ANSWERS TO END-OF-CHAPTER QUESTIONS

Oxidation half-reaction:
__CH4(g) + __OH – (aq) ⎯→ __CO2(g) + __H2O(l) + __e –
Reduction half-reaction:
__O2(g) + __H2O(l) + __e – ⎯→ __OH – (aq)
Answer:
Oxidation half-reaction:
CH4(g) + 8 OH – (aq) ⎯→ CO2(g) + 6 H2O(l) + 8 e –
Reduction half-reaction:
2 O2(g) + 4 H2O(l) + 8 e – ⎯→ 8 OH – (aq)
Overall reaction:
CH4(g) + 2 O2(g) ⎯→ CO2(g) + 2 H2O(l)
19. The reactions in a hydrogen-fueled solid oxide fuel cell (SOFC) are shown in equations
8.19–8.21. This is the skeleton equation for the oxidation half-reaction if CO, rather than H2,
is the fuel.
a. Balance by adding electrons as needed.
b. Combine the balanced equation for oxidation with that for reduction.
c. Write the overall equation for a carbon monoxide-based SOFC.
Answer:
The reactions in a hydrogen-fueled solid oxide fuel cell (SOFC) are shown in equations
8.19–8.21.
a. CO(g) + O 2– ⎯→ CO2(g) + 2 e –
b. The reduction half reaction is: ½ O2(g) + 2 e – ⎯→ O 2–
Combined, this gives: CO(g) + O 2– + ½ O2(g) + 2 e – ⎯→ CO2(g) + 2 e – + O 2–
c. CO(g) + ½ O2(g) ⎯→ CO2(g)
20. a. Potassium, a Group 1A metal, reacts with H2 to form potassium hydride, KH. Write the
chemical equation for the reaction.
b. Potassium hydride reacts with water to release H2 and form potassium hydroxide. Write
the chemical equation.
c. Offer a possible reason that potassium is not used to store H2 for use in fuel cells.
Answer:
a. K(s) + ½ H2(g) ⎯→ 2 KH(s)
b. KH(s) + H2O(l) ⎯→ H2(g) + KOH(s)
c. Potassium is a highly reactive metal that can react explosively with water. The risks would
outweigh the benefits.
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