10.3 STOICHIOMETRY OF CELL REACTIONS
PRACTICE
(Page 748)
Understanding Concepts
1. I = 1.5 A = 1.5 C/s
t = 30 s
q = It
C
= 1.5 30 s
s
q = 45 C
The charge transferred is 45 C.
2. q = 87.6 C
t = 22.5 s
q = It
q
I = t
87.6 C
= 22.5 s
= 3.89 C/s
I = 3.89 A
The average electric current is 3.89 A.
3. I = 250 mA = 250 10–3 C/s
t = 28.5 s
q = It
C
= 250 10–3 28.5 s
s
q = 7.13 C
The charge transferred is 7.13 C.
4. I = 1.60 A = 1.60 C/s
q = 375 C
q = It
q
t = I
375 C
= C
1.60 s
= 234 s
1 min
= 234 s 60 s
t = 3.91 min
The charge is transferred in a time of 3.91 min.
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Chapter 10
Copyright © 2003 Nelson
PRACTICE
(Page 749)
Understanding Concepts
5. I 1.9 A 1.9 C/s
60 s
t 35 min 2.1 103 s
1
min
F 9.65 104 C/mol
q
ne– F
It
F
C
1.9 2.1 103 s
s
C
9.65 104 mol
ne– 0.041 mol 41 mmol
The amount of electrons transferred is 0.041 mol or 41 mmol.
6. I 1.24 A 1.24 C/s
ne– 0.146 mol
F 9.65 104 C/mol
q
ne– F
It
ne– F
n –F
t e
I
C
0.146 mol
9.65 104 mol
C
1.24 s
1h
t 1.13 104 s 3.16 h
3600 s
The time of cell operation is 3.16 h.
60 s
7. t 20 min
1.2 103 s
1
min
ne– 0.015 mol
F 9.65 104 C/mol
q
ne– F
It
ne– F
ne–F
I t
C
9.65 104 0.015 mol
mol
1.2 103 s
I 1.2 C/s 1.2 A
The current required is 1.2 A.
Copyright © 2003 Nelson
Electrolytic Cells
431
PRACTICE
(Page 751)
Understanding Concepts
8.
Zn(s)
→
Zn2+
(aq)
+
m
65.38 g/mol
I 0.500 A 0.500 C/s
60 s
t 10.0 min 600 s
1
min
q
ne– F
It
F
C
0.500 600 s
s
C
9.65 104 mol
2 e–
0.500 A
10.0 min
9.65 104 C/mol
ne– 0.00311 mol
1 mol Zn
nZn 0.00311 mol
e– –
2 m
ol e
nZn 0.00155 mol
g
mZn 0.00155 mol 65.38 mol
mZn 0.102 g
The mass of zinc oxidized is 0.102 g.
3 e–
→
54 A
45 min 30 s
9.65 104 C/mol
I 54 A 54 C/s
60 s
t 45 min 30 s
1
min
2700 s 30 s
9. (a) Cr3+
(aq)
+
Cr(s)
m
52.00 g/mol
t 2730 s
Note: Because time was measured to a precision of seconds, this time value (2730 s) has a certainty of 4 significant
digits.
q
ne– F
It
F
C
54 2730 s
s
C
9.65 104 mol
ne– 1.5 mol
1 mol Cr
nCr 1.5 mol
e– –
3 m
ol e
nCr 0.51 mol
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Chapter 10
Copyright © 2003 Nelson
g
mCr 0.51 mol 52.00 mol
mCr 26 g
The mass of chromium produced is 26 g.
(b) Ni2+
(aq)
+
2 e–
→
0.540 A
9.65 104 C/mol
t
Ni(s)
0.250 g
58.69 g/mol
I 0.540 A 0.540 C/s
1 mol
nNi 0.250 g 58.69 g
nNi 0.00426 mol
2 mol e–
ne– 0.00426 mol
Ni 1 m
ol Ni
ne– 0.00852 mol
q
ne– F
It
ne– F
n –F
t e I
C
9.65 104 0.00852 mol
mol
C
0.540 s
1
m
in
t 1.52 103 s 60 s
t 25.4 min
The time to plate the nickel is 25.4 min.
→ Al ]
10. (cathode) 2 [Al3+
(l) + 3 e
(s)
(anode)
3 [2 Cl
(l) → Cl2(g) + 2 e ]
____________________________________________________
(net)
2 Al3+
+
3 Cl2(g)
(l) + 6 Cl(l) → 2 Al(s)
5.40 g
m
26.98 g/mol
70.90 g/mol
1 mol
nAl 5.40 g 26.98 g
nAl 0.200 mol
3 mol Cl
nCl 0.200 mol
Al 2
2
2 m
ol Al
nCl 0300 mol
2
g
mCl 0300 mol 70.90 2
mol
mCl 21.3 g
2
The mass of chlorine gas produced at the anode is 21.3 g.
+
11. Cr2O72
(aq) + 14 H(aq) + 12 e
→
2.20 h
9.65 104 C/mol
I
Copyright © 2003 Nelson
2 Cr(s) + 7 H2O(l)
17.8 g
52.00 g/mol
Electrolytic Cells
433
3600 s
t 2.20 h 1 h
t 7.92 103 s
1 mol
nCr 17.8 g 52.00 g
nCr 0.342 mol
12 mol e–
ne– 0.342 mol
Cr 2 m
ol Cr
ne– 2.05 mol
q
ne– F
It
ne– F
n –F
I e t
C
2.05 mol
9.65 104 mol
7.92 103 s
I 25.0 C/s 25.0 A
The current required is 25.0 A.
Applying Inquiry Skills
Prediction
12. (a) According to the stoichiometry of the known half-reaction and Faraday’s law, 0.766 g of tin should form as shown
below.
Sn2+
+
2 e–
→
Sn(s)
(aq)
3.46 A
m
6.00 min
118.69 g/mol
9.65 104 C/mol
I 3.46 A 3.46 C/s
60 s
t 6.00 min 360 s
1
min
q
ne– F
It
ne– F
C
3.46 360 s
s
C
9.65 104 mol
ne– 0.0129 mol
1 mol Sn
nSn 0.0129 mol
e– –
2 m
ol e
nSn 0.00645 mol
g
mSn 0.00645 mol 118.69 mol
mSn 0.766 g
Analysis
(b) The mass of tin produced is (118.05 – 117.34) g 0.71 g.
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Chapter 10
Copyright © 2003 Nelson
Evaluation
experimental value predicted value 
(c) % difference 100%
predicted value
0.71 g 0.766 g 
100%
0.766 g
0.06 g
100%
0.766 g
% difference 7%
The percentage difference in this experimental result is 7%.
(The percentage difference is properly obtained by not rounding any numbers in any of the calculations. If
rounded values from previous steps are used, a slightly different percentage will be obtained. Note that the
subtraction of two very close values in the % difference produces only one significant digit in the answer.)
(d) The prediction is judged to be verified because its value agrees fairly well with the experimental value; and the
difference can be accounted for by typical experimental errors or uncertainties. The stoichiometric method based
on the tin half-reaction is judged acceptable because the prediction was verified.
Making Connections
13. (a) A car battery ampere-hour rating tells you how much total charge the battery can deliver, which is proportional
to the quantity of energy the battery can provide.
I = 125 A = 125 C/s
t = 1 h 3600 s
q = It
C
= 125 3600 s
s
q = 4.50 105 C
The maximum charge that can be transferred by this battery is 4.50 105 C.
(b) This is a useful way to rate batteries because cost can then be compared to how much energy a battery will supply,
allowing an informed choice to be made.
(c) Pb(s)
→
Pb2+
+
2 e–
(aq)
m
125 A
207.20 g/mol
1 h 3600 s
9.65 104 C/mol
I 125 A 125 C/s
q
ne– F
It
ne– F
C
125 3600 s
s
C
9.65 104 mol
ne– 4.66 mol
1 mol Pb
nPb 4.66 mol
e– –
2 m
ol e
nPb 2.33 mol
g
mPb 2.33 mol 207.20 mol
mPb 483 g
The mass of lead oxidized as the battery discharges is 483 g.
Copyright © 2003 Nelson
Electrolytic Cells
435
(d) Al3+
(aq)
+
3 e–
→
Al(s)
26.98 g/mol
ne– 4.66 mol
1 mol Al
nAl 4.66 mol
e– –
3 m
ol e
nAl 1.55 mol
g
mol 26.98 mAl 1.55 mol
mAl 41.9 g
The mass of aluminum oxidized would be 41.9 g.
(e) Obviously, there would be a huge weight advantage to using an aluminum–oxygen fuel cell; the mass of metal
required for discharge is less than 9% as much, if the charge delivered is equal.
SECTION 10.3 QUESTIONS
(Page 753)
Understanding Concepts
1. I 0.300 A 0.300 C/s
60 s
t 15.0 min 900 s
1
min
F 9.65 104 C/mol
q
ne– F
It
F
C
0.300 900 s
s
C
9.65 104 mol
ne– 0.00280 mol 2.80 mmol
The amount of electrons transferred is 2.80 mmol.
2. 2 Cl(aq) →
Cl2(g)
+
m
70.90 g/mol
I 55 kA 5.5 104 C/s
3600 s
t 8.0 h 2.9 104 s
1 h
2 e
55 kA
8.0 h
9.65 104 C/mol
q
ne– F
It
F
C
5.5 104 2.9 104 s
s
C
9.65 104 mol
ne– 1.6 104 mol
1 mol Cl2
nCl 1.6 104 mol
e– 2 m
ol e–
2
3
nCl 8.2 10 mol
2
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Chapter 10
Copyright © 2003 Nelson
g
mCl 8.2 103 mol
70.90 2
mol
mCl 5.8 105 g 0.58 Mg
2
The mass of chlorine produced is 0.58 Mg or 0.58 t.
3. Ag+(aq) +
e–
→
1.80 A
9.65 104 C/mol
t
Ag(s)
10.00 g
107.87 g/mol
I 1.80 A 1.80 C/s
1 mol
nAg 10.00 g 107.87 g
nAg 0.0927 mol
1 mol e –
ne– 0.0927 mol
Ag 1 m
ol Ag
0.0927 mol
q
ne– F
It
ne– F
n –F
t e I
C
0.0927 mol
9.65 104 mol
t C
1.80 s
1 min
t 4.97 103 s 60 s
t 82.8 min
The time to silverplate the teapot is 82.8 min.
4. Al3+
(aq) +
3 e
→
24.0 h
9.65 104 C/mol
I
Al(l)
425 kg
26.98 g/mol
3600 s
t 24.0 h 1 h
8.64 104 s
1 mol
nAl 425 kg
26.98 g
nAl 15.8 kmol
3 mol e –
ne– 15.8 kmol
Al 1 m
ol Al
ne– 47.3 kmol
q
ne– F
It
ne– F
n –F
I e t
Copyright © 2003 Nelson
Electrolytic Cells
437
C
47.3 kmol
9.65 104 mol
I 8.64 104 s
I 52.8 kC/s 52.8 kA
The average current required is 52.8 kA.
5. (a) Mg2+
(l)
+
2 e
→
2.0 105 A
18.0 h
9.65 104 C/mol
Mg(s)
m
24.31 g/mol
I 2.0 105 A 2.0 105 C/s
3600 s
t 18.0 h 6.48 104 s
1 h
q
ne– F
It
ne– F
C
2.0 105 6.48 104 s
s
C
9.65 104 mol
ne– 134 kmol
1 mol Mg
nMg 134 kmol
e– 2 m
ol e–
nMg 67.2 kmol
g
mMg 67.2 kmol
24.31 mol
mMg 1.63 103 kg 1.63 Mg
The mass of magnesium produced is 1.63 Mg or 1.63 t.
(b) 2 Cl–(aq)
→
Cl2(g)
+
m
70.90 g/mol
2 e–
134 kmol
1 mol Cl2
nCl 134 kmol
e– 2
2 m
ol e–
nCl 67.2 kmol
2
g
mCl 67.2 kmol
70.90 2
mol
mCl 4.76 103 kg 4.76 Mg
2
The mass of chlorine produced is 4.76 Mg or 4.76 t.
6. Co2+
(aq)
250.0 mL
+
2 e–
→
1.14 A
2.05 h
9.65 104 C/mol
Co(s)
m
58.93 g/mol
I 1.14 A 1.14 C/s
3600 s
t 2.05 h 7.38 103 s
1 h
q
ne– F
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Chapter 10
Copyright © 2003 Nelson
It
ne– F
C
1.14 7.38 103 s
s
C
9.65 104 mol
ne– 0.0872 mol
1 mol Co2+
nCo2+ 0.0872 mol
e– 2 m
ol e–
nCo2+ 0.0436 mol
1 mol CoSO4
nCoSO 0.0436 mol
Co2+ 4
1 mo
l Co2+
nCoSO 0.0436 mol
4
0.0436 mol
CCoSO 4
0.2500 L
CCoSO 0.174 mol/L
4
The minimum molar concentration of cobalt(II) sulfate required is 0.174 mol/L.
7. Cu(s) →
Cu2+
+
(aq)
mi = 25.72 g
mf
63.55 g/mol
I 0.876 A 0.876 C/s
60 s
t 75.0 min 4.50 103
1
min
2 e
0.876 A
75.0 min
9.65 104 C/mol
s
q
ne– F
It
ne– F
C
0.876 4.50 103 s
s
C
9.65 104 mol
ne– 0.0408 mol
1 mol Cu
nCu 0.0408 mol
e– 2 m
ol e–
nCu 0.0204 mol
g
mCu 0.0204 mol 63.55 mol
mCu 1.30 g (mass loss at the anode)
The final mass of the copper anode will be (25.72 1.30) g 24.42 g.
Making Connections
8. In a chlor-alkali cell, Faraday’s law is critical to the process design because the quantities of electrical current required
must be determined by the desired production of chemical products. The cost of power, size of conductors, and time
of operation are all matters of concern to industrial processes. Process control just involves varying the quantity of
charge through the cell, which will proportionally vary the amount of product.
9. (a) Chemical technicians prepare solutions; order inventory; maintain and clean equipment; set up bays/areas for
experimentation; perform many kinds of analyses with all types of equipment; and maintain records of lab stocks,
policies, and procedures.
(b) Chemical technician education requirements vary, but usually include certification in a two- or three-year instruction program at a technical school or college.
(c) (Educational facilities listed may be local facilities or those farther afield.)
(d) (Job listings will, of course, vary with the newspaper selected and the date.)
Copyright © 2003 Nelson
Electrolytic Cells
439