Download Chapter 9 Review, pages 628–633

Chapter 9 Review, pages 628–633
Knowledge
1. (a)
2. (c)
3. (a)
4. (c)
5. (b)
6. (d)
7. (a)
8. (b)
9. (a)
10. False. In the carbonate ion, CO32–(aq), carbon has an oxidation number of +4.
11. True
12. False. In the combustion of methane, oxygen is the oxidizing agent.
13. True
14. True
15. False. Redox reactions can be balanced using the oxidation numbers method or the
half-reactions method.
16. False. The reaction of Cu2+(aq) and I2(aq) is non-spontaneous.
17. False. Zinc sulfate, ZnSO4(aq), is a weaker oxidizing agent than iron(II) chloride,
FeCl2(aq).
18. True
19. (a) (iii)
(b) (ii)
(c) (v)
(d) (vii)
(e) (i)
(f) (vi)
(g) (iv)
20. (a) The oxidation and reduction half-reaction equations for the reaction are
Oxidation: Cr2+(aq) → Cr3+(aq) + e–
Reduction: Cu+(aq) + e– → Cu(s)
(b) The oxidation and reduction half-reaction equations for the reaction are
Oxidation: Fe(s) → Fe3+(aq) + 3 e–
Reduction: Au3+(aq) + 3 e– → Au(s)
(c) The oxidation and reduction half-reaction equations for the reaction are
Oxidation: 2 I – (aq) → I2(s) + 2 e–
Reduction: Br2(l) + 2 e– → 2 Br–(aq)
21. (a) The oxidation number of the Ag+ ion, a monatomic ion, is +1. Since the sum of
the oxidation numbers of all the atoms in the compound is 0, the oxidation number of Br
in AgBr(s) is –1.
(b) Since the oxidation number of oxygen in its compounds is –2, the oxidation number
of O in NaOH(s) is –2.
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Chapter 9: Oxidation–Reduction Reactions
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(c) In a molecule of HCN(g), the carbon and nitrogen atoms share 3 pairs of electrons.
The nitrogen atom is more electronegative, so it has a greater attraction for the electrons
than the carbon atom. We can assume that the nitrogen atom has taken the 3 electrons
from the carbon atom. Thus the oxidation number of N in HCN(g) is –3.
(d) The oxidation number of hydrogen in its compounds is +1, and the oxidation number
of oxygen in its compounds is –2. Since there are 2 hydrogen atoms in H2C2O4(aq), the
total contribution of the hydrogen atoms is +2. Since there are 4 oxygen atoms, the
contribution of the oxygen atoms is –8. The 2 carbon atoms must have an oxidation
number of +6 to give a sum of 0. Therefore, the oxidation number of C in H2C2O4(aq)
is +3.
(e) Since there are 2 hydrogen atoms in H2SO3(aq), the total contribution of the hydrogen
atoms is +2. The polyatomic ion SO32– must have an overall charge of –2. Since each
SO32– ion contains 3 oxygen atoms, each of which has an oxidation number of –2, the
total contribution due to the oxygen atoms is –6. Therefore, the oxidation number of S in
H2SO3(aq) is +4.
22. (a) The oxidation number of Cl–, a monatomic ion, is –1. Since there are 3 chlorine
atoms in NCl3, the total contribution of chlorine is –3. Therefore, the oxidation number of
N in NCl3 is +3, and the oxidation number of Cl in NCl3 is –1.
(b) The oxidation number of oxygen in its compounds is –2. Since there are 2 oxygen
atoms in SeO2, the total contribution of oxygen is –4. Therefore, the oxidation number of
Se in SeO2 is +4, and the oxidation number of O in SeO2 is –2.
(c) In the compound SiS2, sulfur can be treated as a monatomic ion, S2–, with an
oxidation number of –2. Since there are 2 sulfur atoms in SiS2, the total contribution of
sulfur is –4. Therefore, the oxidation number of Si in SiS2 is +4, and the oxidation
number of S in SiS2 is –2.
(d) The oxidation number of fluorine in its compounds is –1. Since there are 6 fluorine
atoms in SF6, the total contribution of fluorine is –6. Therefore, the oxidation number of
S in SF6 is +6, and the oxidation number of F in SF6 is –1.
(e) The oxidation number of Cl–, a monatomic ion, is –1. Since there are 2 chlorine atoms
in SCl2, the total contribution of chlorine is –2. Therefore, the oxidation number of S in
SCl2 is +2, and the oxidation number of Cl in SCl2 is –1.
23. (a) Since the OH– ion has a net charge of –1, the sum of the oxidation numbers of all
the atoms in the OH– ion must equal –1. The oxidation number of hydrogen in its
compounds is +1. Therefore, the oxidation number of oxygen must be –2.
(b) Since the ClO3– ion has a net charge of –1, the sum of the oxidation numbers of all the
atoms in the ClO3– ion must equal –1. The oxidation number of oxygen in its compounds
is –2. Since there are 3 oxygen atoms in ClO3–, the total contribution of oxygen is –6.
Therefore, the oxidation number of Cl in ClO3– is +5, and the oxidation number of O in
ClO3– is –2.
(c) Since the ClO2– ion has a net charge of –1, the sum of the oxidation numbers of all the
atoms in the ClO2– ion must equal –1. The oxidation number of oxygen in its compounds
is –2. Since there are 2 oxygen atoms in ClO2–, the total contribution of oxygen is –4.
Therefore, the oxidation number of Cl in ClO2– is +3 and the oxidation number of O in
ClO2– is –2.
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Chapter 9: Oxidation–Reduction Reactions
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(d) Since the SO32– ion has a net charge of –2, the sum of the oxidation numbers of all the
atoms in the SO32– ion must equal –2. The oxidation number of oxygen in its compounds
is –2. Since there are 3 oxygen atoms in SO32–, the total contribution of oxygen is –6.
Therefore, the oxidation number of S in SO32– is +4 and the oxidation number of O in
SO32– is –2.
(e) Since the S2O82– ion has a net charge of –2, the sum of the oxidation numbers of all
the atoms in the S2O82– ion must equal –2. The oxidation number of oxygen in its
compounds is –2. Since there are 8 oxygen atoms in S2O82–, the total contribution of
oxygen is –16. Therefore, the total contribution from the 2 sulfur atoms must be +14.
That is, the oxidation number of each sulfur atom is +7. Therefore, the oxidation number
of S in S2O82– is +7, and the oxidation number of O in S2O82– is –2.
(f) Since the C2H3O2– ion has a net charge of –1, the sum of the oxidation numbers of all
the atoms in the C2H3O2– ion must equal –1. The oxidation number of hydrogen in its
compounds is +1, and the oxidation number of oxygen in its compounds is –2. Since
there are 3 hydrogen atoms in C2H3O2–, the total contribution of the hydrogen atoms is
+3. Since there are 2 oxygen atoms, the contribution of the oxygen atoms is –4. The 2
carbon atoms must have an oxidation number of 0 to give a sum of –1. Therefore, the
oxidation number of C in C2H3O2– is 0, the oxidation number of H is +1, and the
oxidation number of O is –2.
24. Assign oxidation numbers to the elements to determine how the oxidation number on
each element changes. Then identify these changes as either oxidation or reduction.
(i) The oxidation numbers of the elements are
0
+1
0
+2
Cu(s) + 2 Ag+(aq) → 2 Ag(s) + Cu2+(aq)
The oxidation number of copper increases from 0 to +2, so it is oxidized.
The oxidation number of silver decreases from +1 to 0, so it is reduced.
Therefore, reaction (i) is a redox reaction.
(ii) The oxidation numbers of the elements are
+2
+6 –2
+2 +6 –2
Pb2+(aq) + SO42–(aq) → PbSO4(s)
Since there is no change in oxidation numbers, reaction (ii) is not a redox reaction.
(iii) The oxidation numbers of the elements are
+2
+5 –2
+2 +5 –2
Ca2+(aq) + 2 NO3–(aq) → Ca(NO3)2(s)
Since there is no change in oxidation numbers, reaction (iii) is not a redox reaction.
(iv) The oxidation numbers of the elements are
0
+2
0
+2
Ca(s) + Fe2+(aq) → 2 Fe(s) + Ca2+(aq)
The oxidation number of calcium increases from 0 to +2, so it is oxidized.
The oxidation number of iron decreases from +2 to 0, so it is reduced.
Therefore, reaction (iv) is a redox reaction.
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Chapter 9: Oxidation–Reduction Reactions
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(v) Write a chemical equation representing the reaction. The oxidation numbers of the
elements are
0
0
+1
–1
2 Na(s) + Cl2(g) → NaCl(aq)
The oxidation number of sodium increases from 0 to +1, so it is oxidized.
The oxidation number of chlorine decreases from 0 to –1, so it is reduced.
Therefore, reaction (v) is a redox reaction.
(vi) Write a chemical equation representing the reaction. The oxidation numbers of the
elements are
+1
–2
+2
+5 –2
+2 –2
+1 +5 –2
Na2S(aq) + Cu(NO3)2(aq) → CuS(s) + 2 NaNO3(aq)
Since there is no change in oxidation numbers, reaction (vi) is not a redox reaction.
25. (a) Separate the equation H2O(l) + Au3+(aq) → O2(g) + Au(s) into two half-reactions.
H2O(l) → O2(g) (oxidation)
Au3+(aq) → Au(s) (reduction)
For the oxidation half-reaction, first balance oxygen.
2 H2O(l) → O2(g)
Balance hydrogen by adding hydrogen ions.
2 H2O(l) → O2(g) + 4 H+(aq)
Balance the charge by adding 4 electrons to the product side of the equation.
2 H2O(l) → O2(g) + 4 H+(aq) + 4 e−
For the reduction half-reaction, balance the charge by adding 3 electrons to the reactant
side of the equation.
Au3+(aq) + 3 e− → Au(s)
Therefore, the half-reaction equations for the reaction are
Oxidation: 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e−
Reduction: Au3+(aq) + 3 e− → Au(s)
(b) Equalize the electron transfer in the two half-reaction equations. Multiply the
oxidation half-reaction equation by 3 and the reduction half-reaction equation by 4.
6 H2O(l) → 3 O2(g) + 12 H+(aq) + 12 e−
4 Au3+(aq) + 12 e− → 4 Au(s)
Add the two half-reaction equations.
The balanced net ionic equation is
6 H2O(l) + 4 Au3+(aq) → 3 O2(g) + 4 Au(s) + 12 H+(aq)
26. (a) Separate Cu2+(aq) + Pb(s) → Pb2+(aq) + Cu(s) into two half-reactions.
Cu2+(aq) → Cu(s) (reduction)
Pb(s) → Pb2+(aq) (oxidation)
The elements have been balanced. Use electrons to balance the charge.
Therefore, the half-reaction equations for the redox reaction are
Oxidation: Pb(s) → Pb2+(aq) + 2 e−
Reduction: Cu2+(aq) + 2 e− → Cu(s)
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Chapter 9: Oxidation–Reduction Reactions
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(b) Separate 2 HNO2(aq) + 4 H+(aq) + 2 Fe(s) → 2 Fe2+(aq) + N2O(g) + 3 H2O(l) into
two half-reactions.
2 HNO2(aq) + 4 H+(aq) → N2O(g) + 3 H2O(l) (reduction)
2 Fe(s) → 2 Fe2+(aq) (oxidation)
Divide the coefficients of the above equation by 2 to obtain the lowest possible
coefficients.
Fe(s) → Fe2+(aq)
The elements have been balanced. Use electrons to balance the charge.
Therefore, the half-reaction equations for the redox reaction are
Oxidation: Fe(s) → Fe2+(aq) + 2 e−
Reduction: 2 HNO2(aq) + 4 H+(aq) + 4 e− → N2O(g) + 3 H2O(l)
(c) Separate O2(g) + 2 H2O(l) + Co(s) → Co2+(aq) + 4 OH−(aq) into two half-reactions.
O2(g) + 2 H2O(l) → 4 OH−(aq) (reduction)
Co(s) → Co2+(aq) (oxidation)
The elements have been balanced. Use electrons to balance the charge.
Therefore, the half-reaction equations for the redox reaction are
Oxidation: Co(s) → Co2+(aq) + 2 e−
Reduction: O2(g) + 2 H2O(l) + 4 e− → 4 OH−(aq)
(d) Separate Br2(l) + Ag(s) → Ag+(aq) + 2 Br−(aq) into two half-reactions.
Br2(l) → 2 Br−(aq) (reduction)
Ag(s) → Ag+(aq) (oxidation)
The elements have been balanced. Use electrons to balance the charge.
Therefore, the half-reaction equations for the redox reaction are
Oxidation: Ag(s) → Ag+(aq) + e−
Reduction: Br2(l) + 2 e− → 2 Br−(aq)
(e) Separate Fe2+(aq) + Al(s) → Al3+(aq) + Fe(s) into two half-reactions.
Al(s) → Al3+(aq) (oxidation)
Fe2+(aq) → Fe(s) (reduction)
The elements have been balanced. Use electrons to balance the charge.
Therefore, the half-reaction equations for the redox reaction are
Oxidation: Al(s) → Al3+(aq) + 3 e−
Reduction: Fe2+(aq) + 2 e− → Fe(s)
(f) Separate H2S(aq) + Na+(aq) → S(s) + Na(s) into two half-reactions.
H2S(aq) → S(s) (oxidation)
Na+(aq) → Na(s) (reduction)
For the oxidation half-reaction, balance hydrogen by adding hydrogen ions.
H2S(aq) → S(s) + 2 H+(aq)
The elements have been balanced. Use electrons to balance the charge.
Therefore, the half-reaction equations for the redox reaction are
Oxidation: H2S(aq) → S(s) + 2 H+(aq) + 2 e−
Reduction: Na+(aq) + e− → Na(s)
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Chapter 9: Oxidation–Reduction Reactions
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(g) Separate MnO4−(aq) + Zn(s) + 2 H2O(l) → MnO2(s) + Zn2+(aq) + 4 OH−(aq) into two
half-reactions.
MnO4−(aq) + 4 H+(aq) → MnO2(s) + 2 H2O(l) (reduction)
Zn(s) → Zn2+(aq) (oxidation)
The elements have been balanced. Use electrons to balance the charge.
Therefore, the half-reaction equations for the redox reaction are
Oxidation: Zn(s) → Zn2+(aq) + 2 e−
Reduction: MnO4−(aq) + 2 H2O(l) + 3 e− → MnO2(s) + 4 OH−(aq)
27. (a) Write an unbalanced equation.
Ni(s) + Cl2(g) → NiCl2(g)
Determine the oxidation numbers for each element in the equation.
0
0
+2 –1
Ni(s) + Cl2(g) → NiCl2(g)
The oxidation number of nickel changes from 0 to +2, so each nickel loses 2 electrons.
(b) The oxidation number of chlorine changes from 0 to –1, so each chlorine gains 1
electron.
(c) Separate Ni(s) + Cl2(g) → NiCl2(g) into two half-reactions.
Each nickel loses 2 electrons.
Ni(s) → Ni2+(g) + 2 e−
Since there are 2 chlorine atoms in chlorine gas, each chlorine molecule gains 2 electrons.
Cl2(g) + 2 e− → 2 Cl−(g)
The number of electrons transferred is equal in the two half-reactions.
Therefore, the balanced chemical equation for this reaction is
Ni(s) + Cl2(g) → NiCl2(g)
28. Write an unbalanced chemical equation for the reaction.
Ca(s) + H2O(l) → Ca(OH)2(s) + H2(g)
Separate into two half-reactions.
Ca(s) → Ca2+(aq) (oxidation)
H2O(l) → H2(g) (reduction)
For the reduction half-reaction, balance oxygen by adding OH−(aq) ions.
2 H2O(l) → 2 OH−(aq) + H2(g)
The elements have been balanced. Use electrons to balance the charge.
Therefore, the half-reaction equations for the redox reaction are
Oxidation: Ca(s) → Ca2+(aq) + 2 e−
Reduction: 2 H2O(l) + 2 e− → 2 OH−(aq) + H2(g)
Add the two half-reactions to obtain the net ionic equation.
The net ionic equation for the reaction is
Ca(s) + 2 H2O(l) → Ca2+(aq) + H2(g) + 2 OH−(aq)
29. From a redox table, the entities in order of increasing strength as oxidizing agents are
K+(aq) < H2O(l) < Cd2+(aq) < I2(s) < IO3–(aq) < AuCl4–(aq)
30. From a redox table, the entities in order of increasing strength as reducing agents are
F–(aq) < H2O(l) < I2(s) < Cu+(aq) < K(s)
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Chapter 9: Oxidation–Reduction Reactions
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31. (a) Use a redox table to check if a line from the oxidizing agent to the reducing agent
would form a downward diagonal to the right on the table.
(i) In a redox table, a line from the oxidizing agent, K+(aq), to the reducing agent, Na(s),
does not form a downward diagonal to the right on the table. Therefore, the combination
of K+(aq) and Na(s) would not result in a spontaneous reaction.
(ii) In a redox table, a line from the oxidizing agent, Cr2+(aq), to the reducing agent, K(s),
forms a downward diagonal to the right on the table. Therefore, the combination of
Cr2+(aq) and K(s) would result in a spontaneous reaction.
(iii) In a redox table, a line from the oxidizing agent, Pb2+(aq), to the reducing agent,
Fe(s), forms a downward diagonal to the right on the table. Therefore, the combination of
Pb2+(aq) and Fe(s) would result in a spontaneous reaction.
(iv) In a redox table, a line from the oxidizing agent, Sn2+(aq), to the reducing agent,
AgI(s), does not form a downward diagonal to the right on the table. Therefore, the
combination of Sn2+(aq) and AgI(s) would not result in a spontaneous reaction.
(b) A balanced chemical equation for the predicted reaction in (ii) is
Cr2+(aq) + 2 K(s) → 2 K+(aq) + Cr(s)
A balanced chemical equation for the predicted reaction in (iii) is
Pb2+(aq) + Fe(s) → Fe2+(aq) + Pb(s)
32. Substances that would be able to react spontaneously with a solution of aluminum
ions, Al3+(aq), are all reducing agents below aluminum in the redox table: Li(s), K(s),
Ba(s), Ca(s), Na(s), La(s), Mg(s), and H–(aq).
Understanding
33. Answer may vary. Sample answer: A redox reaction is also called an oxidation–
reduction reaction. During the reaction, electrons are transferred from one entity to
another. The oxidation number of the entity determines how many electrons can be
transferred. In the oxidation part of the reaction, an element loses electrons and becomes
oxidized; the substance that loses electrons is called the reducing agent. In the reduction
part of the reaction, an element gains electrons and is reduced; the substance that gains
electrons is called the oxidizing agent.
34. Answers may vary. Sample answer: A hydrocarbon contains only carbon and
hydrogen. To combust a hydrocarbon, gaseous oxygen is required; the products are
carbon dioxide and water vapour. In a combustion reaction, the oxidation number of
oxygen in gaseous oxygen, O2(g), is 0, since O2(g) is an element. The oxidation number
of oxygen in water vapour, H2O(g) is –2, since water is a compound. Since there is a
change in the oxidation number of oxygen in the reaction, a hydrocarbon combustion
reaction is a redox reaction.
For example, in the combustion reaction of methane, the oxidation numbers of the
elements are
–4 +1
0
+4 –2
+1 –2
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
The oxidation number of carbon increases from –4 to +4, so it is oxidized.
The oxidation number of oxygen decreases from 0 to –2, so it is reduced.
Therefore, the combustion reaction is a redox reaction.
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Chapter 9: Oxidation–Reduction Reactions
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35. Solution: Use the zero-sum rule to assign oxidation numbers to the nitrogen in each
compound: The sum of the oxidation numbers of all atoms in each of the electrically
neutral compounds HNO3, NH4Cl, N2O, NO2, and NaNO2 must be 0.
For HNO3: The oxidation number of hydrogen in its compounds is +1, and the oxidation
number of oxygen in its compounds is –2. Since there is 1 hydrogen atom in HNO3, the
charge due to hydrogen is +1. Since there are 3 oxygen atoms, the total charge due to
oxygen is –6. Therefore, the nitrogen atom in HNO3 must have an oxidation number of
+5 to give a sum of 0.
For NH4Cl: The oxidation number of hydrogen in its compounds is +1. Since the
oxidation number of a monatomic ion is the same as its charge, the oxidation number of
chlorine is –1. Since there are 4 hydrogen atoms in NH4Cl, the charge due to hydrogen is
+4. Since there is 1 chlorine atom, the charge due to chlorine is –1. Therefore, the
nitrogen atom in NH4Cl must have an oxidation number of –3 to give a sum of 0.
For N2O: The oxidation number of oxygen in its compounds is –2. Since there is
1 oxygen atom in N2O, the charge due to oxygen is –2. Therefore, the total charge due
to the 2 nitrogen atoms must be +2. Therefore, the oxidation number of each nitrogen
atom is +1.
For NO2: The oxidation number of oxygen in its compounds is –2. Since there are 2
oxygen atoms in NO2, the total charge due to oxygen is –4. Therefore, the nitrogen atom
in NO2 must have an oxidation number of +4 to give a sum of 0.
For NaNO2: Since the oxidation number of a monatomic ion is the same as its charge,
the oxidation number of sodium is +1. The oxidation number of oxygen in its compounds
is –2. Since there are 2 oxygen atoms in NaNO2, the total charge due to oxygen is –4.
Therefore, the nitrogen atom in NaNO2 must have an oxidation number of +3 to give a
sum of 0.
Statement: The following compounds are listed from highest to lowest oxidation number
of the nitrogen atom.
HNO3 > NO2 > NaNO2 > N2O > NH4Cl
36. (a) Answer may vary. Sample answer:
Reducing agent
Oxidizing agent
• loses electrons in a redox reaction
• gains electrons in a redox reaction
• causes the reduction (a decrease in
• causes the oxidation (an increase in
oxidation number) of another entity
oxidation number) of another entity
• itself is oxidized; its oxidation number • itself is reduced; its oxidation number
increases
decreases
(b) Answers may vary. Sample answer: Oxygen gas is likely an oxidizing agent. The
oxidation number of oxygen in oxygen gas is 0. For redox reactions involving oxygen
gas, the oxidation number of oxygen in the compound formed is usually –2. This means
that there is a decrease in oxidation number, so the oxygen gas is reduced and is therefore
an oxidizing agent.
37. A reaction is a redox reaction if one or more electrons are transferred between
chemical entities during the reaction. If the oxidation numbers of entities change during
the reaction, the reaction is a redox reaction. If the oxidation numbers of the entities
remain the same during the reaction, the reaction is not a redox reaction.
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Chapter 9: Oxidation–Reduction Reactions
9-8
38. The chemical equation is not balanced because electron transfer is not equalized.
During the reaction, iron metal loses 2 electrons and the silver ion gains 1 electron. To
balance the equation, the coefficients of silver ion and silver metal should be 2.
39. The chemical equation is not balanced because electron transfer is not equalized.
Silver ion gains 1 electron, and copper metal loses 2 electrons. To balance the equation,
the coefficient of silver ion, and of silver metal, should be 2. The balanced equation is
2 AgNO3(aq) + Cu(s) → 2 Ag(s) + Cu(NO3)2(aq)
40. (a) The unbalanced equation for the reaction is
NH3(g) + O2(g) → NO2(g) + H2O(g)
(b) Solution:
For NH3(g): The oxidation number of hydrogen in NH3(g) is +1. Since there are
3 hydrogen atoms, the oxidation number of nitrogen must be –3.
For O2(g): The oxidation number of oxygen in O2(g) is 0.
For NO2(g): The oxidation number of oxygen in NO2(g) is –2. Since there are 2 oxygen
atoms, the oxidation number of nitrogen must be +4.
For H2O(g): The oxidation number of oxygen in H2O(g) is –2. Since there are
2 hydrogen atoms, the oxidation number of hydrogen is +1.
Statement: The oxidation numbers of all the elements in this reaction are
Reactants: nitrogen: −3; hydrogen: +1; oxygen: 0
Products: nitrogen: +4; oxygen: –2; hydrogen: +1; oxygen: –2
(c) Solution:
Step 1: Write unbalanced equations for oxidation and reduction half-reactions.
Oxygen gains 2 electrons, so it is reduced. The reduction half-reaction equation is
O2(g) → H2O(g)
Nitrogen loses 7 electrons, so it is oxidized. The oxidation half-reaction equation is
NH3(g) → NO2(g)
Step 2: Balance each half-reaction equation.
For the reduction half-reaction, balance hydrogen by adding hydrogen ions.
O2(g) + 4 H+(aq) → 2 H2O(g)
For the oxidation half-reaction, balance oxygen by adding water.
NH3(g) + 2 H2O(l) → NO2(g) + 7 H+(aq)
Step 3: Use electrons to balance the charge in each half-reaction equation.
O2(g) + 4 H+(aq) + 4 e− → 2 H2O(g)
NH3(g) + 2 H2O(l) → NO2(g) + 7 H+(aq) + 7 e−
Statement: The balanced half-reaction equations are
Oxidation: NH3(g) + 2 H2O(l) → NO2(g) + 7 H+(aq) + 7 e−
Reduction: O2(g) + 4 H+(aq) + 4 e− → 2 H2O(g)
(d) Solution:
Step 1: Equalize the electron transfer in the two half-reaction equations. Multiply the
oxidation half-reaction by 4 and the reduction half-reaction by 7.
4 NH3(g) + 8 H2O(l) → 4 NO2(g) + 28 H+(aq) + 28 e−
7 O2(g) + 28 H+(aq) + 28 e− → 14 H2O(g)
Step 2: Add the two-half-reaction equations.
4 NH3(g) + 7 O2(g) → 4 NO2(g) + 6 H2O(g)
Statement: The balanced equation for the redox reaction is
4 NH3(g) + 7 O2(g) → 4 NO2(g) + 6 H2O(g)
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Chapter 9: Oxidation–Reduction Reactions
9-9
41. Use these steps for determining the oxidation numbers for elements in sodium sulfate,
Na2SO4.
Step 1: Assign oxidation numbers to elements as listed in Table 1. For example, the
oxidation number of the sodium ion, a monatomic ion, is +1, and the oxidation number of
oxygen in its compounds is –2.
Step 2: Use the zero-sum rule to assign oxidation numbers of other elements.
The sum of the oxidation numbers of the atoms in the compound must be 0.
Since each sodium ion has an oxidation number of +1 and there are 2 sodium ions, the
polyatomic sulfate ion must have an overall oxidation number of –2.
Since each oxygen atom has an oxidation number of –2 and there are 4 oxygen atoms, the
total contribution of the oxygen atoms is –8. Therefore, the oxidation number for the
sulfur atom must be +6.
+1
+6 –2
for each atom
Na2SO4
Step 3: Check that the sum of the oxidation numbers is equal to 0.
2(+1) + 1(+6)
+ 4(–2) = 0
↑
↑
↑
number of Na atoms number of S atoms
number of O atoms
In the Na2SO4 molecule, the oxidation number of each sodium ion is +1; the oxidation
number of each sulfate atom is +6; and the oxidation number of each oxygen atom is –2.
42. (a) Separate Cu2+(aq) + Pb(s) → Pb2+(aq) + Cu(s) into two half-reactions.
Cu2+(aq) → Cu(s) (reduction)
Pb(s) → Pb2+(aq) (oxidation)
The elements are balanced. Use electrons to balance the charge.
Therefore, the half-reaction equations for the redox reaction are
Oxidation: Pb(s) → Pb2+(aq) + 2 e−
Reduction: Cu2+(aq) + 2 e− → Cu(s)
The number of electrons transferred is equalized. So, add the two half-reaction equations.
The balanced net ionic equation is:
Cu2+(aq) + Pb(s) → Pb2+(aq) + Cu(s)
(b) Separate HNO2(aq) + H+(aq) + Fe(s) → Fe2+(aq) + N2O(g) + H2O(l) into two halfreactions.
HNO2(aq) + H+(aq) → N2O(g) + H2O(l) (reduction)
Fe(s) → Fe2+(aq) (oxidation)
Balance each half-reaction equation.
For the reduction half-reaction, balance nitrogen.
2 HNO2(aq) + H+(aq) → N2O(g) + H2O(l)
Balance oxygen by adding water.
2 HNO2(aq) + H+(aq) → N2O(g) + 3 H2O(l)
Balance hydrogen by adding hydrogen ions.
2 HNO2(aq) + 4 H+(aq) → N2O(g) + 3 H2O(l)
The oxidation half-reaction is balanced.
Fe(s) → Fe2+(aq)
Use electrons to balance the charge.
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-10
Therefore, the half-reaction equations for the redox reaction are
Oxidation: Fe(s) → Fe2+(aq) + 2 e−
Reduction: 2 HNO2(aq) + 4 H+(aq) + 4 e− → N2O(g) + 3 H2O(l)
Equalize the electron transfer in the two half-reaction equations by multiplying the
oxidation half-reaction by 2.
2 Fe(s) → 2 Fe2+(aq) + 4 e−
Add the two half-reaction equations.
The balanced net ionic equation is:
2 HNO2(aq) + 4 H+(aq) + 2 Fe(s) → 2 Fe2+(aq) + N2O(g) + 3 H2O(l)
(c) Separate O2(g) + H2O(l) + Co(s) → Co2+(aq) + OH−(aq) into two half-reactions.
O2(g) + H2O(l) → OH−(aq) (reduction)
Co(s) → Co2+(aq) (oxidation)
Balance each half-reaction equation.
For the reduction half-reaction, balance hydrogen.
O2(g) + H2O(l) → 2 OH−(aq)
Balance oxygen by adding water.
O2(g) + 2 H2O(l) → 4 OH−(aq)
The oxidation half-reaction is balanced.
Co(s) → Co2+(aq)
Use electrons to balance the charge.
Therefore, the half-reaction equations for the redox reaction are:
Oxidation: Co(s) → Co2+(aq) + 2 e−
Reduction: O2(g) + 2 H2O(l) + 4 e− → 4 OH−(aq)
Equalize the electron transfer in the two half-reaction equations by multiplying the
oxidation half-reaction by 2.
2 Co(s) → 2 Co2+(aq) + 4 e−
Add the two half-reaction equations.
The balanced net ionic equation is
O2(g) + 2 H2O(l) + 2 Co(s) → 2 Co2+(aq) + 4 OH−(aq)
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-11
43. (a) Steps for balancing a redox reaction equation using the oxidation numbers method
are
Step 1: Write the unbalanced chemical equation from the given information. Determine
the oxidation numbers for each element in the equation and identify the elements for
which the oxidation numbers change.
!
Step 2: Adjust the values of the coefficients to balance the electrons transferred.
!
Step 3: Balance the rest of the equation by inspection. If necessary, balance oxygen by
adding water.
!
Step 4: If necessary, balance hydrogen by adding H+(aq) and/or OH−(aq).
!
Step 5: Check your answer.
!
Step 6: Write the balanced equation.
(b) Solution:
Step 1: Use the unbalanced equation to determine the oxidation numbers for each
element in the equation and identify the elements for which the oxidation numbers
change.
The oxidation number of nitrogen changes from +5 to +2, so nitrogen gains 3 electrons.
The oxidation number of arsenic changes from +3 to +5, so arsenic loses 2 electrons.
Step 2: To balance electrons, adjust the coefficients by multiplying electrons gained by
2 and electrons lost by 3.
2 NO3−(aq) + 3 AsO33−(aq) → 2 NO(g) + 3 AsO43−(aq)
Step 3: Balance the rest of the equation by inspection.
Balance oxygen by adding water.
2 NO3−(aq) + 3 AsO33−(aq) → 2 NO(g) + 3 AsO43−(aq) + H2O(l)
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-12
Step 4: Balance hydrogen by adding H+(aq) and/or OH−(aq).
2 NO3−(aq) + 3 AsO33−(aq) + 2 H+(aq) → 2 NO(g) + 3 AsO43−(aq) + H2O(l)
Assume 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 NO3−(aq) + 3 AsO33−(aq) + 2 H+(aq) + 2 OH−(aq) →
2 NO(g) + 3 AsO43−(aq) + H2O(l) + 2 OH−(aq)
Subtract 1 water molecule from each side to eliminate redundant water molecules.
2 NO3−(aq) + 3 AsO33−(aq) + H2O(l) → 2 NO(g) + 3 AsO43−(aq) + 2 OH−(aq)
Step 5: Check the answer.
Step 6: Write the balanced equation.
2 NO3−(aq) + 3 AsO33−(aq) + H2O(l) → 2 NO(g) + 3 AsO43−(aq) + 2 OH−(aq)
44. (a) Solution:
Step 1: Assign oxidation numbers to all entities.
+2
2+
0
+3
3+
0
Ni (aq) + Al(s) → Al (aq) + Ni(s)
The oxidation number of nickel changes from +2 to 0, so nickel gains 2 electrons.
The oxidation number of aluminum changes from 0 to +3, so aluminum loses 3 electrons.
Step 2: To balance electrons, adjust the coefficients by multiplying electrons gained by
3 and electrons lost by 2.
3 Ni2+(aq) + 2 Al(s) → 2 Al3+(aq) + 3 Ni(s)
Statement: The balanced equation is
3 Ni2+(aq) + 2 Al(s) → 2 Al3+(aq) + 3 Ni(s)
(b) Solution:
Step 1: Assign oxidation numbers to all entities.
+7
–2
0
+4
–2
+2
MnO4−(aq) + Ni(s) → MnO2(s) + Ni2+(aq)
The oxidation number of manganese changes from +7 to +4, so manganese gains
3 electrons.
The oxidation number of nickel changes from 0 to +2, so nickel 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 Ni(s) → 2 MnO2(s) + 3 Ni2+(aq)
Step 3: Balance the rest of the equation by inspection.
Balance oxygen by adding water.
2 MnO4−(aq) + 3 Ni(s) → 2 MnO2(s) + 3 Ni2+(aq) + 4 H2O(l)
Balance hydrogen by adding hydrogen ions.
2 MnO4−(aq) + 3 Ni(s) + 8 H+(aq) → 2 MnO2(s) + 3 Ni2+(aq) + 4 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 Ni(s) + 8 H+(aq) + 8 OH−(aq) →
2 MnO2(s) + 3 Ni2+(aq) + 4 H2O(l) + 8 OH−(aq)
Subtract 4 water molecules from each side to eliminate redundant water molecules.
2 MnO4−(aq) + 3 Ni(s) + 4 H2O(l) → 2 MnO2(s) + 3 Ni2+(aq) + 8 OH−(aq)
Statement: The balanced equation is
2 MnO4−(aq) + 3 Ni(s) + 4 H2O(l) → 2 MnO2(s) + 3 Ni2+(aq) + 8 OH−(aq)
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-13
(c) Solution:
Step 1: Assign oxidation numbers to all entities.
+6 –2
+2
+4 –2
+4
SO42−(aq) + Sn2+(aq) → SO32−(aq) + Sn4+(aq)
The oxidation number of sulfur changes from +6 to +4, so sulfur gains 2 electrons.
The oxidation number of tin changes from +2 to +4, so tin loses 2 electrons.
Step 2: The electrons have been balanced. Balance the rest of the equation by inspection.
Balance oxygen by adding water.
SO42−(aq) + Sn2+(aq) → SO32−(aq) + Sn4+(aq) + H2O(l)
Balance hydrogen by adding hydrogen ions.
SO42−(aq) + Sn2+(aq) + 2 H+(aq) → SO32−(aq) + Sn4+(aq) + H2O(l)
Statement: The balanced equation is:
SO42−(aq) + Sn2+(aq) + 2 H+(aq) → SO32−(aq) + Sn4+(aq) + H2O(l)
(d) Solution:
Step 1: Assign oxidation numbers to all entities.
0
0
–1
−
+1
+3
0
Br2(l) + Ag(s) → 2 Br (aq) + Ag+(aq)
The oxidation number of bromine changes from 0 to –1, so each bromine atom gains
1 electron; so each bromine molecule gains 2 electrons.
The oxidation number of silver changes from 0 to +1, so silver loses 1 electron.
Step 2: To balance electrons, adjust the coefficients by multiplying electrons lost by 2.
Br2(l) + 2 Ag(s) → 2 Br−(aq) + 2 Ag+(aq)
Statement: The balanced equation is
Br2(l) + 2 Ag(s) → 2 Br−(aq) + 2 Ag+(aq)
(e) Solution:
Step 1: Assign oxidation numbers to all entities.
+2
0
Fe2+(aq) + Al(s) → Al3+(aq) + Fe(s)
The oxidation number of iron changes from +2 to 0, so iron gains 2 electrons.
The oxidation number of aluminum changes from 0 to +3, so aluminum loses 3 electrons.
Step 2: To balance electrons, adjust the coefficients by multiplying electrons gained by
3 and electrons lost by 2.
3 Fe2+(aq) + 2 Al(s) → 2 Al3+(aq) + 3 Fe(s)
Statement: The balanced equation is
3 Fe2+(aq) + 2 Al(s) → 2 Al3+(aq) + 3 Fe(s)
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-14
(f) Solution:
Step 1: Assign oxidation numbers to all entities.
+1 –2
+1
0
0
H2S(aq) + Na+(aq) → S(s) + Na(s)
The oxidation number of sulfur changes from –2 to 0, so sulfur loses 2 electrons.
The oxidation number of sodium changes from +1 to 0, so sodium gains 1 electron.
Step 2: To balance electrons, adjust the coefficients by multiplying electrons gained by 2.
H2S(aq) + 2 Na+(aq) → S(s) + 2 Na(s)
Step 3: Balance the rest of the equation by inspection.
Balance hydrogen by adding hydrogen ions.
H2S(aq) + 2 Na+(aq) → S(s) + 2 Na(s) + 2 H+(aq)
Statement: The balanced equation is
H2S(aq) + 2 Na+(aq) → S(s) + 2 Na(s) + 2 H+(aq)
(g) Solution:
Step 1: Assign oxidation numbers to all entities.
+7
–2
−
0
+4
–2
+2
MnO4 (aq) + Zn(s) → MnO2(s) + Zn2+(aq)
The oxidation number of manganese changes from +7 to +4, so manganese gains
3 electrons.
The oxidation number of zinc changes from 0 to +2, so zinc 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 Zn(s) → 2 MnO2(s) + 3 Zn2+(aq)
Step 3: Balance the rest of the equation by inspection.
Balance oxygen by adding water.
2 MnO4−(aq) + 3 Zn(s) → 2 MnO2(s) + 3 Zn2+(aq) + 4 H2O(l)
Balance hydrogen by adding hydrogen ions.
2 MnO4−(aq) + 3 Zn(s) + 8 H+(aq) → 2 MnO2(s) + 3 Zn2+(aq) + 4 H2O(l)
Since the reaction takes place in a basic solution, eliminate H+(aq) by adding an equal
number of OH−(aq) to both sides of the equation.
2 MnO4−(aq) + 3 Zn(s) + 8 H+(aq) + 8 OH−(aq) →
2 MnO2(s) + 3 Zn2+(aq) + 4 H2O(l) + 8 OH−(aq)
Subtract 4 water molecules from each side to eliminate redundant water molecules.
2 MnO4−(aq) + 3 Zn(s) + 4 H2O(l) → 2 MnO2(s) + 3 Zn2+(aq) + 8 OH−(aq)
Statement: The balanced equation is
2 MnO4−(aq) + 3 Zn(s) + 4 H2O(l) → 2 MnO2(s) + 3 Zn2+(aq) + 8 OH−(aq)
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-15
45. (a) Solution:
Step 1: Assign oxidation numbers to all entities.
+7
–2
+3 –2
+7
–2
+2–3
+2
+4 –2
+1 –2
MnO4−(aq) + C2O42−(aq) → Mn2+(aq) + CO2(g) + H2O(l)
The oxidation number of manganese changes from +7 to +2, so manganese gains
5 electrons.
The oxidation number of carbon changes from +3 to +4, so carbon loses 1 electron; so
each C2O42− ion gains 2 electrons.
Step 2: To balance electrons, adjust the coefficients by multiplying electrons gained by
2 and electrons lost by 5.
2 MnO4−(aq) + 5 C2O42−(aq) → 2 Mn2+(aq) + 5 CO2(g) + H2O(l)
Step 3: Balance the rest of the equation by inspection.
Balance carbon.
2 MnO4−(aq) + 5 C2O42−(aq) → 2 Mn2+(aq) + 10 CO2(g) + H2O(l)
Balance oxygen by adding water.
2 MnO4−(aq) + 5 C2O42−(aq) → 2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l)
Balance hydrogen by adding hydrogen ions.
2 MnO4−(aq) + 5 C2O42−(aq) + 16 H+(aq) → 2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l)
Statement: The balanced equation is
2 MnO4−(aq) + 5 C2O42−(aq) + 16 H+(aq) → 2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l)
(b) Solution:
Step 1: Assign oxidation numbers to all entities.
+4
–2
–2+4–3
MnO4−(aq) + CN−(aq) → MnO2(s) + OCN−(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 OCN−(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 OCN−(aq) + H2O(l)
Balance hydrogen by adding hydrogen ions.
2 MnO4−(aq) + 3 CN−(aq) + 2 H+(aq) → 2 MnO2(s) + 3 OCN−(aq) + H2O(l)
Since the reaction takes place in a 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 OCN−(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 OCN−(aq) + 2 OH−(aq)
Statement: The balanced equation is
2 MnO4−(aq) + 3 CN−(aq) + H2O(l) → 2 MnO2(s) + 3 OCN−(aq) + 2 OH−(aq)
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-16
(c) Solution:
Step 1: Assign oxidation numbers to all entities.
0
+4 –2
+2
Pb(s) + PbO2(s) → 2 Pb2+(aq)
Step 2: Write unbalanced equations for oxidation and reduction half-reactions.
Pb(s) → Pb2+(aq) (oxidation)
PbO2(s) → Pb2+(aq) (reduction)
Step 3: Balance each half-reaction equation.
The oxidation half-reaction is already balanced.
Pb(s) → Pb2+(aq)
Use H2O(l) to balance oxygen and H+(aq) to balance hydrogen.
PbO2(s) + 4 H+(aq) → Pb2+(aq) + 2 H2O(l)
Step 4: Use electrons to balance the charge in each half-reaction equation.
Pb(s) → Pb2+(aq) + 2 e−
PbO2(s) + 4 H+(aq) + 2 e− → Pb2+(aq) + 2 H2O(l)
Step 5: Electron transfer is equalized. Add the half-reaction equations.
Pb(s) + PbO2(s) + 4 H+(aq) → 2 Pb2+(aq) + 2 H2O(l)
Statement: The balanced equation is
Pb(s) + PbO2(s) + 4 H+(aq) → 2 Pb2+(aq) + 2 H2O(l)
46. Solution:
Step 1: List all of the entities present.
The reaction mixture contains Cu2+(aq), NO3–(aq), Ni(s), Cu(s), Ni2+(aq), and H2O(l).
Step 2: Use a redox table to identify the stronger oxidizing agent in the reaction mixture.
Ni2+(aq) + 2 e– → Ni(s)
E°r = −0.23 V
Cu2+(aq) + 2 e– → Cu(s) E°r = 0.34 V
Since copper occurs higher in the table, Cu2+(aq) is the stronger oxidizing agent, while
Ni(s) is the stronger reducing agent.
Step 3: Predict whether the reaction will occur spontaneously.
Since the relative positions of Cu2+(aq) and Ni(s) form a downward diagonal to the right
on the redox table, the reaction will occur spontaneously.
Statement: The following reaction can occur spontaneously:
Cu(NO3)2(aq) + Ni(s) → Cu(s) + Ni(NO3)2(aq)
47. In a redox table, the reducing agents are listed on the right-hand side of the table with
the strongest reducing agent at the bottom. The positions of the metals from the bottom of
the redox table are as follows: Ca(s), Al(s), Cr(s), Cd(s), Pb(s), and Cu(s).
Therefore, the metals in order of strongest reducing agent to weakest reducing agent are
Ca(s) > Al(s) > Cr(s) > Cd(s) > Pb(s) > Cu(s)
48. The ions of strong oxidizing agents are likely to react with the metals of weaker
oxidizing agents. The ions of weak oxidizing agents will not react with the metals of
stronger oxidizing agents. So, the metal ion that reacted with all the other metals it was
tested with would be the strongest oxidizing agent.
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-17
49. Answers may vary. Sample answer:
Analysis and Application
50. (a) Solution:
Step 1: Use the zero-sum rule to assign oxidation numbers to each atom and ion in the
chemical equation.
The oxidation number of an atom in an element is 0, so the oxidation number of iron in
Fe(s) and the oxidation number of oxygen in O2(g) is 0. The oxidation number of oxygen
in its compounds is –2. Since there are 3 oxygen atoms in Fe2O3, the charge due to
oxygen is –6. Therefore, the 2 iron atoms in Fe2O3 must have an oxidation number of +6
to give a sum of 0. That means the oxidation number of each iron atom is +3.
0
0
+3 –2
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
Step 2: Determine how the oxidation number on each element changes.
The oxidation number of iron increases from 0 to +3, so it is oxidized.
The oxidation number of oxygen decreases from 0 to –2, so it is reduced.
Step3: Determine the number of electrons transferred.
Since there are 4 iron atoms and each loses 3 electrons, the total number of electrons lost
is 12. Since there are 6 oxygen atoms and each gains 2 electrons, the total number of
electrons gained is 12.
Statement: The oxidation numbers of the atoms are
Reactants: iron: 0; oxygen: 0
Products: iron: +3; oxygen: –2
The number of electrons transferred is 12.
Since oxygen in O2 gains electrons, the oxidizing agent is O2(g). Since iron loses
electrons, the reducing agent is Fe(s).
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-18
(b) Solution:
Step 1: Use the zero-sum rule to assign oxidation numbers to each atom/ion in the
chemical equation.
The oxidation number of an atom in an element is 0, so the oxidation number of
hydrogen in H2(g) and the oxidation number of nickel in Ni(s) is 0.
The oxidation number of hydrogen in its compounds is +1 and the oxidation number of
oxygen in its compounds is –2. Since there is 1 oxygen atom in NiO, the charge due to
oxygen is –2. Therefore, the nickel atom in NiO must have an oxidation number of +2 to
give a sum of 0.
+2 –2
0
0
+1 –2
NiO(s) + H2(g) → Ni(s) + H2O(l)
Step 2: Determine how the oxidation number on each element changes.
The oxidation number of hydrogen increases from 0 to +1, so it is oxidized.
The oxidation number of nickel decreases from +2 to 0, so it is reduced.
The oxidation number of oxygen does not change.
Step3: Determine the number of electrons transferred.
Since there is 1 nickel atom and it gains 2 electrons, the total number of electrons gained
is 2.
Since there are 2 hydrogen atoms and each loses 1 electron, the total number of electrons
lost is 2.
Statement: The oxidation numbers of the atoms are:
Reactants: nitrogen: +2; oxygen: –2; hydrogen: 0
Products: nickel: 0; hydrogen: +1; oxygen: –2
The number of electrons transferred is 2.
Since nickel in NiO gains electrons, the oxidizing agent is NiO(s). Since hydrogen in H2
loses electrons, the reducing agent is H2(g).
(c) Solution:
Step 1: Use the zero-sum rule to assign oxidation numbers to each atom/ion in the
chemical equation.
The oxidation number of an atom in an element is 0, so the oxidation number of
hydrogen in H2(g) and the oxidation number of iron in Fe(s) is 0.
The oxidation number of oxygen in its compounds is –2, so the charge due to oxygen in
Fe2O3 is –2. Since there are 3 oxygen atoms in Fe2O3, the charge due to oxygen is –6.
Therefore, the 2 iron atoms in Fe2O3 must have an oxidation number of +6 to give a sum
of 0. That means the oxidation number of each iron atom is +3. Since there are 2 oxygen
atoms in CO2, the charge due to oxygen is –4. Therefore, the carbon atom in CO2 must
have an oxidation number of +4 to give a sum of 0.
+3
–2
+2 –2
0
+4 –2
Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 2 CO2(g)
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-19
Step 2: Determine how the oxidation number on each element changes.
The oxidation number of carbon increases from +2 to +4, so it is oxidized.
The oxidation number of iron decreases from +3 to 0, so it is reduced.
The oxidation number of oxygen does not change.
Step 3: Determine the number of electrons transferred.
Since there are 2 iron atoms and each gains 3 electrons, the total number of electrons
gained is 6. Since there are 3 carbon atoms and each loses 2 electrons, the total number of
electrons lost is 6.
Statement: The oxidation numbers of the atoms are
Reactants: iron: +3; oxygen: –2; carbon: +2; oxygen in CO: –2
Products: iron: 0; carbon: +4; oxygen: –2
The number of electrons transferred is 6.
Since iron in Fe2O3 gains electrons, the oxidizing agent is Fe2O3(s). Since carbon in CO
loses electrons, the reducing agent is CO(g).
(d) Solution:
Step 1: Use the zero-sum rule to assign oxidation numbers to each atom and ion in the
chemical equation.
The oxidation number of an atom in an element is 0, so the oxidation number of copper in
Cu(s) is 0.
The oxidation number of hydrogen in its compounds is +1 and the oxidation number of
oxygen in its compounds is –2.
In HNO3, the charge due to the 1 hydrogen atom is +1 and the charge due to the 3 oxygen
atoms is –6. Therefore, the nitrogen atom in HNO3 must have an oxidation number of +5
to give a sum of 0.
In Cu(NO3)2, the charge of Cu2+, a monatomic ion, is +2. The charge due to the 6 oxygen
atoms is –12. Therefore, the 2 nitrogen atoms in Cu(NO3)2 must have an oxidation
number of +10 to give a sum of 0. That means the oxidation number of each nitrogen
atom is +5.
In H2O, the oxidation number of each hydrogen atom is +1 and the oxidation number of
the oxygen atom is –2.
In NO2, the total charge due to the 2 oxygen atoms is –4. Therefore, the nitrogen atom in
NO2 must have an oxidation number of +4 to give a sum of 0.
0
+1+5 –2
+2
+5 –2
+1 –2
+4 –2
Cu(s) + 4 HNO3(aq) → Cu(NO3)2(aq) + 2 H2O(l) + 2 NO2(g)
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-20
Step 2: Determine how the oxidation number on each element changes. There is no
change in the oxidation numbers of hydrogen and oxygen.
The oxidation number of copper increases from 0 to +2, so it is oxidized.
The oxidation number of nitrogen decreases from +5 to +4, so it is reduced.
Step3: Determine the number of electrons transferred.
Since there are 2 nitrogen atoms and each gains 1 electron, the total number of electrons
gained is 2. Since there is 1 copper atom and it loses 2 electrons, the total number of
electrons lost is 2.
Statement: The oxidation numbers of the atoms are
Reactants: copper: 0; hydrogen: +1; nitrogen: +5; oxygen: –2
Products: copper: +2; hydrogen: +1; nitrogen: +5; oxygen: –2; hydrogen in H2O: +1;
oxygen in H2O: –2; nitrogen in NO2: +4; oxygen in NO2: –2
The number of electrons transferred is 2.
Since nitrogen in HNO3 gains electrons, the oxidizing agent is HNO3(aq). Since copper
loses electrons, the reducing agent is Cu(s).
51. (a) On Earth, the atmosphere contains oxygen gas, which is an oxidizing agent, and
nitrogen gas, which is a reducing agent, for many redox reactions to take place. The
atmosphere of Saturn contains hydrogen gas, which is an oxidizing agent.
(b) On Earth, methanol, CH3OH(l), might undergo a combustion reaction with oxygen
gas as the oxidizing agent. The reaction can be represented by this equation:
–2 +1 –2 +1
0
+4 –2
+1 –2
2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(g)
The oxidation state of carbon changes from –2 to +4, so carbon in CH3OH(l) is oxidized.
On Saturn, methanol, CH3OH(l), might undergo a reduction reaction with hydrogen gas.
The reaction can be represented by this equation:
–2 +1 –2 +1
0
–4 +1
+1 –2
CH3OH(l) + H2(g) → CH4(g) + H2O(g)
The oxidation state of carbon changes from –2 to –4, so carbon in CH3OH(l) is reduced.
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-21
52. Solution:
Step 1: Write a balanced chemical equation for the photosynthesis reaction.
6 CO2(g) + 6 H2O(l) → C6H12O6(s) + 6 O2(g)
Step 2: Use the zero-sum rule to assign oxidation numbers to each atom and ion in the
equation.
The oxidation number of an atom in an element is 0, so the oxidation number of oxygen
in O2(g) is 0.
The oxidation number of hydrogen in its compounds is +1 and the oxidation number of
oxygen in its compounds is –2.
In CO2, the total charge due to the 2 oxygen atoms is –4. Therefore, the carbon atom in
CO2 must have an oxidation number of +4 to give a sum of 0.
In H2O, the oxidation number of each hydrogen atom is +1 and the oxidation number of
the oxygen atom is –2.
In C6H12O6, the total charge due to the 12 hydrogen atoms is +12, and the total charge
due to the 6 oxygen atoms is –12. Therefore, the total charge due to the 6 carbon atoms
must be 0. That is, the oxidation number of each carbon in C6H12O6 is 0.
The oxidation numbers of the elements are
Reactants: C: +4; H: +1; O: –2
Products: C: 0; H: +1; O: –2; O in O2: 0
Step 3: Determine how the oxidation number on each element changes.
The oxidation number of oxygen increases from –2 to 0.
The oxidation number of carbon decreases from +4 to 0.
Statement: Since the oxidation numbers of the atoms changes, which means a transfer of
electron occurs, the process of photosynthesis is a redox reaction.
53. (a) Assign oxidation numbers to each atom/ion in the chemical equation.
+1 –1
+2
0
–1
−
+3
AgI(s) + Cr (aq) → Ag(s) + I (aq) + Cr3+(aq)
Since Cr2+(aq) loses 1 electron, Cr2+(aq) is oxidized. Since silver in AgI(s) gains
1 electron, AgI(s) is reduced.
(b) Assign oxidation numbers to each atom/ion in the chemical equation.
+2
2+
0
+3
0
Ni2+(aq) + Al(s) → Al3+(aq) + Ni(s)
Since Ni2+(aq) gains 2 electrons, Ni2+(aq) is reduced. Since Al(s) loses 3 electrons, Al(s)
is oxidized.
(c) Assign oxidation numbers to each atom/ion in the chemical equation.
+7
–2
0
+4
–2
0
MnO4 (aq) + Ni(s) → MnO2(s) + Ni2+(aq)
Since maganese in MnO4−(aq) gains 3 electrons, MnO4−(aq) is reduced. Since Ni(s) loses
2 electrons, Ni(s) is oxidized.
(d) Assign oxidation numbers to each atom/ion in the chemical equation.
+6 –2
−
+2
+4 –2
+4
SO42−(aq) + Sn2+(aq) → SO32−(aq) + Sn4+(aq)
Since sulfur in SO42−(aq) gains 2 electrons, SO42−(aq) is reduced. Since Sn2+(aq) loses
2 electrons, Sn2+(aq) is oxidized.
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-22
54. (a) Write an unbalanced chemical equation for the reaction.
Ca(s) + H2O(l) → Ca(OH)2(s) + H2(g)
Separate into two half-reactions.
Ca(s) → Ca2+(aq) (oxidation)
H2O(l) → H2(g) (reduction)
For the reduction half-reaction, balance oxygen by adding OH−(aq) ions.
2 H2O(l) → 2 OH−(aq) + H2(g)
The elements have been balanced. Use electrons to balance the charge.
Therefore, the half-reaction equations for the redox reaction are:
Oxidation: Ca(s) → Ca2+(aq) + 2 e−
Reduction: 2 H2O(l) + 2 e− → 2 OH−(aq) + H2(g)
Add the two half-reactions to obtain the net ionic equation.
The net ionic equation for the reaction is:
Ca(s) + 2 H2O(l) → Ca2+(aq) + H2(g) + 2 OH−(aq)
(b) In the reaction, Ca(s) is oxidized, so it is the reducing agent. Since the relative
positions of H2O(l) and Ca(s) form a downward diagonal to the right on the redox table,
the reaction will occur spontaneously.
(c) In the reaction, hydrogen gas is produced, so I would expect to observe bubbles.
(d) To confirm my prediction, I could perform a pH test to see if hydroxide ions are
produced in the reaction.
55. (a) Answers may vary. Sample answers: A laboratory procedure, I could conduct
could be as follows:
(i) Place a piece of copper metal into silver nitrate solution.
(ii) Place a piece of zinc metal into hydrochloric acid.
(iii) Place a piece of Fe3O4(s) into a container of carbon monoxide gas.
(iv) Place a piece of copper metal into nitric acid solution.
(b) (i) Solution:
Step 1: Assign oxidation numbers to each atom/ion in the chemical equation.
+1 +5 –2
0
+2
+5 –2
0
AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + Ag(s)
The oxidation number of silver changes from +1 to 0, so silver gains 1 electron.
The oxidation number of copper changes from 0 to +2, so copper loses 2 electrons.
Step 2: To balance electrons, adjust the coefficients by multiplying electrons gained by 2.
2 AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2 Ag(s)
Statement: The balanced equation is
2 AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2 Ag(s)
(ii) Solution:
Step 1: Assign oxidation numbers to each atom/ion in the chemical equation.
+1 –1
0
+2 –1
0
HCl(aq) + Zn(s) → ZnCl2(aq) + H2(g)
The oxidation number of hydrogen changes from +1 to 0, so hydrogen gains 1 electron.
The oxidation number of zinc changes from 0 to +2, so zinc loses 2 electrons.
Step 2: To balance electrons, adjust the coefficients by multiplying electrons gained by 2.
2 HCl(aq) + Zn(s) → ZnCl2(aq) + H2(g)
Statement: The balanced equation is
2 HCl(aq) + Zn(s) → ZnCl2(aq) + H2(g)
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-23
(iii) Solution:
Step 1: Assign oxidation numbers to each atom/ion in the chemical equation.
(+8) –2
+2 –2
+1+5 –2
0
+4 –2
0
Fe3O4(s) + CO(g) → CO2(g) + Fe(s)
The oxidation number of each of the 4 oxygen atoms in Fe3O4 is –2, so the combined
oxidation number of the 3 iron atoms must equal +8. The oxidation number of the 3 iron
atoms changes from +8 to 0, so the 3 iron atoms gain 8 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 lost by 4.
Fe3O4(s) + 4 CO(g) → 4 CO2(g) + Fe(s)
Step 3: Balance the rest of the equation by inspection.
Balance iron.
Fe3O4(s) + 4 CO(g) → 4 CO2(g) + 3 Fe(s)
Statement: The balanced equation is
Fe3O4(s) + 4 CO(g) → 4 CO2(g) + 3 Fe(s)
(iv) Solution:
Step 1: Assign oxidation numbers to each atom/ion in the chemical equation.
+2 –2
+2
+5 –2
+1 –2
HNO3(aq) + Cu(s) → NO(g) + Cu(NO3)2(aq) + H2O(l)
The oxidation number of nitrogen changes from +5 to +2, so nitrogen gains 3 electrons.
The oxidation number of copper changes from 0 to +2, so copper loses 2 electrons.
Step 2: To balance electrons, adjust the coefficients by multiplying electrons gained by 2
and electrons lost by 3.
2 HNO3(aq) + 3 Cu(s) → 2 NO(g) + 3 Cu(NO3)2(aq) + H2O(l)
Step 3: Balance the rest of the equation by inspection.
Balance nitrogen.
8 HNO3(aq) + 3 Cu(s) → 2 NO(g) + 3 Cu(NO3)2(aq) + H2O(l)
Balance oxygen by adding water.
8 HNO3(aq) + 3 Cu(s) → 2 NO(g) + 3 Cu(NO3)2(aq) + 4 H2O(l)
Statement: The balanced equation is
8 HNO3(aq) + 3 Cu(s) → 2 NO(g) + 3 Cu(NO3)2(aq) + 4 H2O(l)
56. (a) For the first chemical equation, assign oxidation numbers to each atom and ion.
+1–2 +1
–3 +1
–3 +1 +1
+1 –2
HOCl(aq) + NH3(aq) → NHCl2(g) + H2O(l)
The oxidation numbers of all the atoms in the first reaction are:
Reactants: hydrogen: +1; oxygen: –2; chlorine: +1; nitrogen: –3; hydrogen: +1
Products: nitrogen: –3; hydrogen: +1; chlorine: +1; hydrogen: +1; oxygen: –2
There is no change in oxidation numbers.
Balance chlorine.
2 HOCl(aq) + NH3(aq) → NHCl2(g) + H2O(l)
Balance oxygen by adding water.
2 HOCl(aq) + NH3(aq) → NHCl2(g) + 2 H2O(l)
Therefore, the balanced equation for the first reaction is
2 HOCl(aq) + NH3(aq) → NHCl2(g) + 2 H2O(l)
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-24
For the second chemical equation, assign oxidation numbers to each atom and ion.
+1–2 +1
–3 +1
+3 –1
+1 –2
HOCl(aq) + NH3(aq) → NCl3(g) + H2O(l)
The oxidation numbers of all the atoms in the second reaction are
Reactants: hydrogen: +1; oxygen: –2; chlorine: +1; nitrogen: –3; hydrogen: +1
Products: nitrogen: +3; chlorine: –1; hydrogen: +1, oxygen: –2
The oxidation number of chlorine changes from +1 to –1, so chlorine gains 2 electrons
per atom.
The oxidation number of nitrogen changes from –3 to +3, so nitrogen loses 6 electrons.
To balance electrons, adjust the coefficients by multiplying electrons gained by 3.
3 HOCl(aq) + NH3(aq) → NCl3(g) + H2O(l)
Balance oxygen by adding water.
3 HOCl(aq) + NH3(aq) → NCl3(g) + 3 H2O(l)
Therefore, the balanced equation for the second reaction is
3 HOCl(aq) + NH3(aq) → NCl3(g) + 3 H2O(l)
(b) When equal volumes of household bleach and ammonia are mixed, the first reaction
is more likely to occur because less HOCl is needed than in the second reaction. That is,
more NHCl2 will form.
57. (a) Solution: Determine whether any reaction will occur spontaneously in each
container.
(i) Step 1: The entities present in the container are Fe3+(aq), SO42–(aq), Cu(s), and
H2O(l).
Step 2: Using a redox table, the oxidizing agents in the container are:
Fe3+(aq) + e– → Fe2+(s)
E°r = 0.77 V
SO42–(aq) + 4 H+(aq) + 2 e– → H2SO3(aq) + H2O(l)
–
E° r = 0.20 V
–
2 H2O(l) + e → H2(g) + 2 OH (aq)
E°r = −0.83 V
3+
Since Fe (aq) occurs highest in the table, Fe3+(aq) is the strongest oxidizing agent.
The reducing agents in the container are:
O2(g) + 4 H+(g) + 2 e– → 2 H2O(l)
E°r = 1.23 V
Cu2+(aq) + 2 e– → Cu(s)
E°r = 0.34 V
Since Cu(s) occurs lower in the table, Cu(s) is the stronger reducing agent.
Step 3: Since the relative positions of Fe3+(aq) and Cu(s) form a downward diagonal to
the right on the redox table, a reaction will occur spontaneously.
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-25
(ii) Step 1: The entities present in the container are Cu2+(aq), Cl–(aq), Fe(s), and H2O(l).
Step 2: Using a redox table, the oxidizing agents in the container are:
Cu2+(aq) + 2 e– → Cu(s)
E°r = 0.34 V
2 H2O(l) + 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 container are:
Cl2(g) + 2 e– → 2 Cl–(aq)
E°r = 1.36 V
O2(g) + 4 H+(g) + 2 e– → 2 H2O(l)
2+
–
E°r = 1.23 V
E°r = −0.44 V
Fe (aq) + 2 e → Fe(s)
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) form a downward diagonal to
the right on the redox table, a reaction will occur spontaneously.
(iii) Step 1: The entities present in the container are Sn2+(aq), Cl–(aq), Cu(s), and H2O(l).
Step 2: Using a redox table, the oxidizing agents in the container are:
Sn2+(aq) + 2 e– → Sn(s)
E°r = −0.14 V
2 H2O(l) + e– → H2(g) + 2 OH–(aq)
E°r = −0.83 V
2+
Since Sn (aq) occurs higher in the table, Sn2+(aq) is the stronger oxidizing agent.
The reducing agents in the container are:
Cl2(g) + 2 e– → 2 Cl–(aq)
E°r = 1.36 V
O2(g) + 4 H+(g) + 2 e– → 2 H2O(l)
2+
–
E°r = 1.23 V
E°r = 0.34 V
Cu (aq) + 2 e → Cu(s)
Since Cu(s) occurs lowest in the table, Cu(s) is the strongest reducing agent.
Step 3: Since the relative positions of Sn2+(aq) and Cu(s) do not form a downward
diagonal to the right on the redox table, a reaction will not occur spontaneously.
(iv) Step 1: The entities present in the container are Ag+(aq), NO3–(aq), Cu(s), and
H2O(l).
Step 2: Using a redox table, the oxidizing agents in the container are:
Ag+(aq) + e– → Ag(s)
E°r = 0.80 V
2 H2O(l) + e– → H2(g) + 2 OH–(aq)
E°r = −0.83 V
+
Since Ag (aq) occurs higher in the table, Ag +(aq) is the stronger oxidizing agent.
The reducing agents in the container are:
O2(g) + 4 H+(g) + 2 e– → 2 H2O(l)
E°r = 1.23 V
Cu2+(aq) + 2 e– → Cu(s)
E°r = 0.34 V
Since Cu(s) occurs lower in the table, Cu(s) is the stronger reducing agent.
Step 3: Since the relative positions of Ag+(aq) and Cu(s) do form a downward diagonal
to the right on the redox table, a reaction will occur spontaneously.
Statement: Since there will be spontaneous reactions occurring in solutions (i), (ii), and
(iv), the technician should notice some changes in these solutions a day later.
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-26
(b) Answers may vary. Sample answers:
For solution (i), the technician would observe a change in colour of the solution from
yellow-brown Fe2(SO4)3(aq) to blue CuSO4(aq).
For solution (ii), the technician would observe a change in colour of the solution from
blue-green CuCl2(aq) to pale green FeCl2(aq) and some red-brown copper deposits on the
surface of the iron drum.
For solution (iv), the technician would observe a change in colour of the solution from
colourless AgCl(aq) to blue-green CuCl2(aq) and some silvery deposits on the surface of
the copper flask.
(c) A balanced chemical equation for the predicted reaction in solution (i) is
Fe2(SO4)3(aq) + Cu(s) → CuSO4(aq) + 2 FeSO4(aq)
A balanced chemical equation for the predicted reaction in solution (ii) is
CuCl2(aq) + Fe(s) → Cu(s) + FeCl2(aq)
A balanced chemical equation for the predicted reaction in solution (iv) is
2 AgNO3(aq) + Cu(s) → 2 Ag(s) + Cu(NO3)2(aq)
58. (a) According to the evidence, metal X has the greatest tendency to lose electrons
because it reacted with all three metal ions. Therefore, metal X is the strongest reducing
agent.
(b) According to the evidence, Ag+(aq) has the greatest tendency to gain electrons
because it reacted with all three metals. Therefore, Ag+(aq) is the strongest oxidizing
agent.
59. (a) Solution:
In the series of reactions, the nitrogen-containing compounds are nitrogen gas, N2(g),
nitrogen monoxide, NO(g), Nitrogen dioxide, NO2(g), nitric acid, HNO3(aq), and nitrous
acid, HNO2(aq).
Use the zero-sum rule to assign oxidation numbers to the nitrogen in each compound:
The sum of the oxidation numbers of all atoms in each of the electrically neutral
compounds N2, NO, NO2, HNO3, and HNO2 must be 0.
For N2: Since the oxidation number of an atom in an element is 0, the oxidation number
of nitrogen in N2(g) is 0.
For NO: The oxidation number of oxygen in its compounds is –2. Therefore, the
nitrogen atom in NO must have an oxidation number of +2 to give a sum of 0.
For NO2: The oxidation number of oxygen in its compounds is –2. Since there are
2 oxygen atoms in NO2, the total charge due to oxygen is –4. Therefore, the nitrogen
atom in NO2 must have an oxidation number of +4 to give a sum of 0.
For HNO3: The oxidation number of hydrogen in its compounds is +1 and the oxidation
number of oxygen in its compounds is –2. Since there is 1 hydrogen atom in HNO3, the
charge due to hydrogen is +1. Since there are 3 oxygen atoms, the total charge due to
oxygen is –6. Therefore, the nitrogen atom in HNO3 must have an oxidation number of
+5 to give a sum of 0.
For HNO2: The oxidation number of hydrogen in its compounds is +1 and the oxidation
number of oxygen in its compounds is –2. Since there is 1 hydrogen atom in HNO2, the
charge due to hydrogen is +1. Since there are 2 oxygen atoms, the total charge due to
oxygen is –4. Therefore, the nitrogen atom in HNO2 must have an oxidation number of
+3 to give a sum of 0.
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-27
Statement: The oxidation number of nitrogen is 0 in N2(g), +2 in NO(g), +4 in NO2(g),
+5 in HNO3(aq), and +3 in HNO2(aq).
(b) In the reaction of N2(g) with gaseous oxygen to from NO(g), the oxidation number of
nitrogen increases from 0 to +2. This means nitrogen in N2(g) is oxidized.
In the reaction of NO(g) with more oxygen gas to produce NO2(g), the oxidation number
of nitrogen increases from +2 to +4. This means that nitrogen in NO(g) is oxidized.
In the reaction of NO2(g) with water vapour to produce HNO3(aq) and HNO2(aq), the
oxidation number of nitrogen increases from +4 to +5 in HNO3, and decreases from +4 to
+3 in HNO2. This means that nitrogen in NO2(g) is oxidized for the production of
HNO3(aq) and is reduced for the production of HNO2(aq).
(c) In the reaction of N2(g) with oxygen to from NO(g), the oxidizing agent that acts on
the nitrogen atom is gaseous oxygen.
In the reaction of NO(g) with more oxygen gas to produce NO2(g), the oxidizing agent
that acts on the nitrogen atom is also gaseous oxygen.
In the reaction of NO2(g) with water vapour to produce HNO3(aq) and HNO2(aq), NO2(g)
itself acts as the oxidizing agent as well as the reducing agent.
(d) Answers may vary. Sample answer: Gardeners often use chemical or organic
fertilizers to add nitrates and nitrites to soil because nitrates and nitrites are synthesized
naturally in the environment: by the oxidation reaction of gaseous nitrogen by gaseous
oxygen in the atmosphere, and by the redox reaction of nitrogen dioxide produced that
produces nitric acid and nitrous acid. Nitrates and nitrites are sources of nitrogen that
plants can take up and use.
60. (a) Answers may vary. Students’ investigation procedure to distinguish magnesius,
aluminum, and nickel could include using solutions of metal salts to identify the metal.
Two of the metal salts chosen should have metal ions below Ni2+(aq) in the redox table so
that Ni(s) cannot displace the metals from the salts. To distinguish between Al(s) and
Mg(s), one of the metal salts could be a salt containing the Al3+(aq) ion. Mg(s) will be
able to displace Al(s) from the solution, whereas Al(s) will show no reaction.
The observation table below would then help to identify the metal:
Reactions between metals and solutions of salts
Metal ion in salt
Ni(s)
Al(s)
Mg(s)
2+
Fe (aq)
no
yes
yes
Al3+(aq)
no
no
yes
Equipment and materials should include chemical safety goggles and lab apron (for
safety), pieces of metal to be identified, well plates, and dropper bottles for containing
solutions of metal salts. The manipulated variable would the metal ion in solution and the
responding variable would be a spontaneous reaction. The procedure should be a series of
numbered steps with any precautions listed at the beginning of the experiment.
(b) Answers may vary. Possible equations from part (a):
3 Fe2+(aq) + 2 Al(s) → 2 Al3+(aq) + 3 Fe(s)
Fe2+(aq) + Mg(s) → Mg2+(aq) + Fe(s)
2 Al3+(aq) + 3 Mg(s) → 3 Mg2+(aq) + 2 Al(s)
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-28
61. (a) According to the redox table, gold is the weakest reducing agent among metals.
This means that gold has the least tendency to lose electrons and be oxidized. Therefore,
pure gold would be the metal most resistant to corrosion or oxidation.
(b) Pure gold would be too expensive to be used as roofing material. Gold is also very
soft, so it would not be durable enough to protect a roof.
Evaluation
62. According to the activity series of metals, the metal ions listed in an increasing order
of being reduced to metals are as follows:
Na+(aq) < Al3+(aq) < Zn2+(aq) < Co2+(aq) < Ni2+(aq) < Cu2+(aq). Therefore, CuCl2(aq) is
the strongest oxidizing agent and NaCl(aq) is the weakest oxidizing agent. The order of
the compounds down the left side of Table 2 should be as follows: CuCl2(aq), NiCl2(aq)
CoCl2(aq) ZnCl2(aq), AlCl3(aq), and NaCl(aq)
Therefore, the compounds are not all in the correct place in Table 2.
63. (a) According to the redox table, the equations relating to the chemicals as reducing
agents are the following:
Co2+(aq) + 2 e– → Co(s)
E°r = −0.28 V
Sn2+(aq) + 2 e– → Sn(s)
2+
–
Ca (aq) + 2 e → Ca(s)
Cu2+(aq) + 2 e– → Cu(s)
2 H+(aq) + 2 e– → H2(g)
E°r = −0.14 V
E°r = −2.76 V
E°r = 0.34 V
E°r = 0 V
SO42–(aq) + 4 H+(aq) + 2 e– → H2SO3(aq) + H2O(l) E°r = 0.20 V
Since Ca(s) occurs lowest in the table, Ca(s) has the greatest tendency to lose electrons
and be oxidized, so it is the strongest reducing agent.
(b) Answers may vary. Students could base their procedure on the prediction in part (a).
If Ca(s) is the strongest reducing agent, the observations should show Ca(s) being the
only chemical that reacts spontaneously with all the solutions containing the metal ions
Co2+(aq), Sn2+(aq), and Cu2+(aq), and H+(g) and SO42–(aq). Safety precautions should
include chemical safety goggles and lab apron because some solutions may be irritants or
toxic. Other precautions should be listed at the beginning of the procedure.
64. When metals react with oxygen dissolved in water, the metals are oxidized to their
metal ions. Therefore, oxygen in water is the oxidizing agent and the metal is the
reducing agent.
From the table of standard reduction potentials in Appendix B7:
O2(g) + 2 H2O(l) + 4 e– → 4 OH–(aq)
E°r = 0.40 V
Ag+(aq) + e– → Ag(s)
E°r = 0.80 V
2+
–
Cd (aq) + 2 e → Cd(s)
E°r = −0.40 V
Al3+(aq) + 3 e– → Al(s)
2+
–
E°r = −1.66 V
E°r = −0.44 V
Fe (aq) + 2 e → Fe(s)
Since the relative positions of the oxidizing agent, O2(g) + 2 H2O(l), and the reducing
agent Ag(s) do not form a downward diagonal to the right on the redox table, Ag(s) will
not react spontaneously with oxygen when dissolved in water.
Therefore, the student is incorrect.
Copyright © 2012 Nelson Education Ltd.
Chapter 9: Oxidation–Reduction Reactions 9-29
Reflect on Your Learning
65. Answers may vary. Sample answer: Before I studied this chapter, I knew that rust was
a form of oxidation, but I thought that oxygen was involved in all oxidation reactions.
The information in this chapter helped me understand oxidation as gaining electrons, and
that those electrons could come from oxygen or from another entity. It also helped me see
that oxidation really only describes half of a reaction—it is always paired with reduction.
This is because when one entity gains electrons, another entity must lose them.
66. Answers may vary. Students should include descriptions of oxidation–reduction
reactions that occur in everyday life, such as corrosion, the transformation of nitrogen to
nitrites and nitrates, and bleaching clothes. They should discuss the potential societal,
economic, and environmental effects of each reaction and explain how an understanding
of the reaction can help make decisions about it. For example, knowing that metal is
oxidized more quickly in the presence of water can help motivate people to dry metal
tools and bicycles before storing them.
67. Answers may vary. Reasons for finding a topic interesting could include recognizing
its relevance in daily life or recognizing a relationship between it and a topic studied
previously in this course or in another course.
68. Answers may vary: Sample answer: Three questions I still have about the information
in this chapter are the following:
(1) How does the relative strength of oxidizing agents change at different temperatures or
pressures?
(2) What redox reactions occur naturally, for examples as part of the carbon or nitrogen
cycles?
(3) What catalysts are used in industrial applications to make common redox reactions
more likely to occur?
I could find answers to the first question by conducting the same set of experiments at
different temperatures and pressures and comparing the results. I could answer the second
question by reading books or online resources about ecology. I could answer the third
question by sending an email to a manufacturing company that uses a particular reaction
and ask about catalysts.
Research
69. Answers may vary. Sample answer: The common oxidation numbers of iron are +2
and +3. According to the formula unit Fe3O4, there are 3 iron atoms and 4 oxygen atoms
in the compound. Since the oxidation number of oxygen is –2, the total contribution due
to oxygen is –8. From the zero-sum rule, the contribution from iron must be +8. To
obtain this number, the oxidation number of two of the iron atoms must be +3, while that
of the remaining iron atom must be +2. Therefore, this compound consists of iron atoms
with different oxidation numbers.
70. (a) To convert Cr6+ to Cr3+ in industrial waste water, the hexavalent chromium must
gain 3 electrons. Therefore, the industrial waste water will need to be treated with a
reducing agent.
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Chapter 9: Oxidation–Reduction Reactions 9-30
(b) Answers may vary. Sample answer: Hexavalent chromium is commonly found in the
form of chromate (CrO42–) and dichromate (Cr2O72–). One way of achieving the
conversion from hexavalent chromium to trivalent chromium is by reduction with sulfur
dioxide (SO2), sodium metabisulfite (Na2S2O5), or other reduced sulfur species:
3 SO2 + 2 H2CrO4 + 3 H2O → Cr2(SO4)3 + 5 H2O
The reaction proceeds most rapidly at an acidic pH (< 3) and is typically carried out in
the presence of sulfuric acid. This method is suitable for treatment of waste water
recovered directly from industrial processes, where the pH can be monitored and
controlled.
Another method involves reduction with ferrous iron:
3 Fe2+ + HCrO4– + 7H+ → 3 Fe3+ + Cr3+ + 4 H2O
The reaction is carried out in the presence of sodium dithionite to prevent precipitation of
Fe2+ before it can react. This method is effective for treating soil and groundwater that
has been contaminated with hexavalent chromium.
71. Answers may vary. Sample answer: In a battery, an electric current is created by a
redox reaction. A battery contains two electrodes—one where oxidation occurs (the
anode) and one where reduction occurs (the cathode); it also contains appropriate
electrolytes (ionic solutions) in contact with each electrode. When the battery is put into a
functioning device, the circuit between the electrodes is completed, the electrons are
allowed to flow, and the redox reaction is able to proceed. An alkaline battery contains a
zinc metal anode, a manganese dioxide cathode, and a potassium hydroxide electrolyte. It
is a common, general-purpose battery used in devices such as flashlights and remote
controls. The chemistry of lithium ion batteries is complex and can vary. One type of
lithium ion battery contains a carbon anode, a cobalt oxide cathode, and an organic
solvent electrolyte in which lithium ions move between electrodes. Lithium ion batteries
are more expensive, but carry a higher charge and have a smaller mass, so they are used
in portable
Advantages and Disadvantages of Alkaline Batteries and Lithium Ion Batteries
Alkaline Batteries
Lithium Ion Batteries
Advantages
relatively
inexpensive
widely available
Disadvantages
Advantages
most are not
rechargeable
contain toxic and
hazardous materials,
so require proper
disposal
rechargeable
Disadvantages
relatively expensive
contain non-toxic
materials
high amount of
charge for their mass
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Chapter 9: Oxidation–Reduction Reactions 9-31
72. Answers may vary. At least six different metals and alloys should be discussed,
including which one the student thinks is the best choice for use in coinage. Sample
answer:
• Copper: Strong, malleable, and slow to corrode. Pure copper was used in coins in the
past, but has been replaced by copper alloys, which are more resistant to wear.
• Nickel: Very strong, malleable, and slow to corrode. However, contact with nickel may
cause an allergic reaction in some people.
• Bronze (copper-tin alloy): Improved hardness compared to pure copper, making it more
resistant to wear.
• Aluminum bronze (copper-aluminum alloy): Improved strength and corrosion resistance
compared to bronze.
• Cupronickel (copper-nickel alloy): Strong, malleable, and corrosion-resistant. Because
of its silvery appearance and desirable properties, it was used as an inexpensive
replacement for silver metal in coins.
• Steel (iron-carbon alloy): Stronger and harder than pure iron. Due to its relatively low
cost, it is being used to replace more expensive coinage metals such as nickel. However,
steel corrodes readily and must be plated with a more corrosion-resistant metal.
Based on the different advantages of each type of metal used in coins, I think aluminumbronze is the best choice. The copper in the alloy would make aluminum-bronze
malleable and slow to corrode, while the aluminum would make the coins light and
strong. Cupronickel also has some of these properties, but the nickel in the alloy may
cause reactions in some people.
73. Answers may vary. Sample answer: One metal that is important to society is gold.
The method of processing raw ore into a purified metal depends on the composition of
the ore. The following flowchart outlines the processing of low-grade ore:
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Chapter 9: Oxidation–Reduction Reactions 9-32
74. (a) In the first reaction, the oxidation numbers of the reactants are +3 for iron, –2 for
oxygen, +1 for hydrogen, and –2 for sulfur. These oxidation numbers are the same after
the reaction; therefore, it is not a redox reaction. In the second reaction, the oxidation
number for sulfur changes from –2 to 0, and the oxidation number for oxygen changes
from 0 to –2 (the oxidation number for iron is +3 and remains the same); therefore, it is a
redox reaction.
(b) The balanced equations are as follows:
Fe2O3(s) + H2O(l) + 3 H2S(g) → Fe2S3(s) + 4 H2O(l)
2 Fe2S3(s) + 3 O2(g) + H2O(l) → 2 Fe2O3(s) + H2O(l) + 6 S(s)
(c) This is an economical way of removing hydrogen sulfide from natural gas because the
elemental sulfur obtained from the reaction can be sold for use in other industries.
(d) The effects of different concentrations of hydrogen sulfide gas on humans may vary,
depending on length of exposure. In general, the unpleasant odour of hydrogen sulfide
gas is perceived at a concentration of 0.2 ppm (parts per million), but may be detectable
at much lower concentrations. At 10 ppm the gas begins to cause eye irritation. The
effects of higher gas concentrations include respiratory tract irritation and eye injury at
50–200 ppm; loss of smell at 100–200 ppm; pulmonary edema at 250–600 ppm; and
unconsciousness, respiratory system paralysis, and death at 500–2000 ppm.
Concentrations in excess of 1000 ppm lead to immediate collapse.
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Chapter 9: Oxidation–Reduction Reactions 9-33