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Transcript
Unit 5 Review, pages 688–695
Knowledge
1. (a)
2. (a)
3. (a)
4. (c)
5. (a)
6. (d)
7. (d)
8. (b)
9. (c)
10. (d)
11. (a)
12. (c)
13. (d)
14. (d)
15. (a)
16. (c)
17. (b)
18. (c)
19. (d)
20. (b)
21. (c)
22. True
23. False. In a redox reaction, electrons are gained in the reduction half-reaction.
24. False. Oxidation causes an increase in oxidation number.
25. False. Substances in their elemental form have oxidation numbers of 0.
26. False. The sum of the oxidation numbers in a polyatomic ion must equal the overall
charge of the ion.
27. True
28. False. In the synthesis reaction that produces sulfur dioxide from elemental sulfur
and oxygen gas, sulfur is oxidized.
29. True
30. False. Iodine, I2(s), is able to oxidize zinc metal to Zn2+ ions.
31. True
32. True
33. True
34. False. A galvanic cell is a device in which chemical energy is converted to another
form of energy.
35. True
36. False. A secondary cell is designed with a redox reaction that can be reversed by
running a current through the cell.
37. True
38. True
39. True
40. False. Electric current must be supplied to carry out the electrolysis of water.
Copyright © 2012 Nelson Education Ltd.
Unit 5: Electrochemistry
U5-7
41. (a) (iii)
(b) (vii)
(c) (vi)
(d) (iv)
(e) (viii)
(f) (i)
(g) (v)
(h) (ii)
42. (a) In P2O5, the oxidation number of O is –2 and the oxidation number of P is +5.
(b) In NO2, the oxidation number of O is –2 and the oxidation number of N is +4.
(c) In Na2SO4, the oxidation number of Na is +1, the oxidation number of O is –2, and
the oxidation number of S is +6.
(d) In Cu(NO3)2, the oxidation number of O is –2, the oxidation number of Cu is +2,
and the oxidation number of N is +5.
(e) In KMnO4, the oxidation number of K is +1, the oxidation number of O is –2, and
the oxidation number of Mn is +7.
(f) In Na3Fe(OH)6, the oxidation number of Na is +1, the oxidation number of H is +1,
the oxidation number of O is –2, and the oxidation number of Fe is +3.
(g) In XeOF4, the oxidation number of F is –1, the oxidation number of O is –2, and the
oxidation number of Xe is +6.
43. (a) In Na2SO4, since the oxidation number of Na is +1 and the oxidation number of
O is –2, the oxidation number of S is +6.
(b) In K2S2O8, since the oxidation number of K is +1 and the oxidation number of O is
–2, the oxidation number of S is +7.
(c) In S2O32–, since the oxidation number of O is –2 and the overall charge of the ion is
–2, the oxidation number of S is +2.
(d) In SO2, since the oxidation number of O is –2, the oxidation number of S is +4.
44. Redox reactions involve the transfer of electrons. The substance that gains electrons
is the oxidizing agent. The oxidizing agent is reduced in a redox reaction. The substance
that loses electrons is the reducing agent. The reducing agent is oxidized in a redox
reaction.
45. During oxidation, an entity, the oxidizing agent, gains electrons. The electrons
gained have to come from another entity, the reducing agent, which loses electrons—a
reduction reaction. Therefore, oxidation must always occur with reduction.
46. (a) Separate H2(g) + Cl2(g) → 2 HCl(g) into two half-reactions, one for each
element.
H2(g) → 2 H+(g) + 2 e–
Cl2(g) + 2 e– → 2 Cl–(g)
Since Cl2(g) gains electrons, Cl2(g) is reduced.
(b) Separate H2(g) + O2(g) → 2 H2O(l) into two half-reactions, one for each element.
H2(g) → 2 H+(l) + 2 e–
O2(g) + 4 e– → 2 O2–(l)
Since O2(g) gains electrons, O2(g) is reduced.
Copyright © 2012 Nelson Education Ltd.
Unit 5: Electrochemistry
U5-8
(c) Separate S(s) + Cl2(g) → SCl2(l) into two half-reactions, one for each element.
S(s) → S2+(l) + 2 e–
Cl2(g) + 2 e– → 2 Cl–(l)
Since Cl2(g) gains electrons, Cl2(g) is reduced.
(d) Separate 2 Li(s) + F2(g) → 2 LiF(s) into two half-reactions, one for each element.
Li(s) → Li+(s) + e–
F2(g) + 2 e– → 2 F–(s)
Since F2(g) gains electrons, F2(g) is reduced.
47. (a) Assign oxidation numbers to each atom/ion in the chemical equation.
+4 –2
+1 –1
0
+2 –1
+1 –2
+5 –2
0
PbO2(aq) + HI(aq) → I2(aq) + PbI2(s) + 2 H2O(l)
The oxidation number of iodine increases from –1 to 0, so iodine is oxidized.
The oxidation number of lead decreases from +4 to +2, so lead is reduced.
Since Pb in PbO2 gains electrons, the oxidizing agent is PbO2(aq). Since I in HI loses
electrons, the reducing agent is HI(aq).
(b) Assign oxidation numbers to each atom/ion in the chemical equation.
0
+2
+5 –2
+2
Mg(s) + Cu(NO3)2(aq) → Mg(NO3)2(aq) + Cu(s)
The oxidation number of magnesium increases from 0 to +2, so magnesium is oxidized.
The oxidation number of copper decreases from +2 to 0, so copper is reduced.
Since Cu in Cu(NO3)2(aq) gains electrons, the oxidizing agent is Cu(NO3)2(aq). Since
Mg loses electrons, the reducing agent is Mg(s).
(c) Assign oxidation numbers to each atom/ion in the chemical equation.
+3
–2
0
+1 –2
+1 +5 –2
+1 –1
As2O3(s) + Cl2(g) + H2O(l) → H3AsO4(aq) + HCl(aq)
The oxidation number of arsenic increases from +3 to +5, so arsenic is oxidized.
The oxidation number of chlorine decreases from 0 to –1, so chlorine is reduced.
Since Cl in Cl2 gains electrons, the oxidizing agent is Cl2(g). Since As in As2O3 loses
electrons, the reducing agent is As2O3(s).
(d) Assign oxidation numbers to each atom/ion in the chemical equation.
+5 –2
I2O5(s)
+2 –2
0
+4 –2
+ CO(g) → I2(s) + CO2(g)
The oxidation number of carbon increases from +2 to +4, so carbon is oxidized.
The oxidation number of iodine decreases from +5 to 0, so iodine is reduced.
Since I in I2O5 gains electrons, the oxidizing agent is I2O5(s). Since C in CO loses
electrons, the reducing agent is CO(g).
48. (a) The half-reaction for the reduction of chlorine, Cl2, is
Cl2(g) + 2 e– → 2 Cl–(aq)
(b) The half-reaction for the reduction of sulfur, S8, is
S8(s) + 16 e– → 8 S2–(aq)
(c) The half-reaction for the oxidation of zinc is
Zn(s) → Zn2+(aq) + 2 e–
Copyright © 2012 Nelson Education Ltd.
Unit 5: Electrochemistry
U5-9
(d) The half-reaction for the oxidation of sodium is
Na(s) → Na+(aq) + e–
49. First, write a chemical equation representing the reaction.
Mg(s) + CuSO4(aq) → Cu(s) + MgSO4(aq)
The oxidation and reduction half-reaction equations for the reaction are:
Oxidation: Mg(s) → Mg2+(aq) + 2 e–
Reduction: Cu2+(aq) + 2 e– → Cu(s)
50. (a) Write an unbalanced equation for the reaction.
Na(s) + F2(g) → NaF(s)
Determine the oxidation numbers for each element in the equation.
0
0
+1 –1
Na(s) + F2(g) → NaF(s)
The oxidation number of sodium changes from 0 to +1, so each sodium atom loses
1 electron. The oxidation number of fluorine changes from 0 to –1, so each fluorine
atom gains 1 electron.
(b) Each electron gained by a fluorine atom comes from one sodium atom, which loses
1 electron. Since there are 2 fluorine atoms in one fluorine molecule, 2 sodium atoms
are needed to react with one fluorine molecule.
(c) Since 2 sodium atoms are needed to react with one fluorine molecule, multiply
electrons lost by 2.
2 Na(s) + F2(g) → NaF(s)
Balance sodium.
2 Na(s) + F2(g) → 2 NaF(s)
Therefore, the balanced equation for this reaction is:
2 Na(s) + F2(g) → 2 NaF(s)
51. (a) In the reaction Ca(s) + Fe2+(aq) → Ca2+(aq) + Fe(s), Fe2+(aq) is the oxidizing
agent and Ca(s) is the reducing agent. Since the relative positions of Fe2+(aq) and Ca(s)
form a downward diagonal to the right on the redox table, the reaction will occur
spontaneously.
(b) In the reaction Ca2+(aq) + Fe(s) → Ca(s) + Fe2+(aq), Ca2+(aq) is the oxidizing agent
and Fe(s) is the reducing agent. Since the relative positions of Ca2+(aq) and Fe(s) do not
form a downward diagonal to the right on the redox table, the reaction will not occur
spontaneously.
(c) In the reaction 3 Ca(s) + 2 Au3+(aq) → 3 Ca2+(aq) + 2 Au(s), Au3+(aq) is the
oxidizing agent and Ca(s) is the reducing agent. Since the relative positions of Au3+(aq)
and Ca(s) form a downward diagonal to the right on the redox table, the reaction will
occur spontaneously.
(d) In the reaction 2 Au3+(aq) + 3 Fe(s) → 2 Au(s) + 3 Fe2+(aq), Au3+(aq) is the
oxidizing agent and Fe(s) is the reducing agent. Since the relative positions of Au3+(aq)
and Fe(s) form a downward diagonal to the right on the redox table, the reaction will
occur spontaneously.
Copyright © 2012 Nelson Education Ltd.
Unit 5: Electrochemistry
U5-10
(e) In the reaction Fe(s) + 2 H+(aq) → Fe2+(aq) + H2(g), H+(aq) is the oxidizing agent
and Fe(s) is the reducing agent. Since the relative positions of H+(aq) and Fe(s) form a
downward diagonal to the right on the redox table, the reaction will occur
spontaneously.
(f) In the reaction 3 Mg(s) + 2 Cr3+(aq) → 3 Mg2+(aq) + 2 Cr(s), Cr3+(aq) is the
oxidizing agent and Mg(s) is the reducing agent. Since the relative positions of Cr3+(aq)
and Mg(s) form a downward diagonal to the right on the redox table, the reaction will
occur spontaneously.
(g) In the reaction 3 Cu(s) + 2 NO3–(aq) + 8 H+(aq) → 3 Cu2+(aq) + 2 NO(g) + 4 H2O(l),
NO3–(aq) is the oxidizing agent and Cu(s) is the reducing agent. Since the relative
positions of NO3–(aq) and Cu(s) form a downward diagonal to the right on the redox
table, the reaction will occur spontaneously.
52. (a) In the magnesium–iron galvanic cell shown in Figure 1, iron is the cathode
because it has the more positive reduction potential. Magnesium is the anode.
(b)
53. The salt bridge in a galvanic cell is a porous barrier between two half-cells with
different electrolytes. The salt bridge prevents the solutions in the two half-cells from
mixing but allows ions to flow in both directions, preventing charge buildup and
keeping the solution in each half-cell electrically neutral. The salt bridge is a U-shaped
tube filled with a non-reactive electrolyte. The ends of the tube are plugged with cotton
balls which prevent the solution from falling out of the tube, but are porous enough to
permit some fluid and ions to flow in and out of the tube.
54. In a galvanic cell, the cathode is the electrode where reduction occurs. The anode is
the electrode where oxidation occurs.
55. For the cell for which the line notation is Mg(s) | Mg2+(aq) || Fe2+(aq) | Fe:
(a) Anode half-reaction equation: Mg(s) → Mg2+(aq) + 2 e−
Cathode half-reaction equation: Fe2+(aq) + 2 e− → Fe(s)
Add the two half-reactions to obtain the net ionic equation.
Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s)
Copyright © 2012 Nelson Education Ltd.
Unit 5: Electrochemistry
U5-11
(b) From Table 1 in Appendix B7, E°r ( Mg) = !2.37 V and E°r ( Fe) = !0.44 V. Iron is the
cathode because it has the more positive reduction potential; this is also indicated by its
position on the right in the line notation.
!E°r (cell) = E°r (cathode) " E°r (anode)
= "0.44 V " ("2.37 V)
!E°r (cell) = 1.9 V
56. For a nickel–zinc galvanic cell:
(a) From Table 1 in Appendix B7, E°r ( Ni) = !0.23 V and E°r ( Zn) = !0.76 V. Nickel is
the cathode because it has the more positive reduction potential, so nickel ion is reduced
at the cathode.
Ni2+(aq) + 2 e− → Ni(s)
Zinc metal is oxidized at the anode.
Zn(s) → Zn2+(aq) + 2 e−
Add the two half-reactions to obtain the net ionic equation.
Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s)
(b) The cell potential is:
!E°r (cell) = E°r (cathode) " E°r (anode)
= "0.23 V " ("0.76 V)
!E°r (cell) = 0.53 V
(c) The line notation for the galvanic cell is:
Zn(s) | Zn2+(aq) | | Ni2+(aq) | Ni(s)
57. For the galvanic cell that uses the reaction represented by the equation
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
the line notation is
Zn(s) | Zn2+(aq) | | Cu2+(aq) | Cu(s)
!!"
58. 2 H2O(l) + 2 e− #!! H2(g) + 2 OH−(aq)
E° = !0.83 V
r
!!"
Ni2+(aq) + 2 e− #!! Ni(s)
!!"
Au3+(aq) + 3 e− #!! Au(s)
E°r = !0.23 V
E°r = +1.50 V
!!" 1
IO3−(aq) + 6 H+(aq) + 5 e− #!! 2 I2(s) + 3 H2O(l) E°r = +1.20 V
!!"
Li+(aq) + e− #!! Li(s)
E°r = !3.05 V
The reducing agents, from weakest to strongest, are Au(s), I2(s) + H2O(l), Ni(s),
H2(g) + OH−(aq), and Li(s).
59. Batteries consist of cells connected in series.
Copyright © 2012 Nelson Education Ltd.
Unit 5: Electrochemistry
U5-12
60. A dry cell battery is an alkaline dry cell that consists of an electrolyte paste
containing zinc, which is the anode, surrounded by a cathode that contains solid
manganese dioxide.
61. Rusting is a type of corrosion and refers to ferric oxides that form as steel and iron
containing alloys corrode. Corrosion is a result of the oxidation of various metals.
62. 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
63. In an electrolytic cell used for plating chromium onto an object, the chromium
should be present as ions so that is can be reduced and plated onto the object. Chromium
in elemental form cannot undergo reduction and plating.
64. (a) Cathode: Ni2+(l) + 2 e– → Ni(s);
Anode: 2 Br-(l) → Br2(g) + 2 e–
(b) Cathode: Al3+(l) + 3 e– → Al(s);
Anode: 2 F-(l) → F2(g) + 2 e–
(c) Cathode: Mg2+(l) + 2 e– → Mg(s);
Anode: 2 I–(l) → I2 (s) + 2 e–
65. Answers may vary. Sample answer: Electrolytic cells are used in plating coinage, in
the production of aluminum, and in rechargeable batteries.
66. In the electrolysis of water, the electrolyte, for example, sodium sulfate, functions to
allow electrons to flow freely in solution. The free-flowing electrons allow redox
reactions to occur.
Understanding
67. The oxidation number of a monatomic ion is the same as its ionic charge.
68. Oxidation is the process in which one or more electrons is lost by a chemical entity.
Reduction the process in which one or more electrons is gained by a chemical entity. In
an oxidation–reduction reaction, one entity in the reaction gains electrons (reduction)
and another entity loses electrons (oxidation). Therefore, the processes of oxidation and
reduction occur as a pair in the same redox reaction.
69. The position of Cl2(g) is close to the top of the reduction potential table, which
means that it is a strong oxidizing agent. Since the oxidizing agent is reduced in a redox
reaction, Cl2(g) is more likely to be reduced.
70. The half-reactions for a redox reaction represent the oxidation and reduction parts of
the reaction. Since in a redox reaction, the number of electrons lost by oxidation must
always be equal to the number of electrons gained by reduction, so the half-reactions
can be used to balance the redox reaction equation by making the numbers of electrons
lost and gained in the half-reactions equal.
71. Answers may vary. Sample answer:
(a)
(b)
(c)
(d)
Change in
oxidation numbers?
yes
no
no
yes
Element
oxidized
copper
Element
reduced
silver
Oxidizing
agent
Ag+(aq)
Reducing
agent
Cu(s)
magnesium
silicon
SiCl4(l)
Mg(s)
Copyright © 2012 Nelson Education Ltd.
Unit 5: Electrochemistry
U5-13
72. (a) Solution:
Step 1. Assign oxidation numbers to all entities.
–3 +1
0
0
+1
+4 –2
+1 –2
C2H6(g) + O2(g) → CO2(g) + H2O(g)
The oxidation number of carbon changes from –3 to +4, so carbon loses 7 electrons.
The oxidation number of oxygen changes from 0 to –2, so oxygen gains 2 electrons.
Step 2. To balance electrons, adjust the coefficients by multiplying electrons lost by 2
and electrons gained by 7.
2 C2H6(g) + 7 O2(g) → 2 CO2(g) + 7 H2O(g)
Step 3. Balance the rest of the equation by inspection.
Balance carbon.
2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 7 H2O(g)
Balance oxygen by adding water.
2 C2H6(g) + 7 O2(g) + H2O(g) → 4 CO2(g) + 7 H2O(g)
Subtract 1 water molecule from each side to eliminate redundant water molecules.
2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)
Statement: The balanced equation is
2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)
(b) Solution:
Step 1. Write the unbalanced equation with its ionic compounds dissociated.
Mg(s) + H+(aq) + Cl−(aq) → Mg2+(aq) + 2 Cl−(aq) + H2(g)
Step 2. Write the unbalanced net ionic equation, without the spectator ion, Cl−(aq).
Mg(s) + H+(aq) → Mg2+(aq) + H2(g)
Step 3. Assign oxidation numbers to all entities.
+2
0
Mg(s) + H+(aq) → Mg2+(aq) + H2(g)
The oxidation number of magnesium changes from 0 to +2, so magnesium loses 2
electrons.
The oxidation number of hydrogen changes from +1 to 0, so hydrogen gains 1 electron.
Step 4. To balance electrons, adjust the coefficients by multiplying electrons gained by
2.
Mg(s) + 2 H+(aq) → Mg2+(aq) + H2(g)
Step 5. Equation is balanced. Include the spectator ion, Cl−(aq), and balance the charge.
Mg(s) + 2 HCl(aq) → Mg2+(aq) + 2 Cl−(aq) + H2(g)
Statement: The balanced equation is
Mg(s) + 2 HCl(aq) → Mg2+(aq) + 2 Cl−(aq) + H2(g)
Copyright © 2012 Nelson Education Ltd.
Unit 5: Electrochemistry
U5-14
(c) Solution:
Step 1. Write the unbalanced equation with its ionic compounds dissociated.
K+(aq) + IO4−(aq) + K+(aq) + I−(aq) + H+(aq) + Cl−(aq) →
K+(aq) + Cl−(aq) + I2(s) + H2O(l)
Step 2. Write the unbalanced net ionic equation, without the spectator ions, K+(aq) and
Cl−(aq).
IO4−(aq) + I−(aq) + H+(aq) → I2(s) + H2O(l)
Step 3. Assign oxidation numbers to all entities.
+7–2
–1
+4
0
+1 –2
IO4−(aq) + I−(aq) + H+(aq) → I2(s) + H2O(l)
The oxidation number of iodine in IO4−(aq) changes from +7 to 0, so iodine gains
7 electrons.
The oxidation number of iodine in I−(aq) changes from –1 to 0, so iodine loses
1 electron.
Step 4. To balance electrons, adjust the coefficients by multiplying electrons gained
by 7.
IO4−(aq) + 7 I−(aq) + H+(aq) → I2(s) + H2O(l)
Step 5. Balance the rest of the equation by inspection.
Balance iodine.
IO4−(aq) + 7 I−(aq) + H+(aq) → 4 I2(s) + H2O(l)
Balance oxygen by adding water.
IO4−(aq) + 7 I−(aq) + H+(aq) → 4 I2(s) + 4 H2O(l)
Balance hydrogen by adding hydrogen ions.
IO4−(aq) + 7 I−(aq) + 8 H+(aq) → 4 I2(s) + 4 H2O(l)
Step 6. Include the spectator ions, K+(aq) and Cl−(aq).
KIO4(aq) + 7 KI(aq) + 8 HCl(aq) → 8 KCl(aq) + 4 I2(s) + 4 H2O(l)
Statement: The balanced equation is
KIO4(aq) + 7 KI(aq) + 8 HCl(aq) → 8 KCl(aq) + 4 I2(s) + 4 H2O(l)
(d) Solution:
Step 1. Write the unbalanced equation with its ionic compounds dissociated.
Cl2(g) + K+(aq) + OH−(aq) → K+(aq) + ClO3−(aq) + K+(aq) + Cl−(aq) + H2O(l)
Step 2. Write the unbalanced net ionic equation, without the spectator ion, K+(aq).
Cl2(g) + OH−(aq) → ClO3−(aq) + Cl−(aq) + H2O(l)
Step 3. Assign oxidation numbers to all entities.
0
–2 +1
+5 –2
–1
+1 –2
Cl2(g) + OH−(aq) → ClO3−(aq) + Cl−(aq) + H2O(l)
The oxidation number of chlorine in ClO3−(aq)changes from 0 to +5, so chlorine loses
5 electrons.
The oxidation number of chlorine in Cl−(aq)changes from 0 to –1, so chlorine gains
1 electron.
Step 4. To balance electrons, adjust the coefficients by multiplying electrons gained
by 5.
Cl2(g) + OH−(aq) → ClO3−(aq) + 5 Cl−(aq) + H2O(l)
Copyright © 2012 Nelson Education Ltd.
Unit 5: Electrochemistry
U5-15
Step 5. Balance the rest of the equation by inspection.
Balance chlorine.
3 Cl2(g) + OH−(aq) → ClO3−(aq) + 5 Cl−(aq) + H2O(l)
Balance oxygen by adding OH−(aq).
3 Cl2(g) + 4 OH−(aq) → ClO3−(aq) + 5 Cl−(aq) + H2O(l)
Balance hydrogen by adding hydrogen ions.
3 Cl2(g) + 4 OH−(aq) → ClO3−(aq) + 5 Cl−(aq) + H2O(l) + 2 H+(aq)
Eliminate H+(aq) by adding an equal number of OH−(aq) to both sides of the equation.
3 Cl2(g) + 4 OH−(aq) + 2 OH−(aq) →
ClO3−(aq) + 5 Cl−(aq) + H2O(l) + 2 H+(aq) + 2 OH−(aq)
Step 6. The equation becomes
3 Cl2(g) + 6 OH−(aq) → ClO3−(aq) + 5 Cl−(aq) + 3 H2O(l)
Based on the given equation, include the spectator ion, K+(aq).
3 Cl2(g) + 6 KOH(aq) → KClO3(aq) + 5 KCl(aq) + 3 H2O(l)
Statement: The balanced equation is
3 Cl2(g) + 6 KOH(aq) → KClO3(aq) + 5 KCl(aq) + 3 H2O(l)
(e) Solution:
Step 1. Write the unbalanced equation with its ionic compounds dissociated.
2 K+(aq) + Cr2O72−(aq) + HCl(aq) →
K+(aq) + Cl−(aq) + Cr3+(aq) + 3 Cl−(aq) + Cl2(g) + H2O(l)
Step 2. Write the unbalanced net ionic equation, without the spectator ions, K+(aq) and
Cl−(aq).
Cr2O72−(aq) + HCl(aq) → Cr3+(aq) + Cl2(g) + H2O(l)
Step 3. Assign oxidation numbers to all entities.
+6
–2
+1 –1
+3
0
+1 –2
Cr2O72−(aq) + HCl(aq) → Cr3+(aq) + Cl2(g) + H2O(l)
The oxidation number of chromium changes from +6 to +3, so chromium gains 3
electrons, and Cr2O72−(aq) ion gains 6 electrons.
The oxidation number of chlorine changes from –1 to 0, so chlorine loses 1 electron.
Step 4. To balance electrons, adjust the coefficients by multiplying electrons lost by 6.
Cr2O72−(aq) + 6 HCl(aq) → 2 Cr3+(aq) + 3 Cl2(g) + H2O(l)
Step 5. Balance the rest of the equation by inspection.
Balance oxygen by adding water.
Cr2O72−(aq) + 6 HCl(aq) → 2 Cr3+(aq) + 3 Cl2(g) + 7 H2O(l)
Balance hydrogen by adding hydrogen ions.
Cr2O72−(aq) + 6 HCl(aq) + 8 H+(aq) → 2 Cr3+(aq) + 3 Cl2(g) + 7 H2O(l)
Step 6. Based on the given equation, include the spectator ions, K+(aq) and Cl−(aq).
K2Cr2O7(aq) + 14 HCl(aq) → 2 KCl(aq) + 2 CrCl3(aq) + 3 Cl2(g) + 7 H2O(l)
Statement: The balanced equation is
K2Cr2O7(aq) + 14 HCl(aq) → 2 KCl(aq) + 2 CrCl3(aq) + 3 Cl2(g) + 7 H2O(l)
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Unit 5: Electrochemistry
U5-16
(f) Solution:
Step 1. Write the unbalanced equation with its ionic compounds dissociated.
MnO2(s) + H2C2O4(aq) + 2 H+(aq) + SO42−(aq) →
Mn2+(aq) + SO42−(aq) + CO2(g) + H2O(l)
Step 2. Write the unbalanced net ionic equation, without the spectator ion, SO42−(aq).
MnO2(s) + H2C2O4(aq) + 2 H+(aq) → Mn2+(aq) + CO2(g) + H2O(l)
Step 3. Assign oxidation numbers to all entities.
+4
–2
+4
–2
+1 +3 –2
+1
+2
+4 –2
+1 –2
MnO2(s) + H2C2O4(aq) + 2 H+(aq) → Mn2+(aq) + CO2(g) + H2O(l)
The oxidation number of manganese changes from +4 to +2, so manganese gains
2 electrons.
The oxidation number of carbon changes from +3 to +4, so carbon loses 1 electron, and
H2C2O4(aq) loses 2 electrons
Step 4. Electrons are balanced. Balance the rest of the equation by inspection.
Balance carbon.
MnO2(s) + H2C2O4(aq) + 2 H+(aq) → Mn2+(aq) + 2 CO2(g) + H2O(l)
Balance oxygen by adding water.
MnO2(s) + H2C2O4(aq) + 2 H+(aq) → Mn2+(aq) + 2 CO2(g) + 2 H2O(l)
Step 5. Based on the given equation, include the spectator ion, SO42−(aq).
MnO2(s) + H2C2O4(aq) + H2SO4(aq) → MnSO4(aq) + 2 CO2(g) + 2 H2O(l)
Statement: The balanced equation is
MnO2(s) + H2C2O4(aq) + H2SO4(aq) → MnSO4(aq) + 2 CO2(g) + 2 H2O(l)
73. When balancing redox equations in acidic or basic solutions, it is possible to add
hydrogen ions, H+(aq), or hydroxide ions, OH–(aq), to half-reactions because the added
H+(aq) ions or OH−(aq) ions are not likely to take part in the reaction, so they do not
change the equation. You can add H+(aq) ions to balance hydrogen or add OH−(aq) ions
to balance oxygen.
74. (a) Solution:
Step 1. Assign oxidation numbers to all entities.
+4 –2
+7
–2
+2
MnO(s) + PbO2(s) → MnO4−(aq) + Pb2+(aq)
The oxidation number of manganese changes from +2 to +7, so manganese loses
5 electrons.
The oxidation number of lead changes from +4 to +2, so lead gains 2 electrons.
Step 2. To balance electrons, adjust the coefficients by multiplying electrons lost by 2
and electrons gained by 5.
2 MnO(s) + 5 PbO2(s) → 2 MnO4−(aq) + 5 Pb2+(aq)
Step 3. Balance the rest of the equation by inspection.
Balance oxygen by adding water.
2 MnO(s) + 5 PbO2(s) → 2 MnO4−(aq) + 5 Pb2+(aq) + 4 H2O(l)
Balance hydrogen by adding hydrogen ions.
2 MnO(s) + 5 PbO2(s) + 8 H+(aq) → 2 MnO4−(aq) + 5 Pb2+(aq) + 4 H2O(l)
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Unit 5: Electrochemistry
U5-17
Statement: The balanced equation is:
2 MnO(s) + 5 PbO2(s) + 8 H+(aq) → 2 MnO4−(aq) + 5 Pb2+(aq) + 4 H2O(l)
(b) Solution:
Step 1. Assign oxidation numbers to all entities.
–1
−
+1 –2
–1
I3−(aq)
–1
I (aq) + ClO (aq) →
+ Cl−(aq)
The oxidation number of iodine changes from –1 for an I–(aq) ion to –1 for an I3−(aq)
ion, so I3−(aq) loses 2 electrons.
The oxidation number of chlorine changes from +1 to –1, so chlorine gains 2 electrons.
Step 2. Electrons are balanced. Balance the rest of the equation by inspection.
Balance iodine.
3 I−(aq) + ClO−(aq) → I3−(aq) + Cl−(aq)
Balance oxygen atoms by adding water.
3 I−(aq) + ClO−(aq) → I3−(aq) + Cl−(aq) + H2O(l)
Balance hydrogen by adding hydrogen ions.
3 I−(aq) + ClO−(aq) + 2 H+(aq) → I3−(aq) + Cl−(aq) + H2O(l)
Statement: The balanced equation is:
3 I−(aq) + ClO−(aq) + 2 H+(aq) → I3−(aq) + Cl−(aq) + H2O(l)
(c) Solution:
Step 1. Assign oxidation numbers to all entities.
+3
–2
−
+5 –2
+1 +5 –2
+2 –2
As2O3(s) + NO3−(aq) → H3AsO4(aq) + NO(g)
The oxidation number of arsenic changes from +3 to +5, so arsenic loses 2 electrons,
and As2O3(s) loses 4 electrons.
The oxidation number of nitrogen changes from +5 to +2, so nitrogen gains 3 electrons.
Step 2. To balance electrons, adjust the coefficients by multiplying electrons lost by 3
and electrons gained by 4.
3 As2O3(s) + 4 NO3−(aq) → 6 H3AsO4(aq) + 4 NO(g)
Step 3. Balance the rest of the equation by inspection.
Balance oxygen atoms by adding water.
3 As2O3(s) + 4 NO3−(aq) + 7 H2O(l) → 6 H3AsO4(aq) + 4 NO(g)
Balance hydrogen by adding hydrogen ions.
3 As2O3(s) + 4 NO3−(aq) + 7 H2O(l) + 4 H+(aq) → 6 H3AsO4(aq) + 4 NO(g)
Statement: The balanced equation is:
3 As2O3(s) + 4 NO3−(aq) + 7 H2O(l) + 4 H+(aq) → 6 H3AsO4(aq) + 4 NO(g)
(d) Solution:
Step 1. Assign oxidation numbers to all entities.
–1
−
+7
–2
−
0
+2
Br (aq) + MnO4 (aq) → Br2(l) + Mn2+(aq)
The oxidation number of bromine changes from –1 to 0, so Br−(aq) loses 1 electron, and
Br2(l) loses 2 electrons.
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Unit 5: Electrochemistry
U5-18
The oxidation number of manganese changes from +7 to +2, so manganese gains 5
electrons.
Step 2. To balance electrons, adjust the coefficients by multiplying electrons lost by 2
bromine atoms by 5 and electrons gained by 2.
10 Br−(aq) + 2 MnO4−(aq) → 5 Br2(l) + 2 Mn2+(aq)
Step 3. Balance the rest of the equation by inspection.
Balance oxygen atoms by adding water.
10 Br−(aq) + 2 MnO4−(aq) → 5 Br2(l) + 2 Mn2+(aq) + 8 H2O(l)
Balance hydrogen by adding hydrogen ions.
10 Br−(aq) + 2 MnO4−(aq) + 16 H+(aq) → 5 Br2(l) + 2 Mn2+(aq) + 8 H2O(l)
Statement: The balanced equation is:
10 Br−(aq) + 2 MnO4−(aq) + 16 H+(aq) → 5 Br2(l) + 2 Mn2+(aq) + 8 H2O(l)
(e) Solution:
Step 1. Assign oxidation numbers to all entities.
–2 +1 –2+1
+6
–2
0 +1 –2
+3
CH3OH(aq) + Cr2O72−(aq) → CH2O(aq) + Cr3+(aq)
The oxidation number of carbon changes from –2 to 0, so carbon loses 2 electrons.
The oxidation number of chromium changes from +6 to +3, so chromium gains
3 electrons, and Cr2O72−(aq) ion gains 6 electrons.
Step 2. To balance electrons, adjust the coefficients by multiplying electrons lost by 3.
3 CH3OH(aq) + Cr2O72−(aq) → 3 CH2O(aq) + 2 Cr3+(aq)
Step 3. Balance the rest of the equation by inspection.
Balance oxygen atoms by adding water.
3 CH3OH(aq) + Cr2O72−(aq) → 3 CH2O(aq) + 2 Cr3+(aq) + 7 H2O(l)
Balance hydrogen by adding hydrogen ions.
3 CH3OH(aq) + Cr2O72−(aq) + 8 H+(aq) → 3 CH2O(aq) + 2 Cr3+(aq) + 7 H2O(l)
Statement: The balanced equation is:
3 CH3OH(aq) + Cr2O72−(aq) + 8 H+(aq) → 3 CH2O(aq) + 2 Cr3+(aq) + 7 H2O(l)
75. (a) The unbalanced net ionic equation for the reaction is
Mn(s) + Au3+(aq) → Au(s) + Mn2+(aq)
(b) Since Mn(s) loses electrons, Mn(s) is oxidized, so Au3+(aq) is the oxidizing agent.
(c) Since Au3+(aq) gains electrons, Au3+(aq) is reduced, so Mn(s) is the reducing agent.
(d) The half-reaction equations are:
Oxidation: Mn(s) → Mn2+(aq) + 2 e−
Reduction: Au3+(aq) + 3 e− → Au(s)
(e) To balance electrons, multiply the oxidation half-reaction equation by 3 and the
reduction half-reaction equation by 2.
3 Mn(s) → 3 Mn2+(aq) + 6 e−
2 Au3+(aq) + 6 e− → 2 Au(s)
Add the two half-reaction equations. The balanced equation is:
3 Mn(s) + 2 Au3+(aq) → 2 Au(s) + 3 Mn2+(aq)
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Unit 5: Electrochemistry
U5-19
76. Solution:
Step 1. Assign oxidation numbers to all entities.
+7
–2
+2–3
+4
–2
+4–3–2
MnO4−(aq) + CN−(aq) → MnO2(s) + CNO−(aq)
The oxidation number of manganese changes from +7 to +4, so manganese gains
3 electrons.
The oxidation number of carbon changes from +2 to +4, so carbon loses 2 electrons.
Step 2. To balance electrons, adjust the coefficients by multiplying electrons gained by
2 and electrons lost by 3.
2 MnO4−(aq) + 3 CN−(aq) → 2 MnO2(s) + 3 CNO−(aq)
Step 3. Balance the rest of the equation by inspection.
Balance oxygen by adding water.
2 MnO4−(aq) + 3 CN−(aq) → 2 MnO2(s) + 3 CNO−(aq) + H2O(l)
Balance hydrogen by adding hydrogen ions.
2 MnO4−(aq) + 3 CN−(aq) + 2 H+(aq) → 2 MnO2(s) + 3 CNO−(aq) + H2O(l)
Since the reaction takes place in basic solution, eliminate H+(aq) by adding an equal
number of OH−(aq) to both sides of the equation.
2 MnO4−(aq) + 3 CN−(aq) + 2 H+(aq) + 2 OH−(aq) →
2 MnO2(s) + 3 CNO−(aq) + H2O(l) + 2 OH−(aq)
Subtract 1 water molecule from each side to eliminate redundant water molecules.
2 MnO4−(aq) + 3 CN−(aq) + H2O(l) → 2 MnO2(s) + 3 CNO−(aq) + 2 OH−(aq)
Statement: The balanced equation is:
2 MnO4−(aq) + 3 CN−(aq) + H2O(l) → 2 MnO2(s) + 3 CNO−(aq) + 2 OH−(aq)
77. (a) Given: crystals of iodine added to a solution of sodium chloride
Solution:
Step 1. The entities present are I2(s), Na+(aq), Cl–(aq), and H2O(l).
Step 2. Use a redox table. The equations relating to chemicals in the reaction mixture as
oxidizing agents are
I2(s) + 2 e– → 2 I–(aq) E°r = 0.54 V
Na+(aq) + e– → Na(s)
E°r = −2.71 V
2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq)
E°r = −0.83 V
Since I2(s) occurs highest in the table, I2(s) is the strongest oxidizing agent.
The equations relating to chemicals in the reaction mixture as reducing agents are
Cl2(g) + 2 e– → 2 Cl–(aq)
E°r = 1.36 V
O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l)
E°r = 1.23 V
Since H2O(l) occurs lower in the table, H2O(l) is the stronger reducing agent.
Statement: Since the relative positions of I2(s) and H2O(l) do not form a downward
diagonal to the right on the redox table, a reaction will not occur spontaneously.
Copyright © 2012 Nelson Education Ltd.
Unit 5: Electrochemistry
U5-20
(b) Given: chlorine gas bubbled into a solution of sodium iodide
Solution:
Step 1. The entities present are Cl2(g), Na+(aq), I–(aq), and H2O(l).
Step 2. Use a redox table. The equations relating to chemicals in the reaction mixture as
oxidizing agents are
Cl2(g) + 2 e– → 2 Cl–(aq)
E°r = 1.36 V
Na+(aq) + e– → Na(s)
E°r = −2.71 V
2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq)
E°r = −0.83 V
Since Cl2(g) occurs highest in the table, Cl2(g) is the strongest oxidizing agent.
The equations relating to chemicals in the reaction mixture as reducing agents are
I2(s) + 2 e– → 2 I–(aq) E°r = 0.54 V
O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l)
–
E°r = 1.23 V
–
Since I (aq) occurs lower in the table, I (aq) is the stronger reducing agent.
Statement: Since the relative positions of Cl2(s) and I–(aq) form a downward diagonal
to the right on the redox table, a reaction will occur spontaneously. A balanced chemical
equation for the reaction is: Cl2(g) + 2 NaI(aq) → 2 NaCl(aq) + I2(s)
(c) Given: a silver wire placed in a solution of copper(II) chloride
Solution:
Step 1. The entities present are Ag(s), Cu2+(aq), Cl–(aq), and H2O(l).
Step 2. Use a redox table. The equations relating to chemicals in the reaction mixture as
oxidizing agents are
Ag+(aq) + e– → Ag(s)
E°r = 0.80 V
Cu2+(aq) + 2 e– → Cu(s)
E°r = 0.34 V
–
2 H2O(l) + 2 e → H2(g) + 2 OH–(aq)
E°r = −0.83 V
Since Ag+(aq) occurs highest in the table, Ag+(aq) is the strongest oxidizing agent.
The equations relating to chemicals in the reaction mixture as reducing agents are
Cl2(g) + 2 e– → 2 Cl–(aq)
E°r = 1.36 V
O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l)
E°r = 1.23 V
Since H2O(l) occurs lower in the table, H2O(l) is the stronger reducing agent.
Statement: Since the relative positions of Ag+(aq) and H2O(l) do not form a downward
diagonal to the right on the redox table, a reaction will not occur spontaneously.
Copyright © 2012 Nelson Education Ltd.
Unit 5: Electrochemistry
U5-21
(d) Given: an acidic solution of iron(II) sulfate exposed to air
Solution:
Step 1. The entities present are Fe2+(aq), SO42–(aq), H+(aq), O2(g), and H2O(l).
Step 2. Use a redox table. The equations relating to chemicals in the reaction mixture as
oxidizing agents are
Fe2+(aq) + 2 e– → Fe(s)
E°r = −0.44 V
O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l)
O2(g) + 2 H+(aq) + 2 e– → H2O2(l)
O2(g) + 2 H2O(l) + 4 e– → 4 OH–(aq)
E°r = 1.23 V
E°r = 0.68 V
E°r = 0.40 V
SO42–(aq) + 4 H+(aq) + 2 e– → H2SO3(aq) + H2O(l)
2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq)
E°r = 0.20 V
E°r = −0.83 V
Since O2(g) + 4 H+(aq) occurs highest in the table, O2(g) + 4 H+(aq) is the strongest
oxidizing agent.
The equations relating to chemicals in the reaction mixture as reducing agents are
Fe3+(aq) + e– → Fe2+(aq)
E°r = 0.77 V
O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l)
E°r = 1.23 V
Since Fe2+(aq) occurs lower in the table, Fe2+(aq) is the stronger reducing agent.
Statement: Since the relative positions of O2(g) and Fe2+(aq) form a downward
diagonal to the right on the redox table, a reaction will occur spontaneously. A balanced
chemical equation for the reaction is
O2(g) + 4 H+(aq) + 4 Fe2+(aq) → 4 Fe3+(aq) + 2 H2O(l)
78. (a) Given: aqueous potassium permanganate used to titrate an acidic solution of
iron(II) sulfate
Solution:
Step 1. The entities present are K+(aq), MnO4–(aq), Fe2+(aq), SO42–(aq), H+(aq), and
H2O(l).
Step 2. Use a redox table. The equations relating to chemicals in the reaction mixture as
oxidizing agents are
K+(aq) + e– → K(s)
E°r = −2.92 V
Fe2+(aq) + 2 e– → Fe(s)
E°r = −0.44 V
MnO4–(aq) + 8 H+(aq) + 5 e– → Mn2+(aq) + 4 H2O(l)
SO42–(aq) + 4 H+(aq) + 2 e– → H2SO3(aq) + H2O(l)
2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq)
E°r = 1.68 V
E°r = 0.20 V
E°r = −0.83 V
–
Since MnO4 (aq) occurs highest in the table, MnO4–(aq) is the strongest oxidizing agent.
The equations relating to chemicals in the reaction mixture as reducing agents are
Fe3+(aq) + e– → Fe2+(aq)
E°r = 0.77 V
O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l)
E°r = 1.23 V
Since Fe2+(aq) occurs lower in the table, Fe2+(aq) is the stronger reducing agent.
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Unit 5: Electrochemistry
U5-22
Step 3. Since the relative positions of MnO4–(aq) and Fe2+(aq) form a downward
diagonal to the right on the redox table, a reaction will occur spontaneously.
Statement: The half-reaction equations for the reaction are
Oxidation: Fe2+(aq) → Fe3+(aq) + e–
Reduction: MnO4–(aq) + 8 H+(aq) + 5 e– → Mn2+(aq) + 4 H2O(l)
The overall balanced chemical equation is:
MnO4–(aq) + 8 H+(aq) + 5 Fe2+(aq) → Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l)
(b) Given: strip of copper placed in a beaker of hydrochloric acid
Solution:
Step 1. The entities present are Cu(s), H+(aq), Cl–(aq), and H2O(l).
Step 2. Use a redox table. The equations relating to chemicals in the reaction mixture as
oxidizing agents are
2 H+(aq) + 2 e– → H2(g)
E°r = 0 V
2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq)
+
E°r = −0.83 V
+
Since H (aq) occurs higher in the table, H (aq) is the stronger oxidizing agent.
The equations relating to chemicals in the reaction mixture as reducing agents are
Cl2(g) + 2 e– → 2 Cl–(aq)
E°r = 1.36 V
Cu2+(aq) + 2 e– → Cu(s)
E°r = 0.34 V
O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l)
E°r = 1.23 V
Since Cu(s) occurs lowest in the table, Cu(s) is the strongest reducing agent.
Step 3. Since the relative positions of H+(aq) and Cu(s) do not form a downward
diagonal to the right on the redox table, a reaction will not occur spontaneously: no
reaction.
Statement: No reaction would occur spontaneously.
(c) Given: iron pipe exposed to wind and rain
Solution:
Step 1. The entities present are Fe(s), O2(g), and H2O(l).
Step 2. Use a redox table. The equations relating to chemicals in the reaction mixture as
oxidizing agents are
O2(g) + 2 H2O(l) + 4 e– → 4 OH–(aq)
E°r = 0.40 V
2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq)
E°r = −0.83 V
Since O2(g) occurs higher in the table, O2(g) is the stronger oxidizing agent.
The equations relating to chemicals in the reaction mixture as reducing agents are
O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l)
E°r = 1.23 V
Fe2+(aq) + 2 e– → Fe(s)
E°r = −0.44 V
Since Fe(s) occurs lower in the table, Fe(s) is the stronger reducing agent.
Copyright © 2012 Nelson Education Ltd.
Unit 5: Electrochemistry
U5-23
Step 3. Since the relative positions of O2(g) and Fe(s) do form a downward diagonal to
the right on the redox table, a reaction will occur spontaneously.
Statement: The half-reaction equations for the reaction are
Oxidation: Fe(s) → Fe2+(aq) + 2 e–
Reduction: O2(g) + 2 H2O(l) + 4 e– → 4 OH–(aq)
The overall balanced chemical equation is
2 Fe(s) + O2(g) + 2 H2O(l) → 2 Fe2+(aq) + 4 OH–(aq)
79. Solution:
Step 1. The entities present in the solution in the steel can are Cu2+(aq), NO3–(aq), Fe(s),
and H2O(l).
Step 2. Using a redox table, the oxidizing agents in the steel can are
Cu2+(aq) + 2 e– → Cu(s)
E°r = 0.34 V
2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq)
E°r = −0.83 V
2+
Since Cu (aq) occurs higher in the table, Cu2+(aq) is the stronger oxidizing agent.
The reducing agents in the steel can are
O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l)
E°r = 1.23 V
Fe2+(aq) + 2 e– → Fe(s)
E°r = −0.44 V
Since Fe(s) occurs lower in the table, Fe(s) is the stronger reducing agent.
Step 3. Since the relative positions of Cu2+(aq) and Fe(s) do form a downward diagonal
to the right on the redox table, a reaction will occur spontaneously. The overall balanced
chemical equation for the reaction is
Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)
Statement: If a solution of copper(II) nitrate is poured into an empty steel can, a
spontaneously reaction will occur, producing red-brown copper deposits on the surface
of the steel can.
80. (a) When magnesium metal is added to a beaker of hydrochloric acid, HCl(aq), and
a gas is produced, the oxidizing agent is hydrochloric acid and the reducing agent is
magnesium metal.
(b) The balanced chemical equation for the reaction is
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
(c) Two electrons are transferred per magnesium atom reacting.
(d) The reaction does not produce useful electrical energy when magnesium is added
directly to the beaker of hydrochloric acid because the chemical reaction creates thermal
energy.
(e) To produce an electric current, the reactants need to be separated into half-cells.
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Unit 5: Electrochemistry
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81. Diagram of a galvanic cell constructed using magnesium metal and 1.0 mol/L
magnesium nitrate solution along with copper metal and 1.0 mol/L copper(II) nitrate
solution: The half-reaction equations are
Mg(s) → Mg2+(aq) + 2 e−
Cu2+(aq) + 2 e− → Cu(s)
The net ionic equation is
Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)
82. The electrochemical process that occurs in a galvanic cell is as follows: A metal that
is the stronger reducing agent is oxidized at the anode of one half-cell. Electrons from
the anode migrate through the wire to the cathode in the other half-cell, where they
reduce metal cations in solution. Anions from the cathodic half-cell migrate to the
anodic half-cell. Cations from the anodic half-cell migrate to the cathodic half-cell
through the salt bridge to keep the electrolytes neutral so the spontaneous chemical
reaction can continue to produce electricity.
83. Diagram of a galvanic cell constructed using magnesium metal and 1.0 mol/L
magnesium nitrate solution along with copper metal and 1.0 mol/L copper(II) nitrate
solution:
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Unit 5: Electrochemistry
U5-25
The half-reaction equations are
Mg(s) → Mg2+(aq) + 2 e−
Cu2+(aq) + 2 e− → Cu(s)
The net ionic equation is
Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)
84. The double vertical lines in the line notation used to describe galvanic cells
represent the salt bridge.
85. As a galvanic cell operates, the anode loses mass as metal atoms lose electrons and
becomes ions in solution. The cathode gains mass as metal ions from solution gain
electrons and become neutral atoms that are deposited onto the cathode.
86. (a) Diagram of a galvanic cell with one half-cell constructed from a zinc electrode
and a 1.0 mol/L solution of zinc nitrate, and the other half-cell constructed from a nickel
electrode and a 1.0 mol/L solution of nickel(II) nitrate.
(b) Anode half-reaction equation: Zn(s) → Zn2+(aq) + 2 e−
Cathode half-reaction equation: Ni2+(aq) + 2 e− → Ni(s)
(c) Net ionic equation: Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s)
(d) Line notation: Zn(s) | Zn2+(aq) | | Ni2+(aq) | Ni(s)
87. Given: aluminum foil in electrolyte solution, silverware
Required: whether aluminum in solution can restore lustre to silverware
Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard
cell potential and determine whether the reaction is spontaneous.
Copyright © 2012 Nelson Education Ltd.
Unit 5: Electrochemistry
U5-26
Solution: From Table 1 in Appendix B7, the half-reaction equations and reduction
potentials for aluminum and silver are
!!
Al3+(aq) + 3 e− #
!"
! Al(s)
E°r = !1.66 V
!!
Ag+(aq) + e− #
!"
! Ag(s)
E°r = +0.80 V
Silver has the more positive reduction potential, so silver ions are reduced at the
cathode.
Ag+(aq) + e− → Ag(s)
Aluminum is oxidized at the anode. This half-reaction equation is written as an
oxidation reaction:
Al(s) → Al3+(aq) + 3 e−
To balance electrons, multiply the equation for the silver half-reaction equation by 3.
3 Ag+(aq) + 3 e− → 3 Ag(s)
Add the two half-reaction equations to obtain the net ionic equation.
Al(s) + 3 Ag+(aq) → Al3+(aq) + 3 Ag(s)
The cell potential for this reaction is
!E°r (cell) = E°r (cathode) " E°r (anode)
= 0.80 V " ("1.66 V) !E°r (cell) = 2.46 V
The cell potential is positive; therefore, the reaction is spontaneous.
Statement: The cell potential for the reaction of aluminum foil in electrolyte with
silverware is +2.46 V, according to the equation
Al(s) + 3 Ag+(aq) → Al3+(aq) + 3 Ag(s)
The reaction in spontaneous; therefore, aluminum foil and an electrolyte can be used to
restore lustre to silverware.
88. The half-reaction equations and standard reduction potentials for silver, hydrogen
ion, and magnesium are
!!
Ag+(aq) + e− #
!"
! Ag(s)
E°r = +0.80 V
!!
2 H+(aq) + 2 e− #
!"
! H2(g)
E°r = 0.00 V
!!
Mg2+(aq) + 2 e− #
!"
! Mg(s) E°r = !2.37 V
If hydrogen ions are the oxidizing agent, the cell potential for the reaction of silver
metal and hydrogen ions is
!E°r (cell) = E°r (cathode) " E°r (anode)
= 0 V " (+0.80 V)
!E°r (cell) = –0.80 V
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Unit 5: Electrochemistry
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Since !E°r (cell) < 0, a reaction between silver metal and hydrogen ions is not
spontaneous. However, if hydrogen ions are the oxidizing agent, the cell potential for
the reaction of magnesium metal and hydrogen ions is
!E°r (cell) = E°r (cathode) " E°r (anode)
= 0 V " (–2.37 V)
!E°r (cell) = +2.37 V
Since !E °r (cell) > 0 , magnesium metal and hydrogen ions undergo a spontaneous
reaction. Therefore, no reaction occurs when silver metal is placed in hydrochloric acid
solution, whereas when magnesium metal is placed in hydrochloric acid solution, the
following reaction takes place:
2 HCl(aq) + 2 e− → H2(g) + 2 Cl−(aq)
Mg(s) → Mg2+(aq) + 2 e−
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
89. The scientific definition of a battery is two or more galvanic cells connected in
series. Portable sources of electrical energy are usually alkaline dry cells. Therefore, use
of the term “battery” to refer to a portable source of electrical energy is incorrect, unless
it refers to several cells in series circuit.
90. Illustration showing how four cells should be connected in series to provide energy
to light an electric bulb:
91. A fuel cell differs from a conventional galvanic cell in that in a fuel cell the reactants
are continuously supplied.
92. Answers may vary. Sample answer: Two advantages of lithium-ion batteries over
hydrogen fuel cells are lithium-ion batteries are smaller, and they require no additional
fuel, so they are maintenance free. Two advantages of hydrogen fuel cells over lithiumion batteries are hydrogen fuel cells have an indefinite lifespan, as long as fuel is
available and they are more efficient at converting energy to useful forms.
93. (a) Since gold appears higher on the standard reduction potentials table than oxygen
does, gold is a stronger oxidizing agent than oxygen and will not be oxidized by oxygen.
(b) Most elements are found in oxidized form because most metals are stronger
reducing agents than oxygen and are oxidized by oxygen.
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Unit 5: Electrochemistry
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94. The sacrificial anodes prevent or reduce the rusting of steel hulls by providing the
steel hull with electrons produced through the oxidation of the sacrificial anode. These
electrons take part in the oxidation–reduction reaction instead of electrons in the steel in
the hull.
95. Producing 1kg of silver from its ions requires much less electricity than producing 1
kg of aluminum from its ions because silver ions ( E°r = 0.80 V) are stronger oxidizing
agents and are more readily reduced compared to aluminum ions ( E°r = –1.66 V).
96.
97. Answers may vary. Sample answer: Three practical applications of electrolysis are
galvanizing nails to make them weather resistant, plating precious metals like gold and
silver onto decorative metal objects, and aluminum recycling.
98. (a) Cl2(g) will be formed at the anode.
(b) Sn(s) will be formed at the cathode.
(c) Visible evidence that would confirm the predictions include: the formation of a
bleach-like gas at the anode and the buildup of Sn(s) on the cathode.
(d) Net ionic equation: Sn2+(aq) + 2 Cl–(aq) → Sn(s) + Cl2(g)
(e) Sn2+(aq) + 2 e– → Sn(s)
E°r = –0.14 V
2 Cl–(aq) → 2 e– + Cl2(g) E°r = –1.36 V
Total
E°r = –1.50 V
The minimum voltage that must be supplied to the cell is 1.50 V.
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99. Electroplating is a form of electrolysis in which electricity is applied to a cell
containing an object to be plated and metal ions of the metal being plated. The metal
ions are reduced by the electricity and form a solid coating on the object.
100. When aqueous sodium chloride undergoes electrolysis, the pH of the solution near
the cathode may be increased because of the production of hydroxide ions from water.
The pH of the solution near the anode will likely be unaffected.
Analysis and Application
101. Solution:
Step 1. Write a chemical equation for the reaction.
2 NaF(s) → 2 Na(s) + F2(g)
Step 2. In this reaction, the oxidizing agent is Na+(s). The relating equation is
Na+(s) + e– → Na(s) E°r = −2.71 V
In the same reaction, the reducing agent is F–(s). The relating equation is
F2(g) + 2 e– → 2 F–(aq) E°r = 2.87 V
Step 3. Since the relative positions of the oxidizing agent, Na+(s), and the reducing
agent, F–(aq), do not form a downward diagonal to the right on the redox table, the
above reaction will not occur spontaneously.
Statement: Fluorine gas cannot be prepared by the chemical oxidation of sodium
fluoride. Because of the relative positions of the oxidizing agent and reducing agent on
a redox table, I know that the reaction will not occur spontaneously.
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102. (a) The disproportionation of hydrogen peroxide into water and oxygen can be
represented by this chemical equation:
2 H2O2(l) → 2 H2O(l) + O2(g)
(b) Solution:
Step 1. Assign oxidation numbers to each atom/ion in the chemical equation.
+1 –1
+1 –2
0
2 H2O2(l) → 2 H2O(l) + O2(g)
Since the oxidation number of oxygen changes from –1 to –2 and from –1 to 0, oxygen
is both oxidized and reduced.
Step 2. Determine how the oxidation number on each element changes to identify these
changes as either oxidation or reduction.
Since the oxidation number of oxygen increases from –1 to 0, it is oxidized.
Since the oxidation number of oxygen decreases from –1 to –2, it is reduced.
Statement: Oxygen is both oxidized and reduced.
(c) The half-reaction equation for the oxidation reaction is
O22–(l) → O2(g) + 2 e–
Therefore, the number of electrons lost by a H2O2 molecule in the oxidation reaction
is 2.
(d) The half-reaction equation for the reduction reaction is
O22–(l) + 2 e– → 2 O2–(l)
Therefore, the number of electrons gained by a H2O2 molecule in the reduction reaction
is 2.
103. (a) Answers may vary. Sample answer: Oxidizing agents are stored separately from
other substances because they can cause vigorous reactions when they come into contact
with reducing agents. These reactions often result in damages. For example, many
metals are reducing agents, and they should not be used as containers for oxidizing
agents. Many oxidizing agents are flammable and should be kept away from ignition
sources. Some are corrosive so they need to be stored in corrosion-resistant containers.
(b) When working with strong oxidizing agents, wear protective gloves, aprons, boots,
hoods, or other protective clothing, depending on the risk of skin contact. To protect the
eyes and face, avoid ordinary safety glasses. Use chemical safety goggles instead. In
some cases, you should wear a face shield to protect your face from splashes.
(c) Strong oxidizing agents should be stored in a cool, dry place, away from direct
sunlight and other sources of thermal energy and away from moisture, because many of
these oxidizing agents are dangerously reactive: they decompose at temperatures only a
little above normal room temperatures, or they react vigorously with water vapour.
Some of these reactions can lead to serious damages or even explosions in a warm
environment or in the presence of water vapour.
104. (a) Assign oxidation numbers to each atom/ion.
–2 +1 –2 +1 –2 +1
+6
–2
0 +1 0 –2
+3
CH3CH2OH(aq) + Cr2O72−(aq) → CH3CO2H(aq) + Cr3+(aq)
The oxidation numbers of all elements in the reaction are
Reactants: C: –2; H: +1; O: –2; Cr: +6
Products: C: 0; H: +1; O: –2; Cr: +3
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Unit 5: Electrochemistry
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(b) Solution: Determine the changes in oxidation numbers.
Step 1. Identify the elements for which the oxidation numbers change.
The oxidation number of carbon changes from –2 to 0, so carbon loses 2 electrons, and
CH3CH2OH(aq) loses 4 electrons.
The oxidation number of chromium changes from +6 to +3, so chromium gains
3 electrons, and Cr2O72−(aq) gains 6 electrons.
Step 2. To balance electrons, adjust the coefficients by multiplying electrons lost by 3
and electrons gained by 2.
3 CH3CH2OH(g) + 2 Cr2O72−(aq) → 3 CH3CO2H(aq) + 4 Cr3+(aq)
Step 3: Balance the rest of the equation by inspection.
Balance oxygen by adding water.
3 CH3CH2OH(g) + 2 Cr2O72−(aq) → 3 CH3CO2H(aq) + 4 Cr3+(aq) + 11 H2O(l)
Balance hydrogen by adding hydrogen ions.
3 CH3CH2OH(g) + 2 Cr2O72−(aq) + 16 H+(aq) → 3 CH3CO2H(aq) + 4 Cr3+(aq) + 11
H2O(l)
Statement: The balanced equation is
3 CH3CH2OH(g) + 2 Cr2O72−(aq) + 16 H+(aq) → 3 CH3CO2H(aq) + 4 Cr3+(aq) + 11
H2O(l)
(c) The colour of the solution containing only dichromate ions is yellowish-orange. If a
person exhales air containing alcohol into this solution, chromium ion, which is green in
colour, would be formed. So, the colour of the solution would change from yellowishorange to green.
105. The reaction that produces silver ions and chlorine gas from silver metal and
chloride ions must be a redox reaction, because there are changes in oxidation numbers
of the reactants. That is, an oxidation and a reduction have to occur at the same time.
The half-reaction equation for the production of silver ion from silver metal is
Ag(s) → Ag+(s) + e–
The half-reaction equation for the production of chlorine gas from chloride ions is
2 Cl–(aq) → Cl2(g) + e–
Since both of these are oxidation reactions, silver metal in a solution of chloride ions
will not produce silver ions and chlorine gas.
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Unit 5: Electrochemistry
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106. Solution: Determine if any spontaneous reaction occurs between the iron spoon,
Fe(s), and the solution of copper (II) chloride, CuCl2(aq).
Step 1. The entities present are Cu2+(aq), Cl–(aq), Fe(s), and H2O(l).
Step 2. Use a redox table. The equations relating to the chemicals as oxidizing agents
are
Cu2+(aq) + 2 e– → Cu(s)
E°r = 0.34 V
2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq)
E°r = −0.83 V
Since Cu2+(aq) occurs higher in the redox table, Cu2+(aq) is the stronger oxidizing agent.
The equations relating to the chemicals as reducing agents are
Cl2(g) + 2 e– → 2 Cl–(aq)
E°r = 1.36 V
O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l)
2+
–
Fe (aq) + 2 e → Fe(s)
E°r = 1.23 V
E°r = −0.44 V
Since Fe(s) occurs lowest in the table, Fe(s) is the strongest reducing agent.
Step 3. Since the relative positions of Cu2+(aq) and Fe(s) do form a downward diagonal
to the right on the redox table, a reaction will occur spontaneously.
Statement: When an iron spoon is used to stir a solution of copper(II) chloride, a
spontaneous reaction occurs. There would be a change in colour of the solution from
blue-green CuCl2(aq) to pale green FeCl2(aq), and some red-brown copper may deposit
on the surface of the spoon.
107. To determine experimentally which electrode in a galvanic cell is the anode and
which is the cathode, connect the two electrodes to a voltmeter. A positive reading on
the voltmeter means that the positive terminal of the voltmeter is connected to the
cathode of the cell and the negative terminal is connected to the anode. A negative
reading means that the connections are reversed.
108. (a) Diagram of the galvanic cell in which the redox reaction represented by the
following net ionic equation takes place:
Ni2+(aq) + Fe(s) → Ni(s) + Fe2+(aq)
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(b) From Table 1 in Appendix B7, the half-cell reaction equations and standard
reduction potentials for this cell are
Fe2+(aq) + 2 e− → Fe(s) E°r = !0.44 V
Ni2+(aq) + 2 e− → Ni(s) E°r = !0.23 V
Nickel ion has the more positive reduction potential, so nickel ions are reduced at the
cathode.
The standard cell potential is
!E°r (cell) = E°r (cathode) " E°r (anode)
= "0.23 V " ("0.44 V)
!E°r (cell) = 0.21 V
109. (a) Diagram of an electrochemical cell constructed using zinc and silver metals,
and solutions of zinc nitrate and silver nitrate (both 0.1 mol/L):
(b) Line notation for the cell in (a): Zn(s) | Zn2+(aq) | | Ag+(aq) | Ag(s)
(c) Anode half-reaction equation: Zn(s) → Zn2+(aq) + 2 e−
(d) Cathode half-reaction equation: Ag+(aq) + e− → Ag(s)
(e) From Table 1 in Appendix B7, the standard reduction potential for silver ion is
E°r = +0.80 V, and the standard reduction potential for zinc ion is E°r = !0.76 V. The
standard cell potential is
!E°r (cell) = E°r (cathode) " E°r (anode)
= 0.80 V " ("0.76 V)
!E°r (cell) = 1.6 V
(f) The potential of this cell would be less than that predicted in (e) because electrolyte
solution concentrations for measuring standard cell potential are 1.0 mol/L. The
electrolyte solution concentrations in this cell are only 0.1 mol/L.
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110. If a silver battery uses the same anode half-reaction as the alkaline battery whose
cathode half-reaction is represented by the following equation,
Ag2O(s) + H2O(l) + 2 e− → 2 Ag(s) + 2 OH−(aq)
then the equation for the anode half-reaction of the silver battery will be
Zn(s) + 2 OH−(aq) → ZnO(s) + H2O(l) + 2 e−
Add the two half-reaction equations to obtain the net ionic equation:
Ag2O(s) + Zn(s) → ZnO(s) + 2 Ag(s)
111. Answers may vary. Sample answer: Two advantages of using fuel cells to generate
electricity instead of using fossil fuels are that fuel cells, such as hydrogen fuel cells, do
not consume nonrenewable resources and fuel cells do not produce harmful emissions as
by-products.
112. (a) cathode: Cd2+(aq) + 2 e– → Cd(s), where E°r = –0.40 V
(b) A negative sign in the value of E°r indicates that the redox reaction is
non-spontaneous.
113. Cathodic protection is a form of corrosion protection that provides electrons to a
protected metal to prevent oxidation. Copper metal would not be a good cathodic
protector to prevent iron from rusting because it is a stronger oxidizing agent than iron
is, so it would likely increase the rate of corrosion.
114. (a) Reactant oxidation numbers: Zn: +2; S in ZnS: –2;
S in H2SO4: –2; O: –2; O in O2: 0; H: +1
Product oxidation numbers: Zn: +2; S in ZnSO4: +6
Elemental S: 0; H: +1; O: –2
(b) 2 ZnS(s) + 2 H2SO4(aq) + O2(g) → 2 ZnSO4(aq) + 2 S(s) + 2 H2O(l)
(c) The products of electrolysis will be zinc metal and sulfuric acid.
(d) The minimum potential difference that must be supplied to this cell is 0.96 V.
115. (a) Using a flashlight involves a chemical process that produce electricity.
(b) Producing aluminum form its ore involves the use of electricity to drive a chemical
process.
(c) Electroplating jewellery involves the use of electricity to drive a chemical process.
(d) Galvanizing metal involves the use of electricity to drive a chemical process.
116. (a) The reaction at the anode is
2 I–(aq) → I2(s) + 2 e–
so the product at the anode will be solid iodine.
The reaction at the cathode is
Cu2+(aq) + 2 e– → Cu(s)
so the product will be solid copper.
(b) I would expect to see the following evidence. The cathode will be coated in copper.
The anode may be surrounded by a slight brown colour caused by the production of
iodine.
117. No, letting a solution trickle over scrap iron is the not an efficient process because
the copper will be plated to the iron and will have to be purified. Also, the scrap metal
may be oxidized, decreasing scrap iron’s ability to reduce the copper.
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Unit 5: Electrochemistry
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118. C cells are better suited than AA cells for use in remote-controlled cars because
they have a greater capacity and will last longer.
119. The ground jumper should be attached to a remote part of the engine block because
when the dead battery is being charged, gaseous hydrogen and oxygen can build up. If
the jumper cables are placed too close to the battery, sparks produced when the cables
are disconnected may ignite these gases.
120. The spoon should be the cathode. The reaction at the cathode will be
Ag+(aq) + e– → Ag(s)
A silver coin could be the anode. The reaction at the anode will be
Ag(s) → Ag+(aq) + e–
121. (a) 2 H2O(l) + 2 Br–(aq) → H2(g) + 2 OH–(aq) + Br2(g)
(b) Before the electrolysis, the cell contained potassium bromide, KBr(aq), and water,
H2O(l).
Evaluation
122. Answers may vary. Sample answer: The statement: “Oxidation numbers were
developed simply for bookkeeping. Except for simple ions like Na+ of F–, the charges
are fictitious.” is true because for elements such as sulfur in SF6 and manganese in
KMnO4, the oxidation numbers +6 and +7 do not represent an actual charge on an atom.
However, the way oxidation number is defined—the apparent net electric charge that
the atom would have if electron pairs in covalent bonds belonged entirely to the more
electronegative atom—is a useful way to keep track of electrons in a reaction system.
By assigning oxidation numbers to elements in compounds and ions, and by
determining how the oxidation number on each element changes, the oxidation and
reduction parts of a redox reaction can be identified.
123. Answers may vary. Sample answer: Since a redox reaction is defined in terms of
the transfer of electrons, defining oxidation as the loss of electrons and reduction as the
gain of electrons is an advantage because it describes the behaviour of electrons as they
are transferred from one entity to another. However, It may seem odd to say that a
sodium atom “loses an electron” while it is “oxidized,” with its charge going from 0 to
+1. So, the advantage of defining oxidation as an increase in oxidation number (the
apparent net electric charge) would relate oxidation better to the behaviour of the
entities involved in the reaction. Similarly, defining reduction as a decrease in oxidation
number would indicate that Ag+ ion is reduced to silver metal as the oxidation number
decreases from +1 to 0.
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Unit 5: Electrochemistry
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124. Answers may vary. Sample answer:
The oxidation number of phosphorus in each of the compounds is:
PH3: –3
PCl3: +3
P2H4: –2
H3PO4: +5
Na3PO3: +3
From the oxidation numbers of phosphorus in these compounds, I can infer that the
oxidation number of phosphorus can be a positive or negative number. However, the
oxidation numbers do not really represent the charge on the atom, because it is not
likely that phosphorus will form a positive ion of charge +5 by losing 5 electrons or
form a negative ion of charge –3 by gaining 3 electrons.
125. Answers may vary. Sample answer: Both primary and secondary cells are small,
lightweight cells that are used to power portable electronic devices for several hours.
The main difference between them is that primary cells are not rechargeable whereas
secondary cells are rechargeable. I think secondary cells are better because they can be
reused and thus reduce the number of cells that are discarded and also save on raw
materials needed for the production of new cells.
126. Answers may vary. Sample answer: Some advantages of hydrogen fuel cells over
other common sources of energy for transportation are the following:
– hydrogen fuel cells do not consume non-renewable resources, such as fossil fuels
– hydrogen fuel cells do not produce any toxic by-products
– hydrogen fuel cells transform energy into a usable form more efficiently than other
common sources of energy.
Some disadvantages of hydrogen fuel cells are the following:
– hydrogen, which is required by hydrogen fuel cells, usually requires fossil fuels to
synthesize
– hydrogen, which must be transported in the vehicle as fuel, is extremely explosive
127. Answers may vary. Sample answer: Cathodic protection of iron can be
accomplished by the impressed current method, in which electricity is applied to the
structure to prevent oxidation; or by using a sacrificial anode like zinc to provide
electrons to the protected structure. Iron corrosion can also be halted by plating the iron
in an inert metal or airtight layer such as paint or oil, to prevent exposure to moisture
and water. The impressed current method is ideal for large steel bridges because it
would be too costly to provide enough sacrificial anode material to use the sacrificial
anode method effectively. Painting the bridge could be difficult and would likely not
provide complete protection.
128. Trying to prevent iron corrosion is not practical on a large scale because the paint
or oil continuously needs to be continuously reapplied to maintain the oxygen-barrier
and moisture-barrier. This increases labour costs for maintenance and requires constant
monitoring.
129. (a) The other product of the electrolysis of molten sodium is Cl2(g).
(b) It would not be possible to produce sodium by the electrolysis of a sodium chloride
solution because the sodium and chloride ions will act as an electrolyte in the
electrolysis of water to produce hydrogen and oxygen gas.
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130. Answers may vary. Sample answer: No, aluminum is not an infinitely recyclable
material. While it can be recycled many times, it has diverse applications and is
incorporated into many products that are difficult to recycle or are by current cultural
standards currently sent to a landfill. This reduces the supply of new aluminum.
Reflect on Your Learning
131. Answers may vary. Concept maps should show the relationship between oxidation
and reduction, as well as the following:
– the roles of oxidation numbers, oxidation tables, half-reactions, types of cell potential,
and standard hydrogen half-cells in describing and making calculations related to
electrochemical processes
– the difference between galvanic cells and electrolytic cells, and the roles of an anode,
a cathode, and a salt bridge in each
– the difference between primary cells and secondary cells
– corrosion and methods of preventing it
– all the key terms in the unit
132. Answers may vary. Sample answer: I found determining which electrode would be
the anode and which would be the cathode confusing. In their plans for learning,
students may include talking with a classmate who does understand that concept,
reviewing the relevant sections in the textbook, reviewing earlier chemistry that will
help them lay the groundwork to understand this concept, doing online or library
research, developing a graphic organizer to help understand the concept, asking a
teacher for help, and/or other strategies.
133. Answers may vary. Students should elaborate on one of the scientists’
contributions to redox chemistry , either directly or indirectly, according to question
guidelines. Sample answer: Robert Boyle produced hydrogen gas by reacting iron with
acids (a redox reaction). He also performed experiments that showed that metals such as
tin gain weight when heated (oxidized).
134. Terms such as “galvanizing,” “galvanic,” and “galvanism” are named after the
scientist Luigi Galvani. Galvani conducted experiments involving electrically
stimulated muscle contraction in frog legs, which was called galvanism. Today, the term
galvanism is often used to refer to a chemically induced electric current in general.
Galvanizing, or coating iron or steel with a thin layer of zinc to protect it from
corrosion, is a process that is also based on electrochemical principles.
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Unit 5: Electrochemistry
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135. Answers may vary. Sample answer: One example of a biological process that
involves a redox reaction is the decomposition of hydrogen peroxide to water and
oxygen by the enzyme catalase. Hydrogen peroxide, a reactive oxygen intermediate, is a
by-product of metabolic reactions. The breakdown of hydrogen peroxide into water and
oxygen prevents it from participating in reactions that would damage the cell. The
active site of catalase contains an Fe3+ ion. The decomposition of hydrogen peroxide by
catalase proceeds in two steps:
H2O2(aq) + 2 Fe3+(aq) → O2(g) + 2 Fe2+(aq) + 2 H+(aq) (reduction of Fe3+ to Fe2+)
2 H2O2(aq) + 2Fe2+(aq) + 2 H+(g) → 2 H2(l) + 2 Fe3+(aq) (oxidation of Fe2+ to Fe3+)
136. Answers may vary, and should include three examples of electrochemical
processes that affect living organisms or the environment, including their chemical
equations, and their social and economic implications. Sample answer:
One example of an electrochemical process that affects living organisms and the
environment is photosynthesis, in which plants and certain microorganisms capture and
convert light energy into chemical energy, which is then stored in energy-rich sugars.
Many individual electrochemical reactions occur during photosynthesis, but the
following redox reaction represents the net process:
6 CO2(g) + 6 H2O(l) → C6H12O6(aq) + 6 O2(g)
This reaction is of environmental and economic importance because it is a source of
energy for living things and thus the basis for the production of food and biomass. In
addition, consumption of carbon dioxide and production of oxygen affects the
composition of the atmosphere and provide oxygen for aerobic organisms.
Another environmentally significant electrochemical reaction is corrosion, in which
metal deteriorates in a redox reaction. The rusting of iron, demonstrated by the
following reactions, is an example of corrosion:
2 Fe(s) + O2(g) + 2 H2O(l) → 2 Fe(OH)2(s) (oxidation of Fe to Fe2+)
4 Fe(OH)2(s) + O2(g) + 2 H2O(l) → 4 Fe(OH)3(s) (oxidation of Fe2+ to Fe3+)
Further dehydration and hydration reactions (2 Fe(OH)3 → Fe2O3(s) + 3 H2O(l)) occur
to yield hydrated iron(III) oxide (Fe2O3·nH2O(s)), or rust. Corrosion has significant
environmental and economic impact because rusted metal must be replaced, a process
that consumes additional resources and energy.
A third example of an electrochemical process that affects living organisms is alcohol
fermentation. In the absence of oxygen, certain bacteria and yeasts use alcohol
fermentation to extract energy from sugars. Fermentation occurs in several steps, but the
following redox reaction gives a simplified representation of the net process:
C6H12O6(aq) → 2 CH3CH2OH(aq) + 2 CO2(g)
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Unit 5: Electrochemistry
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This fermentation occurs during the brewing of wine and beer, making it a chemical
process of significant economic and social importance.
137. Answers may vary, and should be presented according to question guidelines.
Answers should include information such as the following:
Single-use batteries:
Zinc carbon batteries
Cylindrical batteries that can be used in a variety of portable devices, similarly to
alkaline batteries. They have a lower cost but shorter lifespan compared to alkaline
batteries.
Alkaline batteries
Cylindrical, general-purpose batteries. They are commonly used, widely available, and
have a longer lifespan than zinc carbon batteries but are more expensive.
Silver oxide batteries
Button cells often used in watches and calculators. They are small and lightweight, but
can be relatively expensive. They may contain small amounts of mercury, which is an
environmental concern.
In general, the recycling rates for non-rechargeable batteries are low, and the nonrechargable batteries cannot be reused. This contributes to their environmental impact
even if they do not contain highly toxic chemicals.
Rechargeable batteries:
Nickel cadmium batteries
Used in devices that require frequent charging, such as cordless power tools and phones,
and as a rechargeable alternative to alkaline batteries. They recharge quickly and are
less expensive than other rechargeable batteries. However, their capacity tends to
diminish with repeated use, and because they contain toxic cadmium, they are
environmentally unfriendly
Lithium-ion batteries
Used in portable devices such as laptops and cameras, and in electric cars. Lithium-ion
batteries are lightweight, more environmentally friendly than other types of battery, and
provide more energy. However they are expensive.
Lead storage (lead-acid) batteries
Used in vehicles. They are durable and relatively inexpensive, but large and heavy.
Their lead content raises environmental concerns; however, most of them are recycled.
138. Answers may vary, and should be presented according to question guidelines.
Photographic film contains light-sensitive silver halide crystals. Absorption of light
energy causes an electron to be released from a halide ion (oxidation). This electron
then reduces a silver ion to form metallic silver. The production of silver metal forms a
“latent image” on the film. A visible image is formed when the film is developed.
During development, a reducing agent is used to convert silver ions to silver metal. The
rate of reaction is greater in the areas surrounding the metallic silver formed in the light
reaction, such that even more black silver metal is produced and a dark image forms.
Subsequent steps involve stopping the reaction by adding an acid and removing the
remaining silver halide by forming a soluble complex with thiosulphate that can be
washed away. Processing the film in this way gives rise to a black and white negative.
Copyright © 2012 Nelson Education Ltd.
Unit 5: Electrochemistry
U5-33
139. Answers may vary, and should be presented according to question guidelines.
Metal coatings used to prevent corrosion in automobile body panels include zinc, zincnickel, and zinc-iron. Pure zinc coatings are widely used and inexpensive but offer less
protection and require a greater thickness. Zinc alloys provide better protection and thus
coating thickness can be reduced. Zinc-iron is less expensive than other alloys, but
performs poorly at higher temperatures. Zinc-nickel coating provides the best corrosion
protection but is the most expensive process.
140. Answers may vary, and should describe and evaluate at least two different methods
of copper production including their financial and environmental costs. After copper is
mined, it must be smelted to extract the copper from the ore. This involves oxidation
reactions and produces sulfur dioxide as a by-product.
Two main methods of smelting copper are used today in Canada. Both use less energy
than traditional smelting in a blast furnace. One method is bath smelting. In bath
smelting, the copper concentrate is reacted in a bath of matte (Cu2S and FeS). This
results in the concentrate reaching the required temperatures quickly, so it is energy
efficient and requires smaller furnaces. Sulfur dioxide gas, SO2(g), is released and must
be scrubbed from the exhaust before it is released into the atmosphere.
The other main method of smelting copper used in Canada today is flash smelting. In
flash smelting, some of the energy for melting the ore comes from the oxidation of iron
and sulfur in the ore, so less fuel is used than in traditional smelters. As the copper is
formed, it drops to the bottom of the reaction chamber as fine-grained particles. The
sulfur released by the process precipitates as a solid, which is much easier to collect
than sulfur dioxide gas. For these reasons, flash copper smelting can be more
environmentally friendly than bath smelting. Because the particles of copper are easier
to collect, flash smelting can also be more economical than bath smelting.
Copyright © 2012 Nelson Education Ltd.
Unit 5: Electrochemistry
U5-34