Tải bản đầy đủ - 0trang
18.5 Electrochemistry: An Introduction



599



Salt bridge

Porous disk



a



Figure 18.4

A salt bridge or a porous-disk

connection allows ions to flow,

completing the electric circuit.



The name galvanic cell honors

Luigi Galvani (1737–1798),

an Italian scientist generally

credited with the discovery

of electricity. These cells are

sometimes called voltaic

cells after Alessandro Volta

(1745–1827), another Italian,

who first constructed cells of

this type around 1800.

Anode: The electrode where

oxidation occurs. Cathode: The

electrode where reduction occurs.



b



The salt bridge contains a strong

electrolyte either as a gel or as

a solution; both ends are covered

with a membrane that allows

only ions to pass.



The porous disk allows ion flow

but does not permit overall mixing

of the solutions in the two

compartments.



idizing agent, and ions in the two aqueous solutions flow from one compartment to the other to keep the net charge zero.

Thus an electrochemical battery, also called a galvanic cell, is a

device powered by an oxidation–reduction reaction where the oxidizing

agent is separated from the reducing agent so that the electrons must travel

through a wire from the reducing agent to the oxidizing agent (Figure 18.5).

Notice that in a battery, the reducing agent loses electrons (which flow

through the wire toward the oxidizing agent) and so is oxidized. The electrode where oxidation occurs is called the anode. At the other electrode, the

oxidizing agent gains electrons and is thus reduced. The electrode where reduction occurs is called the cathode.

We have seen that an oxidation–reduction reaction can be used to generate an electric current. In fact, this type of reaction is used to produce elec-



e–

e–



e–



Cathode

(reduction)



Anode

(oxidation)



–



Ions



Corbis-Bettmann



Oxidizing

agent



Figure 18.5

Alessandro Volta.



Schematic of a battery (galvanic cell).



+



Reducing

agent



600 Chapter 18 Oxidation–Reduction Reactions and Electrochemistry

tric currents in many space vehicles. An oxidation–reduction reaction that

can be used for this purpose is hydrogen and oxygen reacting to form water.

2H2(g) ϩ O2(g) n 2H2O(l)

Oxidation states:



c

0



c

0



a

Q

ϩ1

Ϫ2

(each H)



Notice from the changes in oxidation states that in this reaction, hydrogen

is oxidized and oxygen reduced. The opposite process can also occur. We can

force a current through water to produce hydrogen and oxygen gas.

Electrical



2H2O(l) 87777777n

2H2(g) ϩ O2(g)

energy

This process, where electrical energy is used to produce a chemical change, is

called electrolysis.

In the remainder of this chapter, we will discuss both types of electrochemical processes. In the next section we will concern ourselves with the

practical galvanic cells we know as batteries.



18.6 Batteries

OBJECTIVE:



To learn about the composition and operation of commonly used batteries.

In the previous section we saw that a galvanic cell is a device that uses an

oxidation–reduction reaction to generate an electric current by separating

the oxidizing agent from the reducing agent. In this section we will consider

several specific galvanic cells and their applications.



▲

Remember: The oxidizing agent

accepts electrons and the

reducing agent furnishes

electrons.



Lead Storage Battery

Since about 1915, when self-starters were first used in automobiles, the lead

storage battery has been a major factor in making the automobile a practical means of transportation. This type of battery can function for several

years under temperature extremes from Ϫ30 ЊF to 100 ЊF and under incessant

punishment from rough roads. The fact that this same type of battery has

been in use for so many years in the face of all of the changes in science and

technology over that span of time attests to how well it does its job.

In the lead storage battery, the reducing agent is lead metal, Pb, and the

oxidizing agent is lead(IV) oxide, PbO2. We have already considered a simplified version of this reaction in Example 18.6. In an actual lead storage battery, sulfuric acid, H2SO4, furnishes the Hϩ needed in the reaction; it also furnishes SO42Ϫ ions that react with the Pb2ϩ ions to form solid PbSO4. A

schematic of one cell of the lead storage battery is shown in Figure 18.6.

In this cell the anode is constructed of lead metal, which is oxidized. In

the cell reaction, lead atoms lose two electrons each to form Pb2ϩ ions, which

combine with SO42Ϫ ions present in the solution to give solid PbSO4.

The cathode of this battery has lead(IV) oxide coated onto lead grids.

Lead atoms in the ϩ4 oxidation state in PbO2 accept two electrons each (are

reduced) to give Pb2ϩ ions that also form solid PbSO4.

In the cell the anode and cathode are separated (so that the electrons

must flow through an external wire) and bathed in sulfuric acid. The halfreactions that occur at the two electrodes and the overall cell reaction are

shown on the following page.



18.6 Batteries



601



Figure 18.6

In a lead storage battery each

cell consists of several lead grids

that are connected by a metal

bar. These lead grids furnish

electrons (the lead atoms lose

electrons to form Pb2ϩ ions,

which combine with SO42Ϫ ions

to give solid PbSO4). Because the

lead is oxidized, it functions as

the anode of the cell. The

substance that gains electrons is

PbO2; it is coated onto lead grids,

several of which are hooked

together by a metal bar. The

PbO2 formally contains Pb4ϩ,

which is reduced to Pb2ϩ, which

in turn combines with SO42Ϫ to

form solid PbSO4. The PbO2

accepts electrons, so it functions

as the cathode.



e– flow



H2SO4

electrolyte

solution



Pb metal grid

(anode)



PbO2 coated

onto a lead grid

(cathode)



Anode reaction:

Pb ϩ H2SO4 S PbSO4 ϩ 2Hϩ ϩ 2eϪ oxidation

Cathode reaction:

PbO2 ϩ H2SO4 ϩ 2eϪ ϩ 2Hϩ S PbSO4 ϩ 2H2O reduction

Overall reaction:

Pb(s) ϩ PbO2(s) ϩ 2H2SO4(aq) S 2PbSO4(s) ϩ 2H2O(l)

The tendency for electrons to flow from the anode to the cathode in a

battery depends on the ability of the reducing agent to release electrons and

on the ability of the oxidizing agent to capture electrons. If a battery consists

of a reducing agent that releases electrons readily and an oxidizing agent

with a high affinity for electrons, the electrons are driven through the connecting wire with great force and can provide much electrical energy. It is

useful to think of the analogy of water flowing through a pipe. The greater

the pressure on the water, the more vigorously the water flows. The “pressure” on electrons to flow from one electrode to the other in a battery is

called the potential of the battery and is measured in volts. For example,

each cell in a lead storage battery produces about 2 volts of potential. In an

actual automobile battery, six of these cells are connected to produce about

12 volts of potential.



▲



Dry Cell Batteries

The calculators, electronic watches, CD players, and MP3 players that are so

familiar to us are all powered by small, efficient dry cell batteries. They

are called dry cells because they do not contain a liquid electrolyte. The common dry cell battery was invented more than 100 years ago by George

Leclanché (1839–1882), a French chemist. In its acid version, the dry cell battery contains a zinc inner case that acts as the anode and a carbon (graphite)

rod in contact with a moist paste of solid MnO2, solid NH4Cl, and carbon



602 Chapter 18 Oxidation–Reduction Reactions and Electrochemistry

Cathode

(graphite rod)



Cathode (steel)

Insulation



Anode (zinc container)



Paste of HgO (oxidizing

agent) in a basic medium

of KOH and Zn(OH)2



Figure 18.8

A mercury battery of the type used in small

calculators.



Anode (zinc

inner case)

Paste of MnO2 ,

NH4Cl, and carbon



that acts as the cathode (Figure 18.7). The half-cell reactions are complex but

can be approximated as follows:

Anode reaction:



Zn S Zn2ϩ ϩ 2eϪ oxidation



Cathode reaction:



Figure 18.7

A common dry cell battery.



Ϫ

reduction

2NHϩ

4 ϩ 2MnO2 ϩ 2e S Mn2O3 ϩ 2NH3 ϩ H2O



This cell produces a potential of about 1.5 volts.

In the alkaline version of the dry cell battery, the NH4Cl is replaced with

KOH or NaOH. In this case the half-reactions can be approximated as follows:

Anode reaction:



Zn ϩ 2OHϪ S ZnO(s) ϩ H2O ϩ 2eϪ oxidation



Cathode reaction: 2MnO2 ϩ H2O ϩ 2eϪ S Mn2O3 ϩ 2OHϪ reduction

The alkaline dry cell lasts longer, mainly because the zinc anode corrodes less

rapidly under basic conditions than under acidic conditions.

Other types of dry cell batteries include the silver cell, which has a Zn anode and a cathode that employs Ag2O as the oxidizing agent in a basic environment. Mercury cells, often used in calculators, have a Zn anode and a cathode involving HgO as the oxidizing agent in a basic medium (see Figure 18.8).

An especially important type of dry cell is the nickel–cadmium battery, in

which the electrode reactions are

Anode reaction:



Cd ϩ 2OHϪ S Cd(OH)2 ϩ 2eϪ oxidation



Cathode reaction: NiO2 ϩ 2H2O ϩ 2eϪ S Ni(OH)2 ϩ 2OHϪ reduction

In this cell, as in the lead storage battery, the products adhere to the electrodes. Therefore, a nickel–cadmium battery can be recharged an indefinite

number of times, because the products can be turned back into reactants by

the use of an external source of current.



18.7 Corrosion

OBJECTIVE:



To understand the electrochemical nature of corrosion and to learn some

ways of preventing it.

Most metals are found in nature in compounds with nonmetals such as oxygen and sulfur. For example, iron exists as iron ore (which contains Fe2O3

and other oxides of iron).



C H E M I S T R Y I N F OCUS

Stainless Steel: It’s the Pits



O



Built in New York in

1929, the Chrysler

Building’s stainless

steel pinnacle has

been cleaned only a

few times. Despite

the urban setting, the

material shows few

signs of corrosion.



Image not available due to copyright restrictions



Dreamtime



ne of New York’s giants, the Chrysler Building, boasts a much admired art-deco stainless

steel pinnacle that has successfully resisted corrosion since it was built in 1929. Stainless steel is

the nobility among steels. Consisting of iron,

chromium (at least 13%), and nickel (with molybdenum and titanium added to more expensive

types), stainless steel is highly resistant to the

rusting that consumes regular steel. However,

the cheaper grades of stainless steel have an

Achilles heel—pit corrosion. In certain environments, pit corrosion can penetrate several millimeters in a matter of weeks.

Metallurgy, the science of producing useful

metallic materials, almost always requires some

kind of compromise. In the case of stainless steel,

inclusions of MnS make the steel easier to machine

into useful parts, but such inclusions are also the

source of pit corrosion. Recently a group of British

researchers analyzed stainless steel using a highenergy beam of ions that blasted atoms loose

from the steel surface. Studies of the resultant

atom vapor revealed the source

of the problem. It

turns out that

when the stainless

steel is cooling,

the MnS inclusions

“suck’’ chromium

atoms from the

surrounding area,

leaving behind a



chromium-deficient region. The corrosion occurs

in this region, as illustrated in the accompanying

diagram. The essential problem is that to resist corrosion steel must contain at least 13% Cr atoms.

The low-chromium region around the inclusion is

not stainless steel—so it corrodes just like regular

steel. This corrosion leads to a pit that causes deterioration of the steel surface.

Now that the reason for the pit corrosion is

understood, metallurgists should be able to develop methods of stainless steel formulation that

avoid this problem. One British scientist, Mary

P. Ryan, suggests that heat treatment of the

stainless steel may solve the problem by allowing

Cr atoms to diffuse from the inclusion back into

the surrounding area. Because corrosion of regular steel is such an important issue, finding ways

to make cheaper stainless steel will have a significant economic impact. We need stainless without the pits.



603



604 Chapter 18 Oxidation–Reduction Reactions and Electrochemistry

Corrosion can be viewed as the process of returning metals to their

natural state—the ores from which they were originally obtained. Corrosion

involves oxidation of the metal. Because corroded metal often loses its

strength and attractiveness, this process causes great economic loss. For example, approximately one-fifth of the iron and steel produced annually is

used to replace rusted metal.

Because most metals react with O2, we might expect them to corrode so

fast in air that they wouldn’t be useful. It is surprising, therefore, that the

problem of corrosion does not virtually prevent the use of metals in air. Part

of the explanation is that most metals develop a thin oxide coating, which

tends to protect their internal atoms against further oxidation. The best example of this is aluminum. Aluminum readily loses electrons, so it should be

very easily oxidized by O2. Given this fact, why is aluminum so useful for

building airplanes, bicycle frames, and so on? Aluminum is such a valuable

structural material because it forms a thin adherent layer of aluminum oxide, Al2O3, which greatly inhibits further corrosion. Thus aluminum protects

itself with this tough oxide coat. Many other metals, such as chromium,

nickel, and tin, do the same thing.

Iron can also form a protective oxide coating. However, this oxide is

not a very effective shield against corrosion, because it scales off easily, exposing new metal surfaces to oxidation. Under normal atmospheric conditions, copper forms an external layer of greenish copper sulfate or carbonate

called patina. Silver tarnish is silver sulfide, Ag2S, which in thin layers gives

the silver surface a richer appearance. Gold shows no appreciable corrosion

in air.

Preventing corrosion is an important way of conserving our natural

supplies of metals and energy. The primary means of protection is the application of a coating—most often paint or metal plating—to protect the metal

from oxygen and moisture. Chromium and tin are often used to plate steel

because they oxidize to form a durable, effective oxide coating.

Alloying is also used to prevent corrosion. Stainless steel contains

chromium and nickel, both of which form oxide coatings that protect the

steel.

Cathodic protection is the method most often employed to protect

steel in buried fuel tanks and pipelines. A metal that furnishes electrons

more easily than iron, such as magnesium, is connected by a wire to the

pipeline or tank that is to be protected (Figure 18.9). Because the magnesium

is a better reducing agent than iron, electrons flow through the wire from the

magnesium to the iron pipe. Thus the electrons are furnished by the magnesium rather than by the iron, keeping the iron from being oxidized. As oxidation of the magnesium occurs, the magnesium dissolves, so it must be replaced periodically.



Image not

available due

to copyright

restrictions



18.8 Electrolysis

OBJECTIVE:



To understand the process of electrolysis and learn about the commercial

preparation of aluminum.

Unless it is recharged, a battery “runs down” because the substances in it that

furnish and accept electrons (to produce the electron flow) are consumed.

For example, in the lead storage battery (see Section 18.6), PbO2 and Pb are

consumed to form PbSO4 as the battery runs.

PbO2(s) ϩ Pb(s) ϩ 2H2SO4(aq) S 2PbSO4(s) ϩ 2H2O(l)



C H E M I S T R Y I N F OCUS

Water-Powered Fireplace



H



An electrolytic cell uses

electrical energy to produce a

chemical change that would not

otherwise occur.



fuel. The Aqueon fireplace

uses electrolysis to decompose the water to H2(g) and

O2(g); the hydrogen is then

burned to furnish heat for

the home. The 31,000-Btu

fireplace features copper and

stainless steel and has a contemporary design (see accompanying photo). To operate,

the fireplace is simply hooked

up to the water and electrical

supplies for the home.



Courtesy, Hearth & Home Technologies



ydrogen gas is being touted

as an environmentally friendly

fuel because, unlike fossil fuels,

it does not produce the greenhouse gas carbon dioxide. The

only product of combustion of

H2 is water. As a result, hydrogen is being investigated as a

possible fuel for cars, trucks,

and buses. Now comes a manufacturer, Heat & Glo, that is

showcasing an in-home fireplace that uses water as the



However, one of the most useful characteristics of the lead storage battery is

that it can be recharged. Forcing current through the battery in the direction

opposite to the normal direction reverses the oxidation–reduction reaction.

That is, PbSO4 is consumed and PbO2 and Pb are formed in the charging

process. This recharging is done continuously by the automobile’s alternator, which is powered by the engine.

The process of electrolysis involves forcing a current through a cell to

produce a chemical change that would not otherwise occur.

One important example of this type of process is the electrolysis of water. Water is a very stable substance that can be broken down into its elements by using an electric current (Figure 18.10).

Forced

electric



Charles D. Winters/Photo Researchers, Inc.



2H2O(l) 87777777n

2H2(g) ϩ O2(g)

current



Figure 18.10

The electrolysis of water

produces hydrogen gas at the

cathode (on the left) and oxygen

gas at the anode (on the right).

A nonreacting strong electrolyte

such as Na2SO4 is needed to

furnish ions to allow the flow

of current.



The electrolysis of water to produce hydrogen and oxygen occurs

whenever a current is forced through an aqueous solution. Thus, when the

lead storage battery is charged, or “jumped,” potentially explosive mixtures

of H2 and O2 are produced by the current flow through the solution in the

battery. This is why it is very important not to produce a spark near the battery during these operations.

Another important use of electrolysis is in the production of metals

from their ores. The metal produced in the greatest quantities by electrolysis

is aluminum.

Aluminum is one of the most abundant elements on earth, ranking

third behind oxygen and silicon. Because aluminum is a very reactive metal,

it is found in nature as its oxide in an ore called bauxite (named after Les

Baux, France, where it was discovered in 1821). Production of aluminum

metal from its ore proved to be more difficult than the production of most

other metals. In 1782, Lavoisier, the pioneering French chemist, recognized

aluminum as a metal “whose affinity for oxygen is so strong that it cannot

be overcome by any known reducing agent.” As a result, pure aluminum



605



606 Chapter 18 Oxidation–Reduction Reactions and Electrochemistry



Charles Martin Hall (1863–1914) was a

student at Oberlin College in Ohio when

he first became interested in aluminum.

One of his professors commented that

anyone who could find a way to

manufacture aluminum cheaply would

make a fortune, and Hall decided to give

it a try. The 21-year-old worked in a

wooden shed near his house with an iron

frying pan as a container, a blacksmith’s

forge as a heat source, and galvanic cells

constructed from fruit jars. Using these

crude galvanic cells, Hall found that he

could produce aluminum by passing a

current through a molten mixture of Al2O3

and Na3AIF6. By a strange coincidence,

Paul Heroult, a French chemist who was

born and died in the same years as Hall,

made the same discovery at about the

same time.



The Granger Collection



Figure 18.11



Table 18.2 The Price of

Aluminum,

1855–1990

Date



Price of Aluminum

($/lb)*



1855



$100,000



1885



100



1890



2



1895



0.50



1970



0.30



1980



0.80



1990



0.74



*Note the precipitous drop in price after

the discovery of the Hall–Heroult process

in 1886.



C H A P T E R



metal remained unknown. Finally, in 1854, a process was found for producing metallic aluminum by using sodium, but aluminum remained a very expensive rarity. In fact, it is said that Napoleon III served his most honored

guests with aluminum forks and spoons, while the others had to settle for

gold and silver utensils!

The breakthrough came in 1886 when two men, Charles M. Hall in the

United States (Figure 18.11) and Paul Heroult in France, almost simultaneously discovered a practical electrolytic process for producing aluminum,

which greatly increased the availability of aluminum for many purposes.

Table 18.2 shows how dramatically the price of aluminum dropped after this

discovery. The effect of the electrolysis process is to reduce Al3ϩ ions to neutral Al atoms that form aluminum metal. The aluminum produced in this

electrolytic process is 99.5% pure. To be useful as a structural material, aluminum is alloyed with metals such as zinc (for trailer and aircraft construction) and manganese (for cooking utensils, storage tanks, and highway

signs). The production of aluminum consumes about 4.5% of all electricity

used in the United States.



18



REVIEW



Key Terms

oxidation–reduction

(redox) reactions (18.1)

oxidation (18.1, 18.3)

reduction (18.1, 18.3)

oxidation states (18.2)

oxidizing agent (electron

acceptor) (18.3)



reducing agent (electron

donor) (18.3)

half-reactions (18.4)

electrochemistry (18.5)

electrochemical battery

(galvanic cell) (18.5)



anode (18.5)

cathode (18.5)

electrolysis (18.5, 18.8)

lead storage battery (18.6)



potential (18.6)

dry cell battery (18.6)

corrosion (18.7)

cathodic protection (18.7)



Chapter Review



F



VP



directs you to the Chemistry in Focus feature in the chapter

indicates visual problems

interactive versions of these problems are assignable in OWL



607



1. Sketch a galvanic cell, and explain how it works. Look

at Figures 18.1 and 18.5. Explain what is occurring in

each container and why the cell in Figure 18.5

“works,” but the one in Figure 18.1 does not.

2. Make a list of nitrogen compounds with as many different oxidation states for nitrogen as you can.



Summary

1. Oxidation–reduction reactions involve a transfer of

electrons. Oxidation states provide a way to keep

track of electrons in these reactions. A set of rules is

used to assign oxidation states.

2. Oxidation is an increase in oxidation state (a loss of

electrons); reduction is a decrease in oxidation state

(a gain of electrons). An oxidizing agent accepts electrons, and a reducing agent donates electrons. Oxidation and reduction always occur together.

3. Oxidation–reduction equations can be balanced by

inspection or by the half-reaction method. This

method involves splitting a reaction into two parts

(the oxidation half-reaction and the reduction halfreaction).

4. Electrochemistry is the study of the interchange of

chemical and electrical energy that occurs through

oxidation–reduction reactions.

5. When an oxidation–reduction reaction occurs with

the reactants in the same solution, the electrons are

transferred directly, and no useful work can be obtained. However, when the oxidizing agent is separated from the reducing agent, so that the electrons

must flow through a wire from one to the other, chemical energy is transformed into electrical energy. The

opposite process, in which electrical energy is used to

produce chemical change, is called electrolysis.

6. A galvanic (electrochemical) cell is a device in which

chemical energy is transformed into useful electrical

VP

energy. Oxidation occurs at the anode of a cell; reduction occurs at the cathode.

7. A battery is a galvanic cell, or group of cells, that

serves as a source of electric current. The lead storage

battery has a lead anode and a cathode of lead coated

with PbO2, both immersed in a solution of sulfuric

acid. Dry cell batteries do not have liquid electrolytes

but contain a moist paste instead.

8. Corrosion involves the oxidation of metals to form

mainly oxides and sulfides. Some metals, such as aluminum, form a thin protective oxide coating that inhibits their further corrosion. Corrosion of iron can

be prevented by a coating (such as paint), by alloying,

and by cathodic protection.



3. Which of the following are oxidation–reduction reactions? Explain.

a.

b.

c.

d.

e.



PCl3 ϩ Cl2 S PCl5

Cu ϩ 2AgNO3 S Cu(NO3)2 ϩ 2Ag

CO2 ϩ 2LiOH S Li2CO3 ϩ H2O

FeCl2 ϩ 2NaOH S Fe(OH)2 ϩ 2NaCl

MnO2 ϩ 4HCl S Cl2 ϩ 2H2O ϩ MnCl2



4. Which of the following statements is (are) true? Explain. (There may be more than one answer.)

a. Oxidation and reduction cannot occur independently of each other.

b. Oxidation and reduction accompany all chemical

changes.

c. Oxidation and reduction describe the loss and

gain of electron(s), respectively.

5. Why do we say that when something gains electrons

it is reduced? What is being reduced?

6. The equation Agϩ ϩ Cu S Cu2ϩ ϩ Ag has equal numbers of each type of element on each side of the equation. This equation, however, is not balanced. Why is

this equation not balanced? Balance the equation.

7. In balancing oxidation–reduction equations, why is

it permissible to add water to either side of the equation?

8. What does it mean for a substance to be oxidized? The

term “oxidation” originally came from substances reacting with oxygen gas. Explain why a substance that

reacts with oxygen gas will always be oxidized.

9. Label the following parts of the galvanic cell.

anode

cathode

reducing agent

oxidizing agent

e–

e–



e–



A



C



Active Learning Questions

These questions are designed to be considered by groups of

students in class. Often these questions work well for introducing a particular topic in class.



–

B



Ions



+

D



608 Chapter 18 Oxidation–Reduction Reactions and Electrochemistry



Questions and Problems

18.1 Oxidation–Reduction Reactions

QUESTIONS



10. Why is fluorine always assigned an oxidation state of

Ϫ1? What oxidation number is usually assigned to the

other halogen elements when they occur in compounds? In an interhalogen compound involving fluorine

(such as ClF), which atom has a negative oxidation state?



1. Give some examples of how we make good use of

oxidation–reduction reactions in everyday life.



11. The sum of all the oxidation states of all the atoms in

H3PO4 is

.



2. How do chemists define the processes of oxidation

and reduction? Write a simple equation illustrating

each of your definitions.



12. The sum of all the oxidation states of all the atoms in

PO43Ϫ is

.



3. For each of the following oxidation–reduction reactions, identify which element is being oxidized and

which is being reduced.

a.

b.

c.

d.



Cl2(g) ϩ I2(g) S 2ICl(g)

Cl2(g) ϩ 2Li(s) S 2LiCl(s)

2Na(s) ϩ 2H2O(l) S 2NaOH(aq) ϩ H2(g)

Cl2(g) ϩ 2NaBr(aq) S 2NaCl(aq) ϩ Br2(l)



4. For each of the following oxidation–reduction reactions, identify which element is being oxidized and

which is being reduced.

a.

b.

c.

d.



4B(s) ϩ 3O2(g) S 2B2O3(s)

N2(g) ϩ 2O2(g) S 2NO2(g)

CaC2(s) ϩ H2(g) S CaH2(g) ϩ 2C(s)

CuSO4(aq) ϩ Mg(s) S MgSO4(aq) ϩ Cu(s)



5. For each of the following oxidation–reduction reactions, identify which element is being oxidized and

which is being reduced.

a.

b.

c.

d.



Ca(s) ϩ 2H2O(l) S Ca(OH)2(s, aq) ϩ H2(g)

H2(g) ϩ F2(g) S 2HF(g)

4Fe(s) ϩ 3O2(g) S 2Fe2O3(s)

2Fe(s) ϩ 3Cl2(g) S 2FeCl3(s)



6. For each of the following oxidation–reduction reactions, identify which element is being oxidized and

which is being reduced.

a.

b.

c.

d.



2Cr2S3(s) ϩ 3O2(g) S 2Cr2O3(s) ϩ 6S(s)

P4(s) ϩ 5O2(g) S P4O10(s)

CO2(g) ϩ H2(g) S CO(g) ϩ H2O(g)

2B(s) ϩ 3H2(g) S B2H6(g)



18.2 Oxidation States

QUESTIONS

7. What is an oxidation state? Why do we define such a

concept?

8. Why is the oxidation state of an element in the uncombined state equal to zero?

9. Explain why, although it is not an ionic compound,

we still assign oxygen an oxidation state of Ϫ2 in water, H2O. Give an example of a compound in which

oxygen is not in the Ϫ2 oxidation state.



PROBLEMS

13. Assign oxidation states to all of the atoms in each of

the following.

a. CBr4

b. HClO4



c. K3PO4

d. N2O



14. Assign oxidation states to all of the atoms in each of

the following.

a. CrCl3

b. Ni(OH)2



c. H2S

d. CS2



15. What is the oxidation state of sulfur in each of the following substances?

a. S8

b. H2SO4



c. NaHSO4

d. Na2S



16. What is the oxidation state of nitrogen in each of the

following substances?

a. N2

b. NH3



c. NO2

d. NaNO3



17. What is the oxidation state of chlorine in each of the

following substances?

a. ClF

b. Cl2



c. HCl

d. HClO



18. What is the oxidation state of manganese in each of

the following substances?

a. MnCl2

b. KMnO4



c. MnO2

d. Mn(C2H3O2)3



19. Assign oxidation states to all of the atoms in each of

the following.

a. CuCl2

b. KClO3



c. KClO4

d. Na2CO3



20. Assign oxidation states to all of the atoms in each of

the following.

a. CaO

b. Al2O3



c. PF3

d. P2O5



21. Assign oxidation states to all of the atoms in each of

the following ions:

a. CO32Ϫ

b. NO3Ϫ



c. PO43Ϫ

d. SO42Ϫ



All even-numbered Questions and Problems have answers in the back of this book and solutions in the Solutions Guide.



Chapter Review

22. Assign oxidation states to all of the atoms in each of

the following ions:

a. HSO4Ϫ

b. MnO4Ϫ



c. ClO3Ϫ

d. BrO4Ϫ



18.3 Oxidation–Reduction Reactions

Between Nonmetals

QUESTIONS

23. Oxidation can be defined as a loss of electrons or as

an increase in oxidation state. Explain why the two

definitions mean the same thing, and give an example to support your explanation.

24. Reduction can be defined as a gain of electrons or as

a decrease in oxidation state. Explain why the two definitions mean the same thing, and give an example

to support your explanation.

25. What is an oxidizing agent? What is a reducing agent?

26. Give an example of a simple oxidation–reduction

equation. Identify the species being oxidized and the

species being reduced. Identify the oxidizing agent

and the reducing agent in your example.

27. Does an oxidizing agent donate or accept electrons?

Does a reducing agent donate or accept electrons?

F 28. The “Chemistry in Focus” segment Do We Age by Oxi-



dation? discusses antioxidants. What does it mean for

a chemical to be an antioxidant? How would it work

chemically?



PROBLEMS

29. In each of the following reactions, identify which element is being oxidized and which is being reduced

by assigning oxidation numbers.

a. Fe(s) ϩ CuSO4(aq) S FeSO4(aq) ϩ Cu(s)

b. Cl2(g) ϩ 2NaBr(aq) S 2NaCl(aq) ϩ Br2(l)

c. 3CuS(s) ϩ 8HNO3(aq) S

3CuSO4(aq) ϩ 8NO(g) ϩ 4H2O(l)

d. 2Zn(s) ϩ O2(g) S 2ZnO(s)

30. In each of the following reactions, identify which element is being oxidized and which is being reduced

by assigning oxidation numbers.

a.

b.

c.

d.



2Al(s) ϩ 3S(s) S Al2S3(s)

CH4(g) ϩ 2O2(g) S CO2(g) ϩ 2H2O(g)

2Fe2O3(s) ϩ 3C(s) S 3CO2(g) ϩ 4Fe(s, l)

K2Cr2O7(aq) ϩ 14HCl(aq) S

2KCl(aq) ϩ 2CrCl3(s) ϩ 7H2O(l) ϩ 3Cl2(g)



31. In each of the following reactions, identify which element is being oxidized and which is being reduced

by assigning oxidation states.

a.

b.

c.

d.



2Cu(s) ϩ S(s) S Cu2S(s)

2Cu2O(s) ϩ O2(g) S 4CuO(s)

4B(s) ϩ 3O2(g) S 2B2O3(s)

6Na(s) ϩ N2(g) S 2Na3N(s)



609



32. In each of the following reactions, identify which element is being oxidized and which is being reduced

by assigning oxidation numbers.

a. 4KClO3(s) ϩ C6H12O6(s) S

4KCl(s) ϩ 6H2O(l) ϩ 6CO2(g)

b. 2C8H18(l) ϩ 25O2(g) S 16CO2(g) ϩ 18H2O(l)

c. PCl3(g) ϩ Cl2(g) S PCl5(g)

d. Ca(s) ϩ H2(g) S CaH2(g)

33. Pennies in the United States consist of a zinc core

that is electroplated with a thin coating of copper.

Zinc dissolves in hydrochloric acid, but copper does

not. If a small scratch is made on the surface of a

penny, it is possible to dissolve away the zinc core,

leaving only the thin shell of copper. Identify which

element is oxidized and which is reduced in the reaction for the dissolving of the zinc by the acid.

Zn(s) ϩ 2HCl(aq) S ZnCl2(aq) ϩ H2(g)

34. Iron ores, usually oxides of iron, are converted to the

pure metal by reaction in a blast furnace with carbon

(coke). The carbon is first reacted with air to form carbon monoxide, which in turn reacts with the iron oxides as follows:

F2O3(s) ϩ 3CO(g) S 2Fe(l) ϩ 3CO2(g)

Identify the atoms that are oxidized and reduced, and

specify the oxidizing and reducing agents.

35. Although magnesium metal does not react with water at room temperature, it does react vigorously with

steam at higher temperatures, releasing elemental hydrogen gas from the water.

Mg(s) ϩ 2H2O(g) S Mg(OH)2(s) ϩ H2(g)

Identify which element is being oxidized and which

is being reduced.

36. Potassium iodide in solution reacts readily with many

reagents. In the following reactions, identify the

atoms that are being oxidized and reduced, and specify the oxidizing and reducing agents.

a. Cl2(g) ϩ KI(aq) S KCl(aq) ϩ I2(s)

b. 2FeCl3(aq) ϩ 2KI(aq) S

2FeCl2(aq) ϩ 2KCl(aq) ϩ I2(s)

c. 2CuCl2(aq) ϩ 4KI(aq) S

2CuI(s) ϩ 4KCl(aq) ϩ I2(s)



18.4 Balancing Oxidation–Reduction Reactions

by the Half-Reaction Method

QUESTIONS

37. In what two respects must oxidation–reduction reactions be balanced?

38. Why is a systematic method for balancing oxidation–

reduction reactions necessary? Why can’t these equations be balanced readily by inspection?



All even-numbered Questions and Problems have answers in the back of this book and solutions in the Solutions Guide.