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

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828 Chapter 19 Electrochemistry

IIIB IVB VB VIB VIIB IB IIB

IIA IIIA IVA VA VIA VIIA

VIIIA

VIIIB

FIGURE 19.4

Oxidation states of the metals.

* How to recognize hydrides:

Hydrides have a metal first in their chemical formula.

For example: CaH

, MgH

, LiAlH

, NaH

Hydrides contain no nonmetals.

For example, these are not hydrides: CH

, NaHCO

, NH

, HCl

Is the chemical an element

or a simple ion?

Assign elements as zero (Cu

= 0),

simple ions as their charge (Cu

= +2).

Assign H in hydrides as –1, and

assign all other H as +1.*

Ye s

Is hydrogen one of

the elements present?

Are any elements present that

show just one oxidation

state? If so assign them.

How many elements remain

to be assigned?

Ye s

2 or

Ye s

Assign them as follows:

All Group IA metals +1

All Group IIA metals +2

Zn, Cd +2

Group IIIA metals +3

O (except H

= –1; K

O = –

) –2

F –1

Subtract the total of the other oxidation

states from the charge on the overall

compound.

For each element, subtract eight (8)

from its group number.

FIGURE 19.5

Decision rules for assigning oxidation

states.

Video Lesson: Balancing Redox

Reactions by the Oxidation

Number Method

FIGURE 19.6

The electrostatic potential

map for water.

19.2 Oxidation States—Electronic Bookkeeping

829

have shifted away from hydrogen toward the oxygen, as shown in the electrosta-

tic density map in Figure 19.6. This is consistent with our understanding of elec-

tronegativity (Chapter 7). The oxidation states do not mean that hydrogen has a

full +1 charge and oxygen a full −2 charge. Remember from Chapter 7 that water

is a covalent molecule. We can talk more comfortably about oxidation states as

representations of the charge of atoms in ionic compounds. Whether in covalent

or ionic compounds, we will use oxidation state as a bookkeeping tool to see

where electrons shift in chemical reactions.

EXERCISE 19.1 Oxidation States

Assign an oxidation state to each of the elements in formic acid (HCOOH). Given

the Lewis structure shown below, which of the electrostatic density maps that follow

represents formic acid?

H C O H

First Thoughts

We have two questions here: a technical one about the oxidation states and a more

conceptual one concerning what the oxidation states imply about the electron

distribution. In both cases, though, the assignment of the oxidation state and

the electrostatic density map should be consistent with our understanding of

electronegativity.

Solution

According to our rules for assigning oxidation states in Figure 19.5, each oxygen is

assigned an oxidation state of −2, and each hydrogen is +1. For the neutral formic

acid molecule, in which the sum of all of the oxidation states must be 0, we have

2H =+1 × 2 =+2

2O =−2 × 2 =−4

The total is +2 + (−4) =−2. The carbon atom must therefore have an oxidation

state of +2. This means that the electron density in this molecule is focused more

on the oxygen atoms than on the carbon or hydrogen atoms. Given the assignments

of the oxidation states, the second electron density map is the most reasonable rep-

resentation of the molecule. In the other two maps the charge density isn’t in the

correct location. In the ﬁrst map, the electron density appears to be opposite of what

we’d expect, given that the red color on the map indicates regions of high electron

density. The third map again shows electron density that doesn’t appear to match

where we have predicted the electrons to reside in the molecule.

Further Insights

Carbon can have several oxidation states, depending on the atoms with which

it bonds. When we talk about “complete combustion” of carbon, we mean the

Oxidizing and Reducing Agents

Reactant What Happens Examples

Oxidizing agent Gains electrons Element: O

, and halogens

Is reduced Compound: H

Ionic species (typically with a large

positive oxidation state): MnO

−

Reducing agent Loses electrons Element: H

and metals

Is oxidized Compound: BH

Ionic species: NaH, LiAlH

TABLE 19.1

conversion to its highest possible oxidation state, +4, as is true in CO

. For example,

when glucose (C

) is burned in air, the carbon atoms can undergo complete

combustion to CO

. However, they can also form partial combustion products,such

as carbon monoxide (CO), in which the oxidation state of the carbon atom differs.

PRACTICE 19.1

Determine the oxidation states of carbon in glucose and carbon monoxide, the

compounds discussed in the “Further Insights” section just above.

See Problems 5–8 and 44.

We began this section discussing the reaction of hydrogen and oxygen gases in

a fuel cell. By comparing the reactants and products, we see that both H and O

show a change in their oxidation state. The change indicated that a transfer of

electrons from one species to another has occurred. Any chemical reaction in

which atoms change their oxidation states is classiﬁed as an oxidation–reduction

reaction, or redox reaction for short. When O

and H

react to form water, O

reduced (it gains electrons) and its oxidation state decreases from 0 to −2.

Similarly, H

is oxidized (it loses electrons), and its oxidation state increases from

0 to +1:

We can look at the compounds in this reaction in a different way. Oxygen

) causes the oxidation of the hydrogen (H

), so it is an oxidizing agent. In fact,

oxygen stands as the premier oxidizing agent on planet Earth, both because of its

abundance and because of its strong ability to accept electrons. We observe oxy-

gen’s effect every day when we metabolize glucose or explore a rusty, old ship-

wreck (rust results from the oxidation of Fe to Fe

). The hydrogen (H

) in the

fuel cell causes the reduction of the oxygen and is therefore a

reducing agent (see

Table 19.1). Looking at this another way, we can say that the oxidizing agent itself

is reduced and the reducing agent itself is oxidized.

Hydrogen peroxide is sometimes used to bleach hair or disinfect

wounds. Is its decomposition into oxygen and water a redox reac-

tion? If so, which species is the oxidizing agent and which is the

reducing agent?

(l) n O

(g) + 2H

O(l)

Occasionally, a chemical reaction such as this one employs a single

reactant as both the oxidizing and the reducing agent. Such a reac-

tion is known as a

disproportionation. Using our decision rules in Fig-

ure 19.5, we can assign the following oxidation states to each atom.

+1 −10+1 −2

(l) n O

(g) + 2H

O(l)

830 Chapter 19 Electrochemistry

(g) + O

(g) 2H

O(g)

Oxidized

ReducedReduced

(l) O

(g) + 2H

O(l)

Oxidation

ReductionReduction

–1 0 –2

Nitrogen Cycle

Atmospheric

nitrogen (N

)

Dead plants and

animals, animal waste

Decomposers

Nitrogen-

fixing

bacteria

Nitrogen

fixation

Nitrifying

bacteria

Nitrification

Denitrification

Plant

protein

Animal

protein

Ammonia (NH

), or

Ammonium (NH

)

Nitrite

(NO

–

)

Nitrate

(NO

–

)

Denitrifying

bacteria

NaOH + HCl NaCl + H

+1 in both compounds

–1 in both compounds–1 in both compounds

–2 in both com

ounds

In this reaction, the oxygen atoms in hydrogen peroxide (like any peroxide) have

an oxidation state of −1. The two products of the reaction contain oxygen atoms

with different oxidation states, O

with a higher oxidation state (zero) and H

with oxygen in a lower oxidation state (−2). We say that the hydrogen peroxide is

both the oxidizing agent and the reducing agent.

Not all reactions require the transfer of electrons. Many of the reactions we’ve

discussed in this text thus far are not redox reactions. For example, the neutral-

ization of sodium hydroxide by hydrochloric acid is not a redox reaction. We can

tell because the oxidation states for each of the atoms remain the same on both

sides of the equation:

+1 −2 +1 +1 −1 +1 −1 +1 −2

NaOH + HCl n NaCl + H

Many elements show a variety of oxidation states, depending on the chemical

species in which they are found. An example of this can be found in the nonmetal

nitrogen, which has a range of possible oxidation states, as shown in Table 19.2 on

page 832. The gases found in our atmosphere, both naturally and as pollutants,

have different positive oxidation states of nitrogen, since the oxygen atom in each

compound is more electronegative than nitrogen and is assigned an oxidation

state of −2. For the same reason, nitrates (NO

−

) and nitrites (NO

−

) have ni-

trogen with positive oxidation states. However, when combined with a less elec-

tronegative element such as hydrogen, nitrogen is assigned a negative oxidation

state. Accordingly, look for a negative oxidation state of N in ammonia and in

compounds containing the ammonium ion, NH

The

nitrogen cycle (Figure 19.7) illustrates how nitrogen moves through its

different oxidation states on our planet. You can ﬁnd similar cycles where the

oxidation states of sulfur and carbon change as these elements form different

19.2 Oxidation States—Electronic Bookkeeping 831

FIGURE 19.7

The nitrogen cycle. The nitrogen in the environ-

ment is found in many different compounds

and with many different oxidation states.

compounds. What is the point? Whereas oxidation states are ﬁxed for a particu-

lar compound at a particular moment in time, chemicals in our world constantly

undergo change. Oxidation states help us keep track of the changes that involve

the important processes of oxidation and reduction.

EXERCISE 19.2 The Electrochemistry of Smog

Although oxygen and nitrogen do not react at low temperatures, they combine in

a hot automobile engine to form the pollutant NO, with subsequent oxidation by

atmospheric oxygen to form the poisonous brown gas NO

, which is largely re-

sponsible for the smog found in urban areas on hot, sunny days:

(g) + O

(g) n 2NO(g)

2NO(g) + O

(g) n 2NO

(g)

Assign oxidation states to all the atoms involved in these two reactions. Which

species are oxidized? Which are reduced? Do your answers make sense in terms of

the relative electronegativity values of nitrogen and oxygen?

First Thoughts

Which of the two atoms, nitrogen or oxygen, is more likely to be reduced (gain elec-

trons)? Oxygen is the more electronegative, so in this system, nitrogen will be

oxidized. That is how we evaluate whether our answers make sense.

Solution

According to Figure 19.5, we may assign all atoms in elements (such as N

and O

)

an oxidation state of zero. In the compounds, we may assign oxygen as −2, because

the compounds are not peroxides or superoxides. We assign nitrogen so that the

sum of the oxidation states is zero, because NO and NO

, as compounds, carry no

charge.

832 Chapter 19 Electrochemistry

The Oxidation State of Nitrogen

Oxidation State Formula Name Produced in Nature

+5 HNO

Nitric acid From NO

−

by nitrifying bacteria

−

Nitrate ion

+4NO

Nitrogen dioxide In air by oxidation of NO

+3 HNO

Nitrous acid From NH

by nitrifying bacteria

−

Nitrite ion

+2 NO Nitrogen monoxide From N

by lightning or volcanoes

(nitric oxide)

+1N

O Nitrous oxide From NO

−

by denitrifying bacteria

Nitrogen From N

O by denitrifying bacteria

−

Azide ion (not found in nature)

−1NH

OH Hydroxylamine (not found in nature)

−2N

Hydrazine (not found in nature)

−3NH

Ammonia From biological decay of proteins

Ammonium ion

OH Ammonium hydroxide —

TABLE 19.2

Saccharine

19.3 Redox Equations 833

Compound Oxidation Number of N Oxidation Number of O

0 −

− 0

NO +2 −2

+4–2

The nitrogen has been successively oxidized going from N

to NO to NO

, and

the oxygen has been reduced as it changes from O

to its products.

Further Insights

We noted in Figure 19.5 that hydrogen, which is normally assigned an oxidation

state of +1 in compounds, is assigned an oxidation state of −1 when combining

with Group 1A metals. This can result in a very reactive reducing agent that is quite

useful in organic chemical reactions. One example is NaH, mentioned earlier, which

is used in the manufacture of many pharmaceuticals, perfumes, and other organic

chemicals.

PRACTICE 19.2

Rank these chemicals in order of increasing oxidation state of sulfur: SO

, and Li

See Problems 9–12.

19.3 Redox Equations

Ira Remsen (1846–1927), one of the co-discoverers of the artiﬁcial sweetener

saccharine, was particularly interested in chemistry as a boy. As an adult, he told

of a childhood visit to the doctor’s ofﬁce where he, when left alone in the exami-

nation room, set out to discover what was meant by something he had read in a

chemistry book: “Nitric acid acts upon copper.” To discover this for himself, he

placed a pure copper penny on the exam table and poured nitric acid from the

doctor’s bottle onto the penny. Remsen continues,

But what was this wonderful thing which I beheld? The cent was already changed,

and it was no small change either. A greenish blue liquid foamed and fumed over

the cent and over the table. The air in the neighborhood of the performance be-

came dark red.A great colored cloud arose. This was disagreeable and suffocating—

how should I stop this? I tried to get rid of the objectionable mess by picking it up

and throwing it out the window, which I had meanwhile opened. I learned an-

other fact—nitric acid not only acts upon copper but it acts upon ﬁngers. The

pain led to another unpremeditated experiment. I drew my ﬁngers across my

trousers and another fact was discovered. Nitric acid also acts upon trousers. Tak-

ing everything into consideration, that was the most impressive experiment, and,

relatively, probably the most costly experiment I have ever performed.

The reaction that Remsen describes, which is illustrated in Figure 19.8 on page

834, is a redox reaction. Shown on the right is the unbalanced equation. How do

we know that it is a redox reaction? Examine the oxidation state of copper. Cop-

per metal (oxidation state =0) is being oxidized to the copper(II) ion.

Simultaneously, nitrogen in nitric acid is being reduced from +5 to

+2 in forming nitrogen monoxide. The NO gas released by the reac-

tion rapidly reacts with O

in the air to make NO

(g), the brown fumes

that so alarmed the budding chemist.

HNO

(aq) + Cu(s)

(aq) + NO(g)

2NO(g) + O

(g) 2NO

(g)

Half-Reactions

Ira Remsen noted that this reaction proceeded spontaneously to generate a cloud

of noxious fumes. Can we predict this spontaneity by examining the reaction

equation, which tells us whether the driving force (the potential) is favorable for

this reaction? To assist us in answering this question, we need to extract the oxi-

dation and reduction reactions from the overall equation. These half-reactions,

like so many half-reactions, are so well known that the potential for each has been

measured and the results collected into a Table of Standard Reduction Potentials,

such as Table 19.3. A more comprehensive table can be found in the Appendix.

What do you notice about these tables? One of the things is that all of the

reactions are written as reduction reactions. That is, the reactions show the con-

sumption of electrons to make products with less positive oxidation states.

Standard Reduction Potentials tables can be used to determine the potential of a

reaction, be it for the silver oxide battery found in a pacemaker or for the action of

nitric acid on a copper penny. Moreover, knowing the potential of the half-

reactionshelps usdetermine the spontaneityof aredoxreaction,aswewill seelater.

Each half-reaction in the table is balanced both atomically and electrically.

Half-reactions are simply what they appear to be: half of an oxidation–reduction

reaction that is occurring in aqueous solution. The half-reaction listed in the

table for the reduction of copper shows the reactants (copper ions and electrons),

the product (copper metal), and the

standard potential (E°) of the half-reaction.

(aq) + 2e

−

n Cu(s) E° =+0.34 V

The value of E° is a measure of how strongly the reduced species on the right-

hand side of the reduction half-reaction pulls electrons toward itself. The

standard potential is measured in

volts, the SI unit of electrical potential. It is

sometimes referred to as the

electromotive force (emf) of the half-cell or, more

commonly, as the

voltage.

The values listed for E° are measured under a speciﬁc set of conditions:

■

Any aqueous ion is present at a concentration (technically, activity) of 1.0 M.

All gases are at a pressure of 1 bar (approximately 1 atm).

■

The temperature is 25°C (298 K).

These conditions are “standard” for half-reactions and are indicated by the “°”

in E°. If the conditions are not standard, the voltage will be different from that

listed in the table (see Section 19.7), and the potential will be equal to E.Keep

in mind that there are several “standards”! For example, standard conditions

of temperature and pressure of gases (STP) refer to 0°C (273 K) as the standard

temperature.

834 Chapter 19 Electrochemistry

FIGURE 19.8

Nitric acid acts on copper. The spontaneous reaction is evident from the generation of a blue solution and

a cloud of noxious brown gas. The gas results from the reaction of NO with oxygen in the air.

Video Lesson: Balancing Redox

Reactions Using the Half-

Reaction Method

Video Lesson: Standard

Reduction Potentials

Video Lesson: Electromotive

Force

We speak of standard reduction potential in terms of how strongly the species

pulls electrons toward itself. Yet just as in a tug-of-war, we must consider what we

are pulling against. We need a commonly used reference half-reaction with which

to compare our reduction. Our reference is called the

standard hydrogen electrode

reaction (SHE)

. This half-reaction, which is assigned the potential of zero volts, is

also shown in the table as the reduction of H

to H

. To say that the reduction of

to Cu

has a voltage of +0.34 V, as in Table 19.3, really is to say that it has

this voltage compared to the reduction of H

described by the SHE reaction.All po-

tentials that we use in our discussions will be compared to the SHE reaction.

The potential of a half-reaction can be used to assess the spontaneity of the

half-reaction. For instance, ﬂuorine has a strong attraction for electrons. Recall-

ing our discussion of ionization energy and electron afﬁnity from Chapter 7, we

might predict that adding electrons to F

should be more thermodynamically

favorable than adding electrons to a Group IA metal cation such as Li

. The

half-reaction potentials for each reduction bear this out. Fluorine has a large

positive reduction potential (+2.87 V), and lithium has a large negative reduc-

tion potential (−3.04 V). Michael Faraday (1791−1867), an English electro-

chemist, worked hard to illustrate how a favorable reaction could be related to the

potential. Gibbs later was able to show this relationship mathematically as

∆G° = −nFE°

where n is the number of moles of electrons transferred in the reaction, and F is

called Faraday’s constant, which we’ll discuss in a moment. A key feature of this

equation is that the change in free energy, ∆G

°, and the cell potential, E°, have

19.3 Redox Equations 835

Selected Standard Reduction Potentials

The selected potentials shown here were obtained under standard conditions (in aqueous solution, 25°C, all

solutions 1.0 M, all gases 1.0 atm).

Shorthand Notation Half-Cell Reaction Standard Potential, E°(V)

(aq) | Li(s)Li

(aq) + e

–

n Li(s) –3.04

(aq) | Na(s)Na

(aq) + e

–

n Na(s) –2.71

(aq) | Mg(s)Mg

(aq) + 2e

–

n Mg(s) –2.38

(aq) | Al(s)Al

(aq) + 3e

–

n Al(s) –1.66

O(l) | H

(g)2H

O(l) + 2e

–

n H

(g) + 2OH

−

(aq) –0.83

Cd(OH)

(s) | Cd(s) Cd(OH)

(s) + 2e

–

n Cd(s) + 2OH

–

(aq) –0.81

(aq) | Fe(s)Fe

(aq) + 2e

–

n Fe(s) –0.44

(aq) | H

(g)2H

(aq) + 2e

–

n H

(g) 0.00

(aq) | Fe(s)Fe

(aq) + 3e

–

n Fe(s) +0.04

(aq) | Cu(s)Cu

(aq) + 2e

–

n Cu(s) +0.34

(g) | OH

−

(aq)O

(g) + 2H

O(l) + 4e

−

n 4OH

–

(aq) +0.40

NiO

(s) | Ni(OH)

(s)NiO

(s) +2H

O(l) +2e

−

n Ni(OH)

(s) +2OH

−

(aq) +0.49

(aq) | Ag(s)Ag

(aq) + e

−

n Ag(s) +0.80

HNO

(aq) | NO(g)3H

(aq) + HNO

(aq) + 3e

–

n NO(g) + 2H

O(l) +0.96

(l) | Br

−

(aq)Br

(g) + 2e

−

n 2Br

–

(aq) +1.07

(g) | H

O(l)O

(g) + 4H

(aq) + 4e

−

n 4H

O(l) +1.23

(g) | Cl

−

(aq)Cl

(g) + 2e

−

n 2Cl

–

(aq) +1.36

(aq) | Au(s)Au

(aq) + 3e

−

n Au(s) +1.50

(g) | F

−

(aq)F

(g) + 2e

−

n 2F

−

(aq) +2.87

TABLE 19.3

Video Lesson: Using Standard

Reduction Potentials

opposite signs (n and F are always positive). Because a negative value of free en-

ergy indicates a spontaneous process, a positive value of cell potential must also

indicate spontaneity.

Faraday’s constant is a unit of electric charge equal to the magnitude of charge

on a mole of electrons:

1 faraday = 1F= 96,485

coulombs

mol

= 9.6485

mol

On the basis of the relationship shown by Faraday, we can determine that 1 joule

equals 1 coulomb·volt, and 1 volt = 1

joule

coulomb

1 J = 1 C · V

1V=

EXERCISE 19.3 Spontaneity and Potential

Copper ions undergo reduction according to the following half-reaction:

(aq) + 2e

−

n Cu(s) E° =+0.34 V

What is the free energy change associated with this process? Is this a spontaneous

half-reaction?

Solution

The free energy change is negative; the half-reaction is spontaneous. However,

this is only half of a redox reaction.

G

◦

=−nFE

◦

=−(2 mol e

−

)



96485 C

mol e

−



+0.34 J



=−65609.8J=−66 kJ

PRACTICE 19.3

The silver cell battery used in pacemakers utilizes the following reaction with a

measured potential of 1.86 V. What is

G

for this reaction? Is this reaction

spontaneous?

O(s) + Zn(s)

→

2Ag(s) + ZnO(s)

See Problems 25 and 26.

Balancing Redox Reactions

To balance a redox reaction such as the one describing the action of nitric acid on

copper, we ﬁrst determine the identity of the half-reactions. It can be hard (if not

seemingly impossible!) to balance a redox equation correctly using a trial-and-

error approach (Chapter 3), so we often use a series of steps to accomplish the job

(see Figure 19.9). To be fair, this method is just a device to make the balancing go

more quickly, rather than a representation of what actually happens at the mole-

cular level. In the nanoworld, electron transfer processes not only occur simulta-

neously rather than sequentially but also occur in a fairly complex way, with

charges building up at the phase changes (the so-called interfaces) in solution.

Here we will focus just on the technical aspects of balancing equations.

836 Chapter 19 Electrochemistry

Let’s follow the algorithm in Figure 19.9 as we balance the copper–nitric acid

redox reaction:

HNO

(aq) + Cu(s) n Cu

(aq) + NO(g)

Step 1 in Figure 19.9 indicates that we should determine whether we have a

redox reaction. That is, does our oxidation state bookkeeping indicate that one

species is undergoing oxidation and another is undergoing reduction? We deter-

mine that the equation describes a redox process by noting that the oxidation

state of copper increases (from 0 to +2) while that of nitrogen decreases (from

+5 to +2).

Step 2 indicates that we should separate the redox reaction into the two half-

reactions. These half-reactions show the oxidation (copper to copper ion) and

the reduction (nitric acid to nitrogen monoxide).

Cu(s) n Cu

(aq)

HNO

(aq) n NO(g)

The numbers of atoms (not including H’s and O’s) are then balanced on both

sides of each half-reaction in Step 3. No modiﬁcation of our reactions is needed

for this step.

19.3 Redox Equations 837

Step 8: Add the reactions together and simplify.

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

that both half-reactions will have the same number of electrons.

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

Step 5: Balance the hydrogen atoms by adding H

to one side in each half-reaction.

Step 4: Balance the oxygen atoms by adding H

O to one side of each half-reaction.

Step 3: Balance all atoms that are neither oxygen nor hydrogen.

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

reaction. Include only the compounds that have a change in the oxidation state.

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

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

FIGURE 19.9

Algorithm for balancing redox reactions

in acidic solution.

Visualization: Copper Metal in

Nitric Acid