Chapter 19 Practice Problems, Review, and Assessment
Section 1 Oxidation and Reduction: Practice Problems
1. Identify each of the following changes as either oxidation or reduction. Recall that e− is the symbol for an electron.
a. I2 + 2e− → 2I−
b. K → K+ + e−
c. Fe2+ → Fe3+ + e−
d. Ag+ + e− → Ag
SOLUTION:
a. reduction
b. oxidation
c. oxidation
d. reduction
ANSWER:
a. reduction
b. oxidation
c. oxidation
d. reduction
2. Identify what is oxidized and what is reduced in the following processes.
a. 2Br− + Cl2 → Br2 + 2Cl−
b. 2Ce + 3Cu2+ → 3Cu + 2Ce3+
c. 2Zn + O2 → 2ZnO
d. 2Na + 2H+ → 2Na+ + H2
SOLUTION:
a. Br− is oxidized, Cl is reduced
b. Ce is oxidized, Cu2+ is reduced
c. Zn is oxidized, O2 is reduced
d. H+ is reduced, Na is oxidized
ANSWER:
a. Br− is oxidized, Cl is reduced
b. Ce is oxidized, Cu2+ is reduced
c. Zn is oxidized, O2 is reduced
d. H+ is reduced, Na is oxidized
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Chapter 19 Practice Problems, Review, and Assessment
3. Identify the oxidizing agent and the reducing agent in the following equation. Explain your answer.
Fe(s) + 2Ag+(aq) → Fe2+(aq) + 2Ag(s)
SOLUTION:
Ag+ is the oxidizing agent, Fe is the reducing agent; Ag+ is reduced, Fe is oxidized
ANSWER:
Ag+ is the oxidizing agent, Fe is the reducing agent; Ag+ is reduced, Fe is oxidized
4. Challenge Identify the oxidizing agent and the reducing agent in each reaction.
a. Mg + I2 → MgI2
b. H2S + Cl2 → S + 2HCl
SOLUTION:
a. I2 is the oxidizing agent, Mg is the reducing agent
b. Cl2 is the oxidizing agent, H2S is the reducing agent
ANSWER:
a. I2 is the oxidizing agent, Mg is the reducing agent
b. Cl2 is the oxidizing agent, H2S is the reducing agent
5. Determine the oxidation number of the boldface element in the following formulas for compounds.
a. NaClO4
b. AlPO4
c. HNO2
SOLUTION:
a. +7
b. +5
c. +3
ANSWER:
a. +7
b. +5
c. +3
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Chapter 19 Practice Problems, Review, and Assessment
6. Determine the oxidation number of the boldface element in the following formulas for ions.
a. NH4+
b. AsO43−
c. CrO42−
SOLUTION:
a. −3
b. +5
c. +6
ANSWER:
a. −3
b. +5
c. +6
7. Determine the oxidation number of nitrogen in each of these molecules or ions.
a. NH3
b. KCN
c. N2H4
SOLUTION:
a. −3
b. −3
c. −2
ANSWER:
a. −3
b. −3
c. −2
8. Challenge Determine the net change of oxidation number of each of the elements in these redox equations.
a. C + O2 → CO2
b. Cl2 + ZnI2 → ZnCl2 + I2
c. CdO + CO → Cd + CO2
SOLUTION:
a. C, + 4; O, −2
b. I, +1; Cl, −1; Zn, no change
c. C, +2; Cd, −2; O, no change
ANSWER:
a. C, + 4; O, −2
b. I, +1; Cl, −1; Zn, no change
c. C, +2; Cd, −2; O, no change
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Chapter 19 Practice Problems, Review, and Assessment
Section 1 Oxidation and Reduction: Review
9. Explain why oxidation and reduction must always occur together.
SOLUTION:
If an atom or ion loses an electron, some other species must gain the electron.
ANSWER:
If an atom or ion loses an electron, some other species must gain the electron.
10. Describe the roles of oxidizing agents and reducing agents in a redox reaction. How is each changed in the reaction?
SOLUTION:
An oxidizing agent causes another species to be oxidized by gaining the electrons from it. A reducing
agent causes another species to be reduced by losing electrons to that element.
ANSWER:
An oxidizing agent causes another species to be oxidized by gaining the electrons from it. A reducing
agent causes another species to be reduced by losing electrons to that element.
11. Write the equation for the reaction of iron metal with hydrobromic acid to form aqueous iron(III) bromide and
hydrogen gas. Determine which element is reduced and which element is oxidized in this reaction.
SOLUTION:
2Fe(s) + 6HBr (aq) → 2FeBr3 (aq) + 3H2(g); Fe is oxidized, H is reduced
ANSWER:
2Fe(s) + 6HBr (aq) → 2FeBr3 (aq) + 3H2(g); Fe is oxidized, H is reduced
12. Determine the oxidation number of the boldface element in these compounds.
a. HNO3
b. Ca3N2
c. Sb2O5
d. CuWO4
SOLUTION:
a. +5
b. −3
c. +5
d. +6
ANSWER:
a. +5
b. −3
c. +5
d. +6
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Chapter 19 Practice Problems, Review, and Assessment
13. Determine the oxidation number of the boldface element in these ions.
a. IO4−
b. MnO4−
c. B 4O72−
d. NH2−
SOLUTION:
a. +7
b. +7
c. +3
d. −3
ANSWER:
a. +7
b. +7
c. +3
d. −3
14. Make and Use Graphs Alkali metals are strong reducing agents. Make a graph showing how the reducing abilities
of the alkali metals would increase or decrease as you move down the family from sodium to francium.
SOLUTION:
In general, as you move down the periodic table within a family, the tendency to lose electrons increases
so the reducing ability increases.
ANSWER:
In general, as you move down the periodic table within a family, the tendency to lose electrons increases
so the reducing ability increases.
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Chapter 19 Practice Problems, Review, and Assessment
Section 2 Balancing Redox Equations: Practice Problems
Use the oxidation-number method to balance these redox equations.
15. HCl + HNO3 → HOCl + NO + H2O
SOLUTION:
3HCl + 2HNO3 → 2HOCl + 2NO + H2O
oxidation number of Cl increases from −1 to +1; oxidation number of N decreases from +5 to +2
ANSWER:
3HCl + 2HNO3 → 2HOCl + 2NO + H2O
16. SnCl4 + Fe → SnCl2 + FeCl3
SOLUTION:
3SnCl4 + 2Fe → 3SnCl2 + 2FeCl3
oxidation number of Fe increases from 0 to +3; oxidation number of Sn decreases from +4 to +2
ANSWER:
3SnCl4 + 2Fe → 3SnCl2 + 2FeCl3
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Chapter 19 Practice Problems, Review, and Assessment
17. NH3(g) + NO2(g) → N2(g) + H2O(l)
SOLUTION:
8NH3(g) + 6NO2(g) → 7N2(g) + 12H2O(l)
oxidation number of N increases from −3 to 0; oxidation number of N decreases from +4 to 0
ANSWER:
8NH3(g) + 6NO2(g) → 7N2(g) + 12H2O(l)
18. Challenge SO2 + Br2 + H2O → HBr + H2SO4
SOLUTION:
SO2 + Br2 + 2H2O → 2HBr + H2SO4
oxidation number of S increases from +4 to +6; oxidation number of Br decreases from 0 to −1
ANSWER:
SO2 + Br2 + 2H2O → 2HBr + H2SO4
19. H2S(g) + NO3−(aq) → S(s) + NO(g) (in acid solution)
SOLUTION:
2H+(aq) + 3H2S(g) + 2NO3−(aq) → 3S(s) + 2NO(g) + 4H2O(l)
oxidation number of S increases from −2 to 0; oxidation number of N decreases from +5 to +2
ANSWER:
2H+(aq) + 3H2S(g) + 2NO3−(aq) → 3S(s) + 2NO(g) + 4H2O(l)
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Chapter 19 Practice Problems, Review, and Assessment
20. Cr2O72−(aq) + I−(aq) → Cr3+(aq) + I2(s) (in acid solution)
SOLUTION:
14H+(aq) + Cr2O72−(aq) + 6I− → 2Cr3+(aq) + 3I2(s) + 7H2O(l)
oxidation number of I increases from −1 to 0; oxidation number of Cr decreases from +6 to +3
ANSWER:
14H+(aq) + Cr2O72−(aq) + 6I− → 2Cr3+(aq) + 3I2(s) + 7H2O(l)
21. Zn + NO3− → Zn2+ + NO2 (in acid solution)
SOLUTION:
Zn + 2NO3− + 4H+ → Zn2+ + 2NO2 + 2H2O
oxidation number of Zn increases from 0 to +2; oxidation number of N decreases from +5 to +4
ANSWER:
Zn + 2NO3− + 4H+ → Zn2+ + 2NO2 + 2H2O
22. Challenge I−(aq) + MnO4−(aq) → I2(s) + MnO2(s) (in basic solution)
SOLUTION:
6I−(aq) + 2MnO4−(aq) + 4H2O(l) → 3I2(s) + 2MnO2(s) + 8OH−(aq)
oxidation number of I increases from −1 to 0; oxidation number of Mn decreases from +7 to +4
ANSWER:
6I−(aq) + 2MnO4−(aq) + 4H2O(l) → 3I2(s) + 2MnO2(s) + 8OH−(aq)
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Chapter 19 Practice Problems, Review, and Assessment
Use the half-reaction method to balance the redox equations. Begin by writing the oxidation and reduction
half-reactions. Leave the balanced equation in ionic form.
23. Cr2O72−(aq) + I−(aq) → Cr3+(aq) + I2(s) (in acid solution)
SOLUTION:
2I−(aq) → I2(s) + 2e−(oxidation)
14H+(aq) + 6e− + Cr2O72−(aq) → 2Cr3+(aq) + 7H2O(l) (reduction)
Multiply oxidation half-reaction by 3 and add to reduction half-reaction
14H+(aq) + 6e− + Cr2O72−(aq) + 6I− (aq) → 3I2(s) + 2Cr3+(aq) + 7H2O(l) + 6e−
14H+(aq) + Cr2O72−(aq) + 6I−(aq) → 3I2(s) + 2Cr3+(aq) + 7H2O(l)
ANSWER:
14H+(aq) + Cr2O72−(aq) + 6I−(aq) → 3I2(s) + 2Cr3+(aq) + 7H2O(l)
24. Mn2+(aq) + BiO3−(aq) → MnO4(aq) + Bi2+(aq) (in acid solution)
SOLUTION:
Mn2+(aq) + 4H2O(l) → MnO4−(aq) + 5e− + 8H+(aq) (oxidation)
BiO3−(aq) + 3e− + 6H+(aq) → Bi2+(aq) + 3H2O(l) (reduction)
Multiply oxidation half-reaction by 3. Multiply reduction half-reaction by 5 and add to oxidation halfreaction.
3Mn2+(aq) + 12H2O(l) + 5BiO3−(aq) + 15e− + 30H+(aq) → 3MnO4−(aq) + 15e− + 24H+(aq) + 5Bi2+(aq) +
15H2O(l)
3Mn2+(aq) + 5BiO3−(aq) + 6H+(aq) → 3MnO4−(aq) + 5Bi2+(aq) + 3H2O(l)
ANSWER:
3Mn2+(aq) + 5BiO3−(aq) + 6H+(aq) → 3MnO4−(aq) + 5Bi2+(aq) + 3H2O(l)
25. Challenge N2O(g) + ClO−(aq) → NO2−(aq) + Cl−(aq) (in basic solution). Hint: Add O and H in the form of OH−
ions and H2O molecules.
SOLUTION:
6OH−(aq) + N2O(g) → 2NO2−(aq) + 4e− + 3H2O(l) (oxidation)
ClO−(aq) + 2e− + H2O(l) → Cl−(aq) + 2OH−(aq) (reduction)
Multiply reduction half-reaction by 2 and add to oxidation half-reaction.
6OH−(aq) + N2O(g) + 2ClO−(aq) + 4e− + 2H2O(l) → 2NO2−(aq) + 4e− + 3H2O(l) + 2Cl−(aq) + 4OH−(aq)
N2O(g) + 2ClO−(aq) + 2OH−(aq) → 2NO2−(aq) + 2Cl−(aq) + H2O(l)
ANSWER:
N2O(g) + 2ClO−(aq) + 2OH−(aq) → 2NO2−(aq) + 2Cl−(aq) + H2O(l)
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Chapter 19 Practice Problems, Review, and Assessment
Section 2 Balancing Redox Equations: Review
26. Explain how changes in oxidation number are related to the electrons transferred in a redox reaction. How are the
changes related to the processes of oxidation and reduction?
SOLUTION:
Because the nucleus (specifically, number of protons) never changes during this type of reaction,
whenever there is a transfer of electrons to or from a chemical species, there is a change in the net
charge of that species. Oxidation increases the oxidation number; reduction reduces it.
ANSWER:
Because the nucleus (specifically, number of protons) never changes during this type of reaction,
whenever there is a transfer of electrons to or from a chemical species, there is a change in the net
charge of that species. Oxidation increases the oxidation number; reduction reduces it.
27. Describe why it is important to know the conditions under which an aqueous redox reaction takes place in order to
balance the ionic equation for the reaction.
SOLUTION:
It is important to know whether H2O and either H+ or OH− ions are available to balance the equation.
ANSWER:
It is important to know whether H2O and either H+ or OH− ions are available to balance the equation.
28. Explain the oxidation-number method of balancing equations.
SOLUTION:
Assign an oxidation number of each element in the equation. Identify the atoms or ions that are oxidized
and reduced. Adjust the coefficients so that the total increase in oxidation number is equal to the total
decrease in oxidation number. Use the conventional method to balance the remainder of the equation.
ANSWER:
Assign an oxidation number of each element in the equation. Identify the atoms or ions that are oxidized
and reduced. Adjust the coefficients so that the total increase in oxidation number is equal to the total
decrease in oxidation number. Use the conventional method to balance the remainder of the equation.
29. State what an oxidation half-reaction shows. What does a reduction half-reaction show?
SOLUTION:
An oxidation half-reaction shows the loss of electrons from an atom or ion that undergoes an increase in
oxidation number. A reduction half-reaction shows the combining of electrons with an atom or ion that
undergoes a decrease in oxidation number.
ANSWER:
An oxidation half-reaction shows the loss of electrons from an atom or ion that undergoes an increase in
oxidation number. A reduction half-reaction shows the combining of electrons with an atom or ion that
undergoes a decrease in oxidation number.
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Chapter 19 Practice Problems, Review, and Assessment
30. Write the oxidation and reduction half-reactions for this redox equation:
Pb(s) + Pd(NO3)2(aq) → Pb(NO3)2(aq) + Pd(s)
SOLUTION:
oxidation: Pb → Pb2+ + 2e−
reduction: Pd2+ + 2e− → Pd
ANSWER:
oxidation: Pb → Pb2+ + 2e−
reduction: Pd2+ + 2e− → Pd
31. Determine The oxidation half-reaction of a redox reaction is Sn2+ → Sn4+ + 2e− and the reduction half-reaction is
Au3+ + 3e− → Au. What minimum numbers of tin(II) ions and gold(III) ions would have to react in order to have no
electrons left over?
SOLUTION:
Sn2+ → Sn4+ + 2e−
Au3+ + 3e− → Au
Three Sn2+ ions; two Au3+ ions
ANSWER:
Three Sn2+ ions; two Au3+ ions
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Chapter 19 Practice Problems, Review, and Assessment
32. Apply Balance the following equations
a. HClO3(aq) → ClO2(g) + HClO4(aq) + H2O(l)
b. H2SeO3(aq) + HClO3(aq) → H2SeO4(aq) + Cl2(g) + H2O(l)
c. Cr2O72−(aq) + Fe2+(aq) → Cr3+(aq) + Fe3+(aq) (in acid solution)
SOLUTION:
a.
3HClO3 → 2ClO2 + HClO4 + H2O
b.
5H2SeO3 + 2HClO3 → 5H2SeO4 + Cl2 + H2O
c. Cr2O72− + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O
ANSWER:
a. 3HClO3 → 2ClO2 + HClO4 + H2O
b. 5H2SeO3 + 2HClO3 → 5H2SeO4 + Cl2 + H2O
c. Cr2O72− + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O
Chapter 19 Assessment
Section 1 Oxidation and Reduction: Mastering Concepts
33. What is the main characteristic of oxidation-reduction reactions?
SOLUTION:
All oxidation–reduction reactions involve the transfer of electrons.
ANSWER:
All oxidation–reduction reactions involve the transfer of electrons.
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Chapter 19 Practice Problems, Review, and Assessment
34. Explain why not all oxidation reactions involve oxygen.
SOLUTION:
The word oxidation originally referred only to reactions that involved oxygen, but today it is defined as the
complete or partial loss of electrons from a reacting substance.
ANSWER:
The word oxidation originally referred only to reactions that involved oxygen, but today it is defined as the
complete or partial loss of electrons from a reacting substance.
35. In terms of electrons, what happens when an atom is oxidized? When an atom is reduced?
SOLUTION:
Electrons are lost; electrons are gained.
ANSWER:
Electrons are lost; electrons are gained.
36. Define oxidation number.
SOLUTION:
It is a number assigned to an atom or ion to indicate its degree of oxidation or reduction.
ANSWER:
It is a number assigned to an atom or ion to indicate its degree of oxidation or reduction.
37. Metals What is the oxidation number of alkaline earth metals in their compounds? Of alkali metals?
SOLUTION:
alkaline earth metals = +2; alkali metals = +1
ANSWER:
alkaline earth metals = +2; alkali metals = +1
38. How does the oxidation number in an oxidation process relate to the number of electrons lost? How does the change
in oxidation number in a reduction process relate to the number of electrons gained?
SOLUTION:
The change in oxidation number equals the number of electrons lost in oxidation, and electrons gained in
reduction.
ANSWER:
The change in oxidation number equals the number of electrons lost in oxidation, and electrons gained in
reduction.
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Chapter 19 Practice Problems, Review, and Assessment
39. What probably accounts for the different forms of copper shown in Figure 9?
SOLUTION:
Copper has different oxidation states in the two forms.
ANSWER:
Copper has different oxidation states in the two forms.
40. Copper and air Copper statues like the Statue of Liberty begin to appear green rather than copper after they have
been exposed to air. In this redox process, copper metal reacts with oxygen to form solid copper oxide, which forms
the green coating. Write the reaction for this redox process and identify what atom is oxidized and what is reduced in
this process.
SOLUTION:
2Cu(s) + O2(g) → 2CuO(s); Cu is oxidized, O is reduced
ANSWER:
2Cu(s) + O2(g) → 2CuO(s); Cu is oxidized, O is reduced
Chapter 19 Assessment
Section 1 Oxidation and Reduction: Mastering Problems
41. Identify the species oxidized and the species reduced in each of these redox equations.
a. 3Br2 + 2Ga → 2GaBr3
b. 2HCl + Zn → ZnCl2 + H2
c. 3Mg + N2 → Mg3N2
SOLUTION:
a. Ga is oxidized, Br2 is reduced.
b. Zn is oxidized, H is reduced.
c. Mg is oxidized, N2 is reduced.
ANSWER:
a. Ga is oxidized, Br2 is reduced.
b. Zn is oxidized, H is reduced.
c. Mg is oxidized, N2 is reduced.
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Chapter 19 Practice Problems, Review, and Assessment
42. Identify the oxidizing agent and the reducing agent in each of these redox equations.
a. N2 + 3H2 → 2NH3
b. 2Na + I2 → 2NaI
SOLUTION:
a. N2 is the oxidizing agent, H2 is the reducing agent.
b. I2 is the oxidizing agent, Na is the reducing agent.
ANSWER:
a. N2 is the oxidizing agent, H2 is the reducing agent.
b. I2 is the oxidizing agent, Na is the reducing agent.
43. What is the reducing agent in the following balanced equation?
8H+ + Sn + 6Cl− + 4NO31− → SnCl62− + 4NO2 + 4H2O
SOLUTION:
Sn
ANSWER:
Sn
44. What is the oxidation number of manganese in KMnO4?
SOLUTION:
+7
ANSWER:
+7
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Chapter 19 Practice Problems, Review, and Assessment
45. Determine the oxidation number of the bold element in these substances and ions.
a. CaCrO4
b. NaHSO4
c. NO2−
d. BrO3−
SOLUTION:
a. +6
b. +6
c. +3
d. +5
ANSWER:
a. +6
b. +6
c. +3
d. +5
46. Identify each of these half-reactions as either oxidation or reduction.
a. Al → Al3+ + 3e−
b. Cu2+ + e− → Cu+
SOLUTION:
a. oxidation
b. reduction
ANSWER:
a. oxidation
b. reduction
47. Which of these equations does not represent a redox reaction? Explain your answer.
a. LiOH + HNO3 → LiNO3 + H2O
b. MgI2 + Br2 → MgBr2 + I2
SOLUTION:
Choice a is not redox because none of the atoms or ions in the reaction undergoes a change in oxidation
number.
ANSWER:
Choice a is not redox because none of the atoms or ions in the reaction undergoes a change in oxidation
number.
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Chapter 19 Practice Problems, Review, and Assessment
48. Determine the oxidation number of nitrogen in each of these molecules or ions.
a. NO3−
b. N2O
c. NF3
SOLUTION:
a. +5
b. +1
c. +3
ANSWER:
a. +5
b. +1
c. +3
49. Determine the oxidation number of each element in these compounds or ions.
a. Au2(SeO4)3 (gold(III) selenate)
b. Ni(CN)2 (nickel(II) cyanide)
SOLUTION:
a. Au, +3; Se, +6; O, −2
b. Ni, +2; C, +2; N, −3
ANSWER:
a. Au, +3; Se, +6; O, −2
b. Ni, +2; C, +2; N, −3
50. Explain how the sulfite ion (SO32−) differs from sulfur trioxide (SO3), shown in Figure 10.
SOLUTION:
SO32− is a polyatomic ion and the oxidation number of sulfur is +4. SO3 is a compound and the oxidation
number of S in this compound is +6.
ANSWER:
SO32− is a polyatomic ion and the oxidation number of sulfur is +4. SO3 is a compound and the oxidation
number of S in this compound is +6.
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Chapter 19 Practice Problems, Review, and Assessment
Chapter 19 Assessment
Section 2 Balancing Redox Equations: Mastering Concepts
51. Compare and contrast balancing redox equations in acidic and basic solutions.
SOLUTION:
In a redox reaction that takes place in an acidic solution, H+ and H2O can participate in the reaction as
either reactants or products. In a basic solution, a redox reaction may involve OH− and H2O as either
reactants or products.
ANSWER:
In a redox reaction that takes place in an acidic solution, H+ and H2O can participate in the reaction as
either reactants or products. In a basic solution, a redox reaction may involve OH− and H2O as either
reactants or products.
52. Explain why writing hydrogen ions as H+ in redox reactions represents a simplification and not how they actually
exist.
SOLUTION:
In aqueous solution, hydrogen ions combine with water in their hydrated form, the hydronium ions (H3O+)
and are never present as H+. However, they are sometimes shown as H+ to simplify the chemical equation
that is written.
ANSWER:
In aqueous solution, hydrogen ions combine with water in their hydrated form, the hydronium ions (H3O+)
and are never present as H+. However, they are sometimes shown as H+ to simplify the chemical equation
that is written.
53. Before you attempt to balance the equation for a redox reaction, why do you need to know whether the reaction takes
place in acidic or basic solution?
SOLUTION:
The type of solution determines whether H+ or OH− ions are available to balance the redox equation.
ANSWER:
The type of solution determines whether H+ or OH− ions are available to balance the redox equation.
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Chapter 19 Practice Problems, Review, and Assessment
54. Explain what a spectator ion is.
SOLUTION:
A spectator ion is one that is present in the same stoichiometry on both sides of a redox reaction
equation. Spectator ions are not changed during a reaction, so they can be eliminated from the equation.
ANSWER:
A spectator ion is one that is present in the same stoichiometry on both sides of a redox reaction
equation. Spectator ions are not changed during a reaction, so they can be eliminated from the equation.
55. Define the term species as it is used in describing redox reactions.
SOLUTION:
A species is any kind of chemical unit involved in the redox process. A species can be an ion, molecule, or
a free atom.
ANSWER:
A species is any kind of chemical unit involved in the redox process. A species can be an ion, molecule, or
a free atom.
56. How can you tell that the equation below is not balanced?
Fe(s) + Ag+(aq) → Fe2+(aq) + Ag(s)
SOLUTION:
The total charge on the left-hand side does not equal the total charge on the right-hand side.
ANSWER:
The total charge on the left-hand side does not equal the total charge on the right-hand side.
57. Does the following equation represent a reduction or an oxidation process? Explain your answer.
Zn2+ + 2e− → Zn
SOLUTION:
reduction; Electrons are gained and the oxidation number for Zn decreases.
ANSWER:
reduction; Electrons are gained and the oxidation number for Zn decreases.
58. Describe what is happening to electrons in each half-reaction of a redox process.
SOLUTION:
Electrons are accepted by a species during the reduction half-reaction, and electrons are lost from species
during an oxidation half-reaction.
ANSWER:
Electrons are accepted by a species during the reduction half-reaction, and electrons are lost from species
during an oxidation half-reaction.
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Chapter 19 Practice Problems, Review, and Assessment
Chapter 19 Assessment
Section 2 Balancing Redox Equations: Mastering Problems
59. Use the oxidation-number method to balance these redox equations.
a. Cl2 + NaOH → NaCl + HOCl
b. HBrO3 → Br2 + H2O + O2
SOLUTION:
a.
Cl2 + NaOH → NaCl + HOCl
b.
4HBrO3 → 2Br2 + 2H2O + 5O2
ANSWER:
a. Cl2 + NaOH → NaCl + HOCl
b. 4HBrO3 → 2Br2 + 2H2O + 5O2
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Chapter 19 Practice Problems, Review, and Assessment
60. Balance these net ionic equations for redox reactions.
a. Au3+(aq) + I−(aq) → Au(s) + I2(s)
b. Ce4+(aq) + Sn2+(aq) → Ce3+(aq) + Sn4+(aq)
SOLUTION:
a.
2Au3+(aq) + 6I−(aq) → 2Au(s) + 3I2(s)
b.
2Ce4+ + Sn2+ → Sn4+ + 2Ce3+
ANSWER:
a. 2Au3+(aq) + 6I−(aq) → 2Au(s) + 3I2(s)
b. 2Ce4+ + Sn2+ → Sn4+ + 2Ce3+
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Chapter 19 Practice Problems, Review, and Assessment
61. Use the oxidation-number method to balance the following ionic redox equations.
a. Al + I2 → Al3+ + I2
b. MnO2 + Br2 → Mn2+ + Br2
SOLUTION:
a.
2Al + 3I2 → 2Al3+ + 6I−
b.
MnO2 + 2Br− + 4H+ → Mn2+ + Br2 + 2H2O
ANSWER:
a. 2Al + 3I2 → 2Al3+ + 6I−
b. MnO2 + 2Br− + 4H+ → Mn2+ + Br2 + 2H2O
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Chapter 19 Practice Problems, Review, and Assessment
62. Use the oxidation-number method to balance these redox equations.
a. PbS + O2 → PbO + SO2
b. NaWO3 + NaOH + O2 → Na2WO4 + H2O
c. NH3 + CuO → Cu + N2 + H2O
d. Al2O3 + C + Cl2 → AlCl3 + CO
SOLUTION:
a.
2PbS + 3O2 → 2PbO + 2SO2
b.
4NaWO3 + O2 + 4NaOH → 4Na2WO4 + 2H2O
c.
2NH3 + 3CuO → 3Cu + N2 + 3H2O
d.
Al2O3 + 3C + 3Cl2 → 2AlCl3 + 3CO
ANSWER:
a. 2PbS + 3O2 → 2PbO + 2SO2
b. 4NaWO3 + O2 + 4NaOH → 4Na2WO4 + 2H2O
c. 2NH3 + 3CuO → 3Cu + N2 + 3H2O
d. Al2O3 + 3C + 3Cl2 → 2AlCl3 + 3CO
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Chapter 19 Practice Problems, Review, and Assessment
63. Sapphire The color of some gems is caused by the exchange of electrons between transition metal ions with different
oxidation states. Sapphire is mostly aluminum oxide, but it contains small amounts of Fe2+ and Ti4+. The color of
sapphire results from an electron transfer from Fe2+ to Ti4+. Based on Figure 11, draw the reaction that occurs
resulting in the mineral on the right. What are the oxidizing and reducing agents?
SOLUTION:
Fe2+ + Ti4+ → Fe3+ + Ti3−; Fe2+ is the reducing agent, Ti4+ is the oxidizing agent
ANSWER:
Fe2+ + Ti4+ → Fe3+ + Ti3−; Fe2+ is the reducing agent, Ti4+ is the oxidizing agent
64. Write the oxidation and reduction half-reactions represented in each of these redox equations. Write the half-reactions
in net ionic form if they occur in aqueous solution.
a. PbO(s) + NH3(g) → N2(g) + H2O(l) + Pb(s)
b. I2(s) + Na2S2O3(aq) → Na2S2O4(aq) + NaI(aq)
c. Sn(s) + 2HCl(aq) → SnCl2(aq) + H2(g)
SOLUTION:
a. NH3(g) → N2(g) + 3e− oxidation
PbO(s) + 2e− → Pb(s) reduction
b. I2(s) + 2e− → 2I−(aq) reduction
S2O32−(aq) → S2O42−(aq) + e− oxidation
c. Sn(s) → Sn2+(aq) + 2e− oxidation
H+(aq) + e− → H2(g) reduction
ANSWER:
a. NH3(g) → N2(g) + 3e− oxidation
PbO(s) + 2e− → Pb(s) reduction
b. I2(s) + 2e− → 2I−(aq) reduction
S2O32−(aq) → S2O42−(aq) + e− oxidation
c. Sn(s) → Sn2+(aq) + 2e− oxidation
H+(aq) + e− → H2(g) reduction
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Chapter 19 Practice Problems, Review, and Assessment
65. Write the two half-reactions that make up the following balanced redox reaction.
3H2C2O4 + 2HAsO2 → 6CO2 + 2As + 4H2O
SOLUTION:
H2C2O4 → 2CO2 + 2H+ + 2e−
HAsO2 + 3H+ + 3e− → As + 2H2O
ANSWER:
H2C2O4 → 2CO2 + 2H+ + 2e−
HAsO2 + 3H+ + 3e− → As + 2H2O
66. Label each half-reaction as a reduction or an oxidation
a. Fe2+(aq) → Fe3+(aq) + e2
b. MnO4− + 5e− + 8H+ → Mn2+ + 4H2O
c. 2H1 + 2e− → H2
d. F2 → 2F− + 2e−
SOLUTION:
a. oxidation
b. reduction
c. reduction
d. oxidation
ANSWER:
a. oxidation
b. reduction
c. reduction
d. oxidation
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Chapter 19 Practice Problems, Review, and Assessment
67. Copper When solid copper pieces are put into a solution of silver nitrate, as shown in Figure 12, silver metal
appears and blue copper(II) nitrate forms. Write the corresponding chemical equation without balancing it. Next,
determine the oxidation state of each element in the equation. Write the two half-reactions, labeling which is oxidation
and which is reduction. Finally, write a balanced equation for the reaction.
SOLUTION:
Unbalanced: AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + Ag(s)
Reactant oxidation states: Ag, +1; N, +5; O, −2; Cu, 0
Product oxidation states: Ag, 0; N, +5; O, −2; Cu, +2
Oxidation half-reaction: Cu → Cu2+ + 2e−
Reduction half-reaction: e− + Ag+ → Ag
2AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2Ag(s)
ANSWER:
Unbalanced: AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + Ag(s)
Reactant oxidation states: Ag, +1; N, +5; O, −2; Cu, 0
Product oxidation states: Ag, 0; N, +5; O, −2; Cu, +2
Oxidation half-reaction: Cu → Cu2+ + 2e−
Reduction half-reaction: e− + Ag+ → Ag
2AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2Ag(s)
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Chapter 19 Practice Problems, Review, and Assessment
68. Use the oxidation-number method to balance these ionic redox equations.
a. MoCl5 + S2− → MoS2 + Cl− + S
b. TiCl62− + Zn → Ti3+ + Cl− + Zn2+
SOLUTION:
a.
2MoCl5 + 5S2− → 2MoS2 + 10Cl− + S
b.
2TiCl62− + Zn → 2Ti3+ + 12Cl− + Zn2+
ANSWER:
a. 2MoCl5 + 5S2− → 2MoS2 + 10Cl− + S
b. 2TiCl62− + Zn → 2Ti3+ + 12Cl− + Zn2+
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Chapter 19 Practice Problems, Review, and Assessment
69. Use the half-reaction method to balance these equations for redox reactions. Add water molecules and hydrogen ions
(in acid solutions) or hydroxide ions (in basic solutions) as needed.
a. NH3(g) + NO2(g) → N2(g) + H2O(l) (in acid solution)
b. Br2 → Br− + BrO3− (in basic solution)
SOLUTION:
a.
b.
ANSWER:
a. 8NH3 + 6NO2 → 7N2 + 12H2O
b. 3Br2 + 6OH−→ 5Br− + BrO3− + 3H2O
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Chapter 19 Practice Problems, Review, and Assessment
70. Balance the following redox chemical equation. Rewrite the equation in full ionic form, then derive the net ionic
equation and balance by the half-reaction method. Give the final answer as it is shown below but with the balancing
coefficients.
KMnO4(aq) + FeSO4(aq) + H2SO4(aq) → Fe2(SO4)3(aq) + MnSO4(aq) + K2SO4(aq) + H2O(l)
SOLUTION:
Full: K+(aq) + MnO4−(aq) + Fe2+(aq) + SO42−(aq) + 2H+(aq) + SO42−(aq) → 2Fe3+(aq) + 3SO42−(aq) +
Mn2+(aq) + SO42−(aq) + 2K+(aq) + SO42 −(aq)
Net: MnO4−(aq) + Fe2+(aq) + H+(aq) → 2Fe3+(aq) + Mn2+(aq)
10Fe2+ → 10Fe3+ + 10e−
2MnO4− + 10e− + 16H+ → 2Mn2+ + 8H2O
10Fe2+ + 2MnO4− + 10e−+16H+ → 10Fe3+ + 10e− + 2Mn2+ + 8H2O
10Fe2+ + 2MnO4− + 16H+ → 10Fe3+ + 2Mn2+ + 8H2O
2KMnO4(aq) + 10FeSO4(aq) + 8H2SO4(aq) → 5Fe2(SO4)3(aq) + 2MnSO4(aq) + K2SO4(aq) + 8H2O(l)
ANSWER:
2KMnO4(aq) + 10FeSO4(aq) + 8H2SO4(aq) → 5Fe2(SO4)3(aq) + 2MnSO4(aq) + K2SO4(aq) + 8H2O(l)
71. Use the oxidation-number method to balance these redox equations.
a. CO + I2O5 → I2 + 5CO2
b. SO2 + Br2 + H2O → HBr + H2SO4
c. Cu + NO3− → Cu2+ + NO (in acidic solution)
d. Zn + NO3− → Zn2+ + NO2 (in acidic solution)
e . Al + OH− + H2O → H2 + AlO2−
SOLUTION:
a.
5CO + I2O5 → I2 + 5CO2
b.
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Chapter 19 Practice Problems, Review, and Assessment
SO2 + Br2 + 2H2O → 2HBr + H2SO4
c.
3Cu + 2NO3− + 8H+ → 3Cu2+ + 2NO + 4H2O
d.
Zn + 2NO3− + 4H+ → Zn2+ + 2NO2 + 2H2O
e.
2Al + 2OH− + 2H2O → 3H2 + 2AlO2−
ANSWER:
a. 5CO + I2O5 → I2 + 5CO2
b. SO2 + Br2 + 2H2O → 2HBr + H2SO4
c. 3Cu + 2NO3− + 8H+ → 3Cu2+ + 2NO + 4H2O
d. Zn + 2NO3− + 4H+ → Zn2+ + 2NO2 + 2H2O
e. 2Al + 2OH− + 2H2O → 3H2 + 2AlO2−
72. Use the half-reaction method to balance these equations. Add water molecules and hydrogen ions (in acid solutions)
or hydroxide ions (in basic solutions) as needed. Keep balanced equations in net ionic form.
a. Cl−(aq) + NO3−(aq) → ClO−(aq) + NO(g) (in acid solution)
b. IO3−(aq) + Br−(aq) → Br2(l) + IBr(s) (in acid solution)
c. I2(s) + Na2S2O3(aq) → Na2S2O4(aq) + NaI(aq) (in acid solution)
SOLUTION:
a.
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Chapter 19 Practice Problems, Review, and Assessment
b.
c.
ANSWER:
a. 3Cl− + 2NO3− + 2H+→ 3ClO− + 2NO + H2O
b. 6H+ + 5Br− + IO3−→ 2Br2 + IBr + 3H2O
c. S2O32− + H2O + I2 → S2O42− + 2H+ + 2I−
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Chapter 19 Practice Problems, Review, and Assessment
Chapter 19 Assessment: Mixed Review
73. Determine the oxidation number of the bold element in each of the following examples.
a. OF2
b. UO22+
c. RuO4
d. Fe2O3
SOLUTION:
a. O, +2
b. U, +6
c. Ru, +8
d. Fe, +3
ANSWER:
a. O, +2
b. U, +6
c. Ru, +8
d. Fe, +3
74. Identify each of the following changes as either oxidation or reduction.
a. 2Cl− → Cl2 + 2e−
b. Na → Na+ + e−
c. Ca2+ + 2e−→ 2Ca
d. O2 + 4e−→ 2O2−
SOLUTION:
a. oxidation
b. oxidation
c. reduction
d. reduction
ANSWER:
a. oxidation
b. oxidation
c. reduction
d. reduction
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Chapter 19 Practice Problems, Review, and Assessment
75. Use the rules for assigning oxidation numbers to complete Table 7.
SOLUTION:
Oxidation Number Assignment
Element
K in KBr
Br in KBr
Cl in Cl2
K in KCl
Cl in KCl
Br in Br2
Oxidation
number
+1
–1
0
+1
–1
0
Rule
7
8
1
7
8
1
ANSWER:
rule 7, −1, 0, +1, rule 8, rule 1
Oxidation Number Assignment
Element
K in KBr
Br in KBr
Cl in Cl2
K in KCl
Cl in KCl
Br in Br2
Oxidation
number
+1
–1
0
+1
–1
0
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Rule
7
8
1
7
8
1
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Chapter 19 Practice Problems, Review, and Assessment
76. Identify the reducing agents in these equations.
a. 4NH3 + 5O2 → 4NO + 6H2O
b. Na2SO4 + 4C → Na2S + 4CO
c. 4IrF5 + Ir → 5IrF4
SOLUTION:
a.
b.
c.
ANSWER:
a. NH3
b. C
c. Ir
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Chapter 19 Practice Problems, Review, and Assessment
77. Write a balanced ionic redox equation using the following pairs of redox half-reactions.
a. Fe → Fe2+ + 2e−
Te2+ + 2e−→ Te
b. IO4− + 2e−→ IO3−
Al → Al3+ + 3e− (in acid solution)
c. I2 + 2e−→ 2I−
N2O → 2NO3− + 8e− (in acid solution)
SOLUTION:
a.
b.
c.
ANSWER:
a. Fe + Te2+ → Fe2+ + Te
b. 3IO4− + 2Al + 6H+→ 3IO3− + 2Al3+ + 3H2O
c. 4I2 + N2O + 5H2O → 8I− + 2NO3− + 10H+
78. What probably accounts for the different forms of chromium shown in Figure 13?
SOLUTION:
Chromium has different oxidation states in the two forms.
ANSWER:
Chromium has different oxidation states in the two forms.
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Chapter 19 Practice Problems, Review, and Assessment
79. Balance these ionic redox equations by any method.
a. Sb3+ + MnO4− → SbO43− + Mn2+ (in acid solution)
b. N2O + ClO− → Cl− + NO2− (in basic solution)
SOLUTION:
a.
5Sb3+ + 2MnO4− + 12H2O → 5SbO43− + 2Mn2+ + 24H+
b. Oxidation: N2O + 6OH− → 2NO2− + 3H2O + 4e−
Reduction: H2O + ClO− + 2e− → Cl− + 2OH−
2H2O + 2ClO− + 4e− → 2Cl− + 4OH−
N2O + 6OH− + 2H2O + 2ClO− + 4e− → 2NO2− + 3H2O + 4e− + 2Cl− + 4OH−
N2O + 2OH− + 2ClO− → 2NO2− + H2O + 2Cl−
ANSWER:
a. 5Sb3+ + 2MnO4− + 12H2O → 5SbO43− + 2Mn2+ + 24H+
b. N2O + 2OH− + 2ClO− → 2NO2− + H2O + 2Cl−
80. Gemstones Rubies are gemstones made up mainly of aluminum oxide. Their red color comes from a small amount of
chromium(III) ions replacing some of the aluminum ions. Write the formula for aluminum oxide and show the reaction
in which an aluminum ion is replaced with a chromium ion. Is this a redox reaction?
SOLUTION:
Al2O3; Al2O3 + 2Cr3+ → Cr2O3 + 2Al3+; it is not a redox reaction—the oxidation numbers stay the same
ANSWER:
Al2O3; Al2O3 + 2Cr3+ → Cr2O3 + 2Al3+; it is not a redox reaction—the oxidation numbers stay the same
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Chapter 19 Practice Problems, Review, and Assessment
81. Balance these ionic redox equations by any method.
a. Mg + Fe3+→ Mg2+ + Fe
b. ClO3− + SO2 → Cl− + SO42− (in acid solution)
SOLUTION:
a.
3Mg + 2Fe3+ → 3Mg2+ + 2Fe
b.
3SO2 + ClO3− + 3H2O → 3SO42− + Cl− + 6H+
ANSWER:
a. 3Mg + 2Fe3+ → 3Mg2+ + 2Fe
b. 3SO2 + ClO3− + 3H2O → 3SO42− + Cl− + 6H+
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Chapter 19 Practice Problems, Review, and Assessment
82. Balance these redox equations by any method.
a. P + H2O + HNO3 → H3PO4 + NO
b. KClO3 + HCl → Cl2 + ClO2 + H2O + KCl
SOLUTION:
a.
3P + 2H2O + 5HNO3 → 3H3PO4 + 5NO
b.
2KClO3 + 4HCl → Cl2 + 2ClO2 + 2H2O + 2KCl
ANSWER:
a. 3P + 2H2O + 5HNO3 → 3H3PO4 + 5NO
b. 2KClO3 + 4HCl → Cl2 + 2ClO2 + 2H2O + 2KCl
Chapter 19 Assessment: Think Critically
83. Apply The following equations show redox reactions that are sometimes used in the laboratory to generate pure
nitrogen gas and pure dinitrogen monoxide gas (nitrous oxide, N2O).
NH4NO2(s) → N2(g) + 2H2O(l)
NH4NO3(s) → N2O(g) + 2H2O(l)
a. Determine the oxidation number of each element in the two equations and then make diagrams showing the
changes in oxidation numbers that occur in each reaction.
b. Identify the atom that is oxidized and the atom that is reduced in each of the two reactions.
c. Identify the oxidizing and reducing agents in each of the two reactions.
d. Write a sentence telling how the electron transfer taking place in these two reactions differs from that taking place
here.
2AgNO3 + Zn → Zn(NO3)2 + 2Ag
SOLUTION:
a.
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Chapter 19 Practice Problems, Review, and Assessment
b. N−3 to N2 loses 3 e− (oxidized); N+3 to N2 gains 3 e− (reduced)
N−3 to N+1 loses 4 e− (oxidized); N+5 to N+1 gains 4 e− (reduced)
c. NO2− and NO3− (oxidizing agents); NH4+ (reducing agent)
d. In the first two reactions, nitrogen is both oxidized and reduced. The third reaction involves redox
between two different elements.
ANSWER:
a.
b. N−3 to N2 loses 3 e− (oxidized); N+3 to N2 gains 3 e− (reduced)
N−3 to N+1 loses 4 e− (oxidized); N+5 to N+1 gains 4 e− (reduced)
c. NO2− and NO3− (oxidizing agents); NH4+ (reducing agent)
d. In the first two reactions, nitrogen is both oxidized and reduced. The third reaction involves redox
between two different elements
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Chapter 19 Practice Problems, Review, and Assessment
84. Analyze Examine the net ionic equation below for the reaction that occurs when the thiosulfate ion (S2O32−) is
oxidized to the tetrathionate ion (S4O62−). Balance the equation using the half-reaction method. Figure 14 will help
you to determine the oxidation numbers to use.
S2O32− + I2 → I− + S4O62− (in acid solution)
SOLUTION:
2S2O32− + I2 → 2I− + S4O62− (in acid solution)
ANSWER:
2S2O32− + I2 → 2I− + S4O62− (in acid solution)
85. Predict Consider the fact that all of the following are stable compounds. What can you infer about the oxidation state
of phosphorus in its compounds?
PH3, PCl3, P2H4, PCl5, H3PO4, Na3PO3
SOLUTION:
Phosphorus has several oxidation states (−3, −2, +3, +5) that make phosphorus very flexible when
combining with nonmetals.
ANSWER:
Phosphorus has several oxidation states (−3, −2, +3, +5) that make phosphorus very flexible when
combining with nonmetals.
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Chapter 19 Practice Problems, Review, and Assessment
86. Solve Potassium permanganate oxidizes chloride ions to chlorine gas. Balance the equation for this redox reaction
taking place in acid solution.
SOLUTION:
2MnO4− (aq) + 10Cl− (aq) + 16H+ (aq) → 2Mn2+ (aq) + 5Cl2 (aq) + 8H2O (l)
ANSWER:
2MnO4− (aq) + 10Cl− (aq) + 16H+ (aq) → 2Mn2+ (aq) + 5Cl2 (aq) + 8H2O (l)
87. In the half reaction NO3− → NH4+, should electrons be added to the side on which NO3− is found or on which NH4+
is found? Add the correct number of electrons to the side on which they are needed and rewrite the equation.
SOLUTION:
The oxidation state of N is reduced from +5 to −3; N must gain 8 electrons.
8 e− to the left side; NO3− + 8e− → NH4+
ANSWER:
The oxidation state of N is reduced from +5 to −3; N must gain 8 electrons.
8 e− to the left side; NO3− + 8e− → NH4+
88. The redox reaction between dichromate ion and iodide ion in acid solution is shown in Figure 15. Use the halfreaction method to balance the equation for this redox reaction.
SOLUTION:
2I−→ I2 + 2e−
6I−→ 3I2 + 6e−
Cr2O72− + 14H+ + 6e−→ 2Cr3+ + 7H2O
6I−(aq) + 14H+(aq) + Cr2O72−(aq) → 3I2(s) + 7H2O(l) + 2Cr3+(aq)
ANSWER:
6I−(aq) + 14H+(aq) + Cr2O72−(aq) → 3I2(s) + 7H2O(l) + 2Cr3+(aq)
Chapter 19 Assessment: Challenge Problem
89. For each reaction described, write the corresponding chemical equation without putting coefficients to balance it.
Next, determine the oxidation state of each element in the equation. Then write the two half-reactions, labeling which
is oxidation and which is reduction. Finally, write a balanced equation for the reaction.
a. Solid mercuric oxide is put into a test tube and gently heated. Liquid mercury forms on the sides and in the bottom
of the tube and oxygen gas bubbles out from the test tube.
b. Solid copper pieces are put into a solution of silver nitrate. Silver metal appears and blue copper(II) nitrate forms in
the solution.
SOLUTION:
a.
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Chapter 19 Practice Problems, Review, and Assessment
Chemical Equation: HgO(s) → O2(g) + Hg(l)
Oxidation States: +2 −2 0 0
Half Reactions:
Oxidation
O−2 → O2 + 2e−
Reduction Hg+2 + 2e−→ Hg
Balanced chemical equation: 2HgO(s) → O2(g) + 2Hg(l)
b. Chemical Equation:
AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + Ag(s)
Oxidation States:
+1 +5 −2 0 +2 +5 −2 0
Half Reactions:
Oxidation
Cu0 → Cu2+ + 2e−
Reduction e− + Ag+→ Ag
Balanced chemical equation: 2AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2Ag(s)
ANSWER:
a. Chemical Equation: HgO(s) → O2(g) + Hg(l)
Oxidation States: +2 −2 0 0
Half Reactions:
Oxidation
O−2 → O2 + 2e−
Reduction Hg+2 + 2e−→ Hg
Balanced chemical equation: 2HgO(s) → O2(g) + 2Hg(l)
b. Chemical Equation:
AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + Ag(s)
Oxidation States:
+1 +5 −2 0 +2 +5 −2 0
Half Reactions:
Oxidation
Cu0 → Cu2+ + 2e−
Reduction e− + Ag+→ Ag
Balanced chemical equation: 2AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2Ag(s)
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Chapter 19 Practice Problems, Review, and Assessment
Chapter 19 Assessment: Cumulative Review
90. A gaseous sample occupies 32.4 mL at −23ºC and 0.75 atm. What volume will it occupy at STP?
SOLUTION:
ANSWER:
V2 = 27 mL
91. When iron(III) chloride reacts in an atmosphere of pure oxygen the following occurs:
4FeCl3(s) + 3O2(g) → 2Fe2O3(s) + 6Cl2(g)
If 45.0 g of iron(III) chloride reacts and 20.5 g of iron(III) oxide is recovered, determine the percent yield.
SOLUTION:
ANSWER:
percent yield = 92.5%
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Chapter 19 Practice Problems, Review, and Assessment
Chapter 19 Assessment: Writing in Chemistry
92. Steel Research the role of oxidation-reduction reactions in the manufacture of steel. Write a summary of your
findings, including appropriate diagrams and equations representing the reactions.
SOLUTION:
Student answers will likely include descriptions and diagrams of some or all of the following: The chief
ores of iron are its oxides, hematite (Fe2O3) and magnetite (Fe3O4), and its carbonate (FeCO3). Most
commonly, iron ores are reduced in a blast furnace. In the blast furnace, the important reactions are the
oxidation of coke to carbon monoxide (2C(s) + O2(g) → 2CO(g)) and the reduction of iron ore by carbon
monoxide, which usually occurs in steps.
CO(g) + 3Fe2O3(s) → 2Fe3O4(s) + CO2(g)
Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g)
FeO(s) + CO(g)Fe(l) + CO2(g)
ANSWER:
Student answers will likely include descriptions and diagrams of some or all of the following: The chief
ores of iron are its oxides, hematite (Fe2O3) and magnetite (Fe3O4), and its carbonate (FeCO3). Most
commonly, iron ores are reduced in a blast furnace. In the blast furnace, the important reactions are the
oxidation of coke to carbon monoxide (2C(s) + O2(g) → 2CO(g)) and the reduction of iron ore by carbon
monoxide, which usually occurs in steps.
CO(g) + 3Fe2O3(s) → 2Fe3O4(s) + CO2(g)
Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g)
FeO(s) + CO(g)Fe(l) + CO2(g)
93. Silverware Practice your technical writing skills by writing (in your own words) a procedure for cleaning tarnished
silverware by a redox chemical process. Be sure to include background information describing the process as well as
logical steps that would enable anyone to accomplish the task.
SOLUTION:
Answers will vary, but students should create a logical procedure based on the concepts learned in the
MiniLab for this chapter.
ANSWER:
Answers will vary, but students should create a logical procedure based on the concepts learned in the
MiniLab for this chapter.
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Chapter 19 Practice Problems, Review, and Assessment
94. Copper was a very useful metal even before iron, silver, and gold metals were extracted and used from their ores and
used as tools, utensils, jewelry, and artwork. Copper was smelted by heating copper ores with charcoal to high
temperatures as early as 8000 years ago. Thousands of pieces of scrap copper have been unearthed in Virginia,
where in the 1600’s the colonists may have traded this material for food. Compare and contrast the processing and
use of copper in those older civilizations and today.
SOLUTION:
Answers will vary.
ANSWER:
Answers will vary.
Chapter 19 Assessment: Document-Based Questions
Glazes The formation of color in ceramic glazes can be influenced by the firing conditions. Metal ions like
copper that have more than one oxidation state can impart different colors to a glaze. In an oxidative firing,
plenty of oxygen is allowed in the kiln, and copper ions present will make the glaze a green to blue color.
Under reducing conditions, oxygen is limited and carbon dioxide is abundant. Copper ions in the glaze
provide a reddish color.
Data obtained from: Denio, Allen A. 2001. The joy of color in ceramic glazes with the help of redox chemistry. Journal of Chemical Education.
78 No 10.
95. Write the equation for what has occurred to pottery shown in Figure 16.
SOLUTION:
Cu2+ + 2e2 → Cu; Cu2+ + e− → Cu+
ANSWER:
Cu2+ + 2e2 → Cu; Cu2+ + e− → Cu+
96. Based on the color of the pottery, what is the oxidation state of the copper that is reduced? Oxidized?
SOLUTION:
Red, reduced: 0 or +1; green, oxidized: Cu+2
ANSWER:
Red, reduced: 0 or +1; green, oxidized: Cu+2
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