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

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838 Chapter 19 Electrochemistry

In Step 4 we balance the number of oxygen atoms by adding water molecules

to the product side of the equation.

Cu(s) n Cu

(aq)

HNO

(aq) n NO(g) + 2H

O(l)

In Step 5 the number of hydrogen atoms is then balanced by adding H

the reactant side. Note that in both steps 4 and 5, we did not need to modify the

copper half-reaction.

Cu(s) n Cu

(aq)

(aq) + HNO

(aq) n NO(g) + 2H

O(l)

The charges are then balanced in Step 6 by adding electrons to the two half-

reactions. We note that the change in the oxidation number of the element being

oxidized or reduced must equal the number of electrons lost or gained in the

half-reaction. The copper half-reaction has electrons added as a product; the ni-

tric acid half-reaction has electrons added as a reactant. In every redox reaction,

one half-reaction gets the electrons on the right, the other on the left.

Cu(s) n Cu

(aq) + 2e

−

+ 3H

(aq) + HNO

(aq) n NO(g) + 2H

O(l)

In Step 7 we make sure that the number of electrons is the same in both half-

reactions by multiplying the half-reactions by an integer. In this case, we multiply

the copper reaction by 3 and the nitric acid reaction by 2. Then, in Step 8, we add

the two reactions together and simplify by eliminating similar items from both

sides of the equation.

3{Cu(s) n Cu

(aq) + 2e

−

}

2 {3e

−

+ 3H

(aq) + HNO

(aq) n NO(g) + 2H

O(l)}

3Cu(s) +6H

(aq) + 2HNO

(aq) n 3Cu

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

O(l)

The result is the balanced redox reaction. One observation and one question can

reasonably arise. The observation is that even though the equation is electrically

balanced, each side is not electrically neutral. That is, we have the same +6 charge

on each side! Now for the question: Where are the negative charges that would

make each side electrically neutral? The answer comes when we remember that

we have written net ionic equations, rather than complete ionic or molecular

equations. In this case, the molecular form comes by adding 6 mol of nitrate ion

(NO

−

) to both sides to give

3Cu(s) +8HNO

(aq) n 3Cu(NO

)

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

O(l)

In other reactions, spectator ions such as Na

or Cl

−

(if they are actually in the

solution!) that are not listed in the net ionic equation make the system electrically

neutral. The key point is that when we write equations in net ionic form, we expect

them to be electrically balanced, not electrically neutral.

3Cu(s) + 6H

(aq) + 2HNO

(aq)

+ 6NO

–

3Cu

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

O(l)

+ 6NO

–

3Cu(s) + 8HNO

(aq) 3Cu(NO

)

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

O(l)

→

EXERCISE 19.4 Balancing Redox Equations in Acidic Solutions

Balance the following equation in acidic solution.

2−

(aq) + NO(g) n Cr

(aq) + NO

−

(aq)

Solution

Step 1: Determine the oxidation state on each element in each compound in

the chemical equation. Do we in fact have a redox reaction?

2−

(aq) + NO(g) n Cr

(aq) + NO

−

(aq)

Oxidation states +6 −2 +2 −2 +3 +5 −2

Nitrogen is being oxidized from +2 to +5. Chromium is being reduced from

+6 to +3. This is a redox reaction.

Step 2: Separate the redox reaction into two parts: an oxidation reaction and a

reduction reaction. Include just the compounds that have a change in the oxi-

dation state.

2−

(aq) n Cr

(aq) (reduction)

NO(g) n NO

−

(aq) (oxidation)

Step 3: Balance all atoms except oxygen and hydrogen.

2−

(aq) n 2Cr

(aq)

NO(g) n NO

−

(aq)

Step 4: Balance the oxygen atoms by adding H

O to one side in each half-

reaction.

2−

(aq) n 2Cr

(aq) + 7H

O(l)

O(l) + NO(g) n NO

−

(aq)

Step 5: Balance the hydrogen atoms by adding H

to one side in each half-

reaction.

14H

(aq) + Cr

2−

(aq) n 2Cr

(aq) + 7H

O(l)

O(l) + NO(g) n NO

−

(aq) + 4H

(aq)

Step 6: Balance the charges in each half-reaction by adding electrons to one

side.

−

+ 14H

(aq) + Cr

2−

(aq) n 2Cr

(aq) + 7H

O(l)

O(l) + NO(g) n NO

−

(aq) + 4H

(aq) + 3e

−

Step 7: As needed, multiply one or both of the half-reactions by some coefﬁ-

cient so that the same number of electrons will appear in both half-reactions.

−

+ 14H

(aq) + Cr

2−

(aq) n 2Cr

(aq) + 7H

O(l)

2{2H

O(l) + NO(g) n NO

−

(aq) + 4H

(aq) + 3e

−

}

Step 8: Add the reactions together and simplify. Note that the electrons on

each side mathematically cancel, indicating the same number of electrons

gained in the reduction as lost in the oxidation.

−

+ 14H

(aq) + Cr

2−

(aq) n 2Cr

(aq) + 7H

O(l)

O(l) + 2NO(g) n 2NO

−

(aq) + 8H

(aq) + 6e

−

+ 14H

(aq) + Cr

2−

(aq) + 4H

O(l) + 2NO(g) n

2Cr

(aq) + 7H

O(l) + 2NO

−

(aq) + 8H

(aq) + 6e

−

= 6H

(aq) + Cr

2−

(aq) + 2NO(g) n 2Cr

(aq) + 2NO

−

(aq) + 3H

O(l)

19.3 Redox Equations 839

We have 8 water molecules on the right and 2 on the left. This leaves an excess of

6 water molecules on the right, to give us our ﬁnal balanced equation in basic

solution:

8OH

−

(aq) + CH

OH(aq) + 6MnO

−

(aq) n CO

2−

(aq) + 6H

O(l) + 6MnO

2−

(aq)

EXERCISE 19.5 Balancing Redox Equations in Basic Solutions

Balance the following equation in basic solution.

−

(aq) + S

2−

(aq) n I

−

(aq) + SO

2−

(aq)

Solution

We may ﬁrst balance the equation in acidic solution, using our multistep procedure

presented in Figure 19.9.

−

(aq) + S

2−

(aq) n I

−

(aq) + SO

2−

(aq)

PRACTICE 19.4

Balance the following reaction in acidic solution.

ClO

−

(aq) + H

(aq) + Cu(s) n Cl

−

(aq) + H

O(l) + Cu

(aq)

See Problems 35–38, 47, 48, and 51.

The process we have used to balance the reaction focuses on the use of acidic

solutions. Chemical reactions can also occur in basic conditions. The method we

present to balance basic reactions requires a little chemical sleight-of-hand but

does work effectively. Essentially, we balance the reaction as though it were in an

acidic solution (by adding H

as necessary) and then add a quantity of hydroxide

ion (OH

−

) as necessary, to both sides of the equation. Mathematically, we still

have our equality. Chemically, we neutralize the H

, getting water on one side

and excess base on the other. Although this does not depict what goes on in the so-

lution (we are starting with a base, not doing a titration!) we correctly end up

with an excess of OH

−

on one side or the other.

We will show the method by looking at the ﬁrst step in a procedure to analyze

methanol (CH

OH) by reaction with the permanganate ion (MnO

−

) in base.

Before balancing, we have

OH(aq) + MnO

−

(aq) n CO

2−

(aq) + MnO

2−

(aq)

To balance the reaction in basic solution, we ﬁrst balance it as though it were in

acidic solution. Using our half-reaction technique, we get the following oxidation

and reduction half-reactions:

O(l) + CH

OH(aq) n CO

2−

(aq) + 8H

(aq) + 6e

−

(oxidation)

−

+ MnO

−

(aq) n MnO

2−

(aq) (reduction)

We can multiply the reduction reaction by 6 to balance electrons and add the

half-reactions to get the ﬁnal equation, balanced in acidic solution:

O(l) + CH

OH(aq) + 6MnO

−

(aq) n CO

2−

(aq) + 8H

(aq) + 6MnO

2−

(aq)

To balance in basic solution, we add an amount of OH

−

to each side equal to the

amount of H

840 Chapter 19 Electrochemistry

The reaction of purple permanganate

ion with methanol.

O(l) + CH

OH(aq) + 6MnO

−

(aq) n CO

2−

(aq) + 8H

(aq) + 6MnO

2−

(aq)

+ 8OH

−

(aq) + 8OH

−

(aq)

8OH

−

(aq) + 2H

O(l) + CH

OH(aq) + 6MnO

−

(aq) n CO

2−

(aq) + 8H

O(l) + 6MnO

2−

(aq)

The iodine changes from an oxidation state of −

⁄3 in I

−

to −1inI

−

. (reduction)

The sulfur changes from an oxidation state of +2 in S

2−

to +6 in SO

2−

(oxidation)

The balanced half-reactions in acidic solution are

−

+ I

−

(aq) n 3I

−

(aq)

O(l) + S

2−

(aq)n 2SO

2−

(aq) + 10H

(aq) + 8e

−

We multiply the reduction equation by 4 to equalize the electrons gained and lost in

the reduction and oxidation half-reactions

4{2e

−

+ I

−

(aq) n 3I

−

(aq)} = 8e

−

+ 4I

−

(aq) n 12I

−

(aq)

O(l) + S

2−

(aq) n 2SO

2−

(aq) + 10H

(aq) + 8e

−

We add the two half-reactions to get the ﬁnal, balanced equation in acidic solution.

O(l) + S

2−

(aq) + 4I

−

(aq) n 2SO

2−

(aq) + 12I

−

(aq) + 10H

(aq)

To balance in base, we add 10OH

−

to both sides.

O(l) + S

2−

(aq) + 4I

−

(aq) n 2SO

2−

(aq) + 12I

−

(aq) + 10H

(aq)

+ 10OH

−

(aq) + 10OH

−

(aq)

We cancel our the extra waters to give the equation that is now balanced in base.

10OH

−

(aq) +S

2−

(aq) + 4I

−

(aq) n 2SO

2−

(aq) + 12I

−

(aq) + 5H

O(l)

Is the equation electrically and atomically balanced? Check to make sure that the

charge is the same on both sides of the equation and that there is the number of

atoms on each side. If one or both of these checks do not work, then we’ve made a

mistake balancing the equation.

PRACTICE 19.5

Balance the following equation in basic solution.

MnO

−

(aq) + Mn

(aq) n MnO

(s)

(Hint: MnO

can be written as the product in both an oxidation and a reduction

half-reaction.)

See Problems 39–42 and 52.

Manipulating Half-Cell Reactions

We have learned to balance redox reactions by recognizing that they include elec-

tron loss and electron gain.Although calculating the spontaneity of an individual

half-reaction may lead to the conclusion that the half-reaction is spontaneous,

this is only half of the picture. Because a redox reaction is simply the sum of an

oxidation half-reaction and a reduction half-reaction, we determine the spon-

taneity of the resulting reaction only by including both. Remember that we are

using the SHE as our reference point in these calculations.

We can then say that for two half-reactions, the sum of their change in free en-

ergy should be equal to the free energy change for the complete redox reaction:

∆

rxn

G° = ∆G

° + ∆G

If we substitute the equivalent expression (−nFE°) for ∆G°, the equation

becomes

−n

rxn

° =−n

° +−n

19.3 Redox Equations 841

Because the value of F is a constant, and because the number of electrons was the

same when we balanced the reaction (n

= n

rxn

), all terms but the cell po-

tentials cancel:

rxn

° =E

° + E

This means that if the redox reactions are written so that one is a reduction and

one is an oxidation, and the cell potential for the oxidation reaction relative to the

SHE has the sign opposite that of its half-reaction potential when written as a re-

duction, then the two half-reaction potentials are additive for a balanced redox

equation.

rxn

° =E

red

° + E

From that conclusion, we can say, for example, that if copper is oxidized in a re-

action, we write it as an oxidation, and just as we reverse the standard free energy

value,

∆G°, when we reverse a reaction, we reverse the E°, as follows:

(aq) + 2e

−

n Cu(s) E° =+0.34 V (reduction)

Cu(aq) n Cu

(aq) + 2e

−

E° =−0.34 V (oxidation)

This notion of reversing the cell potential relative to the SHE when we reverse the

reaction gives a consistent picture of the thermodynamic relationship between

∆G°

and E°. In addition, it is a reminder that until 1953, when IUPAC changed its pro-

tocol to list half-reaction potentials as reductions, they were formerly listed as

oxidations, with opposite signs relative to the SHE.

We know that the nitric acid−copper reaction that we introduced at the be-

ginning of this section is spontaneous, as evidenced by its effect on Ira Remsen’s

lungs, hands, and pants. Let’s see how we can manipulate the half-reactions to

conﬁrm this. In Table 19.3, we notice that one of the reactions is the reverse of the

desired reaction.

From the balancing process steps 8 and 9:

3{Cu(s) n Cu

(aq) + 2e

−

}

2{3e

−

+ 3H

(aq) + HNO

(aq) n NO(g) + 2H

O(l)}

3Cu(s) +6H

(aq) + 2HNO

(aq) n 3Cu

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

O(l)

From Table 19.3:

(aq) + 2e

−

n Cu(s) E° =+0.34 V

(aq) + HNO

(aq) + 3e

−

n NO(g) + 2H

O(l) E° =+0.96 V

To make the half-reactions from the table look like the half-reactions in the redox

reaction, we need to reverse the copper reduction and write it as an oxidation. As

we pointed out above, reversing the reaction also changes the sign on the poten-

tial of the reaction, or, in this case, the half-reaction.

(aq) + 2e

−

n Cu(s) E° =+0.34 V (reduction)

Cu(aq) n Cu

(aq) + 2e

−

E° =−0.34 V (oxidation)

However, no modiﬁcation of the potential is necessary when we double or triple a

half-reaction because the number of electrons involved in the process also doubles or

triples. The free energy, however, does change when we multiply the equation,

because

G

◦

=−nFE

◦

, and if, as in this case, we triple n,

G

◦

will triple as

well. This also is consistent with our discussion of thermodynamics in Chapters 5

and 14.

Cu(aq) n Cu

(aq) + 2e

−

E° =−0.34 V (oxidation)

3Cu(s) n 3Cu

(aq) + 6e

−

E° =−0.34 V (oxidation)

Table 19.4 lists the key aspects of manipulating half-reactions.

842 Chapter 19 Electrochemistry

For the copper–nitric acid reaction, the reaction potential is determined by adding

the nitric acid reduction (+0.96 V) to the copper oxidation (−0.34 V). The resulting

potential is positive (+0.62 V), identifying the reaction as a spontaneous process, as

is consistent with Ira Remsen’s experience.

3{Cu(s) n Cu

(aq) + 2e

−

} E° =−0.34 V

2{3e

−

+ 3H

(aq) + HNO

(aq) n NO(g) + 2H

O(l)} E° =+0.96 V

3Cu(s) +6H

(aq) + 2HNO

(aq) n 3Cu

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

O(l) E°

cell

=+0.62 V

EXERCISE 19.6 Dissolving Gold

Nitric acid can be used to dissolve copper. Can nitric acid dissolve gold at standard

conditions? The unbalanced reaction is shown here.

HNO

(aq) +Au(s) n Au

(aq) + NO(g)

First Thoughts

What we’re really asking in this problem is “Is the reaction shown spontaneous?”

To determine the spontaneity we can use Faraday’s equation, which requires us to

know the potential of the reaction. We can obtain this by combining properly

balanced half-reactions, which have reduction potentials given in Table 19.3.

Solution

The two unbalanced half-reactions of interest are

Au(s) n Au

(aq)

HNO

(aq) n NO(g)

We can balance them in acidic solution (because of the presence of nitric acid) and

combine them to give

Au(s) n Au

(aq) + 3e

−

(oxidation)

−

+ 3H

(aq) + HNO

(aq) n NO(g) + 2H

O(l) (reduction)

(aq) + HNO

(aq) +Au(s) n NO(g) +Au

(aq) + 2H

O(l)

Table 19.3 lists the potential relative to the SHE for each half-reaction written as

a reduction. We can reverse the gold reduction reaction because the metal is

19.3 Redox Equations 843

Important Points for Potentials of Redox Reactions

• Just as free energies can be combined for chemical reactions, an oxidation and a re-

duction half-cell potential can be combined to produce a chemical equation.

• Just as you reverse the sign of ∆G° when you reverse a chemical reaction, you re-

verse the sign of E° when you reverse a half-cell reaction.

• Although ∆G° values depend on the coefﬁcients in the chemical equation (that is,

when you double the coefﬁcients, you double ∆G°), E°values do not. In a half-

reaction, if the coefﬁcients change, the number of electrons, n, will change as well,

in essence canceling the effect of the change to the coefﬁcients.

• Because of the negative sign in

G

◦

=−nFE

◦

, electrochemical cell reactions with a

positive voltage are spontaneous. (Recall from Chapter 15 that the rate is not related

to the spontaneity of the reaction.)

TABLE 19.4

Connective tissue

Nerve

Papillae

Nucleus

Blood supply

Electrically conductive

tissue

Application

HEMICAL

ENCOUNTERS:

The Electric Eel

being oxidized. We can use the potentials of the half-reactions to calculate the

cell voltage.

Au(s) n Au

(aq) + 3e

−

E° =−1.50 V

−

+ 3H

(aq) + HNO

(aq) n NO(g) + 2H

O(l) E° =+0.96 V

(aq) +HNO

(aq) +Au(s) n NO(g) +Au

(aq) +2H

O(l) E°

cell

=−0.54 V

The negative value for the cell voltage indicates that the reaction of nitric acid and

gold is not spontaneous. Nitric acid doesn’t dissolve gold.

Further Insights

You’ve probably noticed that you don’t have to balance a redox reaction in order to

determine the standard voltage of the reaction. This is because the voltage of a reac-

tion is not dependent on the stoichiometry of the equation. Simply adding the half-

reaction potentials (as long as one is an oxidation and the other a reduction) yields

the voltage of the resulting reaction. However, it is good practice to balance the

equation so that we can tell how many electrons are exchanged. We will need this in-

formation if our system is at nonstandard conditions, which is most of the time.

A process for dissolving gold was known during the early days of the alchemists.

Because metals like gold and platinum wouldn’t react with strong acids, such as ni-

tric acid, they were known as the royal metals. However, there existed a solution that

would dissolve gold and platinum. These metals were soluble in a mixture of one

part nitric acid and three parts hydrochloric acid known as aqua regia, or royal

water. Aqua regia is used commercially as a ﬁrst step in separating platinum from

gold that is combined with other metals in their ore mixtures. Both metals are ini-

tially dissolved and then selectively precipitated from solution.

PRACTICE 19.6

Determine the cell voltage at standard conditions for the following redox reactions:

(g) + H

O(g) n CO(g) + H

(g)

Ag(s) + H

(aq) n H

O(l) +Ag

(aq)

See Problems 27 and 28.

19.4 Electrochemical Cells

It may be “shocking” to learn, but it’s true: The electric eel is a formidable oppo-

nent. When startled, stepped on, or hunting for food, the eel can deliver up to

1 ampere (1 coulomb of charge per second) at 600 V, enough electrical energy

to stun or even kill large animals. (To be accurate, the electric eel isn’t

an eel. It lives in fresh water and is really a ﬁsh.) How does the electric

eel generate the electricity used in hunting?

The powerful shock that the eel can deliver is produced by 5000 to

10,000 specialized cells, called electroplates, in its tail (see Figure 19.10).

Using biochemical processes, the eel charges up the electroplates in

much the same manner as muscle and nerve cells are charged. Then,

when a nerve impulse is sent to the tail, the electroplates discharge their

stored potential. Can you think of an electrical storage device similar to

the electric eel’s that we use every day? The electrochemical cell seems to

ﬁt the deﬁnition. We’ll explore the electrochemical cell in this section

844 Chapter 19 Electrochemistry

FIGURE 19.10

Electroplates in the tail of the electric eel develop a potential charge that

can be delivered to shock its prey.

(Drawing by Rick Simonson.)

Video Lesson: Electrochemical

Cells

–

The salt bridge.

FIGURE 19.12

Using the electrochemical cell to light a bulb.

and learn how we can make use of the electrical en-

ergy of chemical reactions.

Electrochemical Cells in the Laboratory

Redox reactions, such as the dissolving penny, can

take place inside a beaker just like other reactions.

However, we can harness the energy from the elec-

trons that are exchanged in the redox reaction if we

modify the experimental setup. Take a look at what

happens when we separate the copper penny from

the nitric acid shown in Figure 19.11. The brown

fumes produced by the subsequent oxidation of NO

to NO

by O

still waft from the beaker, and the cop-

per ions are still generated. The redox reaction is still

working. Examine the experimental setup closely to see why it works. We have

3Cu(s) +6H

(aq) + 2HNO

(aq) n 3Cu

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

O(l) E°

cell

=+0.62 V

followed by the oxidation of NO to the poisonous NO

gas:

2NO(g) + O

(g) n 2NO

(g) ∆H° =−112 kJ

The oxidation half-reaction is in the beaker on the right, and the reduction

half-reaction is on the left. The two beakers are connected with a wire that trans-

fers the electrons from beaker to beaker. At both ends of the wire is an electrode.

The electrode in the oxidation reaction is called the anode, and the electrode in

the reduction reaction is called the cathode. Remember, electrons are one of the

products of the copper oxidation. They need someplace to go. Providing a wire

for them to travel into the nitric acid reduction reaction keeps the reaction run-

ning. If we open the circuit on the wire, the reaction stops. The wire is an essen-

tial part of this electrochemical cell.

The tube labeled

salt bridge in the diagram contains an electrolyte such as

potassium chloride or sodium nitrate and allows ions to pass from beaker to

beaker. The salt bridge is needed because as electrons move from the beaker on

the right (the oxidation reaction) to the beaker on the left (the reduction reac-

tion), a strong positive charge will develop at the anode as more of the copper

penny becomes Cu

. A similar thing happens in the other beaker; the H

being removed to make NO and water, and the cathode becomes negatively

charged. The developing charges attract the electrons as they move away from the

19.4 Electrochemical Cells 845

Copper

electrode

Copper

cell

Electrode

Salt bridge

Nitric acid

cell

FIGURE 19.11

The electrochemical cell. Ira Remsen’s ob-

servations are still valid in this setup. The

nitric acid still acts on the copper. Note

the presence of the salt bridge and the

electrodes.

Visualization: Electrochemical

Half-Reactions in a Galvanic Cell

Visualization: Galvanic (Voltaic)

Cells

beaker. Without the salt bridge, the electrons would stop this type of movement.

We must have a salt bridge in our electrochemical cell design so that the entire

system can remain electrically neutral. The ﬁnished electrochemical cell produces

the same reaction that Ira Remsen observed (nitric acid acts upon copper), but

the movement of the electrons through the wire can be used to light a bulb, as

shown in Figure 19.12. We have harnessed the electrons as electrical energy.

EXERCISE 19.7 Alkaline Batteries

Some electrochemical cells are constructed using an alkaline electrolyte. Here are

two half-cells that can be employed:

Cd(OH)

(s) n Cd(s)

NiO

(s) n Ni(OH)

(s)

a. Based on the information in the standard reduction potentials table (Table 19.3),

write the two oxidation−reduction reactions possible from these half-

reactions, and calculate their cell potentials.

b. Which could be used as the basis of an electrolytic cell? ...ofa galvanic cell?

What conditions are required to obtain these speciﬁc cell potentials? (Recall

our discussion in Section 19.1.)

c. When used as a voltaic cell, this set of reactions makes a useful battery that can

power portable appliances and tools. Judging on the basis of the elements that

make up the galvanic cell, can you identify the name of this type of battery?

What signiﬁcance does the reverse reaction have?

Solution

a. The two half-cell reactions are

Cd(OH)

(s) + 2e

−

n Cd(s) + 2OH

−

(aq) E° =−0.81 V

NiO

(s) + 2H

O(l) + 2e

−

n Ni(OH)

(s) + 2OH

−

(aq) E° =+0.49 V

If the ﬁrst reaction is reversed, E°

cell

= (+0.81 V) + (+0.49 V) =+1.30 V.

Cd(s) + NiO

(s) + 2H

O(l) n Cd(OH)

(s) + Ni(OH)

(s) E° =+1.30 V

If the second reaction is reversed, E°

cell

= (−0.81 V) + (−0.49 V) =−1.30 V.

Cd(OH)

(s) +Ni(OH)

(s) n Cd(s) + NiO

(s) +2H

O(l) E° =−1.30 V

In both cases, the OH

−

(aq) and the electrons cancel when the half-cell reac-

tions are added.

b. To obtain these voltages, the reactions must be run at 25°C at 1 atm pressure

(standard conditions). The concentration of OH

−

(aq) must be 1 M (solids

have a constant concentration). The ﬁrst oxidation–reduction reaction could

form the basis of a galvanic cell, because E°

cell

is positive. The second would be

electrolytic, because E°

cell

is negative.

c. This voltaic cell is better known as a NiCad battery. The reverse electrolytic

reaction proceeds well, because once formed, the reaction products remain in

contact with the electrodes. Therefore NiCad batteries can be recharged by

running the cell in reverse by applying a reverse voltage greater than the cell

potential.

PRACTICE 19.7

Draw the electrochemical cell for each of the redox reactions given in Practice 19.6.

What is the potential of the galvanic cell? ...ofthe electrolytic cell?

See Problems 57–59.

846 Chapter 19 Electrochemistry

Mg(s) Mg

(aq)Cu

(aq) Cu(s)

Bridge

Anode Cathode

Zn(s) ZnO(s)Ag

O(s) Ag(s)

Bridge

Anode Cathode

Cathode

Insulation

Anode

(zinc container)

Paste of Ag

in a medium of

KOH and ZnO

Cutaway view of button battery.

Cell Notation

Because the balanced chemical equations require a lot of time to write, shorthand

notation (called

cell notation) is often used when describing electrochemical cells.

It may look hard to do, but cell notation is as easy as knowing your ABCs—that

is, anode, bridge, cathode. Consider the reaction that takes place in the silver

oxide cell found in pacemaker batteries:

O(s) + Zn(s) n 2Ag(s) + ZnO(s) E° =+1.86 V

A closer look at the half-reactions tells us that silver is being reduced

and zinc is being oxidized. The silver metal must be the cathode, and

the zinc metal must be the anode. We write each half-reaction, and

then, without even balancing them, we construct the overall short-

hand notation.

Zn(s) n ZnO(s) (oxidation, anode) Zn(s) | ZnO(s)

O(s) n Ag(s) (reduction, cathode) Ag

O(s) | Ag(s)

EXERCISE 19.8 Shorthand Notation

Write the shorthand cell notation for the following voltaic cell.

(aq) + Mg(s) n Cu(s) + Mg

(aq)

Solution

Our goal is to write our ABCs—anode, bridge, and cathode. In order to do this, we

need to know which half-reaction is the oxidation and which is the reduction. The

copper ion gains two electrons, so it is reduced. The copper is the cathode. The mag-

nesium loses two electrons, so it is oxidized. It is the anode. We now have enough

information to write the cell notation:

19.4 Electrochemical Cells 847

PRACTICE 19.8

Create a galvanic cell using these half-reactions, and write the cell notation. Check

to make sure your reaction is written as a spontaneous redox reaction.

(aq) + 2e

−

n Fe(s)

(aq) + 3e

−

n Al(s)

See Problems 55 and 56.

Batteries

Commercial batteries come in many shapes and sizes and are based on a wide va-

riety of chemical processes (Table 19.5). The best battery for any application is

usually chosen by considering power output, cost, convenience, size, and whether

the battery will be rechargeable. The chemistry of commercial batteries is often

rather complex compared to the simple electrochemical cells used in the teaching

lab to convey the basic principles.

Visualization: Zinc/Copper Cells

with Lemons

Video Lesson: CIA

Demonstration: The Fruit-

Powered Clock

Video Lesson: Batteries