ANSWERS TO END-OF-CHAPTER QUESTIONS

ANSWERS TO
END-OF-CHAPTER QUESTIONS
CHAPTER 8: ENERGY FROM ELECTRON TRANSFER
Emphasizing Essentials
1. a. Define the terms oxidation and reduction.
b. Why must these processes take place together?
Answer:
a. Oxidation is a process in which an atom, ion, or molecule loses one or more electrons.
Reduction is a process in which an atom, ion, or molecule gains one or more electrons.
b. Electrons must be transferred from the species losing electrons to the species gaining
electrons.
2. Which of these half-reactions represent oxidation and which reduction? Explain your
reasoning.
a. Fe(s) ⎯→ Fe 2+ (aq) + 2 e –
b. Ni 4+ (aq) + 2 e – ⎯→ Ni 2+ (aq)
c. 2 H2O(l) + 2 e – ⎯→ H2(g) + 2 OH – (aq)
Answer:
a. Oxidation. Iron loses two electrons to form the iron(II) ion.
b. Reduction. The nickel(IV) ion gains two electrons to form the nickel(II) ion.
c. Reduction. Water gains two electrons to the hydrogen ion and the hydroxide ion.
3. You have seen several examples of oxidation–reduction reactions in this chapter. Now
examine these equations and decide which are oxidation–reduction reactions and which are
not. Explain your decisions.
a. Zn(s) + 2 MnO2(s) + H2O(l) ⎯→ Zn(OH)2(s) + Mn2O3(s)
b. HCl(aq) + NaOH(aq) ⎯→ NaCl(aq) + H2O(l)
c. CH4(g) + 2 O2(g) ⎯→ CO2(g) + 2 H2O(g)
Answer:
Parts a and c are redox reactions. Electrons must be transferred when an element (Zn in
part a, O2 in part c) reacts to form a compound. Part b is a neutralization reaction in which
ions combine to form a soluble salt and water. Electron transfer does not occur in this case.
4. Comment on the statement: Every combustion reaction is an oxidation–reduction reaction.
Answer:
This is a true statement. Combustion is the rapid combination of a fuel and oxygen to form
products. In the case of burning a hydrocarbon, these products are CO2 and H2O. The oxygen
is reduced, and the hydrogen and the carbon are oxidized to form H2O and CO2, respectively.
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Even when substances that do not contain carbon are burned, the reactions can still be
described in terms of oxidation and reduction.
5. Two common units associated with electricity are the volt and the amp. What does each unit
measure?
Answer:
Electric current (an amount of charge per second) is measured in amps. In contrast, the volt is
a measure of electric potential, that is, the force or “pressure” behind this current.
6. Consider this galvanic cell. A coating of impure silver metal begins to appear on the surface
of the silver electrode as the cell discharges.
a. Identify the anode and write the oxidation half-reaction.
b. Identify the cathode and write the reduction half-reaction.
Answer:
a. The anode is Zn(s) and the oxidation half-reaction is:
Zn(s) ⎯→ Zn 2+ (aq) + 2 e –
b. The cathode is Ag(s) and the reduction half-reaction is:
2 Ag + (aq) + 2 e – ⎯→ 2 Ag(s)
7. In the lithium–iodine cell, Li is oxidized to Li + ; I2 is reduced to 2 I – .
a. Write equations for the two half-reactions that take place in this cell, labeling one as
oxidation and the other as reduction.
b. Write an equation for the overall reaction in this cell.
c. Identify the half-reaction that occurs at the anode and the half-reaction that occurs at the
cathode.
Answer:
a. oxidation half-reaction: Li(s) ⎯→ Li + + e –
reduction half-reaction: I2(s) + 2 e – ⎯→ 2 I –
b. overall reaction: 2 Li(s) + I2(s) ⎯→ 2 LiI(s)
c. The oxidation reaction, Li(s) ⎯→ Li + (aq) + e – , occurs at the anode.
The reduction reaction, I2(s) + 2 e – ⎯→ 2 I – (aq), occurs at the cathode.
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8. a. Is the voltage from a tiny AAA-size alkaline cell the same as that from a large D alkaline
cell? Explain.
b. Will both batteries sustain the flow of electrons for the same time? Why or why not?
Answer:
a. The voltage from both kinds of cells is the same (1.54 V) because voltage depends on the
chemical reaction that is producing the electrical energy and not the size of the electrodes.
The biggest difference in these two kinds of cells is the amount of current each produces. The
amount of current produced by a cell does depend on the size of the cell: Larger cells
generate more current than smaller cells do. The larger D cell will generate a larger current,
but the same voltage as the smaller AAA cells.
b. Larger cells contain more materials and can sustain the transfer of electrons over a longer
period.
9. Identify the type of battery commonly used in each of these consumer electronic products.
Assume none uses solar cells.
a. battery-powered watch c. digital camera
b. MP3 player d. handheld calculator
Answer:
a. battery-powered watch: silver oxide battery
b. MP3 player: Lithium ion battery
c. digital camera: Lithium ion battery
d. calculator: AA or AAA alkaline batteries
10. The mercury battery has been used extensively in medicine and industry. Its overall reaction
can be represented by this equation.
HgO(l) + Zn(s) ⎯→ ZnO(s) + Hg(l)
a. Write the oxidation half-reaction.
b. Write the reduction half-reaction.
c. Why is the mercury battery no longer in common use?
Answer:
a. Oxidation: Zn(s) + 2 OH – (aq) ⎯→ ZnO(s) + H2O(l) + 2 e –
b. Reduction: HgO(s) + H2O(l) + 2 e – ⎯→ Hg(l) + 2 OH – (aq)
c. In both 1970 and 1980, a major use for mercury was in batteries. By 1990, awareness of
the dangers of mercury in urban trash had grown. Mercury is a toxic metal and (in some
forms) can accumulate in the biosphere. Safer batteries and the need to recycle batteries led
to the passage of the Mercury-Containing and Rechargeable Battery Management Act (The
Battery Act) in 1996.
11. a. What is the function of the electrolyte in a galvanic cell?
b. What is the electrolyte in an alkaline cell?
c. What is the electrolyte in a lead–acid storage battery?
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Answer:
a. The electrolyte completes the electrical circuit. It provides a medium for transport of ions,
thus allowing charge to be transferred.
b. KOH paste
c. H2SO4(aq)
12. These are the incomplete equations for the half-reactions in a lead storage battery. They do
not show the electrons either lost or gained.
Pb(s) + SO4 2– (aq) ⎯→ PbSO4(s)
PbO2(s) + 4 H + (aq) + SO4 2– (aq) ⎯→ PbSO4(s) + 2 H2O(l)
a. Balance both equations with respect to charge by adding electrons on either side of the
equations, as needed.
b. Which half-reaction represents oxidation and which represents reduction?
c. One of the electrodes is made of lead, the other of lead dioxide. Which is the anode and
which is the cathode?
Answer:
a. Pb(s) + SO4 2– (aq) ⎯→ PbSO4(s) + 2 e –
PbO2(s) + 4 H + (aq) + SO4 2– (aq) + 2 e – ⎯→ PbSO4(s) + 2 H2O(l)
b. The first half-reaction shows electrons being lost, so it is the half-reaction of oxidation.
The second half-reaction shows electrons being gained, so it is the half-reaction of reduction.
c. Lead is being oxidized, so lead is the anode. Lead dioxide is being reduced, so it is the
cathode.
13. What is meant by the term hybrid car?
Answer:
In current usage, the term hybrid car refers to the combination of a gasoline engine with a
nickel-metal hydride battery, an electric motor, and an electric generator. Other hybrids using
fuel cells are either available or under development.
14. a. What is the role of the electrolyte in a fuel cell?
b. List two advantages fuel cells have over internal combustion engines.
Answer:
a. The electrolyte allows ions to move between the electrodes, thus completing the circuit.
b. Fuel cells using hydrogen gas (H2) as the fuel produce only water as a product, whereas
internal combustion engines produce many products including the greenhouse gas carbon
dioxide and pollutants such as carbon monoxide, soot, VOCs, and nitrogen monoxide. When
a fuel is oxidized in a fuel cell, 45–55% of the energy released is transferred into electricity.
Combustion engines are only 20–30% efficient because combustion releases energy in the
form of heat, which then must be transferred to electrical and/or mechanical energy.
15. Is the conversion of O2(g) to H2O(l) in a fuel cell an example of oxidation or reduction? Use
electron loss or gain to support your answer.
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Answer:
The conversion of O2 to H2O requires a supply of electrons (see equation 8.13) so it is a
reduction reaction. This is the reduction half-reaction:
1/2 O2(g) + 2 H + (aq) + 2 e – ⎯→ H2O(l)
16. Consider this diagram of a hydrogen–oxygen fuel cell used in earlier space missions.
a. How does the reaction between hydrogen and oxygen in a fuel cell differ from the
combustion of hydrogen with oxygen?
b. Write the half-reaction that takes place at the anode in this fuel cell.
c. Write the half-reaction that takes place at the cathode in this fuel cell.
Answer:
a. In the combustion of hydrogen with oxygen, the chemical energy is released in the form of
heat and light, often explosively. In a fuel cell utilizing the same chemical reaction, the
chemical energy is released in a controlled manner in the form of electricity.
b. Anode reaction: H2(g) ⎯→ 2 H + (aq) + 2 e –
c. Cathode reaction: 1/2 O2(g) + 2 H + (aq) + 2 e – ⎯→ H2O(l)
17. a. What is a PEM fuel cell? How does it differ from the fuel cell represented in question #16?
b. What is an SOFC? How does it differ from the fuel cell represented in question #16?
Answer:
a. PEM stands for proton exchange membrane. In a PEM fuel cell, H2 is oxidized to form H + .
These H + move through the membrane to react with O2 (which is reduced) to form water. In
the cell described in question #16, the same oxidation-reduction reactions occur, but the
electrodes and electrolytes are different. In addition, the PEM fuel cell operates at room
temperature and the one used for the space mission does not.
b. SOFC stands for Solid Oxide Fuel Cell. An SOFC uses solid zirconium oxide and yttrium
oxide as the electrolyte instead of the hot KOH solution shown in question #16. Also, oxide
ions (O 2– ) instead of H + ions are transported from one electrode to the other through the
electrolyte.
18. In addition to hydrogen, methane also has been studied for use in PEM fuel cells. Balance the
given oxidation and reduction half-reactions and write the overall equation for a methanebased
fuel cell.
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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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21. a. What is meant by “the hydrogen economy”?
b. Even if methods for producing hydrogen cheaply and in large quantities were to become
available, what problems would still remain for the hydrogen economy?
Answer:
a. The hydrogen economy refers to producing, storing, and using hydrogen gas to produce
energy.
b. Before achieving all the benefits of the hydrogen economy, the problems of the safe
production, transportation, and storage of large quantities of hydrogen will need to be solved.
22. a. How are equations 8.24 and 8.25 the same and how are they different?
b. How will the energy released in the reaction shown in equation 8.24 compare with the
energy released in the reaction represented by equation 8.25? Explain your reasoning.
Answer:
a. Equations 8.24 and 8.25 are alike in that they both represent the same chemical process,
the oxidation of O2 and reduction of H2 to form H2O. The coefficients differ by a factor of 2.
Equation 8.24 represents the reaction of 2 moles of H2 with 1 mole of O2 to form 2 moles of
H2O, while equation 8.25 represents the reaction of 1 mole of H2 with ½ mole of O2 to
produce 1 mole of O2.
b. The energy released in the reaction shown in equation 8.24 is double the amount released
by the reaction shown in equation 8.25 because the first equation represents twice as many
moles of reactants.
23. Given that 286 kJ of energy is released per mole of H2 burned, what is the maximum amount
of energy that can be released when 370 kg of H2 is burned?
Answer:
370 kg H 2 "
1000 g
1 kg " 1 mol H 2
2.0 g H 2
" 286 kJ
1 mol H 2
PAGE 8-7
= 5.3 × 10 7 kJ
24. a. Use bond energies from Table 4.2 to calculate the energy released when 1 mol of H2 burns.
b. Compare your result with the stated value of 286 kJ. Account for any difference.
Answer:
a. First write the chemical equation for the reaction.
Then consider the Lewis structures for the reaction.
Energy needed to break bonds:
436 kJ + 1/2 (498 kJ) = 685 kJ
Energy released as new bonds form:
2(467 kJ) = 2934 kJ
The net energy change is (685 kJ) + (2934 kJ) = 2249 kJ.

. Average bond energies are based on bonds within molecules in the gaseous state. In the
given chemical equation, the H2O formed is present as a liquid rather than as a gas.
Additional energy is released when gaseous water condenses to the liquid state, so the stated
value of 286 kJ is greater than the 249 kJ calculated in part a.
25. Every year, 5.6 × 10 21 kJ of energy comes to Earth from the Sun. Why can’t this energy be
used to meet all of our energy needs?
Answer:
Solar energy is distributed over Earth’s surface. Therefore, it is not very easy to capture,
store, and transform solar energy to meet our energy needs.
26. This unbalanced equation represents the last step in the production of pure silicon for use in
solar cells.
__Mg(s) + __SiCl4(l) ⎯→ __MgCl2(l) + __Si(s)
a. How many electrons are transferred per atom of pure silicon formed?
b. Is the production of pure silicon an oxidation or a reduction reaction? Why do you think
so?
Answer:
a. This is the balanced equation: 2 Mg(s) + SiCl4(l) ⎯→ 2 MgCl2(l) + Si(s)
Each magnesium atom loses 2 electrons, and there are 2 magnesium atoms providing
electrons for one silicon atom. Silicon in SiCl4 picks up the four electrons.
b. The silicon atoms in SiCl4 gain four electrons to form elemental silicon, a reduction
process. The two magnesium atoms each lose two electrons, an oxidation process. Overall,
the production of pure silicon is an oxidation-reduction reaction.
27. The symbol represents an electron and the symbol represents a silicon atom. Does this
diagram represent a gallium-doped p-type silicon semiconductor, or does it represent an
arsenic-doped n-type silicon semiconductor? Explain your answer.
Answer:
Each Si atom is surrounded by 8 electrons, but the atom in the center (the one that is doping
the semiconductor) is surrounded by 9 electrons. Silicon is in Group 4A and has 4 outer
electrons. Thus central atom in the figure must have 5 outer electrons. This is consistent with
an element in Group 5A such as arsenic, so this is an n-type silicon semiconductor.
28. Describe the main reasons why solar cells have solar energy conversion efficiencies
significantly less than the theoretical value of 31%.
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Answer:
The efficiency is reduced because some of the radiant energy is reflected by the cell or
absorbed to produce heat instead of an electric current.
Concentrating on Concepts
29. Explain the significance of the title of this chapter, “Energy from Electron Transfer.”
Answer:
In every electrochemical process described in this chapter, energy is produced through
electron transfer. Chemical reaction, such as those that take place in galvanic cells, batteries
and fuel cells, produce electrons that can do work because the anode and cathode are
physically separated in space. The transfer of electrons also may be initiated when light
strikes a photovoltaic cell.
30. Consider these three sources of light: a candle, a battery-powered flashlight, and an electric
light bulb. For each source, provide:
a. The origin of the light
b. The immediate source of the energy that appears as light
c. The original source of the energy that appears as light. Hint: Trace this back stepwise as
far as possible.
d. The end-products and by-products of using each
e. The environmental costs associated with each
f. The advantages and disadvantages of each light source
Answer:
Candlelight
Origin The hot gases that burn and emit light.
Immediate E The hydrocarbon wax, made either by bees or from petroleum.
source
Original E source Sunlight that drove photosynthesis, which in turn produced the plants
from which bees gathered their food (or years ago died and formed
fossil fuels).
Products Products: CO2, H2O and small amounts of soot and CO.
Environmental
costs
Advantages
Disadvantages
Mainly from the pollutants (and CO2, not classified as a pollutant as of
2007) produced while burning the candles.
Convenient, pretty to look at, produce dirty soot and sometimes start
fires.
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Light in a battery-powered flashlight
Origin A wire that glows when it is heated to a high temperature.
Immediate E Energy to heats the wire comes from a chemical reaction in the battery.
source
Original E source Several possibilities, depending on what energy source was used to
produce the battery. Could have been fossil fuel consumption
(ultimately solar energy) or nuclear power plant (nuclear fission).
Products The end products are different chemicals in the battery while the byproducts
are those that are produced during the manufacture of the
battery, bulb, and flashlight.
Environmental
costs
Advantages
Disadvantages
All those associated with the production and disposal of the battery
materials, as well as the side products during the combustion of fossil
fuels (or nuclear fission).
Portable, convenient, clean for the user.
Somewhat expensive.
Light from an electric light bulb
Origin A wire that glows when it is heated to a high temperature.
Immediate E
source
Several possibilities, depending on what energy source was used to
produce the electricity. Could have been fossil fuel consumption or
nuclear power plant.
Original E source The Sun or the ancient stellar synthesis that produced the uranium and
other metals on our planet.
Products The light bulb is very clean at the site where it is used, but produces
pollutants such as NOx, SO2 and particulate matter at the power plant
(if coal combustion) or spent nuclear fuel (if nuclear).
Environmental
costs
Advantages
Disadvantages
See above.
Convenient, safe, inexpensive.
Few to the user, except that the energy costs are relatively high in
comparison to using a fluorescent bulb.
31. Explain the difference between a rechargeable battery and one that must be discarded. Use a
NiCad battery and an alkaline battery as examples.
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Answer:
These batteries derive their voltage from different sets of chemical reactions. A rechargeable
battery (such as a NiCad battery) is one in which the oxidation-reduction reaction can be
reversed with the input of energy. This recharges the battery. The oxidation-reduction
reaction in a non-rechargeable battery, such as an alkaline battery, cannot easily be reversed.
Since no way exists to recharge alkaline batteries, they are discarded once they stop
producing electrical energy.
32. Is there a difference between a galvanic cell and an electrochemical cell? Explain, giving
examples to support your answer.
Answer:
The chapter defines only one type of cell: galvanic. A galvanic cell (sometimes called a
voltaic cell) is a device that converts the energy released in a spontaneous chemical reaction
into electrical energy. The term electrochemical cell is broader than this. It also includes cells
that require the input of energy to make them run. Batteries are galvanic cells. So are fuel
cells. Electrolysis, a process described in Chapter 5 that splits water into oxygen and
hydrogen, is an electrolytic cell.
33. What is the difference between a storage battery and a fuel cell?
Answer:
A storage battery converts chemical energy into electrical energy by means of a reversible
reaction. No reactants or products leave the “storage” battery and the reactants can be
reformed during the recharging cycle. A fuel cell also converts chemical energy into
electrical energy but the reaction is not reversible. A fuel cell continues to operate only if fuel
and oxidant are continuously added, which is why it is classed as a “flow” battery.
34. Why are electric cars powered by lead–acid storage batteries alone only a short-term solution
to the problem of air pollution emissions from automobiles? Outline your reasoning.
Answer:
The emissions in question are the oxides of sulfur and nitrogen (SO2 and NOx) as well as
particulate matter. Although cars powered by lead storage batteries do not emit these
chemicals, nonetheless all of them are associated with the manufacture of these batteries.
Thus the emissions still exist; they are just one step removed from the automobile. In
addition, lead is a toxic metal with environmental consequences. Because of its high density,
it is heavy (and expensive) to transport.
35. AgZn batteries are replacing lead–acid batteries in small airplanes, such as
Cessna172s.
a. Why are these batteries, although more expensive, preferable to the lead–acid batteries
used previously?
b. Write the half-reaction of oxidation and of reduction.
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Answer:
a. These batteries are lighter and less toxic than the lead-acid batteries.
b. The overall reaction is:
Zn(s) + Ag2O(s) ⎯→ ZnO(s) + 2 Ag(s)
The oxidation half-reaction is: Zn(s) + O 2– (aq) ⎯→ ZnO(s) + 2 e –
The reduction half-reaction is: Ag2O(s) + 2 e – ⎯→ 2 Ag(s) + O 2– (aq)
36. The battery of a cell phone discharges when the phone is in use. A manufacturer, while
testing a new “power boost” system, reported these data.
Time, min.sec Voltage, V
0.00 6.56
1.00 6.31
2.00 6.24
3.00 6.18
4.00 6.12
5.00 6.07
6.35 6.03
8.35 6.00
11.05 5.90
13.50 5.80
16.00 5.70
16.50 5.60
a. Prepare a graph of these data.
b. The manufacturer’s goal was to retain 90% of its initial voltage after 15 minutes of
continuous use. Has that goal been achieved? Justify your answer using your graph.
Answer:
a.
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. The manufacturer’s goal of retaining 90% of the initial battery voltage after 15 minutes of
continuous use has not been achieved. 90% of the initial 6.56 V is 5.90 V, a level that was
reached when the cell phone had been used for only 11 minutes.
37. Assuming that hybrid cars are available in your area, what questions would you ask the car
dealer before deciding to buy or lease a hybrid? Which of these questions do you consider
most important? Offer reasons for your choices.
Answer:
The primary question is likely to be that of fuel economy. How do the new hybrids compare
to other vehicles on the market? Follow-up questions include purchase price, safety record,
maintenance costs, battery lifetime, environmental impacts, and available tax credits. All of
these factors will influence the costs of purchasing and operating the car. Each factor’s
relative importance will depend on personal choices and on the region of the country where
the car is being purchased and driven.
38. You never need to plug in Toyota’s gasoline–battery hybrid car to recharge the batteries.
Explain.
Answer:
The Toyota hybrid car (the Prius) has a gasoline engine sitting side-by-side with nickel-metal
hydride batteries, an electric motor, and a generator. The battery is recharged by regenerative
braking, in which the kinetic energy of the car is transferred to stored electrical energy.
39. Prepare a list of the environmental costs and benefits associated with hybrid vehicles.
Compare that list with the environmental costs and benefits of vehicles powered by gasoline.
On balance, which energy source do you favor, and why?
Answer:
Many hybrid vehicles have greater average fuel economy than their gasoline counterparts.
(Note that many smaller gasoline-powered cars currently achieve better fuel efficiency than
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large hybrid vehicles such as SUVs.) Hybrid vehicles generally have fewer emissions (CO,
CO2, NOx, and particulate matter) than gasoline-powered cars. When hybrid vehicles are
running on the electricity generated by braking, they are not depleting fossil fuel supplies.
However, the greater consumer familiarity and lower price of gasoline-powered vehicles
makes them attractive to many people.
40. William C. Ford, Jr., chairman of the board of Ford Motor Company, is quoted as saying that
going “totally green” with zero-emissions vehicles will be a real challenge. Regular drivers
won’t buy high-tech clean cars, Ford admits, until the industry has a “no-trade-off” vehicle
widely available. What do you think he means by a no-trade-off vehicle? Do you think he is
justified in this opinion?
Answer:
A no-trade-off vehicle will perform just as well as a gasoline vehicle at the same price (and
still have zero emissions). Although Mr. Ford’s words may have made sense when he spoke
them, they are less true today (2007 as the 6 th edition goes to press). Those working in the
industry are seeing the need to “go green” and are finding that “green” practices give them a
competitive edge and that people are buying them.
41. Fuel cells were invented in 1839, but never developed into practical devices for producing
electrical energy until the U.S. space program in the 1960s. What advantages did fuel cells
have over previous power sources?
Answer:
The space program required reliable, relatively lightweight power sources. Fuels cells, when
compared to other types of batteries that were available at that time, met those specifications
and did not “run down” or require recharging. A fuel cell will continue to operate as long as
fuel is available.
42. Hydrogen, H2, and methane, CH4, can each be used with oxygen in a fuel cell. Hydrogen and
methane also can be burned directly. Which has greater heat content when burned, 1.00 g of
H2 or 1.00 g of CH4? Hint: Write the balanced chemical equation for each reaction and use
the bond energies in Table 4.2 to help answer this question.
Answer:
The bond energies listed in Table 4.2 for bonds breaking and forming in gases can be used to
calculate the following heats of combustion. These differ somewhat from the values given in
the beginning of this chapter, where the product water is given in the liquid state.
H2(g) + ½ O2(g) ⎯→ H2O(g) heat of combustion = 249 kJ/mol
CH4(g) + 2 O2(g) ⎯→ CO2(g)+ 2 H2O(g) heat of combustion = 814 kJ/mol
In each case, the units of the calculated heat of combustion can be changed to kJ/gram by
dividing by the molar mass of the fuel.
249 kJ
For hydrogen: " 1 mol H2 = 124 kJ
For methane:
!
!
mol H2 2.01g H2 814 kJ
mol CH4 " 1 mol CH4 16.0 g CH4 g H 2
= 50.9 kJ
PAGE 8-14
g CH 4

43. Engineers have developed a prototype fuel cell that converts gasoline to hydrogen and carbon
monoxide. The carbon monoxide, in contact with a catalyst, then reacts with steam to
produce carbon dioxide and more hydrogen.
a. Write a set of reactions that describes this prototype fuel cell, using octane (C8H18) to
represent the hydrocarbons in gasoline.
b. Speculate as to the future economic success of this prototype fuel cell.
Answer:
a. Conversion of fuel: C8H18(l) + 4 O2(g) ⎯→ 9 H2(g) + 8 CO(g)
catalyst
Elimination of CO: CO(g) + H2O(g) ⎯→ CO2(g) + H2(g)
b. This type of fuel cell is convenient because it runs on a liquid fuel, gasoline, rather than
using gaseous hydrogen. However, the liquid fuel is still petroleum-based and therefore nonrenewable.
It also burns to produce CO2, a greenhouse gas. Therefore, although such fuel
cells may find specialty applications in the near future, their long-term prospects are not
good.
44. At this time, the U.S. Department of Transportation (DOT) prohibits passengers from
carrying flammable fluids aboard aircraft. Explain how this might affect the development of
microfuel cells for use in consumer electronics such as portable computers.
Answer:
While the microcells described in section 8.6 use methanol as a fuel, the small amounts
required would be unlikely to cause a flammability concern. However, the volume of liquid
fuel required to power a laptop or iPod may exceed the three ounces of liquid per container
currently allowed by U.S. regulations.
45. Consider this representation of two water molecules in the liquid state.
a. What bonds are broken when water boils? Are these intermolecular or intramolecular
bonds? Hint: See Chapter 5 for definitions.
b. What bonds are broken when water is electrolyzed? Are these intermolecular or
intramolecular bonds?
Answer:
a. When water boils, the hydrogen bonds between water molecules are broken. These are
intermolecular bonds.
PAGE 8-15

!
!
b. When water is electrolyzed, the covalent bonds within water molecules are broken. These
are intramolecular bonds.
46. Although hydrogen gas can be produced by the electrolysis of water, this reaction is usually
not carried out on a large scale. Suggest a reason for this fact.
Answer:
Electrolysis of water requires an input of 286 kJ of energy per mole of water electrolyzed.
Most of this energy comes from the burning of fossil fuels in conventional power plants. The
inherent inefficiency associated with transforming heat into work limits the usefulness of
large-scale electrolysis and makes the process energy intensive.
47. Small quantities of hydrogen gas can be prepared in the lab by reacting metallic sodium with
water, as shown in this equation.
a. Calculate the grams of sodium needed to produce 1.0 mol of hydrogen gas.
b. Calculate the grams of sodium needed to produce sufficient hydrogen to meet an
American’s daily energy requirement of 1.1 × 10 6 kJ.
c. If the price of sodium were $94/kg, what would be the cost of producing 1.0 mol of
hydrogen? Assume the cost of water is negligible.
Answer:
a. 1.0 mol H2 "
b.
2 mol Na
1 mol H 2
1.1"10 6 kJ " 1 mol H 2
286 kJ
c. Using the result from part a:
" 23.0 g Na
1 mol Na
" 2 mol Na
1 mol H 2
PAGE 8-16
= 46.0 g Na
" 23.0 g Na
46.0 g Na "
1 mol Na
= 1.8 "10 5 g Na
1 kg Na
10 3 g Na "
$94
1 kg Na
= $4.32
48. a. As a fuel, hydrogen has both advantages and disadvantages. Set up parallel lists for the
advantages and the disadvantages of using hydrogen as the fuel for transportation and for
producing electricity. !
b. Do you advocate the use of hydrogen as a fuel for transportation or for the production of
electricity? Explain your position in a short article for your student newspaper.

Answer:
a.
Advantages of H 2
Disadvantages of H 2
Transportation Electricity
• lightweight fuel
• saves fossil fuels if the
hydrogen is derived from
renewable sources
• can be used in fuel cells
• potentially explosive
• difficult to store, handle
PAGE 8-17
• large supply potentially
available from water through
electrolysis
• fuel cells using H 2 are
practical for some
applications
• expensive to extract H 2 from
water
• difficult to transport, store,
and handle
b. Personal decisions about the use of hydrogen as a fuel for transportation and/or the
production of electricity should be based largely on the real advantages and disadvantages of
this fuel in the two different applications.
49. Fossil fuels have been called “. . . Sun’s ancient investment on Earth.” Explain this statement
to a friend who is not enrolled in your course.
Answer:
Fossil fuels were formed as the result of photosynthetic processes that took place hundreds of
millions of years ago. Solar energy was a necessary ingredient in the formation, so the energy
that is stored in chemical bonds in fossil fuels originally was invested from the Sun.
50. The cost of electricity generated by solar thermal power plants currently is greater than that
of electricity produced by burning fossil fuels. Given this economic fact, suggest some
strategies that might be used to promote the use of environmentally cleaner electricity from
photovoltaics.
Answer:
Environmentally cleaner electricity may be required to meet the mandates of the Kyoto
Accord. One strategy to promote the use of environmentally cleaner electricity could be to
provide a tax break or a subsidy for companies that produce electricity without consuming
fossil fuels. Changes in the tax system could be imposed that reflect the true dollar and
environmental costs of burning fossil fuels.
51. Name some of the current applications of photovoltaic cells other than the production of
electricity in remote areas.
Answer:
PV devices have demonstrated their practical utility for satellites, highway signs, security and
safety lighting, navigational buoys, and automobile recharging stations.

Exploring Extensions
52. The aluminum–air battery is being explored for use in automobiles. In this battery, aluminum
metal undergoes oxidation to Al 3+ and forms Al(OH)3. Oxygen from the air undergoes
reduction to OH – ions.
a. Write equations for the oxidation and reduction half-reactions. Use H2O as needed to
balance the number of hydrogen atoms present, and add electrons as needed to balance the
charge.
b. Add the half-reactions to obtain the equation for the overall reaction in this cell.
c. Specify which half-reaction occurs at the anode and which occurs at the cathode in the
battery.
d. What are the benefits of the widespread use of the aluminum–air battery? What are
some of the limitations? Write a brief summary of your findings.
e. What is the current state of development of this battery? Is it in use in any vehicles at
the present time? What is its projected future use?
Answer:
a. Al + 3 H2O ⎯→ Al(OH)3 + 3 H + + 3 e – half-reaction of oxidation
O2 + 2 H2O + 4 e – ⎯→ 4 OH – half-reaction of reduction
b. To make the number of electrons the same, the first half-reaction must be multiplied by 4,
and the second half-reaction must be multiplied by 3. In this way, 12 moles of electrons are
exchanged in the overall reaction.
4 (Al + 3 H2O ⎯→ Al(OH)3 + 3 H + + 3 e – )
3 (O2 + 2 H2O + 4 e – ⎯→ 4 OH – )
The net equation is 4 Al + 3 O2 + 6 H2O ⎯→ 4 Al(OH)3.
c. The half-reaction of oxidation, Al + 3 H2O ⎯→ Al(OH)3 + 3 H + + 3 e – , occurs at the
anode.
The half-reaction of reduction, O2 + 2 H2O + 4 e – ⎯→ 4 OH – , occurs at the cathode.
d. Potential benefits are that aluminum is recyclable, non-combustible, and it is capable of
storing a lot of energy. An aluminum–air battery is quite light and uses air as a reactant. It is a
non-polluting source of power. Some of the limitations are that obtaining aluminum from its
ore is a very energy-intensive process. Even recycling aluminum in a large volume requires
considerable energy, potentially limiting the cost and availability of aluminum. In early
versions of the battery, the electrolyte degraded the aluminum even when the battery was not
in use. Recently, companies have overcome this problem.
e. Despite twenty-five years of research to develop the aluminum–air battery, its use has been
mostly in military applications and no commercial products use the technology. A French
company, Métalectrique (http://www.metalectrique.com), is developing aluminum–air
batteries for use in mobile refrigerators, electric cars, and electric wheelchairs.
53. An iron-based “superbattery” is a promising alternative for delivering more power with
fewer environmental effects than alkaline batteries. Find out how the superbattery is designed
and its state of commercial acceptance.
PAGE 8-18

Answer:
Developed by Stuart Licht of the Technion-Israel Institute of Technology in Haifa, Israel,
this rechargeable battery is designed using less toxic materials than other batteries and has a
50% higher energy advantage. In these batteries the manganese dioxide cathode found in a
traditional battery is replaced with compounds containing iron in a +6 oxidation state. (Thus
the batteries are also called “super-iron” batteries.) The batteries can be used anywhere where
AAA batteries are currently used, but the superbatteries are not yet on the market. Additional
information about Licht’s battery is available at:
http://www.weizmann.ac.il/ICS/booklet/3/pdf3/3.pdf.
54. Although Alessandro Volta is credited with the invention of the first electric battery in
1800, some feel this is a reinvention. Research the “Baghdad battery” to evaluate the merit of
this claim.
Answer:
Clay jars found in the National Museum of Iraq have been described as the oldest known
electric batteries in existence. They were initially attributed to the Parthian Empire, an
ancient Asian culture that ruled most of the Middle East from 247 B.C. to A.D. 228.
However, modern archaeologists think the jars may have been created in the later Sassanid
era (A. D. 224 to 640).
The jars were first described in 1938 by German archaeologist Wilhelm König, and to this
day, it is uncertain whether Konig dug them up himself or found them archived in the
museum.
Those who have examined the jars closely say that there is little else that they can be but a
battery. A typical nondescript earthen jar is only 5½ inches high by 3 inches across. The
opening was sealed with an asphalt plug, which held in place a copper sheet, rolled into a
tube. This tube was capped at the bottom with a copper disc held in place by more asphalt. A
narrow iron rod was stuck through the upper asphalt plug and hung down into the center of
the copper tube — not touching any part of it.
Fill the jar with an acidic liquid, such as vinegar or fermented grape juice, and the jar
becomes a battery capable of generating a small current. The acidic liquid permits a flow of
electrons from the copper tube to the iron rod – an electric flow – when the two metal
terminals are connected on the outside of the jar.
Some archeologists remain skeptical that the people making these batteries understood the
batteries’ electrical properties. It is simple to prove that the batteries can generate an
electrical current today, but it is much more difficult to predict what the batteries were used
for in ancient times.
PAGE 8-19

55. If all of today’s technology presently based on fossil fuel combustion were replaced by
H2–O2 fuel cells, significantly more H2O would be released into the environment. Is this
effect a concern? Find out what other effects might be anticipated from switching to a
hydrogen economy.
Answer:
While water is not a criteria air pollutant, it is a greenhouse gas. Excess water in the
atmosphere could therefore contribute to global warming. Other effects include the release of
excess CO2 from the hydrogen production, and those listed in 56b below.
56. a. Hydrogen is generally considered as an environmentally friendly fuel, only
producing water after reacting with oxygen. What effect could the widespread use of
hydrogen have on urban air quality?
b. Some scientists are reporting concerns that leakage of hydrogen gas from cars, hydrogen
production plants, and fuel transportation could cause problems in Earth’s ozone layer. How
significant are these concerns? What is the mechanism by which hydrogen could destroy
ozone?
Answer:
a. The widespread use of hydrogen fuel could dramatically improve urban air quality by
reducing levels of particulate matter and SOx produced by internal combustion engines.
b. In June of 2003, geochemists at Caltech created a stir in the scientific community when
they published theoretical models in the journal Science that predicted hydrogen could leak
from production plants, transport systems, and hydrogen-powered cars in the future. They
predicted a large amount of hydrogen would reach the stratosphere assuming all current
fossil fuel combustion was replaced by hydrogen fuel cells and 10-20% of the hydrogen was
lost to leakage. The additional hydrogen in the stratosphere would react and significantly
increase the amount of water molecules found as ice in the stratosphere. Because ozone is
destroyed by reactions with chlorine radicals that take place on stratospheric ice, the
researchers proposed that the rate of ozone destruction would be enhanced. However, other
scientists in academic and industrial research claim the Caltech group’s hydrogen leakage
estimate is too high. One automobile executive argues that the industry would not allow such
a large amount of a valuable commodity (hydrogen) to go to waste. Another scientist points
out that the leaking hydrogen can be recovered. Others use the Caltech group’s research to
support the idea of on-site hydrogen production.
57. At the cutting edge of technology the line between science and science fiction often
blurs. Investigate the “futuristic” idea of putting mirrors in orbit around Earth to focus and
concentrate solar energy for use in generating electricity.
Answer:
Mirrors or lenses could be used to concentrate solar energy on a collection array in space and
send the electricity back to Earth. This technology would counteract one of the major
PAGE 8-20

limitations of current solar technology – the light from the Sun spreads out over Earth and is
relatively weak. Unlike Earth-based solar collectors, orbiting collecting satellites would also
have the advantage of being in the light all the time.
http://www.thespacereview.com/article/214/1
58. Although silicon, used to make solar cells, is one of the most abundant elements in
Earth’s crust, extracting it from minerals is costly. The increased demand for solar cells has
some companies worried about a “silicon shortage.” Use the resources of the Web to find out
how silicon is purified and how the PV industry is coping with the rising prices.
Answer:
Crystalline silicon is used for the production of photovoltaic cells. Two of the common
procedures for synthesizing crystalline silicon are Czochralski Crystal Growth and Float
Zone Crystal Growth. Information can be found at:
http://www.siliconfareast.com/crystal.htm.
To cope with the shortage of silicon, the photovoltaic industry is undergoing a series of
changes. Companies are pursuing business models to increase productivity and secure supply
lines. Researchers are developing and testing a number of synthetic molecules for the next
generation of photovoltaic collectors.
http://www.renewableenergyaccess.com/rea/news/story?id=43983
59. Figure 8.27 shows an array of photovoltaic cells installed at the Bavaria Solar Park in
Germany. Where in the United States is the largest photovoltaic power plant? Use the Web to
learn of other large-scale photovoltaic cell installations. What factors help to influence this
approach, one that uses a centralized array rather than individual rooftop solar units?
Answer:
Photovoltaics (PV) are being used in parts of California and in Arizona in large-scale
demonstration projects. For example, Pacific Gas and Electric in Kerman, CA is close to
rapidly growing Fresno, CA. The PV substation experiment is designed to measure the value
to the utility of a 500-kW generating plant that can be quickly (within 6 months) placed
where extra power is needed. The results of this experiment will be widely distributed.
In April 2007, construction began on an array of photovoltaics at an Air Force base in
Nevada. When completed, the array will be the largest in the US and will provide the base
with 15-18 megawatts of power, about 25-30% of its need.
PV generation plants have several characteristics that have slowed their use by utilities.
Under current utility accounting, PV-generated electricity still costs considerably more than
electricity generated by conventional plants, and regulatory agencies require most utilities to
supply electricity for the lowest cash cost. Furthermore, photovoltaic systems produce power
PAGE 8-21

only during daylight hours and their output varies with the weather. Utility planners must
therefore treat a PV power plant differently than they would treat a conventional plant.
Despite the costs, utilities are becoming more involved with PV. An alliance has been formed
between DOE, the Electric Power Research Institute (EPRI), and several utilities. The
program is called Photovoltaics for Utility-Scale Applications (PVUSA) and there are
presently three pilot projects, including the Kerman PV plant. Photovoltaics have found some
use in other parts of the world, including Australia, Japan, and as the text discusses,
Switzerland. Developing countries have been slower to use photovoltaics on a large scale
because of the high cost.
60. The Solar America Initiative program receives so little publicity that most people in the
United States are unaware of its existence. Design a poster to explain and promote some part
of this program to the general public.
Answer:
The Solar America Initiative is coordinated by the U.S. Department of Energy. The program
aims to improve the cost-effectiveness of solar energy to the levels of conventional forms of
energy (coal, nuclear power, hydroelectric power, etc) by 2015. Funding is available to
universities and companies for research and development of solar technologies. The program
also gives grants to local governments to encourage changes in energy regulations and
promotion of solar technology among residents. Thirteen U.S. cities were chosen to be Solar
America Cities in 2007, and a second round of grants will be announced in 2008. A third
grant program supports Solar America Showcases, which are large projects that draw public
attention to the benefits of solar energy. The first showcases funded are powering homes on a
military base in Oahu, Hawaii, large buildings in San Jose, CA, and the Orange County
convention center in Florida. Student posters could focus on any of these aspects of the Solar
America Initiative.
PAGE 8-22