Kelter P., Mosher M., Scott A. Chemistry. The Practical Science

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848 Chapter 19 Electrochemistry

Selected Batteries

Zinc–carbon battery Also known as a standard carbon bat-

tery. Zinc–carbon chemistry is used in all inexpensive AA, C,

and D dry-cell batteries. The electrodes are zinc and carbon,

with an acidic paste between them that serves as the

electrolyte.

Alkaline battery Used in common Duracell and Energizer

batteries. The electrodes are zinc and manganese oxide, with

an alkaline electrolyte.

Lithium photo battery Lithium, lithium iodide, and lead-

iodide are used in cameras because of their ability to supply

power surges.

Lead–acid battery (rechargeable) Used in automobiles. The

electrodes are made of lead and lead oxide with a strong

acidic electrolyte.

Nickel–cadmium battery (rechargeable) The electrodes are

nickel hydroxide and cadmium, with potassium hydroxide

as the electrolyte.

TABLE 19.5

The Chemistry of Some Common Batteries

Nickel metal hydride (NiMH) rechargeable batteries are used in many cellular

phones (Figure 19.13). During the charging phase, an external source of electric-

ity causes water in the electrolyte (often aqueous potassium hydroxide) to react

with a rare earth– or zirconium metal–based alloy at what will be the negative

electrode of the battery when it is in operation. This generates hydrogen atoms

that are absorbed into the alloy, and releases hydroxide ions:

Alloy + H

O(l) + e

−

n Alloy−H(s) + OH

−

(aq) (reduction)

At the other electrode, which will be the positive electrode when the battery is

powering the phone, nickel hydroxide reacts with hydroxide ions to form nickel

oxyhydroxide, which has nickel in what for it is an unusual +3 oxidation state:

Ni(OH)

(s) + OH

−

(aq) n NiOOH + H

O + e

−

(oxidation)

When the battery is in use, the hydrogen atoms that were absorbed into the alloy

at the negative electrode are released, combining with hydroxide ions to form

water and supply the electrons that ﬂow through a circuit to power the phone.

Alloy−H(s) + OH

−

(aq) n Alloy + H

O(l) + e

−

(oxidation)

At the positive electrode, nickel oxyhydroxide is reduced back to nickel hydroxide

by the electrons that arrive through the circuit, having done their work for us:

NiOOH(s) + H

O(l) + e

−

n Ni(OH)

(s) + OH

−

(aq) (reduction)

The cycle of charge and discharge can be repeated many times, to power all

the talking and text messaging on the move that is such a pervasive part of mod-

ern life.

A typical nonrechargeable “alkaline” battery for a ﬂashlight (Figure 19.14)

uses the oxidation of zinc metal into zinc ions to generate the electrons for the

electric current:

Zn(s) n Zn

(aq) + 2e

−

(oxidation)

Negative

electrode

(hydrogen-

absorbing

alloy)

Case (–) Cap (+)

Insulator

Positive

electrode

NiO(OH)

Separator

FIGURE 19.13

Nickel metal hydride cell.

Nickel–metal hydride battery (rechargeable) This bat-

tery is rapidly replacing nickel–cadmium because it

does not suffer from the “voltage depression”that nickel-

cadmiums do, in which repeated charging after only

partial discharges prevents it from fully discharging.

Lithium–ion battery (rechargeable) With a very good

power-to-weight ratio, this is often found in high-end

laptop computers and cell phones.

Zinc–air battery This battery is lightweight and

rechargeable.

Zinc–mercury oxide battery This is often used in hear-

ing aids.

Silver–zinc battery This is used in aeronautical applica-

tions because the power-to-weight ratio is good.

Metal–chloride battery Used in electric vehicles.

Hydrogen fuel cell Used in electric vehicles and to

power the space shuttle.

19.5 Chemical Reactivity Series 849

Anode

(zinc)

Cathode (graphite)

Paste of

MnO

Cl,

and carbon

Separator

FIGURE 19.14

A common dry cell battery.

Aluminum

case

Negative

electrode

Positive

electrode

Positive

terminal

Negative

terminal

Separator

FIGURE 19.15

A lithium ion battery.

When electrons ﬂow back into the battery at the other electrode, they combine

with manganese dioxide:

2MnO

(s) + 2H

O(l) + 2e

−

n 2MnO(OH)(s) + 2OH

−

(aq) (reduction)

Therefore, the indirect reaction of zinc with manganese dioxide is the source of

the energy that lights the bulb.

Your laptop computer may be powered by a “lithium ion battery” (Fig-

ure 19.15). These batteries use lithium oxide mixed with other metal oxides as the

positive electrode, and crystalline graphite with lithium ions intercalated within

it as the negative electrode. Unlike most conventional batteries, however, the

lithium ion battery is not powered by a redox reaction. Instead, lithium ions

move back and forth within the battery during the cycle of charging and recharg-

ing, accompanied by electrons moving in the external circuit. During charging,

the lithium ions are driven into the graphite cathode by application of the exter-

nal electric current. When the battery is used as a source of power, the ions drift

back to the lithium oxide anode. As the ions leave the cathode, electrons must

travel around the external circuit, ensuring that there is no overall transfer of

electric charge from cathode to anode as the battery is discharged.

19.5 Chemical Reactivity Series

Plumbers often use metal pipes to deliver water from the main water line to your

sink faucet. In fact, plumbers used to use lead pipes, but because the lead in the

pipe can leach into the water and cause heavy metal poisoning, that practice isn’t

followed anymore. Even though plastic pipes made from polyvinyl chloride

(PVC) are much cheaper than metal pipes, there is still a demand for copper

pipes. Why do plumbers use copper for pipes? The demand is mostly due to the

durability of copper compared to plastic, but why don’t they use iron, lithium,

calcium, or magnesium pipes?

The Table of Standard Reduction Potentials (Table 19.3) provides not only

cell potentials but also a ranking of reducing agents and oxidizing agents. This

ranking is called a

reactivity series. Table 19.6 shows a relative listing of some of

the more common metals and includes hydrogen as a reference point. The

strongest reducing agents (those elements that are most easily oxidized) are the

most reactive metals and can be found at the top of the table: Li, Na, and Mg.

Copper pipes in a home transport water

but don’t corrode easily.

Because these metals are such good reducing agents, you would not want an ear-

ring made out of them; a better choice would be a relatively unreactive metal (and

a weak reducing agent) such as gold, copper, or silver, found at the bottom of the

table. “Fourteen-carat” gold (which means that

⁄24 of the sample is gold) is an

alloy of gold, copper, and silver. “Twenty-four-carat” gold is pure gold. The reac-

tivity series has social importance beyond its signiﬁcance to earrings. Underwater

steel pipelines, which contain substantial amounts of iron, electrochemically cor-

rode (the metal deteriorates via oxidation) as a consequence of the interaction of

the iron with water, salt, and oxygen dissolved in the water. This

corrosion can be

minimized by putting, for example, magnesium strips in direct contact with the

pipeline, shown in Figure 19.16. The magnesium, which is more chemically active

than iron, will preferentially oxidize, ideally leaving the iron in its elemental form.

The magnesium, in effect, is sacriﬁced for the good of the pipeline and therefore

is known as a

sacriﬁcial anode. Other sacriﬁcial anodes include aluminum

wrapped around steel in hot water heaters and zinc coating the propellers and

rudders of ships.

EXERCISE 19.9 Which Is More Reactive?

Iron, especially in the form of stainless steel, can be used in jewelry, such as in the

post of an earring. Where does iron fall on the activity series? Based on the poten-

tial for the oxidation of iron to iron(III) ion, why is iron suitable (or not suitable)

for jewelry?

Solution

From Table 19.3 we see that the oxidation of metallic iron, Fe(s), has a half-reaction

potential of +0.04 V. Metallic copper (−0.34 V), silver (−0.80 V), and gold

(−1.50 V) have negative potentials. From this information we can conclude that

iron is more reactive than copper, silver, or gold. In this sense pure iron would not

be suitable, and you may know that iron spontaneously reacts to make rust when in

contact with moisture and air. Stainless steel, however, is an alloy formulated to in-

hibit corrosion. Some formulas have as much as 18% chromium and 8% nickel

added to the iron. The half-reaction potential for stainless steel is different, and it is

difﬁcult to predict reactivity from the E° values for its constituent elements.

850 Chapter 19 Electrochemistry

Reactivity Series of the Metals

Ion Atom

Ions difﬁcult to displace Li

Li Metals that react with water

Zn Metals that react with acid

Ions easy to displace H

Ag Metals that are highly unreactive

TABLE 19.6

Steel

pipeline

Sacrificial

anode

FIGURE 19.16

A sacriﬁcial anode on a steel pipeline.

Reaction of copper with water.

Sodium reacting with water.

Video Lesson: Corrosion and the

Prevention of Corrosion

PRACTICE 19.9

Arrange the following in decreasing order of reactivity:

Na Al Ca Cu

See Problems 63–66.

19.6 Not-So-Standard Conditions:

The Nernst Equation

Native copper and other copper objects can be cleaned with relatively dilute solu-

tions of nitric acid. Concentrated nitric acid is too strong for the job, as Ira

Remsen noted, so ancient coins should not be cleaned with concentrated nitric

acid. A very dilute solution in the hands of a professional, however, can transform

a 1600-year-old coin into a masterpiece that looks as new as the day it was

minted.

How does lowering the concentration of nitric acid change the reactivity? It

does reduce the rate of the reaction (see Chapter 15), but is there an effect on the

potential of the reaction as well? What about the temperature of the reaction? We

know that in general, the rate of a reaction increases as the temperature is raised

(also from Chapter 15). Does cold, dilute nitric acid still react with copper? We

can understand these relationships, which are at nonstandard conditions, by con-

sidering a simpler system in which the copper ion reacts with zinc metal.

If we put a zinc strip into a solution of copper(II) chloride, we note that a dark

ﬁlm immediately begins to form on the zinc strip (Figure 19.17a on page 852).

Within an hour, the entire zinc strip has oxidized into the solution, and we are left

with a brown mass of copper metal (Figure 19.17b). We also note the disappear-

ance of the blue color that is characteristic of Cu

in water. This is consistent

with the following half-reactions and overall cell reaction:

(aq) + 2e

−

n Cu(s) E° =+0.34 V

Zn(s) n Zn

(aq) + 2e

−

E° =+0.76 V

(aq) + Zn(s) n Zn

(aq) + Cu(s) E°

cell

=+1.10 V

We know from Section 19.3 that we can relate the free energy of a system to the

cell potential:



G =−nFE

We also discussed in Section 14.6 the relationship between free energy change

and the standard free energy change:



G =



G ° + RT lnQ

in which Q = the reaction quotient, the ratio of the concentrations (more prop-

erly, the activities) of the products over the reactants; R is the universal gas con-

stant; and T is the temperature in kelvins. For our current example,



G =



G° + RT ln

[Zn

]

[Cu

]

We can substitute for



G the expression for the cell potential,



G =−nFE:

−nFE =−nFE ° + RT ln

[Zn

]

[Cu

]

Dividing each term by −nF enables us to solve for the cell potential at nonstan-

dard conditions:

E = E° −

[Zn

]

[Cu

]

19.6 Not-So-Standard Conditions: The Nernst Equation 851

This is one form of the Nernst equation, named after Walther Hermann Nernst

(1864–1941), a German chemist who studied the effect of concentration on the

potential of an electrochemical cell. We can write a more general form of the

equation, taking into account the reaction quotient for any process:

E = E

◦

−

ln Q

Because most reactions are conducted at standard temperature (25°C, 298 K), we

can calculate the term

8.3145 J·mol

−1

·K

−1

(298 K)

96485 C·mol

−1

= 0.0257

which we can substitute into the Nernst equation in this form:

E = E

◦

−

0.0257

ln Q

Although our calculators can easily handle the natural logarithm (“ln”) in the

equation, we typically convert the equation to the more familiar base 10 “log” by

multiplying by 2.3026 (that is,log(10) =1, and ln 10 =2.3026,so log =ln/2.3026).

852 Chapter 19 Electrochemistry

(a) (b)

–

FIGURE 19.17

Zinc metal reacting with copper(II) chloride

solution.

Video Lesson: The Nernst

Equation

FIGURE 19.18

The concentration cell. The voltage observed in the

copper concentration cell is due to the differences in

concentration of Cu

at the anode and cathode.

To account for that, we multiply the coefﬁcient,

0.0257

, by 2.3026, which gives

us the ﬁnal, common form of the Nernst equation:

E = E

◦

−

0.0592

log Q

Using this equation, we can determine the effect of lowering the concentration of

nitric acid on the potential of the copper oxidation.

The Nernst equation can also be used to measure the concentration of a solu-

tion if the standard cell potential and the actual potential are known. Electro-

chemists take advantage of this use of the Nernst equation via ion-selective

electrodes, in which the concentration of an ion such as chloride, ammonium,

cadmium, nitrate, or hydrogen (in a pH electrode) is determined. And there’s

something else that’s interesting about the Nernst equation. For example, let’s

consider the hypothetical redox reaction between copper metal and copper ions

in solution. The potential of the reduction half-reaction and that of the oxidation

half-reaction are the same at standard conditions, and, as we’d expect, no net po-

tential should be noticed for an electrochemical cell containing these reactions.

(aq) + 2e

−

n Cu(s) E°

red

=+0.34 V

Cu n Cu

(aq) + 2e

−

E°

=−0.34 V

Cu(s) + Cu

(aq) n Cu

(aq) + Cu(s) E°

cell

= 0.00 V

However, what would happen if we increased the concentration of the reactant

copper ions to 2.0 M instead of the standard 1.0 M? Doing so changes the distri-

bution of the species in the reaction. Using the Nernst equation, we can calculate

the result of our modiﬁcation.

E = E

−

0.0592

log



1.0

2.0



E = 0.00 −

0.0592

log(0.5)

E = 0.00 − (−0.0089)

E =+0.0089 V

The electrochemical potential of the reaction is a nonzero value. By ad-

justing the concentrations of the products and reactants, we have cre-

ated an electrochemical cell. This type of cell is called a

concentration

cell

because the concentrations are driving the potential of the cell (Fig-

ure 19.18). This is the same type of potential that develops across the

membrane of a muscle cell or nerve cell.

How does our cell notation change to reﬂect the fact that the con-

ditions are not-so-standard? By indicating the concentrations in

parentheses immediately after the species, we can immediately show

how the reaction should be written. For example,

Cu(s) | Cu

(aq) (1.0 M) || Cu

(aq) (2.0 M) | Cu(s)

EXERCISE 19.10 Heart Cell Potential

At the beginning of this chapter, we mentioned the electrical signals in the heart

muscle. Given the differing concentrations of potassium ions inside and outside the

heart cells, what is the electrochemical potential that corresponds to this concentra-

tion gradient? Assume that the electrochemical cell in the body can be represented

by the following cell notation. (In truth, no elemental potassium exists in the

19.6 Not-So-Standard Conditions: The Nernst Equation 853

Cathode

Anode

Potential

Salt bridge

human body. We use this concentration cell as a model merely to estimate the

potential that is obtained in a heart muscle cell.)

K(s) | K

(aq) (0.005 M) || K

(aq) (0.166 M) | K(s)

Solution

This concentration cell is based on the potassium half-reaction. The complete

reaction is

K(s) + K

(aq) n K(s) + K

(aq) E°

cell

= 0.00 V

There is one electron involved in the reaction. Using the information from the

Nernst equation, we get a potential of +0.090 V.

cell

= E

◦

cell

−

0.0592

log



product

]

reactant

]



cell

= 0.00 −

0.0592

log



0.005

0.166



cell

= 0.00 −0.0592 log (0.03012)

cell

= 0.00 −0.0592(−1.521)

cell

= 0.00 −(−0.09005)

cell

=+0.090 V

In the heart cell, the sodium gradient provides a potential of −0.052 V. The net

potential across the membrane of the heart cell is +0.038 V.

PRACTICE 19.10

What is the potential of the following electrochemical cell written in shorthand cell

notation? What is the half-reaction listed on the right-hand side? Is this a voltaic

cell? (Assume that the pressure of hydrogen gas is 1.0 atm in each half-cell.)

(g) | H

(aq) (1.0 M) || H

(aq) (0.10 M) | H

(g)

See Problems 69–72.

When we listen to a battery-powered radio, the sound tends to get softer as

the radio is used. What is happening inside the battery that causes this power

loss? During the progress of the reaction inside the battery, the reactants are being

used up as the products are being formed. According to the Nernst equation, as

the value of Q gets larger, the modiﬁcation to the standard cell potential (E°) gets

more and more negative and closer and closer to zero. What is the result? As the

reaction proceeds, the potential of the battery decreases as the system inches ever

closer to equilibrium, and the music gets softer. At some point, the battery doesn’t

have enough voltage to run the radio. It has not yet reached equilibrium, but it is

below the threshold that will allow the radio to operate. We say that the batteries

are dead. As we have pointed out already, some batteries can be recharged, be-

cause their discharge reactions can be run in reverse. Rechargeable batteries can

be charged only a ﬁnite number of times, typically in the range of 500 to 1000

times for household batteries, because the surfaces of the recharged electrodes do

not form as cleanly as the original surface, and they eventually become too worn

to be useful.

854 Chapter 19 Electrochemistry

The Nernst Equation and the Equilibrium Constant

Measuring the standard cell potential is a very powerful way to solve for the equi-

librium constant of a reaction. Here’s how. When a redox reaction proceeds with-

out any intervention, it will inevitably reach equilibrium. At that point the value

of E

cell

must become zero, and the free energy change for the process also becomes

zero—our thermodynamic deﬁnition of equilibrium, introduced in Chapter 16.

When this occurs, the Q value in the Nernst equation is equal to the equilibrium

constant K. We can show this reasoning by using the following equations:



G =



G° + RT ln Q

At equilibrium,



G = 0

so 0 =



G° + RT ln K (note that “Q” has become “K”)



G° = −RT ln K

We know that



G° = −nFE°

Substituting −nFE°

into the previous equation yields

−RT ln K = −nFE

Solving for K, we get ln K =

nFE

Calculating

at standard conditions (as we did for the Nernst equation),

we ﬁnd that

ln K =

38.92 nE°

Converting from natural ln to base 10 log (by dividing by 2.3026, as we did in

the Nernst equation, yields

log K = 16.9 nE°

To clearly show the connection to the Nernst equation, we will invert 16.9 and

put the result in the denominator.

log K =

nE°

0.0592

We can use the equation in this way. Alternatively, we can take the antilog of

both sides and use it in this way:

K = 10

nE°

0.0592

Which form of the equation you use depends mostly on your comfort level. We

can solve for the equilibrium constant either way. The key point is that our un-

derstanding of the thermodynamic meaning of equilibrium enables us to relate cell

potential to the equilibrium constant. The rest is just manipulating equations to

get where we want to go. Let’s see how we use this relationship to determine

the equilibrium constant for the copper–nitric acid reaction performed by Ira

Remsen. Recall that the reaction is

3Cu(s) +6H

(aq) + 2HNO

(aq) n 3Cu

(aq) + 2NO(g) + 4H

O(l) E °

cell

=+0.62 V

K = 10

nE°

0.0592

K = 10

6(+0.62)

0.0592

= 10

62.8

≈ 10

Alternatively,

log

K =

nE°

0.0592

6(+0.62)

0.0592

= 62.8

Raising both sides to the power of 10 yields

K ≈ 10

19.6 Not-So-Standard Conditions: The Nernst Equation 855

Video Lesson: Electrochemical

Determinants of Equilibria

This is yet another conﬁrmation that the reaction does, in fact, proceed toward

products. As we have this discussion, however, please keep in mind the difference

between thermodynamics and kinetics. Thermodynamics answers the question

“Can a process occur spontaneously?” It says absolutely nothing about speed.

Kinetics addresses the issues of rates and mechanisms. All that our calculations

tell us is that nitric acid can react spontaneously with the copper penny. They

don’t say how fast. That’s a question of kinetics.

EXERCISE 19.11 Equilibrium Constants and Cell Potential

Vanadium(V) ion can be reduced stepwise (that is, to V

and, ﬁnally, V

) by

reaction with a “Jones Reductor,” a zinc–mercury amalgam. The reaction for the

reduction of V

to V

ion includes the following two half-reactions:

(aq) + H

(aq) n VO

(aq) + H

O(l) E° =+1.00 V

(aq) n Zn(s) E° =−0.76 V

Calculate the equilibrium constant for this reaction.

First Thoughts

There are a number of steps to solving this problem. One way to ﬁgure out what we

need to do in working forward is to start by working backward (where do we want

to be, and how do we get there?). We want the value for K. In order to get that, we

need the value for E°. In order to get that, we need to have a balanced redox equa-

tion for the reduction of VO

to VO

, in which V

is reduced to V

, and Zn

oxidized to Zn

. Our order of operations, then, is

1. Balance the redox reaction.

2. Calculate the E° value for the reaction.

3. Calculate the equilibrium constant, knowing the number of electrons

exchanged in the reaction, along with the value for E°.

How might we assess whether the answer we calculate makes sense? At this point,

because we know that the reduction of vanadium ion does occur, our equilibrium

constant should be greater than 1. How much greater will depend on the cell volt-

age and on the number of electrons transferred in the process.

Solution

The balanced half-reactions and overall cell reaction are

2VO

(aq) + 4H

(aq) + 2e

−

n 2VO

(aq) + 2H

O(l) E° =+1.00 V

Zn(s) n Zn

(aq) + 2e

−

E° =+0.76 V

2VO

(aq)+4H

(aq)+Zn(s) n2VO

(aq)+Zn

(aq)+2H

O(l) E°=+1.76V

K = 10

nE°

0.0592

K = 10

2(+1.76)

0.0592

= 10

59.5

≈ 10

Alternatively,

log K =

nE°

0.0592

2(+1.76)

0.0592

= 59.5

Raising both sides to the power of 10 yields

K ≈ 10

856 Chapter 19 Electrochemistry

Zn is added. Time

Time

Further Insights

Does our answer make sense? The cell voltage is very high and positive in this two-

electron transfer, so our equilibrium constant is large, indicating that the reduction

of vanadium proceeds essentially to completion.

Here are pictures of the vanadium solutions, showing each ion from V

(left-

most picture) to V

(rightmost picture). We will discuss the reasons why transition

metal ions in solution have color, and often change color when reduced or oxidized,

in the next chapter.

19.7 Electrolytic Reactions 857

The varying oxidation states of vanadium.

PRACTICE 19.11

What is the equilibrium constant for the reaction of copper metal with zinc ion,

discussed at the beginning of this section?

See Problem 68.

19.7 Electrolytic Reactions

Some metals, including copper, gold, and silver, are found in their pure elemental

state in the environment. On the other hand, aluminum metal, based on its reac-

tivity (Section 19.5) is found only chemically combined in ores such as bauxite

(hydrated aluminum oxide; Al

· H

O or Al

· 3H

O). In fact, aluminum,

largely in the form of the aluminum oxides and silicates, makes up 8.1% of the

Earth’s crust. However, we know that pure aluminum can be produced, because it

is a major component in so many common products: the can in which we store

our soda, the wrap in which we put our ﬁsh for freezing, and the lightweight

bicycle we ride down the street. How is aluminum metal made from bauxite?

The basic process, called electrolysis, entails passing a current through a solu-

tion of metal ions in an electrochemical cell in the direction opposite to the spon-

taneous reaction. Doing so forces the nonspontaneous reaction to occur. This

process, also known as

electrowinning, is responsible for the manufacture and

puriﬁcation not only of aluminum but also of many other metals.

+ 3e

−

n Al

+ 2e

−

n Cu

+ e

−

n Ag etc....

Electrowinning is the most inexpensive method for making aluminum and

magnesium metals. Electrowinning of metals such as aluminum and magnesium

Video Lesson: Electrolytic Cells

Tutorial: Electrolytic Cells