2
0
ε
0.78 V
a 1 mol e
96,500 C
a 1 mol OH
1 mol e
a 0.400 C
1 s
(b)
1.24 V
(c)
0.43 V
(d)
0.47 V
500.0 s
b
b
b
X 2 .
17.21
(a) No
(b) Ye s
(c) Z will reduce
0.00207 3 mol OH
Cu
17.22
disproportionates (reacts with itself) to yield Cu and
(Reverse the equation, add it to the
, and get a positive potential.)
0.00207 3 mol
0.02500 L
Cu 2 .
Cu 2 /Cu
0.0829 M
Cu /Cu
17.42
The system will not experience any net change.
a 0.0592
n
1
[Fe 2 ]
0
17.24
(a)
ε ε
b
log
a
b
17.43
(a) The freezing (melting) point is lowered (Section 15.5).
(b) To save energy in heating the salt.
(c)
a 0.0592
2
1
(0.100)
It cannot be more easily reduced; if it were, the sodium
would not be reduced first.
0.44
b
log
a
b
0.47 V
 
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