Redox Reactions and Electrochemistry

CONFIRMING PAGES

CHAPTER Hydrogen gas generated from an illuminated photoelectrode. Using light energy to produce hydrogen from water can play an important role in the development of fuel cells.

CHAPTER OUTLINE ESSENTIAL CONCEPTS 19.1 Redox Reactions 643 Redox Reactions and Electrochemical Cells Equations repre- Balancing Redox Equations senting redox reactions can be balanced using the half-reaction method. These reactions involve the transfer of electrons from a 19.2 Galvanic Cells 646 reducing agent to an oxidizing agent. Using separate compart- 19.3 Standard Reduction Potentials 648 ments, such a reaction can be used to generate electrons in an 19.4 Thermodynamics of Redox Reactions 654 arrangement called a galvanic cell. 19.5 The Effect of Concentration on Cell Emf 657 Thermodynamics of Galvanic Cells The voltage measured in a The Nernst Equation • Concentration Cells galvanic cell can be broken down into the electrode potentials of the anode (where oxidation takes place) and cathode (where 19.6 Batteries 661 reduction takes place). This voltage can be related to the Gibbs The Dry Cell Battery • The Mercury Battery • The Lead Storage free energy change and the equilibrium constant of the redox Battery • The Lithium-Ion Battery • Fuel Cells process. The Nernst equation relates the cell voltage to the cell 19.7 Corrosion 665 voltage under standard-state conditions and the concentrations of 19.8 Electrolysis 668 reacting species. Electrolysis of Molten Sodium Chloride • Electrolysis of Batteries Batteries are electrochemical cells that can supply Water • Electrolysis of an Aqueous Sodium Chloride direct electric current at a constant voltage. There are many Solution • Quantitative Aspects of Electrolysis different types of batteries used in automobiles, flashlights, and 19.9 Electrometallurgy 673 pacemakers. Fuel cells are a special type of electrochemical cell Production of Aluminum Metal • Purification of Copper Metal that generates electricity by the oxidation of hydrogen or hydro- carbons. Corrosion Corrosion is a spontaneous redox reaction that results in the formation of rust from iron, silver sulfide from silver, and patina (copper carbonate) from copper. Corrosion causes enormous damage to buildings, structures, ships, and cars. Many methods have been devised to prevent or minimize the effect of corrosion. Electrolysis Electrolysis is the process in which electrical energy Activity Summary is used to cause a nonspontaneous redox reaction to occur. The quantitative relationship between the current supplied and 1. Interactivity: Redox Reactions—Oxidation States for the products formed is provided by Faraday. Electrolysis is the Nitrogen (19.1) major method for producing active metals and nonmetals and many essential industrial chemicals. 2. Interactivity: Redox Reactions— Oxidized or Reduced? (19.1) 3. Animation: Galvanic Cells (19.2) cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 643 CONFIRMING PAGES

19.1 Redox Reactions 643

19.1 Redox Reactions Electrochemistry is the branch of chemistry that deals with the interconversion of electrical energy and chemical energy . Electrochemical processes are redox (oxidation- reduction) reactions in which the energy released by a spontaneous reaction is converted to electricity or in which electrical energy is used to cause a nonspontaneous reaction to occur. Although redox reactions were discussed in Chapter 4, it is helpful to review some of the basic concepts that will come up again in this chapter. In redox reactions electrons are transferred from one substance to another. The reaction between magnesium metal and hydrochloric acid is an example of a redox reaction: 0 ϩ1 ϩ2 0 Mg (s) ϩ 2HCl (aq ) ¡ Mg Cl 2(aq ) ϩ H2(g) Recall that the numbers above the elements are the oxidation numbers of the elements. The loss of electrons by an element during oxidation is marked by an increase in the element’s oxidation number. In reduction, there is a decrease in oxidation number resulting from a gain of electrons by an element. In the preceding reaction Mg metal is oxidized and H ϩ ions are reduced; the Cl Ϫ ions are spectator ions.

Balancing Redox Equations

Equations for redox reactions like the preceding one are relatively easy to balance. Interactivity: However, in the laboratory we often encounter more complex redox reactions involv- Redox Reactions—Oxidation 2Ϫ 2Ϫ States for Nitrogen ing oxoanions such as chromate (CrO 4 ), dichromate (Cr 2O7 ), permanganate ARIS, Interactives Ϫ Ϫ 2Ϫ (MnO 4 ), nitrate (NO 3 ), and sulfate (SO 4 ). In principle, we can balance any redox Interactivity: equation using the procedure outlined in Section 3.7, but there are some special tech- Redox Reactions—Oxidized or Reduced? niques for handling redox reactions, techniques that also give us insight into electron ARIS, Interactives transfer processes. Here we will discuss one such procedure, called the half-reaction method. In this approach, the overall reaction is divided into two half-reactions, one for oxidation and one for reduction. The equations for the two half-reactions are balanced separately and then added together to give the overall balanced equation. Suppose we are asked to balance the equation showing the oxidation of Fe 2ϩ ions 3ϩ 2Ϫ 2Ϫ to Fe ions by dichromate ions (Cr 2O7 ) in an acidic medium. As a result, the Cr 2O7 ions are reduced to Cr 3ϩ ions. The following steps will help us balance the equation. Step 1: Write the unbalanced equation for the reaction in ionic form.

2ϩ 2Ϫ 3ϩ 3ϩ Fe ϩ Cr 2O7 ¡ Fe ϩ Cr Step 2: Separate the equation into two half-reactions.

ϩ2 ϩ3 Oxidation: Fe 2ϩ ¡ Fe 3ϩ ϩ6 ϩ3 2Ϫ 3ϩ Oxidation: Cr 2O7 ¡ Cr Step 3: Balance each half-reaction for number and type of atoms and charges. For ϩ reactions in an acidic medium, add H 2O to balance the O atoms and H to balance the H atoms. Oxidation half-reaction: The atoms are already balanced. To balance the charge, we

add an electron to the right-hand side of the arrow: In an oxidation half-reaction, electrons appear as a product; in a reduction half- 2ϩ 3ϩ Ϫ Fe ¡ Fe ϩ e reaction, electrons appear as a reactant. cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 644 CONFIRMING PAGES

644 CHAPTER 19 Redox Reactions and Electrochemistry

Reduction half-reaction: Because the reaction takes place in an acidic medium, we add seven H 2O molecules to the right-hand side of the arrow to balance the O atoms:

2Ϫ 3ϩ Cr 2O7 ¡ 2Cr ϩ 7H 2O To balance the H atoms, we add 14 H ϩ ions on the left-hand side:

ϩ 2Ϫ 3ϩ 14H ϩ Cr 2O7 ¡ 2Cr ϩ 7H 2O

There are now 12 positive charges on the left-hand side and only six positive charges on the right-hand side. Therefore, we add six electrons on the left

ϩ 2Ϫ Ϫ 3ϩ 14H ϩ Cr 2O7 ϩ 6e ¡ 2Cr ϩ 7H 2O Step 4: Add the two half-reactions together and balance the ﬁnal equation by inspection. The electrons on both sides must cancel. If the oxidation and reduction half-reactions contain different numbers of electrons, we need to multiply one or both half-reactions to equalize the number of electrons. Here we have only one electron for the oxidation half-reaction and six electrons for the reduction half-reaction, so we need to multiply the oxidation half-reaction by 6 and write 6(Fe 2ϩ ¡ Fe 3ϩ ϩ eϪ) ϩ 2Ϫ Ϫ 3ϩ 14H ϩ Cr 2O7 ϩ 6e ¡ 2Cr ϩ 7H 2O 2ϩ ϩ 2Ϫ Ϫ 3ϩ 3ϩ Ϫ 6Fe ϩ 14H ϩ Cr 2O7 ϩ 6e ¡ 6Fe ϩ 2Cr ϩ 7H 2O ϩ 6e

The electrons on both sides cancel, and we are left with the balanced net ionic equation:

2ϩ ϩ 2Ϫ 3ϩ 3ϩ This reaction can be carried out by 6Fe ϩ 14H ϩ Cr 2O7 ¡ 6Fe ϩ 2Cr ϩ 7H 2O dissolving potassium dichromate and iron(II) sulfate in a dilute sulfuric acid Step 5: Verify that the equation contains the same type and numbers of atoms and solution. the same charges on both sides of the equation. A ﬁnal check shows that the resulting equation is “atomically” and “electrically” balanced. For reactions in a basic medium, we proceed through step 4 as if the reaction were carried out in a acidic medium. Then, for every H ϩ ion we add an equal number of OH Ϫ ions to both sides of the equation. Where H ϩ and OH Ϫ ions appear on the same side of the equation, we combine the ions to give H 2O. Example 19.1 illus- trates this procedure.

Example 19.1 Write a balanced ionic equation to represent the oxidation of iodide ion (I Ϫ) by perman- Ϫ ganate ion (MnO 4 ) in basic solution to yield molecular iodine (I 2) and manganese(IV) oxide (MnO 2). Strategy We follow the preceding procedure for balancing redox equations. Note that the reaction takes place in a basic medium. Solution Step 1: The unbalanced equation is

Ϫ Ϫ MnO 4 ϩ I ¡ MnO 2 ϩ I2 (Continued ) cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 645 CONFIRMING PAGES

19.1 Redox Reactions 645

Step 2: The two half-reactions are Ϫ Ϫ1 0 Oxidation: I ¡ I2 ϩ ϩ 7 Ϫ 4 Reduction: MnO 4 ¡ MnO 2

Step 3: We balance each half-reaction for number and type of atoms and charges. Oxidation half-reaction: We ﬁrst balance the I atoms:

Ϫ 2I ¡ I2

To balance charges, we add two electrons to the right-hand side of the equation:

Ϫ Ϫ 2I ¡ I2 ϩ 2e

Reduction half-reaction: To balance the O atoms, we add two H 2O molecules on the right: Ϫ MnO 4 ¡ MnO 2 ϩ 2H 2O

To balance the H atoms, we add four H ϩ ions on the left: Ϫ ϩ MnO 4 ϩ 4H ¡ MnO 2 ϩ 2H 2O

There are three net positive charges on the left, so we add three electrons to the same side to balance the charges: Ϫ ϩ Ϫ MnO 4 ϩ 4H ϩ 3e ¡ MnO 2 ϩ 2H 2O

Step 4: We now add the oxidation and reduction half reactions to give the overall reaction. To equalize the number of electrons, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 as follows:

Ϫ Ϫ 3(2I ¡ I2 ϩ 2e ) Ϫ ϩ Ϫ 2(MnO 4 ϩ 4H ϩ 3e ¡ MnO 2 ϩ 2H 2O) Ϫ Ϫ ϩ Ϫ Ϫ 6I ϩ 2MnO 4 ϩ 8H ϩ 6e ¡ 3I 2 ϩ 2MnO 2 ϩ 4H 2O ϩ 6e

The electrons on both sides cancel, and we are left with the balanced net ionic equation:

Ϫ Ϫ ϩ 6I ϩ 2MnO 4 ϩ 8H ¡ 3I 2 ϩ 2MnO 2 ϩ 4H 2O

This is the balanced equation in an acidic medium. However, because the reaction is carried out in a basic medium, for every H ϩ ion we need to add an equal number of OH Ϫ ions to both sides of the equation:

Ϫ Ϫ ϩ Ϫ Ϫ 6I ϩ 2MnO 4 ϩ 8H ϩ 8OH ¡ 3I 2 ϩ 2MnO 2 ϩ 4H 2O ϩ 8OH

Finally, combining the H ϩ and OH Ϫ ions to form water, we obtain

Ϫ Ϫ Ϫ 6I ϩ 2MnO 4 ϩ 4H 2O ¡ 3I 2 ϩ 2MnO 2 ϩ 8OH To carry out this reaction, mix KI and KMnO 4 in a basic medium. Step 5: A ﬁnal check shows that the equation is balanced in terms of both atoms and charges. Similar problems: 19.1, 19.2. Practice Exercise Balance the following equation for the reaction in an acidic medium by the half-reaction method:

2ϩ Ϫ 3ϩ 2ϩ Fe ϩ MnO 4 ¡ Fe ϩ Mn cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 646 CONFIRMING PAGES

646 CHAPTER 19 Redox Reactions and Electrochemistry

19.2 Galvanic Cells

In Section 4.4 we saw that when a piece of zinc metal is placed in a CuSO 4 solution, Zn is oxidized to Zn 2ϩ ions while Cu 2ϩ ions are reduced to metallic copper (see Figure 4.13):

Zn( s) ϩ Cu 2ϩ(aq ) ¡ Zn 2ϩ(aq ) ϩ Cu(s)

The electrons are transferred directly from the reducing agent (Zn) to the oxidizing agent (Cu 2ϩ) in solution. However, if we physically separate the oxidizing agent from the reducing agent, the transfer of electrons can take place via an external conducting medium (a metal wire). As the reaction progresses, it sets up a constant ﬂow of electrons and hence generates electricity (that is, it produces electrical work such as driving an electric motor). Animation: The experimental apparatus for generating electricity through the use of a spon- Galvanic Cells taneous reaction is called a galvanic cell or voltaic cell, after the Italian scientists ARIS, Animations Luigi Galvani and Alessandro Volta, who constructed early versions of the device. Figure 19.1 shows the essential components of a galvanic cell. A zinc bar is immersed

Voltmeter

e– e–

Zinc – + Copper anode Cl K cathode Salt bridge

Cotton plugs Cu 2ϩ

2+ 2– Zn SO 4

ZnSO 4 solution CuSO4 solution 2e– 2e– Cu 2+ Zn Zn 2+

Zn is oxidized Cu 2+ is reduced to Zn 2+ at anode. to Cu at cathode.

Zn( s) Zn2+ (aq ) + 2 e– 2e– + Cu 2+ (aq ) Cu(s) Net reaction

Zn( s) + Cu2+ (aq ) Zn2+ (aq ) + Cu(s)

Figure 19.1 A galvanic cell. The salt bridge (an inverted U tube) containing a KCl solution provides an electrically conducting medium between two solutions. The openings of the U tube are loosely plugged with cotton balls to prevent the KCl solution from ﬂowing into the containers while allowing the anions and cations to move across. Electrons ﬂow externally from the Zn electrode (anode) to the Cu electrode (cathode). cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 647 CONFIRMING PAGES

19.2 Galvanic Cells 647

in a ZnSO 4 solution, and a copper bar is immersed in a CuSO 4 solution. The cell operates on the principle that the oxidation of Zn to Zn 2ϩ and the reduction of Cu 2ϩ to Cu can be made to take place simultaneously in separate locations with the transfer of electrons between them occurring through an external wire. The zinc and copper bars are called electrodes. This particular arrangement of electrodes (Zn and Cu) and solutions (ZnSO 4 and CuSO 4) is called the Daniell cell. By deﬁnition, the anode Alphabetically anode precedes cathode in a galvanic cell is the electrode at which oxidation occurs and the cathode is the and oxidation precedes reduction. Therefore, anode is where oxidation electrode at which reduction occurs. occurs and cathode is where reduction For the Daniell cell, the half-cell reactions, that is, the oxidation and reduction takes place. reactions at the electrodes, are

2ϩ Ϫ Zn electrode (anode): Zn( s) 88n Zn (aq ) ϩ 2e Half-cell reactions are similar to the half- Cu electrode (cathode): Cu 2ϩ(aq ) ϩ 2eϪ 88n Cu( s) reactions discussed earlier.

Note that unless the two solutions are separated from each other, the Cu 2ϩ ions will react directly with the zinc bar:

Cu 2ϩ(aq ) ϩ Zn( s) ¡ Cu(s) ϩ Zn 2ϩ(aq )

and no useful electrical work will be obtained. To complete the electrical circuit, the solutions must be connected by a conducting medium through which the cations and anions can move from one electrode compartment to the other. This requirement is satisfied by a salt bridge, which, in its simplest form, is an inverted U tube containing an inert electrolyte solution, such as KCl or NH 4NO 3, whose ions will not react with other ions in solution or with the electrodes (see Figure 19.1). During the course of the overall redox reaction, electrons flow externally from the anode (Zn electrode) through the wire and voltmeter to the cathode (Cu electrode). In the solution, the cations (Zn 2ϩ, Cu 2ϩ, and K ϩ) move 2Ϫ Ϫ toward the cathode, while the anions (SO 4 and Cl ) move toward the anode. With- out the salt bridge connecting the two solutions, the buildup of positive charge in the anode compartment (due to the formation of Zn 2ϩ ions) and negative charge in the cathode compartment (created when some of the Cu 2ϩ ions are reduced to Cu) would quickly prevent the cell from operating. An electric current flows from the anode to the cathode because there is a difference in electrical potential energy between the electrodes. This flow of electric current is analogous to that of water down a waterfall, which occurs because there is a difference in gravitational potential energy, or the flow of gas from a high-pressure region to a low-pressure region. The voltage across the electrodes of a galvanic cell is called the cell voltage, or cell potential. Experimentally, this is measured by a voltmeter (Figure 19.2). Another common term for the cell potential is the electromotive force or emf (E), which, despite the name, is a measure of voltage, not force. We will see that the voltage of a cell depends not only on the nature of electrodes and the ions, but also on the concentrations of the ions and the temperature at which the cell is operated. The conventional notation for representing galvanic cells is the cell diagram. For the Daniell cell shown in Figure 19.1, if we assume that the concentrations of Zn 2ϩ and Cu 2ϩ ions are 1 M, the cell diagram is

Zn( s)|Zn 2ϩ(1 M)|| Cu 2ϩ(1 M)|Cu(s)

The single vertical line represents a phase boundary. For example, the zinc electrode 2ϩ is a solid and the Zn ions (from ZnSO 4) are in solution. Thus, we draw a line between Zn and Zn 2ϩ to show the phase boundary. The double vertical lines denote cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 648 CONFIRMING PAGES

648 CHAPTER 19 Redox Reactions and Electrochemistry

Figure 19.2 Practical setup of the galvanic cell described in Figure 19.1. Note the U tube (salt bridge) connecting the two beakers. When the concentrations of Salt bridge ZnSO 4 and CuSO 4 are 1 molar (1 M) at 25 ЊC, the cell voltage is 1.10 V.

H2 gas at 1 atm

the salt bridge. By convention, the anode is written ﬁrst, to the left of the double lines Pt electrode and the other components appear in the order in which we would encounter them in moving from the anode to the cathode. 1 M HCl

Figure 19.3 A hydrogen electrode operating 19.3 Standard Reduction Potentials under standard-state conditions. 2ϩ 2ϩ Hydrogen gas at 1 atm is When the concentrations of the Cu and Zn ions are both 1.0 M, we ﬁnd that the bubbled through a 1 M HCl voltage or emf of the Daniell cell is 1.10 V at 25 ЊC (see Figure 19.2). This voltage solution. The platinum electrode must be related directly to the redox reactions, but how? Just as the overall cell reac- is part of the hydrogen electrode. tion can be thought of as the sum of two half-cell reactions, the measured emf of the cell can be treated as the sum of the electrical potentials at the Zn and Cu electrodes. Knowing one of these electrode potentials, we could obtain the other by subtraction The choice of an arbitrary reference for (from 1.10 V). It is impossible to measure the potential of just a single electrode, but measuring electrode potential is if we arbitrarily set the potential value of a particular electrode at zero, we can use analogous to choosing the surface of the ocean as the reference for altitude, it to determine the relative potentials of other electrodes. The hydrogen electrode, calling it zero meters, and then referring shown in Figure 19.3, serves as the reference for this purpose. Hydrogen gas is bub- to any terrestrial altitude as being a Њ certain number of meters above or below bled into a hydrochloric acid solution at 25 C. The platinum electrode has two func- sea level. tions. First, it provides a surface on which the dissociation of hydrogen molecules can take place: ϩ Ϫ H2 ¡ 2H ϩ 2e

Second, it serves as an electrical conductor to the external circuit. Standard states are deﬁned in Table 18.2 Under standard-state conditions (when the pressure of H 2 is 1 atm and the con- (p. 624). centration of the HCl solution is 1 M; see Table 18.2), the potential for the reduction of H ϩ at 25 ЊC is taken to be exactly zero:

ϩ Ϫ 2H (1 M) ϩ 2e ¡ H2(1 atm) E° ϭ 0 V The superscript “ Њ” denotes standard-state conditions, and EЊ is the standard reduction potential, or the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. Thus, the standard reduction potential of the hydrogen electrode is deﬁned as zero. The hydrogen electrode is called the standard hydrogen electrode (SHE). cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 649 CONFIRMING PAGES

19.3 Standard Reduction Potentials 649

Voltmeter Voltmeter e– e– e– e–

Zn H2 gas at 1 atm H2 gas at 1 atm Cu Salt bridge Salt bridge

Pt electrode Pt electrode

1 M ZnSO 4 1 M HCl 1 M HCl 1 M CuSO4 Zinc electrode Hydrogen electrode Hydrogen electrode Copper electrode (a) (b)

Figure 19.4 (a) A cell consisting of a zinc electrode and a hydrogen electrode. (b) A cell consisting of a copper electrode and a hydrogen electrode. Both cells are operating under standard-state conditions. Note that in (a) the SHE acts as the cathode, but in (b) it is the anode.

We can use the SHE to measure the potentials of other kinds of electrodes. For example, Figure 19.4(a) shows a galvanic cell with a zinc electrode and a SHE. In this case the zinc electrode is the anode and the SHE is the cathode. We deduce this fact from the decrease in mass of the zinc electrode during the operation of the cell, which is consistent with the loss of zinc to the solution caused by the oxidation reaction:

Zn( s) ¡ Zn 2ϩ(aq ) ϩ 2eϪ

The cell diagram is

2ϩ ϩ Zn( s)|Zn (1 M)|| H (1 M)|H2(1 atm) |Pt( s)

As mentioned earlier, the Pt electrode provides the surface on which the reduction takes place. When all the reactants are in their standard states (that is, H 2 at 1 atm, Hϩ and Zn 2ϩ ions at 1 M), the emf of the cell is 0.76 V at 25 ЊC. We can write the half-cell reactions as follows:

Anode (oxidation): Zn( s) 88n Zn 2ϩ(1 M) ϩ 2eϪ ϩ Ϫ Cathode (reduction): 2H (1 M) ϩ 2e 88n H2(1 atm) ϩ 2ϩ Overall: Zn( s) ϩ 2H (1 M) 88n Zn (1 M) ϩ H2(1 atm)

-By convention, the standard emf of the cell, E؇cell , which is composed of a contribu tion from the anode and a contribution from the cathode, is given by

E°cell ϭ E°cathode Ϫ E°anode (19.1)

where both E cathodeЊ and EanodeЊ are the standard reduction potentials of the electrodes. For the Zn-SHE cell, we write

ϭ ϩ Ϫ 2ϩ E°cell E°H րH2 E°Zn րZn 0.76 V ϭ 0 Ϫ E°Zn 2ϩրZn

ϩ ϩ Ϫ 2ϩ where the subscript H ͞H2 means 2H ϩ 2e → H2 and the subscript Zn ͞Zn means 2ϩ Ϫ Zn ϩ 2e → Zn. Thus, the standard reduction potential of zinc, E°Zn 2ϩրZn , is Ϫ0.76 V. cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 650 CONFIRMING PAGES

650 CHAPTER 19 Redox Reactions and Electrochemistry

The standard electrode potential of copper can be obtained in a similar fashion, by using a cell with a copper electrode and a SHE [Figure 19.4(b)]. In this case, the copper electrode is the cathode because its mass increases during the operation of the cell, as is consistent with the reduction reaction:

Cu 2ϩ(aq ) ϩ 2eϪ 88n Cu( s)

The cell diagram is

ϩ 2ϩ Pt( s)|H2(1 atm) |H (1 M)|| Cu (1 M)|Cu(s)

and the half-cell reactions are ϩ Ϫ Anode (oxidation): H2(1 atm) 88n 2H (1 M) ϩ 2e Cathode (reduction): Cu 2ϩ(1 M) ϩ 2eϪ 88n Cu( s) 2ϩ ϩ Overall: H 2(1 atm) ϩ Cu (1 M) 88n 2H (1 M) ϩ Cu( s) Under standard-state conditions and at 25 ЊC, the emf of the cell is 0.34 V, so we write

E°cell ϭ E°cathode Ϫ E°anode ϭ 2ϩ Ϫ ϩ 0.34 V E°Cu /Cu E°H /H 2 ϭ E°Cu 2ϩ/Cu Ϫ 0

In this case, the standard reduction potential of copper, E°Cu 2ϩ/Cu , is 0.34 V, where the subscript means Cu 2ϩ ϩ 2eϪ → Cu. For the Daniell cell shown in Figure 19.1, we can now write Anode (oxidation): Zn( s) 88n Zn 2ϩ(1 M) ϩ 2eϪ Cathode (reduction): Cu 2ϩ(1 M) ϩ 2eϪ 88n Cu( s) Overall: Zn( s) ϩ Cu 2ϩ(1 M) 88n Zn 2ϩ(1 M) ϩ Cu( s) The emf of the cell is

E°cell ϭ E°cathode Ϫ E°anode ϭ E°Cu 2ϩ/Cu Ϫ E°Zn 2ϩ/Zn ϭ 0.34 V Ϫ (Ϫ0.76 V) ϭ 1.10 V

As in the case of ⌬GЊ (p. 630), we can use the sign of EЊ to predict the extent of a redox reaction. A positive EЊ means the redox reaction will favor the formation of products at equilibrium. Conversely, a negative EЊ means that more reactants than products will be formed at equilibrium. We will examine the relationships among EЊcell , ⌬GЊ, and K later in this chapter. The activity series in Figure 4.14 is based Table 19.1 lists standard reduction potentials for a number of half-cell reactions. on data given in Table 19.1. By deﬁnition, the SHE has an EЊ value of 0.00 V. Below the SHE the negative standard reduction potentials increase, and above it the positive standard reduction potentials increase. It is important to know the following points about the table in calculations: 1. The EЊ values apply to the half-cell reactions as read in the forward (left to right) direction. 2. The more positive EЊ is, the greater the tendency for the substance to be reduced. For example, the half-cell reaction Ϫ Ϫ ϭ F2(1 atm) ϩ 2e ¡ 2F (1 M) E° 2.87 V cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 651 CONFIRMING PAGES

19.3 Standard Reduction Potentials 651

*TABLE 19.1 Standard Reduction Potentials at 25 ؇C

Half-Reaction EЊ(V)

Ϫ Ϫ F2(g) ϩ 2e 88n 2F (aq ) ϩ2.87 ϩ Ϫ O3(g) ϩ 2H (aq ) ϩ 2e 88n O2(g) ϩ H2O ϩ2.07 Co 3ϩ(aq ) ϩ eϪ 88n Co 2ϩ(aq ) ϩ1.82 ϩ Ϫ H2O2(aq ) ϩ 2H (aq ) ϩ 2e 88n 2H 2O ϩ1.77 ϩ 2Ϫ Ϫ PbO 2(s) ϩ 4H (aq ) ϩ SO 4 (aq ) ϩ 2e 88n PbSO 4(s) ϩ 2H 2O ϩ1.70 Ce 4ϩ(aq ) ϩ eϪ 88n Ce 3ϩ(aq ) ϩ1.61 Ϫ ϩ Ϫ 2ϩ MnO 4 (aq ) ϩ 8H (aq ) ϩ 5e 88n Mn (aq ) ϩ 4H 2O ϩ1.51 Au 3ϩ(aq ) ϩ 3eϪ 88n Au( s) ϩ1.50 Ϫ Ϫ Cl 2(g) ϩ 2e 88n 2Cl (aq ) ϩ1.36 2Ϫ ϩ Ϫ 3ϩ Cr 2O7 (aq ) ϩ 14H (aq ) ϩ 6e 88n 2Cr (aq ) ϩ 7H 2O ϩ1.33 ϩ Ϫ 2ϩ MnO 2(s) ϩ 4H (aq ) ϩ 2e 88n Mn (aq ) ϩ 2H 2O ϩ1.23 ϩ Ϫ O2(g) ϩ 4H (aq ) ϩ 4e 88n 2H 2O ϩ1.23 Ϫ Ϫ Br 2(l) ϩ 2e 88n 2Br (aq ) ϩ1.07 Ϫ ϩ Ϫ NO 3 (aq ) ϩ 4H (aq ) ϩ 3e 88n NO( g) ϩ 2H 2O ϩ0.96 2ϩ Ϫ 2ϩ 2Hg (aq ) ϩ 2e 88n Hg 2 (aq ) ϩ0.92 2ϩ Ϫ Hg 2 (aq ) ϩ 2e 88n 2Hg( l) ϩ0.85 Ag ϩ(aq ) ϩ eϪ 88n Ag( s) ϩ0.80 Fe 3ϩ(aq ) ϩ eϪ 88n Fe 2ϩ(aq ) ϩ0.77 ϩ Ϫ O2(g) ϩ 2H (aq ) ϩ 2e 88n H2O2(aq ) ϩ0.68 Ϫ Ϫ Ϫ MnO 4 (aq ) ϩ 2H 2O ϩ 3e 88n MnO 2(s) ϩ 4OH (aq ) ϩ0.59 Ϫ Ϫ I2(s) ϩ 2e 88n 2I (aq ) ϩ0.53 Ϫ Ϫ O2(g) ϩ 2H 2O ϩ 4e 88n 4OH (aq ) ϩ0.40 Cu 2ϩ(aq ) ϩ 2eϪ 88n Cu( s) ϩ0.34 AgCl( s) ϩ eϪ 88n Ag( s) ϩ Cl Ϫ(aq ) ϩ0.22 2Ϫ ϩ Ϫ SO 4 (aq ) ϩ 4H (aq ) ϩ 2e 88n SO 2(g) ϩ 2H 2O ϩ0.20 Cu 2ϩ(aq ) ϩ eϪ 88n Cu ϩ(aq ) ϩ0.15 Sn 4ϩ(aq ) ϩ 2eϪ 88n Sn 2ϩ(aq ) ϩ0.13 ϩ Ϫ 2H (aq ) ϩ 2e 88n H2(g) 0.00 Pb 2ϩ(aq ) ϩ 2eϪ 88n Pb( s) Ϫ0.13 Sn 2ϩ(aq ) ϩ 2eϪ 88n Sn( s) Ϫ0.14 Ni 2ϩ(aq ) ϩ 2eϪ 88n Ni( s) Ϫ0.25 Increasing strength as oxidizing agent oxidizing as strength Increasing Co 2ϩ(aq ) ϩ 2eϪ 88n Co( s) Ϫ0.28 agent reducing as strength Increasing Ϫ 2Ϫ PbSO 4(s) ϩ 2e 88n Pb( s) ϩ SO 4 (aq ) Ϫ0.31 Cd 2ϩ(aq ) ϩ 2eϪ 88n Cd( s) Ϫ0.40 Fe 2ϩ(aq ) ϩ 2eϪ 88n Fe( s) Ϫ0.44 Cr 3ϩ(aq ) ϩ 3eϪ 88n Cr( s) Ϫ0.74 Zn 2ϩ(aq ) ϩ 2eϪ 88n Zn( s) Ϫ0.76 Ϫ Ϫ 2H 2O ϩ 2e 88n H2(g) ϩ 2OH (aq ) Ϫ0.83 Mn 2ϩ(aq ) ϩ 2eϪ 88n Mn( s) Ϫ1.18 Al 3ϩ(aq ) ϩ 3eϪ 88n Al( s) Ϫ1.66 Be 2ϩ(aq ) ϩ 2eϪ 88n Be( s) Ϫ1.85 Mg 2ϩ(aq ) ϩ 2eϪ 88n Mg( s) Ϫ2.37 Na ϩ(aq ) ϩ eϪ 88n Na( s) Ϫ2.71 Ca 2ϩ(aq ) ϩ 2eϪ 88n Ca( s) Ϫ2.87 Sr 2ϩ(aq ) ϩ 2eϪ 88n Sr( s) Ϫ2.89 Ba 2ϩ(aq ) ϩ 2eϪ 88n Ba( s) Ϫ2.90 Kϩ(aq ) ϩ eϪ 88n K( s) Ϫ2.93 Li ϩ(aq ) ϩ eϪ 88n Li( s) Ϫ3.05

*For all half-reactions the concentration is 1 M for dissolved species and the pressure is 1 atm for gases. These are the standard-state values cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 652 CONFIRMING PAGES

652 CHAPTER 19 Redox Reactions and Electrochemistry

has the highest positive EЊ value among all the half-cell reactions. Thus, F 2 is the strongest oxidizing agent because it has the greatest tendency to be reduced. At the other extreme is the reaction

Li ϩ(1 M) ϩ eϪ ¡ Li( s) E° ϭ Ϫ 3.05 V

which has the most negative EЊ value. Thus, Li ϩ is the weakest oxidizing agent because it is the most difﬁcult species to reduce. Conversely, we say that F Ϫ is the weakest reducing agent and Li metal is the strongest reducing agent. Under standard-state conditions, the oxidizing agents (the species on the left-hand side of the half-reactions in Table 19.1) increase in strength from bottom to top and the reducing agents (the species on the right-hand side of the half-reactions) increase in strength from top to bottom. 3. The half-cell reactions are reversible. Depending on the conditions, any electrode can act either as an anode or as a cathode. Earlier we saw that the SHE is the ϩ cathode (H is reduced to H 2) when coupled with zinc in a cell and that it ϩ becomes the anode (H 2 is oxidized to H ) when used in a cell with copper. 4. Under standard-state conditions, any species on the left of a given half-cell reaction will react spontaneously with a species that appears on the right of any half- cell reaction located below it in Table 19.1. This principle is sometimes called the diagonal rule. In the case of the Daniell cell,

ϩ Ϫ Cu 2 (1 M) ϩ 2e ¡ Cu(s) E° ϭ 0.34 V

ϩ Ϫ Zn 2 (1 M) ϩ 2e ¡ Zn( s) E° ϭ Ϫ 0.76 V

We see that the substance on the left of the ﬁrst half-cell reaction is Cu 2ϩ and the substance on the right in the second half-cell reaction is Zn. Therefore, as we saw earlier, Zn spontaneously reduces Cu 2ϩ to form Zn 2ϩ and Cu. 5. Changing the stoichiometric coefﬁcients of a half-cell reaction does not affect the value of EЊ because electrode potentials are intensive properties. This means that the value of EЊ is unaffected by the size of the electrodes or the amount of solutions present. For example, Ϫ Ϫ I2(s) ϩ 2e ¡ 2I (1 M) E° ϭ 0.53 V

but EЊ does not change if we multiply the half-reaction by 2:

Ϫ Ϫ 2I 2(s) ϩ 4e ¡ 4I (1 M) E° ϭ 0.53 V 6. Like ⌬H, ⌬G, and ⌬S, the sign of EЊ changes but its magnitude remains the same when we reverse a reaction. As Examples 19.2 and 19.3 show, Table 19.1 enables us to predict the outcome of redox reactions under standard-state conditions, whether they take place in a galvanic cell, where the reducing agent and oxidizing agent are physically separated from each other, or in a beaker, where the reactants are all mixed together.

Example 19.2

Predict what will happen if molecular bromine (Br 2) is added to a solution containing NaCl and NaI at 25 ЊC. Assume all species are in their standard states. (Continued ) cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 653 CONFIRMING PAGES

19.3 Standard Reduction Potentials 653

Strategy To predict what redox reaction(s) will take place, we need to compare the standard reduction potentials of Cl 2, Br 2, and I 2 and apply the diagonal rule. Solution From Table 19.1, we write the standard reduction potentials as follows:

Ϫ Ϫ Cl 2(1 atm) ϩ 2e ¡ 2Cl (1 M) E° ϭ 1.36 V Ϫ Ϫ Br 2(l) ϩ 2e ¡ 2Br (1 M) E° ϭ 1.07 V Ϫ Ϫ I 2(s) ϩ 2e ¡ 2I (1 M) E° ϭ 0.53 V

Ϫ Ϫ Applying the diagonal rule we see that Br 2 will oxidize I but will not oxidize Cl . Therefore, the only redox reaction that will occur appreciably under standard-state conditions is

Ϫ Ϫ Oxidation: 2I (1 M) 88n I2(s) ϩ 2e Ϫ Ϫ Reduction: Br 2(l) ϩ 2e 88n 2Br (1 M) Ϫ Ϫ Overall: 2I (1 M) ϩ Br 2(l) 88n I2(s) ϩ 2Br (1 M)

ϩ Check We can conﬁrm our conclusion by calculating EЊcell . Try it. Note that the Na ions are inert and do not enter into the redox reaction. Similar problems: 19.14, 19.17. Practice Exercise Can Sn reduce Zn 2ϩ(aq ) under standard-state conditions?

Example 19.3

A galvanic cell consists of a Mg electrode in a 1.0 M Mg(NO 3)2 solution and a Ag electrode in a 1.0 M AgNO 3 solution. Calculate the standard emf of this cell at 25 ЊC. Strategy At ﬁrst it may not be clear how to assign the electrodes in the galvanic cell. From Table 19.1 we write the standard reduction potentials of Ag and Mg and apply the diagonal rule to determine which is the anode and which is the cathode. Solution The standard reduction potentials are

ϩ Ϫ Ag (1.0 M) ϩ e ¡ Ag( s) E° ϭ 0.80 V ϩ Ϫ Mg 2 (1.0 M) ϩ 2e ¡ Mg( s) E° ϭ Ϫ 2.36 V

Applying the diagonal rule, we see that Ag ϩ will oxidize Mg:

Anode (oxidation): Mg( s) 88n Mg 2ϩ(1.0 M) ϩ 2eϪ Cathode (reduction): 2Ag ϩ(1.0 M) ϩ 2eϪ 88n 2Ag( s) Overall: Mg( s) ϩ 2Ag ϩ(1.0 M) 88n Mg 2ϩ(1.0 M) ϩ 2Ag( s)

Note that in order to balance the overall equation we multiplied the reduction of Ag ϩ by 2. We can do so because, as an intensive property, EЊ is not affected by this procedure. We ﬁnd the emf of the cell by using Equation (19.1) and Table 19.1:

E°cell ϭ E°cathode Ϫ E°anode ϭ E°Ag ϩ/Ag Ϫ E°Mg 2ϩ/Mg ϭ 0.80 V Ϫ (Ϫ2.37 V) ϭ 3.17 V

Check The positive value of EЊ shows that the forward reaction is favored. Similar problems: 19.11, 19.12. Practice Exercise What is the standard emf of a galvanic cell made of a Cd electrode in a 1.0 M Cd(NO 3)2 solution and a Cr electrode in a 1.0 M Cr(NO 3)3 solution at 25 ЊC? cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 654 CONFIRMING PAGES

654 CHAPTER 19 Redox Reactions and Electrochemistry

19.4 Thermodynamics of Redox Reactions

Our next step is to see how EЊcell is related to thermodynamic quantities such as ⌬GЊ and K. In a galvanic cell, chemical energy is converted to electrical energy to do electrical work. Electrical energy in this case is the product of the emf of the cell and the total electrical charge (in coulombs) that passes through the cell:

V ؋ 1 C electrical energy ϭ volts ϫ coulombs 1 ؍ J 1 ϭ joules

The total charge is determined by the number of electrons that pass through the cell, so we have total charge ϭ number of eϪ ϫ charge of one eϪ

In general, it is more convenient to express the total charge in molar quantities. The charge of one mole of electrons is called the Faraday constant (F), after the English chemist and physicist Michael Faraday, where ϭ ϫ 23 Ϫ Ϫ ϫ ϫ Ϫ19 Ϫ In most calculations, we round the 1 F 6.022 10 e /mol e 1.602 10 C/e ؊ Ϫ Faraday constant to 96,500 C/mol e . ϭ 9.647 ϫ 10 4 C/mol e

Therefore, the total charge can now be expressed as nF , where n is the number of moles of electrons exchanged between the reducing agent and the oxidizing agent in the overall redox equation. The measured emf ( Ecell ) is the maximum voltage the cell can achieve. It is given by the electrical work done ( wele ) divided by the total charge; that is,

Ϫwele Ϫwele Ecell ϭ ϭ total charge nF

or wele ϭ Ϫ nFE cell

The sign convention for electrical work is The negative sign indicates that the electrical work is done by the system (galvanic the same as that for P-V work, discussed cell) on the surroundings. In Chapter 18 we deﬁned free energy as the energy avail- in Section 6.3. able to do work. Speciﬁcally, the change in free energy ( ⌬G) represents the maximum amount of useful work that can be obtained in a reaction:

¢G ϭ wmax ϭ wele Therefore, we can write

¢G ϭ Ϫ nFE cell (19.2)

Both n and F are positive quantities and ⌬G is negative for a spontaneous process, so Ecell must be positive. For reactions in which reactants and products are in their standard states, Equation (19.2) becomes

¢G° ϭ Ϫ nFE °cell (19.3)

Now we can relate EЊcell to the equilibrium constant ( K) of a redox reaction. In Section 18.5 we saw that the standard free-energy change ⌬GЊ for a reaction is related to its equilibrium constant as follows [see Equation (18.14)]:

¢G° ϭ Ϫ RT ln K cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 655 CONFIRMING PAGES

19.4 Thermodynamics of Redox Reactions 655

Therefore, if we combine Equations (18.14) and (19.3) we obtain

ϪnFE °cell ϭ Ϫ RT ln K

Solving for EЊcell

RT ϭ E°cell ln K (19.4) nF

When T ϭ 298 K, Equation (19.4) can be simpliﬁed by substituting for R and F:

(8.314 J/K и mol)(298 K) E° ϭ ln K In calculations involving F, we sometimes cell ؊ n(96,500 J/V и mol) omit the symbol e .

0.0257 V E° ϭ ln K (19.5) cell n

Alternatively, Equation (19.5) can be written using the base-10 logarithm of K:

0.0592 V Ecell° ϭ log K (19.6) n

Thus, if any one of the three quantities ⌬GЊ, K, or EЊ is known, the other two cell E° can be calculated using Equation (18.14), Equation (19.3), or Equation (19.4) cell ⌬ Њ Њ (Figure 19.5). We summarize the relationships among G , K, and E cell and charac- E ° cell ° terize the spontaneity of a redox reaction in Table 19.2. For simplicity, we will omit = ln

nFE ___RT the subscript “cell” in EЊ and E. nF

° = – G K ∆ cell

∆G° = – RT ln K Example 19.4 ∆G° K Calculate the equilibrium constant for the following reaction at 25 ЊC:

ϩ ϩ ϩ Figure 19.5 Sn( s) ϩ 2Cu2 (aq ) ∆ Sn 2 (aq ) ϩ 2Cu (aq ) Relationships among EЊcell , K, and ⌬GЊ. Strategy The relationship between the equilibrium constant K and the standard emf is given by Equation (19.5): EЊcell ϭ (0.0257 V͞n)ln K. Thus, if we can determine the standard emf, we can calculate the equilibrium constant. We can determine the EЊcell of a hypothetical galvanic cell made up of two couples (Sn 2ϩ͞Sn and Cu 2ϩ͞Cu ϩ) from the standard reduction potentials in Table 19.1. (Continued )

TABLE 19.2 Relationships Among ⌬G؇, K, and E؇cell

G؇ K E Њcell Reaction Under Standard-State Conditions⌬ Negative Ͼ1 Positive Favors formation of products. 0 ϭ1 0 Reactants and products are equally favored. Positive Ͻ1 Negative Favors formation of reactants. cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 656 CONFIRMING PAGES

656 CHAPTER 19 Redox Reactions and Electrochemistry

Solution The half-cell reactions are

ϩ Ϫ Anode (oxidation): Sn( s) ¡ Sn 2 (aq ) ϩ 2e Cathode (reduction): 2Cu2ϩ(aq ) ϩ 2eϪ ¡ 2Cu ϩ(aq )

E°cell ϭ E°cathode Ϫ E°anode ϭ E°Cu 2ϩ/Cu ϩ Ϫ E°Sn 2ϩ/Sn ϭ 0.15 V Ϫ (Ϫ0.14 V) ϭ 0.29 V Equation (19.5) can be written

nE ° ln K ϭ 0.0257 V

In the overall reaction we ﬁnd n ϭ 2. Therefore,

(2)(0.29 V) ln K ϭ ϭ 22.6 0.0257 V 22.6 9 Similar problems: 19.21, 19.22. K ϭ e ϭ 7 ϫ 10

Practice Exercise Calculate the equilibrium constant for the following reaction at 25 ЊC:

Fe 2ϩ(aq ) ϩ 2Ag( s) ∆ Fe( s) ϩ 2Ag ϩ(aq )

Example 19.5 Calculate the standard free-energy change for the following reaction at 25 ЊC:

2Au( s) ϩ 3Ca2ϩ(1.0 M) ¡ 2Au 3ϩ(1.0 M) ϩ 3Ca( s)

Strategy The relationship between the standard free energy change and the standard emf of the cell is given by Equation (19.3): ⌬GЊ ϭ Ϫ nFE Њcell . Thus, if we can determine EЊcell , we can calculate ⌬GЊ. We can determine the EЊcell of a hypothetical galvanic cell made up of two couples (Au 3ϩ͞Au and Ca 2ϩ͞Ca) from the standard reduction potentials in Table 19.1. Solution The half-cell reactions are

ϩ Ϫ Anode (oxidation): 2Au( s) ¡ 2Au 3 (1.0 M) ϩ 6e ϩ Ϫ Cathode (reduction): 3Ca2 (1.0 M) ϩ 6e ¡ 3Ca( s)

E°cell ϭ E°cathode Ϫ E°anode ϭ E°Ca 2ϩ/Ca Ϫ E°Au 3ϩ/Au ϭ Ϫ 2.87 V Ϫ 1.50 V ϭ Ϫ 4.37 V

Now we use Equation (19.3):

¢G° ϭ Ϫ nFE °

The overall reaction shows that n ϭ 6, so

¢G° ϭ Ϫ (6)(96,500 J/V и mol)( Ϫ4.37 V) ϭ 2.53 ϫ 10 6 J/mol ϭ 2.53 ϫ 10 3 kJ/mol (Continued ) cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 657 CONFIRMING PAGES

19.5 The Effect of Concentration on Cell Emf 657

Check The large positive value of ⌬GЊ tells us that the reaction favors the reactants at equilibrium. The result is consistent with the fact that EЊ for the galvanic cell is negative. Similar problem: 19.24. Practice Exercise Calculate ⌬GЊ for the following reaction at 25 ЊC:

2Al 3ϩ(aq ) ϩ 3Mg( s) ∆ 2Al( s) ϩ 3Mg 2ϩ(aq )

19.5 The Effect of Concentration on Cell Emf So far we have focused on redox reactions in which reactants and products are in their standard states, but standard-state conditions are often difﬁcult, and sometimes impossible, to maintain. Nevertheless, there is a mathematical relationship between the emf of a galvanic cell and the concentration of reactants and products in a redox reaction under nonstandard-state conditions. This equation is derived next.

The Nernst Equation Consider a redox reaction of the type

aA ϩ bB ¡ cC ϩ dD

From Equation (18.13)

¢G ϭ ¢G° ϩ RT ln Q

Because ⌬G ϭ Ϫ nFE and ⌬GЊ ϭ Ϫ nFE Њ, the equation can be expressed as

ϪnFE ϭ Ϫ nFE ° ϩ RT ln Q

Dividing the equation through by ϪnF , we get

RT E ϭ E° Ϫ ln Q (19.7) nF

where Q is the reaction quotient (see Section 15.3). Equation (19.7) is known as the Nernst equation (after the German chemist Walter Nernst). At 298 K, Equation (19.7) can be rewritten as

0.0257 V E ϭ E° Ϫ ln Q (19.8) n

or, expressing Equation (19.8) using the base-10 logarithm of Q:

0.0592 V E ϭ E° Ϫ log Q (19.9) n

During the operation of a galvanic cell, electrons ﬂow from the anode to the cathode, resulting in product formation and a decrease in reactant concentration. Thus, Q increases, which means that E decreases. Eventually, the cell reaches equilibrium. At equilibrium, there is no net transfer of electrons, so E ϭ 0 and Q ϭ K, where K is the equilibrium constant. cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 658 CONFIRMING PAGES

658 CHAPTER 19 Redox Reactions and Electrochemistry

The Nernst equation enables us to calculate E as a function of reactant and product concentrations in a redox reaction. For example, for the Daniell cell in Figure 19.1,

Zn( s) ϩ Cu 2ϩ(aq ) ¡ Zn 2ϩ(aq ) ϩ Cu(s)

The Nernst equation for this cell at 25 ЊC can be written as

[Zn 2ϩ] ϭ Ϫ 0.0257 V Remember that concentrations of pure E 1.10 V ln 2ϩ solids (and pure liquids) do not appear in 2 [Cu ] the expression for Q (see p. 505). If the ratio [Zn 2ϩ]͞[Cu 2ϩ] is less than 1, ln ([Zn 2ϩ]͞[Cu 2ϩ]) is a negative number, so that the second term on the right-hand side of the preceding equation is positive. Under this condition E is greater than the standard emf EЊ. If the ratio is greater than 1, E is smaller than EЊ.

Example 19.6 Predict whether the following reaction would proceed spontaneously as written at 298 K:

Co(s) ϩ Fe 2ϩ(aq ) ¡ Co 2ϩ(aq ) ϩ Fe( s)

given that [Co 2ϩ] ϭ 0.15 M and [Fe 2ϩ] ϭ 0.68 M. Strategy Because the reaction is not run under standard-state conditions (concentrations are not 1 M), we need Nernst’s equation [Equation (19.8)] to calculate the emf ( E) of a hypothetical galvanic cell and determine the spontaneity of the reaction. The standard emf ( EЊ) can be calculated using the standard reduction potentials in Table 19.1. Remember that solids do not appear in the reaction quotient ( Q) term in the Nernst equation. Note that 2 moles of electrons are transferred per mole of reaction, that is, n ϭ 2. Solution The half-cell reactions are

ϩ Ϫ Anode (oxidation): Co( s) ¡ Co 2 (aq ) ϩ 2e ϩ Ϫ Cathode (reduction): Fe 2 (aq ) ϩ 2e ¡ Fe( s)

E°cell ϭ E°cathode Ϫ E°anode ϭ E°Fe 2ϩ/Fe Ϫ E°Co 2ϩ/Co ϭ Ϫ 0.44 V Ϫ (Ϫ0.28 V) ϭ Ϫ 0.16 V

From Equation (19.8) we write

0.0257 V E ϭ E° Ϫ ln Q n ϩ 0.0257V [Co 2 ] ϭ E° Ϫ ln n [Fe 2ϩ] 0.0257 V 0.15 ϭ Ϫ 0.16 V Ϫ ln 2 0.68 ϭ Ϫ 0.16 V ϩ 0.019 V ϭ Ϫ 0.14 V

Similar problems: 19.29, 19.30. Because E is negative, the reaction is not spontaneous in the direction written. Practice Exercise Will the following reaction occur spontaneously at 25 ЊC, given that [Fe 2ϩ] ϭ 0.60 M and [Cd 2ϩ] ϭ 0.010 M?

Cd(s) ϩ Fe 2ϩ(aq ) ¡ Cd 2ϩ(aq ) ϩ Fe( s) cha48518_ch19_642-683.qxd 12/21/06 1:42 PM Page 659 CONFIRMING PAGES

19.5 The Effect of Concentration on Cell Emf 659

Now suppose we want to determine at what ratio of [Co 2ϩ] to [Fe 2ϩ] the reaction in Example 19.6 would become spontaneous. We can use Equation (19.8) as follows: 0.0257 V E ϭ E° Ϫ ln Q n

.K ؍ Q ,0 ؍ We ﬁrst set E equal to zero, which corresponds to the equilibrium situation. When E

0.0257 V [Co 2ϩ] 0 ϭ Ϫ 0.16 V Ϫ ln 2 [Fe 2ϩ] [Co 2ϩ] ln ϭ Ϫ 12.5 [Fe 2ϩ] [Co 2ϩ] ϭ eϪ12.5 ϭ K [Fe 2ϩ] or K ϭ 4 ϫ 10 Ϫ6

Thus, for the reaction to be spontaneous, the ratio [Co 2ϩ]͞[Fe 2ϩ] must be smaller than 4 ϫ 10 Ϫ6 so that E would become positive. As Example 19.7 shows, if gases are involved in the cell reaction, their concentrations should be expressed in atm.

Example 19.7 Consider the galvanic cell shown in Figure 19.4(a). In a certain experiment, the emf ( E) of Њ 2ϩ ϭ ϭ the cell is found to be 0.54 V at 25 C. Suppose that [Zn ] 1.0 M and PH2 1.0 atm. Calculate the molar concentration of H ϩ. Strategy The equation that relates standard emf and nonstandard emf is the Nernst equation. The overall cell reaction is

ϩ 2ϩ Zn( s) ϩ 2H (? M) ¡ Zn (1.0 M) ϩ H2(1.0 atm)

Given the emf of the cell ( E), we apply the Nernst equation to solve for [H ϩ]. Note that 2 moles of electrons are transferred per mole of reaction; that is, n ϭ 2. Solution As we saw earlier (p. 649), the standard emf ( EЊ) for the cell is 0.76 V. From Equation (19.8) we write

0.0257 V E ϭ E° Ϫ ln Q n 2ϩ 0.0257 V [Zn ]PH ϭ E° Ϫ ln 2 n [H ϩ]2 0.0257 V (1.0)(1.0) 0.54 V ϭ 0.76 V Ϫ ln 2 [H ϩ]2 0.0257 V 1 Ϫ0.22 V ϭ ln 2 [H ϩ]2 1 17.1 ϭ ln [H ϩ]2 1 e17.1 ϭ [H ϩ]2 1 [H ϩ] ϭ ϭ 2 ϫ 10 Ϫ4 M B 3 ϫ 10 7 (Continued ) cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 660 CONFIRMING PAGES

660 CHAPTER 19 Redox Reactions and Electrochemistry

Check The fact that the nonstandard-state emf ( E) is given in the problem means that not all the reacting species are in their standard-state concentrations. Thus, because both 2ϩ ϩ Similar problem: 19.32. Zn ions and H 2 gas are in their standard states, [H ] is not 1 M. Practice Exercise What is the emf of a galvanic cell consisting of a Cd 2ϩ͞Cd half- ϩ 2ϩ ϭ ϩ ϭ ϭ cell and a Pt ͞H ͞H2 half-cell if [Cd ] 0.20 M, [H ] 0.16 M, and PH2 0.80 atm?

Example 19.7 shows that a galvanic cell whose cell reaction involves H ϩ ions can be used to measure [H ϩ] or pH. The pH meter described in Section 16.3 is based on this principle. However, the hydrogen electrode (see Figure 19.3) is normally not employed in laboratory work because it is awkward to use. Instead, it is replaced by a glass electrode, shown in Figure 19.6. The electrode consists of a very thin glass membrane that is permeable to H ϩ ions. A silver wire coated with silver chloride is immersed in a dilute hydrochloric acid solution. When the electrode is placed in a solution whose pH is different from that of the inner solution, the potential difference that develops between the two sides of the membrane can be monitored using a reference electrode. The emf of the cell made up of the glass electrode and the reference electrode is measured with a voltmeter that is calibrated in pH units. Ag—AgCl electrode Thin-walled glass membrane Concentration Cells HCl solution Because electrode potential depends on ion concentrations, it is possible to construct a galvanic cell from two half-cells composed of the same material but differing in ion Figure 19.6 concentrations. Such a cell is called a concentration cell. A glass electrode that is used in conjunction with a reference Consider a situation in which zinc electrodes are put into two aqueous solutions electrode in a pH meter. of zinc sulfate at 0.10 M and 1.0 M concentrations. The two solutions are connected by a salt bridge, and the electrodes are joined by a piece of wire in an arrangement like that shown in Figure 19.1. According to Le Châtelier’s principle, the tendency for the reduction

Zn 2ϩ(aq ) ϩ 2eϪ ¡ Zn( s)

increases with increasing concentration of Zn 2ϩ ions. Therefore, reduction should occur in the more concentrated compartment and oxidation should take place on the more dilute side. The cell diagram is

Zn( s)͉Zn 2ϩ(0.10 M)ʈZn 2ϩ(1.0 M)͉Zn( s)

and the half-reactions are

Oxidation: Zn( s) ¡ Zn 2ϩ(0.10 M) ϩ 2eϪ Reduction: Zn 2ϩ(1.0 M) ϩ 2eϪ ¡ Zn( s) Overall: Zn 2ϩ(1.0 M) ¡ Zn 2ϩ(0.10 M)

The emf of the cell is

0.0257 V [Zn 2ϩ] ϭ Ϫ dil E E° ln 2ϩ 2 [Zn ]conc where the subscripts “dil” and “conc” refer to the 0.10 M and 1.0 M concentrations, respectively. The EЊ for this cell is zero (the same electrode and the same type of ions are involved), so cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 661 CONFIRMING PAGES

19.6 Batteries 661

0.0257 V 0.10 E ϭ 0 Ϫ ln 2 1.0 ϭ 0.0296 V

The emf of concentration cells is usually small and decreases continually during the operation of the cell as the concentrations in the two compartments approach each other. When the concentrations of the ions in the two compartments are the same, E becomes zero, and no further change occurs. A biological cell can be compared to a concentration cell for the purpose of calculating its membrane potential. Membrane potential is the electrical potential that exists across the membrane of various kinds of cells, including muscle cells and nerve cells. It is responsible for the propagation of nerve impulses and heartbeat. A membrane potential is established whenever there are unequal concentrations of the same type of ion in the interior and exterior of a cell. For example, the concentrations of ϩ ؊ .؋ 10 3 M 1 ؍ K ions in the interior and exterior of a nerve cell are 400 m M and 15 m M, respec- 1 m M tively. Treating the situation as a concentration cell and applying the Nernst equation for just one kind of ions, we can write 0.0257 V [K ϩ] ϭ Ϫ ex E E° ln ϩ 1 [K ]in 15 ϭ Ϫ (0.0257 V) ln 400 ϭ 0.084 V or 84 mV where “ex” and “in” denote exterior and interior. Note that we have set EЊ ϭ 0 because the same type of ion is involved. Thus, an electrical potential of 84 mV exists across the membrane due to the unequal concentrations of Kϩ ions.

19.6 Batteries A battery is a galvanic cell, or a series of combined galvanic cells, that can be used as a source of direct electric current at a constant voltage. Although the operation of a battery is similar in principle to that of the galvanic cells described in Section 19.2, a battery has the advantage of being completely self-contained and requiring no aux- iliary components such as salt bridges. Here we will discuss several types of batteries that are in widespread use.

The Dry Cell Battery + The most common dry cell, that is, a cell without a fluid component, is the Leclanché Paper spacer cell used in flashlights and transistor radios. The anode of the cell consists of a zinc Moist paste of can or container that is in contact with manganese dioxide (MnO 2) and an electrolyte. ZnCl 2 and NH 4Cl The electrolyte consists of ammonium chloride and zinc chloride in water, to which Layer of MnO 2 starch is added to thicken the solution to a pastelike consistency so that it is less likely Graphite cathode to leak (Figure 19.7). A carbon rod serves as the cathode, which is immersed in the Zinc anode electrolyte in the center of the cell. The cell reactions are 88n 2ϩ ϩ Ϫ Figure 19.7 Anode: Zn( s) Zn (aq ) 2e Interior section of a dry cell of ϩ Ϫ 88n Cathode: 2NH 4(aq ) ϩ 2MnO 2(s) ϩ 2e Mn 2O3(s) ϩ 2NH 3(aq ) the kind used in flashlights and ϩ H2O( l) transistor radios. Actually, the cell is not completely dry, as it ϩ ϩ ϩ 88n 2ϩ ϩ Overall: Zn( s) 2NH 4(aq ) 2MnO 2(s) Zn (aq ) 2NH 3(aq ) contains a moist electrolyte ϩ H2O( l) ϩ Mn 2O3(s) paste. cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 662 CONFIRMING PAGES

662 CHAPTER 19 Redox Reactions and Electrochemistry

Cathode (steel) Actually, this equation is an oversimpliﬁcation of a complex process. The voltage pro- Insulation Anode (Zn can) duced by a dry cell is about 1.5 V.

The Mercury Battery The mercury battery is used extensively in medicine and electronic industries and is more expensive than the common dry cell. Contained in a stainless steel cylinder, Electrolyte solution containing KOH the mercury battery consists of a zinc anode (amalgamated with mercury) in contact and paste of Zn(OH) 2 and HgO with a strongly alkaline electrolyte containing zinc oxide and mercury(II) oxide (Figure 19.8). The cell reactions are Figure 19.8 Interior section of a mercury Ϫ Ϫ Anode: Zn(Hg) ϩ 2OH (aq ) 88n ZnO( s) ϩ H2O( l) ϩ 2e battery. Ϫ Ϫ Cathode: HgO( s) ϩ H2O( l) ϩ 2e 88n Hg( l) ϩ 2OH (aq ) Overall: Zn(Hg) ϩ HgO( s) 88n ZnO( s) ϩ Hg( l)

Because there is no change in electrolyte composition during operation—the overall cell reaction involves only solid substances—the mercury battery provides a more constant voltage (1.35 V) than the Leclanché cell. It also has a considerably higher capacity and longer life. These qualities make the mercury battery ideal for use in pacemakers, hearing aids, electric watches, and light meters.

The Lead Storage Battery The lead storage battery commonly used in automobiles consists of six identical cells joined together in series. Each cell has a lead anode and a cathode made of lead dioxide (PbO 2) packed on a metal plate (Figure 19.9). Both the cathode and the anode are immersed in an aqueous solution of sulfuric acid, which acts as the electrolyte. The cell reactions are

2Ϫ Ϫ Anode: Pb( s) ϩ SO 4 (aq ) 88n PbSO 4(s) ϩ 2e ϩ 2Ϫ Ϫ Cathode: PbO 2(s) ϩ 4H (aq ) ϩ SO 4 (aq ) ϩ 2e 88n PbSO 4(s) ϩ 2H 2O( l) ϩ 2Ϫ Overall: Pb( s) ϩ PbO 2(s) ϩ 4H (aq ) ϩ 2SO 4 (aq ) 88n 2PbSO 4(s) ϩ 2H 2O( l)

Figure 19.9 Removable cap Interior section of a lead Anode Cathode storage battery. Under normal – operating conditions, the concentration of the sulfuric acid solution is about 38 percent by mass. +

H2SO 4 electrolyte

Negative plates (lead grills filled with spongy lead)

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19.6 Batteries 663

Under normal operating conditions, each cell produces 2 V; a total of 12 V from the six cells is used to power the ignition circuit of the automobile and its other electrical systems. The lead storage battery can deliver large amounts of current for a short time, such as the time it takes to start up the engine. Unlike the Leclanché cell and the mercury battery, the lead storage battery is rechargeable. Recharging the battery means reversing the normal electrochemical reaction by applying an external voltage at the cathode and the anode. (This kind of process is called electrolysis, see p. 668.) The reactions that replenish the original materials are

Ϫ 2Ϫ PbSO 4(s) ϩ 2e 88n Pb( s) ϩ SO 4 (aq ) ϩ 2Ϫ Ϫ PbSO 4(s) ϩ 2H 2O( l) 88n PbO 2(s) ϩ 4H (aq ) ϩ SO 4 (aq ) ϩ 2e ϩ 2Ϫ Overall: 2PbSO 4(s) ϩ 2H 2O( l) 88n Pb( s) ϩ PbO 2(s) ϩ 4H (aq ) ϩ 2SO 4 (aq ) The overall reaction is exactly the opposite of the normal cell reaction. Two aspects of the operation of a lead storage battery are worth noting. First, because the electrochemical reaction consumes sulfuric acid, the degree to which the battery has been discharged can be checked by measuring the density of the electrolyte with a hydrometer, as is usually done at gas stations. The density of the ﬂuid in a “healthy,” fully charged battery should be equal to or greater than 1.2 g ͞mL. Sec- ond, people living in cold climates sometimes have trouble starting their cars because the battery has “gone dead.” Thermodynamic calculations show that the emf of many galvanic cells decreases with decreasing temperature. However, for a lead storage battery, the temperature coefﬁcient is about 1.5 ϫ 10 Ϫ4 V/ ЊC; that is, there is a decrease in voltage of 1.5 ϫ 10 Ϫ4 V for every degree drop in temperature. Thus, even allowing for a 40 ЊC change in temperature, the decrease in voltage amounts to only 6 ϫ 10 Ϫ3 V, which is about

6 ϫ 10 Ϫ3 V ϫ 100% ϭ 0.05% 12 V

of the operating voltage, an insigniﬁcant change. The real cause of a battery’s appar- ent breakdown is an increase in the viscosity of the electrolyte as the temperature decreases. For the battery to function properly, the electrolyte must be fully conducting. However, the ions move much more slowly in a viscous medium, so the resist- ance of the ﬂuid increases, leading to a decrease in the power output of the battery. If an apparently “dead battery” is warmed to near room temperature on a frigid day, it recovers its ability to deliver normal power.

The Lithium-Ion Battery Figure 19.10 shows a schematic diagram of a lithium-ion battery. The anode is made of a conducting carbonaceous material, usually graphite, which has tiny spaces in its structure that can hold both Li atoms and Li ϩ ions. The cathode is ϩ made of a transition metal oxide such as CoO 2, which can also hold Li ions. Because of the high reactivity of the metal, nonaqueous electrolyte (organic sol- vent plus dissolved salt) must be used. During the discharge of the battery, the half- cell reactions are

Anode (oxidation): Li( s) ¡ Li ϩ ϩ eϪ ϩ Ϫ Cathode (reduction): Li ϩ CoO2 ϩ e ¡ LiCoO 2(s) ϩ Overall: Li( s) CoO2 ¡ LiCoO 2(s) Ecell ϭ 3.4 V cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 664 CONFIRMING PAGES

664 CHAPTER 19 Redox Reactions and Electrochemistry

Figure 19.10 A lithium-ion battery. Lithium atoms are embedded in the graphite, which serves as the e— e— anode and CoO 2 is the cathode. During operation, Li ϩ ions migrate through the nonaqueous Anode Cathode electrolyte from the anode to the cathode while electrons ﬂow externally from the anode to the Li + cathode to complete the circuit.

Li in CoO2 graphite Nonaqueous electrolyte

+ — + — Li Li + e Li + CoO 2 + e LiCoO 2

The advantage of the battery is that lithium has the most negative standard reduction potential (see Table 19.1) and hence the greatest reducing strength. Furthermore, lithium is the lightest metal so that only 6.941 g of Li (its molar mass) are needed to produce 1 mole of electrons. A lithium-ion battery can be recharged literally hundreds of times without deterioration. These desirable characteristics make it suitable for use in cellular telephones, digital cameras, and laptop computers.

Fuel Cells Fossil fuels are a major source of energy, but conversion of fossil fuel into electrical energy is a highly inefﬁcient process. Consider the combustion of methane:

CH 4(g) ϩ 2O 2(g) ¡ CO 2(g) ϩ 2H 2O( l) ϩ energy

To generate electricity, heat produced by the reaction is first used to convert water to steam, which then drives a turbine that drives a generator. An appreciable fraction of the energy released in the form of heat is lost to the surroundings at each step; even the most efficient power plant converts only about 40 percent of the original chemical energy into electricity. Because combustion reactions are redox reactions, it is more desirable to carry them out directly by electrochemical means, thereby greatly increasing the efficiency of power production. This objective can be accomplished by a device known as a fuel cell, a galvanic cell that requires a continuous supply of reactants to keep functioning . In its simplest form, a hydrogen-oxygen fuel cell consists of an electrolyte solution, such as potassium hydroxide solution, and two inert electrodes. Hydrogen and oxygen gases are bubbled through the anode and cathode compartments (Figure 19.11), where the following reactions take place:

Ϫ Ϫ Anode: 2H 2(g) ϩ 4OH (aq ) ¡ 4H 2O( l) ϩ 4e Ϫ Ϫ Cathode: O2(g) ϩ 2H 2O( l) ϩ 4e ¡ 4OH (aq )

Overall: 2H 2(g) ϩ O2(g) ¡ 2H 2O( l) The standard emf of the cell can be calculated as follows, with data from Table 19.1:

E°cell ϭ E°cathode Ϫ E°anode ϭ 0.40 V Ϫ (Ϫ0.83 V) The positive standard emf means that the process will occur spontaneously. ϭ 1.23 V cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 665 CONFIRMING PAGES

19.7 Corrosion 665

Figure 19.11 A hydrogen-oxygen fuel cell. The Ni and NiO embedded in the porous carbon electrodes e– e– are electrocatalysts. Anode Cathode

H2 O2

Porous carbon electrode Porous carbon electrode containing Ni containing Ni and NiO

Hot KOH solution Oxidation Reduction – – – – 2H 2(g) + 4OH (aq ) 4H 2O( l) + 4 e O2(g) + 2H2O( l) + 4 e 4OH (aq )

Thus, the cell reaction is spontaneous under standard-state conditions. Note that the reaction is the same as the hydrogen combustion reaction, but the oxidation and reduction A car powered by hydrogen fuel are carried out separately at the anode and the cathode. Like platinum in the standard cells manufactured by General hydrogen electrode, the electrodes have a twofold function. They serve as electrical con- Motors. ductors, and they provide the necessary surfaces for the initial decomposition of the molecules into atomic species, prior to electron transfer. They are electrocatalysts. Met- als such as platinum, nickel, and rhodium are good electrocatalysts. In addition to the H 2-O 2 system, a number of other fuel cells have been developed. Among these is the propane-oxygen fuel cell. The half-cell reactions are

ϩ Ϫ Anode: C 3H8(g) ϩ 6H 2O( l) ¡ 3CO2(g) ϩ 20H (aq ) ϩ 20 e ϩ Ϫ Cathode: 5O 2(g) ϩ 20H (aq ) ϩ 20 e ¡ 10H2O( l)

Overall: C 3H8(g) ϩ 5O 2(g) ¡ 3CO2(g) ϩ 4H 2O( l) The overall reaction is identical to the burning of propane in oxygen. Unlike batteries, fuel cells do not store chemical energy. Reactants must be constantly resupplied, and products must be constantly removed from a fuel cell. In this respect, a fuel cell resembles an engine more than it does a battery. Properly designed fuel cells may be as much as 70 percent efficient, about twice as efficient as an internal combustion engine. In addition, fuel-cell generators are free of the noise, vibration, heat transfer, thermal pollution, and other problems normally associated with conventional power plants. Nevertheless, fuel cells are not yet in widespread use. A major problem lies in the lack of cheap electrocatalysts able to function efficiently for long periods of time without contamination. The most successful application of fuel cells to date has been in space vehicles (Figure 19.12).

19.7 Corrosion Figure 19.12 A hydrogen-oxygen fuel cell Corrosion is the term usually applied to the deterioration of metals by an electro- used in the space program. The chemical process. We see many examples of corrosion around us. Rust on iron, tarnish pure water produced by the cell on silver, and the green patina formed on copper and brass are a few of them. Corrosion is consumed by the astronauts. cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 666 CONFIRMING PAGES

666 CHAPTER 19 Redox Reactions and Electrochemistry

causes enormous damage to buildings, bridges, ships, and cars. The cost of metallic corrosion to the U.S. economy has been estimated to be well over 100 billion dollars a year! This section discusses some of the fundamental processes that occur in corrosion and methods used to protect metals against it. By far the most familiar example of corrosion is the formation of rust on iron. Oxygen gas and water must be present for iron to rust. Although the reactions involved are quite complex and not completely understood, the main steps are believed to be as follows. A region of the metal’s surface serves as the anode, where oxidation occurs:

Fe( s) ¡ Fe 2ϩ(aq ) ϩ 2eϪ

The electrons given up by iron reduce atmospheric oxygen to water at the cathode, which is another region of the same metal’s surface:

ϩ Ϫ O2(g) ϩ 4H (aq ) ϩ 4e ¡ 2H 2O( l) The overall redox reaction is

ϩ 2ϩ 2Fe( s) ϩ O2(g) ϩ 4H (aq ) ¡ 2Fe (aq ) ϩ 2H 2O( l) With data from Table 19.1, we ﬁnd the standard emf for this process:

E°cell ϭ E°cathode Ϫ E°anode ϭ 1.23 V Ϫ (Ϫ0.44 V) ϭ 1.67 V

Note that this reaction occurs in an acidic medium; the H ϩ ions are supplied in part by the reaction of atmospheric carbon dioxide with water to form H 2CO 3. The Fe 2ϩ ions formed at the anode are further oxidized by oxygen:

2ϩ # ϩ 4Fe (aq ) ϩ O2(g) ϩ (4 ϩ 2x)H 2O( l) ¡ 2Fe 2O3 xH2O( s) ϩ 8H (aq ) This hydrated form of iron(III) oxide is known as rust. The amount of water associated with the iron oxide varies, so we represent the formula as Fe 2O3 и xH2O. Figure 19.13 shows the mechanism of rust formation. The electrical circuit is completed by the migration of electrons and ions; this is why rusting occurs so rapidly in salt water. In cold climates, salts (NaCl or CaCl 2) spread on roadways to melt ice and snow are a major cause of rust formation on automobiles.

Figure 19.13 Air O2 The electrochemical process involved in rust formation. ϩ Water The H ions are supplied by Fe 2+ Rust H2CO 3, which forms when CO 2 Fe 3+ dissolves in water.

Iron e–

Anode Cathode 2+ – + – Fe( s) Fe (aq ) + 2 e O2(g) + 4H (aq ) + 4 e 2H 2O( l) Fe 2+ (aq ) Fe3+ (aq ) + e – cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 667 CONFIRMING PAGES

19.7 Corrosion 667

Metallic corrosion is not limited to iron. Consider aluminum, a metal used to make many useful things, including airplanes and beverage cans. Aluminum has a much greater tendency to oxidize than iron does; in Table 19.1 we see that Al has a more negative standard reduction potential than Fe. Based on this fact alone, we might expect to see airplanes slowly corrode away in rainstorms, and soda cans transformed into piles of corroded aluminum. These processes do not occur because the layer of insoluble aluminum oxide (Al 2O3) that forms on its surface when the metal is exposed to air serves to protect the aluminum underneath from further corrosion. The rust that forms on the surface of iron, however, is too porous to protect the underlying metal. Coinage metals such as copper and silver also corrode, but much more slowly.

Cu( s) ¡ Cu 2ϩ(aq ) ϩ 2eϪ Ag( s) ¡ Ag ϩ(aq ) ϩ eϪ

In normal atmospheric exposure, copper forms a layer of copper carbonate (CuCO 3), a green substance also called patina, that protects the metal underneath from further corrosion. Likewise, silverware that comes into contact with foodstuffs develops a layer of silver sulﬁde (Ag 2S). A number of methods have been devised to protect metals from corrosion. Most of these methods are aimed at preventing rust formation. The most obvious approach is to coat the metal surface with paint. However, if the paint is scratched, pitted, or dented to expose even the smallest area of bare metal, rust will form under the paint layer. The surface of iron metal can be made inactive by a process called passivation. A thin oxide layer is formed when the metal is treated with a strong oxidizing agent such as concentrated nitric acid. A solution of sodium chromate is often added to cooling systems and radiators to prevent rust formation. The tendency for iron to oxidize is greatly reduced when it is alloyed with certain other metals. For example, in stainless steel, an alloy of iron and chromium, a layer of chromium oxide forms that protects the iron from corrosion. An iron container can be covered with a layer of another metal such as tin or zinc. A “tin” can is made by applying a thin layer of tin over iron. Rust formation is prevented as long as the tin layer remains intact. However, once the surface has been scratched, rusting occurs rapidly. If we look up the standard reduction potentials, according to the diagonal rule, we ﬁnd that iron acts as the anode and tin as the cathode in the corrosion process:

ϩ Ϫ Sn 2 (aq ) ϩ 2e ¡ Sn( s) E° ϭ Ϫ 0.14 V ϩ Ϫ Fe 2 (aq ) ϩ 2e ¡ Fe( s) E° ϭ Ϫ 0.44 V

The protective process is different for zinc-plated, or galvanized, iron. Zinc is more easily oxidized than iron (see Table 19.1):

ϩ Ϫ Zn 2 (aq ) ϩ 2e ¡ Zn( s) E° ϭ Ϫ 0.76 V

So even if a scratch exposes the iron, the zinc is still attacked. In this case, the zinc metal serves as the anode and the iron is the cathode. Cathodic protection is a process in which the metal that is to be protected from corrosion is made the cathode in what amounts to a galvanic cell. Figure 19.14 shows how an iron nail can be protected from rusting by connecting the nail to a piece of Figure 19.14 zinc. Without such protection, an iron nail quickly rusts in water. Rusting of under- An iron nail that is cathodically protected by a piece of zinc ground iron pipes and iron storage tanks can be prevented or greatly reduced by con- strip does not rust in water, necting them to metals such as zinc and magnesium, which oxidize more readily than while an iron nail without such iron (Figure 19.15). protection rusts readily. cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 668 CONFIRMING PAGES

668 CHAPTER 19 Redox Reactions and Electrochemistry

Figure 19.15 Cathodic protection of an iron storage tank (cathode) by magnesium, a more electropositive metal (anode). Because only the Mg e– magnesium is depleted in the electrochemical process, it is sometimes called the sacriﬁcial Iron storage tank anode.

2+ – + – Oxidation: Mg( s) Mg (aq ) + 2 e Reduction: O 2(g) + 4H (aq ) + 4 e 2H 2O( l)

19.8 Electrolysis In contrast to spontaneous redox reactions, which result in the conversion of chemical energy into electrical energy, electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur. An electrolytic cell is an apparatus for carrying out electrolysis. The same principles underlie electrolysis and the processes that take place in galvanic cells. Here we will discuss three examples of electrolysis based on those principles. Then we will look at the quantitative aspects of electrolysis.

Electrolysis of Molten Sodium Chloride In its molten state, sodium chloride, an ionic compound, can be electrolyzed to form sodium metal and chlorine. Figure 19.16(a) is a diagram of a Downs cell, which is used for large-scale electrolysis of NaCl. In molten NaCl, the cations and anions are the Na ϩ and Cl Ϫ ions, respectively. Figure 19.16(b) is a simpli- fied diagram showing the reactions that occur at the electrodes. The electrolytic cell contains a pair of electrodes connected to the battery. The battery serves as an “electron pump,” driving electrons to the cathode, where reduction occurs, and withdrawing electrons from the anode, where oxidation occurs. The reactions at the electrodes are

Figure 19.16 Cl 2 gas Battery (a) A practical arrangement NaCl called a Downs cell for the electrolysis of molten NaCl e – e – ϭ Њ (m.p. 801 C). The sodium Anode Cathode metal formed at the cathodes is in the liquid state. Because Liquid Na Liquid Na liquid sodium metal is less dense than molten NaCl, the sodium ﬂoats to the surface, as shown, and is collected. Chlo- Molten NaCl rine gas forms at the anode and is collected at the top. (b) A simpliﬁed diagram showing the electrode reactions during the Iron cathode Iron cathode Oxidation Reduction electrolysis of molten NaCl. The 2Cl– Cl (g) + 2 e – 2Na + + 2 e – 2Na( l) battery is needed to drive the Carbon anode 2 nonspontaneous reactions. (a) (b) cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 669 CONFIRMING PAGES

19.8 Electrolysis 669

Ϫ Ϫ Anode (oxidation): 2Cl (l) ¡ Cl 2(g) ϩ 2e Cathode (reduction): 2Na ϩ(l) ϩ 2eϪ ¡ 2Na( l) H2 O2 ϩ Ϫ Overall: 2Na (l) ϩ 2Cl (l) ¡ 2Na( l) ϩ Cl 2(g) This process is a major source of pure sodium metal and chlorine gas. Theoretical estimates show that the EЊ value for the overall process is about Ϫ4 V, which means that this is a nonspontaneous process. Therefore, a minimum of 4 V must be supplied by the battery to carry out the reaction. In practice, a higher voltage is necessary because of inefﬁciencies in the electrolytic process and because of overvoltage, to be discussed shortly. Electrolysis of Water Water in a beaker under atmospheric conditions (1 atm and 25 ЊC) will not spontaneously decompose to form hydrogen and oxygen gas because the standard free- energy change for the reaction is a large positive quantity: Figure 19.17 2H 2O( l) ¡ 2H 2(g) ϩ O2(g) ¢G° ϭ 474.4 kJ/mol Apparatus for small-scale electrolysis of water. The volume of However, this reaction can be induced in a cell like the one shown in Figure 19.17. hydrogen gas generated (left This electrolytic cell consists of a pair of electrodes made of a nonreactive metal, such column) is twice that of oxygen as platinum, immersed in water. When the electrodes are connected to the battery, gas (right column). nothing happens because there are not enough ions in pure water to carry much of an electric current. (Remember that at 25 ЊC, pure water has only 1 ϫ 10 Ϫ7 M Hϩ ions and 1 ϫ 10 Ϫ7 M OH Ϫ ions.) On the other hand, the reaction occurs readily in a 0.1 M H2SO 4 solution because there are a sufﬁcient number of ions to conduct electricity. Immediately, gas bubbles begin to appear at both electrodes. Figure 19.18 shows the electrode reactions. The process at the anode is ϩ Ϫ 2H 2O( l) ¡ O2(g) ϩ 4H (aq ) ϩ 4e while at the cathode we have ϩ ϩ Ϫ 1 H (aq ) e ¡ 2H2(g) The overall reaction is given by ϩ Ϫ Anode (oxidation): 2H 2O( l) ¡ O2(g) ϩ 4H (aq ) ϩ 4e ϩ ϩ Ϫ 1 Cathode (reduction): 4[H (aq ) e ¡ 2H2(g)] What is the minimum voltage needed for this electrolytic process? Overall: 2H 2O( l) ¡ 2H 2(g) ϩ O2(g)

Note that no net H 2SO 4 is consumed.

Battery Figure 19.18 A diagram showing the electrode reactions during the electrolysis – – e e of water. Anode Cathode

Dilute H 2SO 4 solution

Oxidation Reduction + – + – 2H 2O( l) O2(g) + 4H (aq ) + 4 e 4H (aq ) + 4 e 2H 2(g) cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 670 CONFIRMING PAGES

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Electrolysis of an Aqueous Sodium Chloride Solution This is the most complicated of the three examples of electrolysis considered here because aqueous sodium chloride solution contains several species that could be oxidized and reduced. The oxidation reactions that might occur at the anode are

Ϫ Ϫ (1) 2Cl (aq ) ¡ Cl 2(g) ϩ 2e ϩ Ϫ (2) 2H 2O( l) ¡ O2(g) ϩ 4H (aq ) ϩ 4e Referring to Table 19.1, we ﬁnd

Ϫ Ϫ Cl 2(g) ϩ 2e ¡ 2Cl (aq ) E° ϭ 1.36 V ϩ Ϫ O2(g) ϩ 4H (aq ) ϩ 4e ¡ 2H 2O( l) E° ϭ 1.23 V

Because Cl 2 is more easily reduced than The standard reduction potentials of (1) and (2) are not very different, but the values O2, it follows that it would be more Ϫ do suggest that H 2O should be preferentially oxidized at the anode. However, by difficult to oxidize Cl than H 2O at the anode. experiment we find that the gas liberated at the anode is Cl 2, not O 2! In studying electrolytic processes, we sometimes find that the voltage required for a reaction is considerably higher than the electrode potential indicates. The overvoltage is the difference between the electrode potential and the actual voltage required to cause electrolysis. The overvoltage for O 2 formation is quite high. Therefore, under normal operating conditions Cl 2 gas is actually formed at the anode instead of O 2. The reductions that might occur at the cathode are

ϩ Ϫ (3) 2H (aq ) ϩ 2e ¡ H2(g) E° ϭ 0.00 V Ϫ Ϫ (4) 2H 2O( l) ϩ 2e ¡ H2(g) ϩ 2OH (aq ) E° ϭ Ϫ 0.83 V (5) Na ϩ(aq ) ϩ eϪ ¡ Na( s) E° ϭ Ϫ 2.71 V Reaction (5) is ruled out because it has a very negative standard reduction potential. Reaction (3) is preferred over (4) under standard-state conditions. At a pH of 7 (as is the case for a NaCl solution), however, they are equally probable. We generally use (4) to describe the cathode reaction because the concentration of H ϩ ions is too low (about 1 ϫ 10 Ϫ7 M) to make (3) a reasonable choice. Thus, the half-cell reactions in the electrolysis of aqueous sodium chloride are Ϫ Ϫ Anode (oxidation): 2Cl (aq ) ¡ Cl 2(g) ϩ 2e Ϫ Ϫ Cathode (reduction): 2H 2O( l) ϩ 2e ¡ H2(g) ϩ 2OH (aq ) Ϫ Ϫ Overall: 2H 2O( l) ϩ 2Cl (aq ) ¡ H2(g) ϩ Cl 2(g) ϩ 2OH (aq ) As the overall reaction shows, the concentration of the Cl Ϫ ions decreases during elec- Ϫ trolysis and that of the OH ions increases. Therefore, in addition to H 2 and Cl 2, the useful by-product NaOH can be obtained by evaporating the aqueous solution at the end of the electrolysis. Keep in mind the following from our analysis of electrolysis: cations are likely to be reduced at the cathode and anions are likely to be oxidized at the anode, and in aqueous solutions water itself may be oxidized and/or reduced. The outcome depends on the nature of other species present.

Example 19.8

An aqueous Na 2SO 4 solution is electrolyzed, using the apparatus shown in Figure 19.17. If the products formed at the anode and cathode are oxygen gas and hydrogen gas, respectively, describe the electrolysis in terms of the reactions at the electrodes. (Continued ) cha48518_ch19_642-683.qxd 12/21/06 12:56 PM Page 671 CONFIRMING PAGES

19.8 Electrolysis 671

2؊ The SO 4 ion is the conjugate base of the ؋ ؊2 ؍ Ϫ Strategy Before we look at the electrode reactions, we should consider the following weak acid HSO 4 (Ka 1.3 10 ). 2؊ However, the extent to which SO 4 facts: (1) Because Na 2SO 4 does not hydrolyze, the pH of the solution is close to 7. 2؊ ϩ 2Ϫ hydrolyzes is negligible. Also, the SO 4 (2) The Na ions are not reduced at the cathode and the SO 4 ions are not oxidized at ion is not oxidized at the anode. the anode. These conclusions are drawn from the electrolysis of water in the presence of sulfuric acid and in aqueous sodium chloride solution, as discussed earlier. Therefore, both the oxidation and reduction reactions involve only water molecules. Solution The electrode reactions are

ϩ Ϫ Anode: 2H 2O( l) ¡ O2(g) ϩ 4H (aq ) ϩ 4e Ϫ Ϫ Cathode: 2H 2O( l) ϩ 2e ¡ H2(g) ϩ 2OH (aq )

The overall reaction, obtained by doubling the cathode reaction coefﬁcients and adding the result to the anode reaction, is

ϩ Ϫ 6H 2O( l) ¡ 2H 2(g) ϩ O2(g) ϩ 4H (aq ) ϩ 4OH (aq )

If the H ϩ and OH Ϫ ions are allowed to mix, then

ϩ Ϫ 4H (aq ) ϩ 4OH (aq ) ¡ 4H 2O( l)

and the overall reaction becomes

2H 2O( l) ¡ 2H 2(g) ϩ O2(g) Similar problem: 19.46(a).

Practice Exercise An aqueous solution of Mg(NO 3)2 is electrolyzed. What are the gaseous products at the anode and cathode?

Electrolysis has many important applications in industry, mainly in the extraction and puriﬁcation of metals. We will discuss some of these applications in Section 19.9.

Quantitative Aspects of Electrolysis The quantitative treatment of electrolysis was developed primarily by Faraday. He observed that the mass of product formed (or reactant consumed) at an electrode is proportional to both the amount of electricity transferred at the electrode and the molar mass of the substance in question. For example, in the electrolysis of molten NaCl, the cathode reaction tells us that one Na atom is produced when one Na ϩ ion accepts an electron from the electrode. To reduce 1 mole of Na ϩ ions, we must supply Avo- gadro’s number (6.02 ϫ 10 23 ) of electrons to the cathode. On the other hand, the sto- ichiometry of the anode reaction shows that oxidation of two Cl Ϫ ions yields one chlorine molecule. Therefore, the formation of 1 mole of Cl 2 results in the transfer of 2 moles of electrons from the Cl Ϫ ions to the anode. Similarly, it takes 2 moles of electrons to reduce 1 mole of Mg 2ϩ ions and 3 moles of electrons to reduce 1 mole of Al 3ϩ ions:

Mg 2ϩ ϩ 2eϪ ¡ Mg Al 3ϩ ϩ 3eϪ ¡ Al

In an electrolysis experiment, we generally measure the current (in amperes, A) that passes through an electrolytic cell in a given period of time. The relationship between charge (in coulombs, C) and current is

1 C ϭ 1 A ϫ 1 s cha48518_ch19_642-683.qxd 12/21/06 12:57 PM Page 672 CONFIRMING PAGES

672 CHAPTER 19 Redox Reactions and Electrochemistry

that is, a coulomb is the quantity of electrical charge passing any point in the circuit Current (amperes) in 1 second when the current is 1 ampere. and time (seconds) Figure 19.19 shows the steps involved in calculating the quantities of substances produced in electrolysis. Let us illustrate the approach by considering molten CaCl 2 Product of in an electrolytic cell. Suppose a current of 0.452 A is passed through the cell for current and time 1.50 h. How much product will be formed at the anode and at the cathode? In solving electrolysis problems of this type, the ﬁrst step is to determine which species will be Charge oxidized at the anode and which species will be reduced at the cathode. Here the in choice is straightforward because we only have Ca 2ϩ and Cl Ϫ ions in molten CaCl . coulombs 2 Thus, we write the half- and overall cell reactions as

Divide by Ϫ Ϫ the Faraday Anode (oxidation): 2Cl (l) ¡ Cl 2(g) ϩ 2e constant Cathode (reduction): Ca 2ϩ(l) ϩ 2eϪ ¡ Ca( l) 2ϩ Ϫ Number Overall: Ca (l) ϩ 2Cl (l) ¡ Ca( l) ϩ Cl 2(g) of moles of electrons The quantities of calcium metal and chlorine gas formed depend on the number of electrons that pass through the electrolytic cell, which in turn depends on current ϫ Use mole ratio in half-cell reaction time, or charge:

3600 s 1 C 3 Moles of ? C ϭ 0.452 A ϫ 1.50 h ϫ ϫ # ϭ 2.44 ϫ 10 C substance reduced 1 h 1 A s or oxidized Because 1 mol eϪ ϭ 96,500 C and 2 mol eϪ are required to reduce 1 mole of Ca 2ϩ Use molar mass or ions, the mass of Ca metal formed at the cathode is calculated as follows: ideal gas equation Ϫ 3 1 mol e 1 mol Ca 40.08 g Ca ? g Ca ϭ 2.44 ϫ 10 C ϫ ϫ Ϫ ϫ ϭ 0.507 g Ca Grams or liters 96,500 C 2 mol e 1 mol Ca of product The anode reaction indicates that 1 mole of chlorine is produced per 2 mol eϪ of electricity. Hence the mass of chlorine gas formed is

Figure 19.19 Ϫ Steps involved in calculating 3 1 mol e 1 mol Cl 2 70.90 g Cl2 ? g Cl 2 ϭ 2.44 ϫ 10 C ϫ ϫ Ϫ ϫ ϭ 0.896 g Cl2 amounts of substances reduced 96,500 C 2 mol e 1 mol Cl 2 or oxidized in electrolysis.

Example 19.9 A current of 1.26 A is passed through an electrolytic cell containing a dilute sulfuric acid solution for 7.44 h. Write the half-cell reactions and calculate the volume of gases generated at STP. Strategy Earlier (see p. 669) we saw that the half-cell reactions for the process are

ϩ Ϫ Anode (oxidation): 2H 2O( l) ¡ O2(g) ϩ 4H (aq ) ϩ 4e ϩ ϩ Ϫ 1 Cathode (reduction): 4[H (aq ) e ¡ 2H2(g)]

Overall: 2H 2O( l) ¡ 2H 2(g) ϩ O2(g)

According to Figure 19.19, we carry out the following conversion steps to calculate the

quantity of O 2 in moles: Ϫ current ϫ time S coulombs S moles of e S moles of O 2

Then, using the ideal gas equation we can calculate the volume of O 2 in liters at STP. A similar procedure can be used for H 2. (Continued ) cha48518_ch19_642-683.qxd 12/21/06 12:57 PM Page 673 CONFIRMING PAGES

19.9 Electrometallurgy 673

Solution First we calculate the number of coulombs of electricity that pass through the cell: 3600 s 1 C ? C ϭ 1.26 A ϫ 7.44 h ϫ ϫ ϭ 3.77 ϫ 10 4 C 1 h 1 A # s

Next, we convert number of coulombs to number of moles of electrons

1 mol eϪ 3.37 ϫ 10 4 C ϫ ϭ 0.349 mol eϪ 96,500 C

Ϫ From the oxidiation half-reaction we see that 1 mol O 2 ϴ 4 mol e . Therefore, the number of moles of O 2 generated is 1 mol O 0.349 mol eϪ ϫ 2 ϭ 0.0873 mol O 4 mol eϪ 2

The volume of 0.0873 mol O 2 at STP is given by

nRT V ϭ P (0.0873 mol)(0.0821 L # atm/K # mol)(273 K) ϭ ϭ 1.96 L 1 atm

The procedure for hydrogen is similar. To simplify, we combine the ﬁrst two steps

to calculate the number of moles of H 2 generated:

1 mol e 1 mol H 3.37 ϫ 10 4C ϫ ϫ 2 ϭ 0.175 mol H 96,500 C 2 mol e 2

The volume of 0.175 mol H 2 at STP is given by

nRT V ϭ P (0.175 mol)(0.0821 L # atm/K # mol)(273 K) ϭ 1 atm ϭ 3.92 L

Check Note that the volume of H 2 is twice that of O 2 (see Figure 19.17), which is what we would expect based on Avogadro’s law (at the same temperature and pressure, volume is directly proportional to the number of moles of gases). Similar problem: 19.49. Practice Exercise A constant current is passed through an electrolytic cell containing 5 molten MgCl 2 for 18 h. If 4.8 ϫ 10 g of Cl 2 are obtained, what is the current in amperes?

Carbon anodes 19.9 Electrometallurgy Electrolysis methods are useful for obtaining a pure metal from its ores or for reﬁning Carbon cathode (purifying) the metal. Collectively, these processes are called electrometallurgy. In the last section we saw how an active metal, sodium, can be obtained by electrolytically reducing its cation in the molten NaCl salt (p. 668). Here we will consider two other examples. Molten aluminum Al 2O3 in Production of Aluminum Metal molten cryolite и Figure 19.20 Aluminum is usually prepared from bauxite ore (Al 2O3 2H 2O). The ore is ﬁrst treated Electrolytic production of to remove various impurities and then heated to obtain the anhydrous Al 2O3. The oxide aluminum based on the Hall is dissolved in molten cryolite (Na 3AlF 6) in a Hall electrolytic cell (Figure 19.20). The process. cha48518_ch19_642-683.qxd 12/21/06 12:57 PM Page 674 CONFIRMING PAGES

674 CHAPTER 19 Redox Reactions and Electrochemistry

Battery cell contains a series of carbon anodes; the cathode is also made of carbon and con- stitutes the lining inside the cell. The solution is electrolyzed to produce aluminum and Impure Pure oxygen gas: copper copper anode cathode 2Ϫ Ϫ Anode: 3[2O ¡ O2(g) ϩ 4e ] Cathode: 4[Al 3ϩ ϩ 3eϪ ¡ Al( l)]

Overall: 2Al 2O3 ¡ 4Al( l) ϩ 3O 2(g) Oxygen gas reacts with the carbon anodes at 1000 ЊC (the melting point of cryolite) Cu 2+ to form carbon monoxide, which escapes as a gas. The liquid aluminum metal (m.p. 2– Њ SO 4 660 C) sinks to the bottom of the vessel, from which it can be drained.

Figure 19.21 Puriﬁcation of Copper Metal Electrolytic puriﬁcation of The copper metal obtained from its ores usually contains a number of impurities such as copper. zinc, iron, silver, and gold. The more electropositive metals are removed by an electrolysis process in which the impure copper acts as the anode and pure copper acts as the cathode in a sulfuric acid solution containing Cu 2ϩ ions (Figure 19.21). The half-reactions are

ϩ Ϫ Anode: Cu(s) ¡ Cu 2 (aq ) ϩ 2e ϩ Ϫ Cathode: Cu 2 (aq ) ϩ 2e ¡ Cu(s)

Reactive metals in the copper anode, such as iron and zinc, are also oxidized at the anode and enter the solution as Fe 2ϩ and Zn 2ϩ ions. They are not reduced at the cathode, however. The less electropositive metals, such as gold and silver, are not oxidized at the anode. Eventually, as the copper anode oxidizes, these metals fall to the bottom of the cell. Thus, the net result of this electrolysis process is the transfer of copper from the anode to the cathode. Copper prepared this way has a purity greater than 99.5 percent. It is interesting to note that the metal impurities (mostly silver and gold) from the copper anode are valuable by-products, the sale of which often pays for the electricity used to drive the electrolysis.

KEY EQUATIONS

EЊcell ϭ EЊcathode Ϫ EanodeЊ (19.1) Calculating the standard emf of a galvanic cell.

⌬G ϭ Ϫ nFE cell (19.2) Relating free-energy change to the emf of the cell.

⌬GЊ ϭ Ϫ nFE Њcell (19.3) Relating the standard free-energy change to the standard emf of the cell. RT E°cell ϭ ln K (19.4) Relating the standard emf of the cell to the nF equilibrium constant. 0.0257 V E°cell ϭ ln K (19.5) Relating the standard emf of the cell to the n equilibrium constant at 298 K. 0.0592 V E°cell ϭ log K (19.6) Relating the standard emf of the cell to the n equilibrium constant at 298 K. RT E ϭ E° Ϫ ln Q (19.7) The Nernst equation. For calculating the emf of a nF cell under non-standard-state conditions. 0.0257 V E ϭ E° Ϫ ln Q (19.8) The Nernst equation. For calculating the emf of a n cell under non-standard-state conditions at 298 K. 0.0592 V E ϭ E° Ϫ log Q (19.9) The Nernst equation. For calculating the emf of a n cell under non-standard-state conditions at 298 K. cha48518_ch19_642-683.qxd 12/21/06 12:57 PM Page 675 CONFIRMING PAGES

Questions and Problems 675

SUMMARY OF FACTS AND CONCEPTS

1. Redox reactions involve the transfer of electrons. Equa- 8. The decrease in free energy of the system in a sponta- tions representing redox processes can be balanced us- neous redox reaction is equal to the electrical work done ing the half-reaction method. by the system on the surroundings, or ⌬G ϭ Ϫ nFE . 2. All electrochemical reactions involve the transfer of 9. The equilibrium constant for a redox reaction can be electrons and therefore are redox reactions. found from the standard electromotive force of a cell. 3. In a galvanic cell, electricity is produced by a sponta- 10. The Nernst equation gives the relationship between the neous chemical reaction. Oxidation and reduction take cell emf and the concentrations of the reactants and place separately at the anode and cathode, respectively, products under nonstandard-state conditions. and the electrons flow through an external circuit. 11. Batteries, which consist of one or more galvanic cells, 4. The two parts of a galvanic cell are the half-cells, and are used widely as self-contained power sources. Some the reactions at the electrodes are the half-cell reac- of the better-known batteries are the dry cell, such as the tions. A salt bridge allows ions to flow between the Leclanché cell, the mercury battery, the lithium-ion bat- half-cells. tery, and the lead storage battery used in automobiles. 5. The electromotive force (emf) of a cell is the voltage Fuel cells produce electrical energy from a continuous difference between the two electrodes. In the external supply of reactants. circuit, electrons flow from the anode to the cathode in 12. The corrosion of metals, such as the rusting of iron, is a galvanic cell. In solution, the anions move toward the an electrochemical phenomenon. anode and the cations move toward the cathode. 13. Electric current from an external source is used to drive 6. The quantity of electricity carried by 1 mole of elec- a nonspontaneous chemical reaction in an electrolytic trons is called a faraday, which is equal to 96,500 C. cell. The amount of product formed or reactant con- 7. Standard reduction potentials show the relative likeli- sumed depends on the quantity of electricity transferred hood of half-cell reduction reactions and can be used to at the electrode. predict the products, direction, and spontaneity of redox 14. Electrolysis plays an important role in obtaining pure reactions between various substances. metals from their ores and in purifying metals.

KEY WORDS

Anode, p. 647 Electrochemistry, p. 643 Faraday constant, p. 654 Overvoltage, p. 670 Battery, p. 661 Electrolysis, p. 668 Fuel cell, p. 664 Standard emf ( EЊ), p. 649 Cathode, p. 647 Electrolytic cell, p. 668 Galvanic cell, p. 646 Standard reduction Cell voltage, p. 647 Electromotive force Half-cell reaction, p. 647 potential, p. 648 Corrosion, p. 665 (emf) ( E), p. 647 Nernst equation, p. 657

QUESTIONS AND PROBLEMS

2ϩ Balancing Redox Equations (b) Cu ϩ HNO 3 ¡ Cu ϩ NO ϩ H2O (in Problems acidic solution) Ϫ Ϫ Ϫ (c) CN ϩ MnO 4 ¡ CNO ϩ MnO 2 (in basic 19.1 Balance the following redox equations by the half- solution) reaction method: (d)Br ¡ BrO Ϫ ϩ Br Ϫ (in basic solution) (a)H O ϩ Fe 2ϩ ¡ Fe 3ϩ ϩ H O (in acidic 2 3 2 2 2 2Ϫ ϩ 88n Ϫ ϩ 2Ϫ solution) (e) S 2O3 I2 I S4O6 (in acidic solution) cha48518_ch19_642-683.qxd 12/21/06 12:57 PM Page 676 CONFIRMING PAGES

676 CHAPTER 19 Redox Reactions and Electrochemistry

19.2 Balance the following redox equations by the half- 19.15 Consider the following half-reactions: reaction method: Ϫ ϩ ϩ ϩ Ϫ 2ϩ MnO 4 (aq ) 8H (aq ) 5e ¡ (a)Mn ϩ H2O2 ¡ MnO 2 ϩ H2O (in basic 2ϩ Mn (aq ) ϩ 4H 2O( l) solution) Ϫ ϩ ϩ ϩ Ϫ 88n ϩ 2Ϫ 2Ϫ NO 3 (aq ) 4H (aq ) 3e NO( g) 2H 2O( l) (b) Bi(OH) 3 ϩ SnO 2 ¡ SnO 3 ϩ Bi Ϫ 2ϩ (in basic solution) Predict whether NO 3 ions will oxidize Mn to Ϫ 2Ϫ 2Ϫ 3ϩ (c) Cr 2O7 ϩ C2O4 ¡ Cr ϩ CO 2 MnO 4 under standard-state conditions. (in acidic solution) 19.16 Predict whether the following reactions would occur Ϫ Ϫ Њ (d)ClO 3 ϩ Cl ¡ Cl 2 ϩ ClO 2 (in acidic spontaneously in aqueous solution at 25 C. Assume solution) that the initial concentrations of dissolved species are all 1.0 M. (a) Ca( s) ϩ Cd 2ϩ(aq ) ¡ Ca 2ϩ(aq ) ϩ Cd( s) Galvanic Cells and Standard Emfs Ϫ 2ϩ (b) 2Br (aq ) ϩ Sn (aq ) ¡ Br 2(l) ϩ Sn( s) Review Questions ϩ ϩ (c) 2Ag( s) ϩ Ni 2 (aq ) ¡ 2Ag (aq ) ϩ Ni( s) 19.3 Define the following terms: anode, cathode, cell (d) Cu ϩ(aq ) ϩ Fe 3ϩ(aq ) ¡ voltage, electromotive force, standard reduction Cu 2ϩ(aq ) ϩ Fe 2ϩ(aq ) potential. 19.17 Which species in each pair is a better oxidizing 19.4 Describe the basic features of a galvanic cell. Why agent under standard-state conditions? (a) Br 2 or 3ϩ ϩ 2ϩ 3ϩ are the two components of the cell separated from Au , (b) H 2 or Ag , (c) Cd or Cr , (d) O 2 in each other? acidic media or O 2 in basic media. 19.5 What is the function of a salt bridge? What kind of 19.18 Which species in each pair is a better reducing agent electrolyte should be used in a salt bridge? under standard-state conditions? (a) Na or Li, (b) H 2 2ϩ Ϫ 2ϩ 19.6 What is a cell diagram? Write the cell diagram for a or I 2, (c) Fe or Ag, (d) Br or Co . galvanic cell consisting of an Al electrode placed in a 1 M Al(NO 3)3 solution and a Ag electrode placed Spontaneity of Redox Reactions in a 1 M AgNO 3 solution. 19.7 What is the difference between the half-reactions Review Questions discussed in redox processes in Chapter 4 and the 19.19 Write the equations relating ⌬GЊ and K to the stan- half-cell reactions discussed in Section 19.2? dard emf of a cell. Deﬁne all the terms. 19.8 After operating a Daniell cell (see Figure 19.1) for a 19.20 Compare the ease of measuring the equilibrium con- few minutes, a student notices that the cell emf be- stant electrochemically with that by chemical means gins to drop. Why? [see Equation (18.14)]. 19.9 Use the information in Table 2.1, and calculate the Faraday constant. Problems 19.10 Discuss the spontaneity of an electrochemical reaction in terms of its standard emf ( EЊ ). 19.21 What is the equilibrium constant for the following cell reaction at 25 ЊC? 2ϩ 2ϩ Problems Mg( s) ϩ Zn (aq ) ∆ Mg (aq ) ϩ Zn( s) 19.11 Calculate the standard emf of a cell that uses the 19.22 The equilibrium constant for the reaction 2ϩ 2ϩ Mg ͞Mg and Cu ͞Cu half-cell reactions at 25 ЊC. Sr( s) ϩ Mg 2ϩ(aq ) ∆ Sr 2ϩ(aq ) ϩ Mg( s) Write the equation for the cell reaction that occurs is 2.69 ϫ 10 12 at 25 ЊC. Calculate EЊ for a cell made under standard-state conditions. 2ϩ 2ϩ ϩ up of Sr ͞Sr and Mg ͞Mg half-cells. 19.12 Calculate the standard emf of a cell that uses Ag ͞Ag and Al ͞Al 3ϩ half-cell reactions. Write the cell reac- 19.23 Use the standard reduction potentials to ﬁnd the tion that occurs under standard-state conditions. equilibrium constant for each of the following reac- ϩ Ϫ tions at 25 ЊC: 19.13 Predict whether Fe 3 can oxidize I to I under 2 ϩ Ϫ Ϫ ϩ standard-state conditions. (a) Br 2(l) 2I (aq ) ∆ 2Br (aq ) I2(s) (b) 2Ce 4ϩ(aq ) ϩ 2ClϪ(aq ) ∆ 19.14 Which of the following reagents can oxidize H 2O to ϩ Cl (g) ϩ 2Ce 3ϩ(aq ) O2(g) under standard-state conditions? H (aq ), 2 Ϫ 2ϩ 2ϩ Ϫ 2ϩ Ϫ ϩ Cl (aq ), Cl 2(g), Cu (aq ), Pb (aq ), MnO 4 (aq ) (c) 5Fe (aq ) ϩ MnO 4 (aq ) ϩ 8H (aq ) ∆ 2ϩ 3ϩ (in acid). Mn (aq ) ϩ 4H 2O( l) ϩ 5Fe (aq ) cha48518_ch19_642-683.qxd 12/21/06 12:57 PM Page 677 CONFIRMING PAGES

Questions and Problems 677

⌬ Њ 19.24 Calculate G and Kc for the following reactions at Batteries and Fuel Cells 25 ЊC: Review Questions (a) Mg( s) ϩ Pb 2ϩ(aq ) ∆ Mg 2ϩ(aq ) ϩ Pb( s) Ϫ Ϫ 19.35 Explain the differences between a primary galvanic (b) Br 2(l) ϩ 2I (aq ) ∆ 2Br (aq ) ϩ I2(s) ϩ 2ϩ cell—one that is not rechargeable—and a storage (c) O2(g) ϩ 4H (aq ) ϩ 4Fe (aq ) ∆ ϩ cell (for example, the lead storage battery), which is 2H O( l) ϩ 4Fe 3 (aq ) 2 rechargeable. (d) 2Al( s) ϩ 3I (s) ∆ 2Al 3ϩ(aq ) ϩ 6I Ϫ(aq ) 2 19.36 Discuss the advantages and disadvantages of fuel 19.25 Under standard-state conditions, what spontaneous cells over conventional power plants in producing reaction will occur in aqueous solution among the electricity. ions Ce 4ϩ, Ce 3ϩ, Fe 3ϩ, and Fe 2ϩ? Calculate ⌬GЊ and Kc for the reaction. Problems Њ ϭ ϩ ϩ 19.26 Given that E 0.52 V for the reduction Cu (aq ) 19.37 The hydrogen-oxygen fuel cell is described in Sec- Ϫ n Њ ⌬ Њ e Cu( s), calculate E , G , and K for the fol- tion 19.6. (a) What volume of H (g), stored at 25 ЊC Њ 2 lowing reaction at 25 C: at a pressure of 155 atm, would be needed to run an 2Cu ϩ(aq ) ¡ Cu 2ϩ(aq ) ϩ Cu( s) electric motor drawing a current of 8.5 A for 3.0 h? (b) What volume (liters) of air at 25 ЊC and 1.00 atm will have to pass into the cell per minute to run the motor? Assume that air is 20 percent O by volume The Effect of Concentration on Cell Emf 2 and that all the O 2 is consumed in the cell. The other Review Questions components of air do not affect the fuel-cell reactions. Assume ideal gas behavior. 19.27 Write the Nernst equation and explain all the terms. 19.38 Calculate the standard emf of the propane fuel cell 19.28 Write the Nernst equation for the following processes discussed on p. 665 at 25 ЊC, given that ⌬GЊf for at some temperature T: propane is Ϫ23.5 kJ͞mol. (a) Mg( s) ϩ Sn 2ϩ(aq ) ¡ Mg 2ϩ(aq ) ϩ Sn( s) ϩ 2ϩ 3ϩ ϩ (b) 2Cr( s) 3Pb (aq ) ¡ 2Cr (aq ) 3Pb( s) Corrosion Problems Review Questions 19.29 What is the potential of a cell made up of Zn ͞Zn 2ϩ 19.39 Steel hardware, including nuts and bolts, is often and Cu ͞Cu 2ϩ half-cells at 25 ЊC if [Zn 2ϩ] ϭ 0.25 M coated with a thin plating of cadmium. Explain the and [Cu 2ϩ] ϭ 0.15 M? function of the cadmium layer. 19.30 Calculate EЊ, E, and ⌬G for the following cell 19.40 “Galvanized iron” is steel sheet that has been coated reactions. with zinc; “tin” cans are made of steel sheet coated with tin. Discuss the functions of these coatings and ϩ 2ϩ ¡ 2ϩ ϩ (a) Mg( s) Sn (aq ) Mg (aq ) Sn( s) the electrochemistry of the corrosion reactions that 2ϩ ϭ 2ϩ ϭ [Mg ] 0.045 M, [Sn ] 0.035 M occur if an electrolyte contacts the scratched surface ϩ ϩ (b) 3Zn( s) ϩ 2Cr 3 (aq ) ¡ 3Zn 2 (aq ) ϩ 2Cr( s) of a galvanized iron sheet or a tin can. [Cr 3ϩ] ϭ 0.010 M, [Zn 2ϩ] ϭ 0.0085 M 19.41 Tarnished silver contains Ag 2S. The tarnish can be 19.31 Calculate the standard potential of the cell consisting removed by placing silverware in an aluminum pan ϩ of the Zn ͞Zn 2 half-cell and the SHE. What will the containing an inert electrolyte solution, such as 2ϩ emf of the cell be if [Zn ] ϭ 0.45 M, PH ϭ 2.0 atm, NaCl. Explain the electrochemical principle for this ϩ 2 and [H ] ϭ 1.8 M? procedure. [The standard reduction potential for the Ϫ Ϫ 2ϩ ϩ → ϩ 2 19.32 What is the emf of a cell consisting of a Pb ͞Pb half-cell reaction Ag 2S( s) 2e 2Ag( s) S (aq ) ϩ 2ϩ Ϫ half-cell and a Pt ͞H ͞H2 half-cell if [Pb ] ϭ 0.10 M, is 0.71 V.] ϩ ϭ ϭ [H ] 0.050 M, and PH2 1.0 atm? 19.42 How does the tendency of iron to rust depend on the 19.33 Referring to the arrangement in Figure 19.1, calcu- pH of solution? late the [Cu 2ϩ]͞[Zn 2ϩ] ratio at which the following reaction is spontaneous at 25 ЊC: Electrolysis Cu( s) ϩ Zn 2ϩ(aq ) ¡ Cu 2ϩ(aq ) ϩ Zn( s) Review Questions 19.34 Calculate the emf of the following concentration cell: 19.43 What is the difference between a galvanic cell (such ϩ ϩ Mg( s)|Mg 2 (0.24 M)|| Mg 2 (0.53 M)|Mg( s) as a Daniell cell) and an electrolytic cell? cha48518_ch19_642-683.qxd 12/21/06 12:57 PM Page 678 CONFIRMING PAGES

678 CHAPTER 19 Redox Reactions and Electrochemistry

19.44 What is Faraday’s contribution to quantitative 19.55 What is the hourly production rate of chlorine gas electrolysis? (in kg) from an electrolytic cell using aqueous NaCl electrolyte and carrying a current of 1.500 ϫ 10 3 A? Problems The anode efficiency for the oxidation of Cl Ϫ is 93.0 percent. 19.45 The half-reaction at an electrode is 19.56 Chromium plating is applied by electrolysis to ob- Mg 2ϩ(molten) ϩ 2eϪ ¡ Mg( s) jects suspended in a dichromate solution, according to the following (unbalanced) half-reaction: Calculate the number of grams of magnesium that 2Ϫ Ϫ ϩ can be produced by supplying 1.00 F to the electrode. Cr 2O7 (aq ) ϩ e ϩ H (aq ) 88n Cr( s) ϩ H2O( l) 19.46 Consider the electrolysis of molten barium chloride, How long (in hours) would it take to apply a Ϫ2 BaCl 2. (a) Write the half-reactions. (b) How many chromium plating 1.0 ϫ 10 mm thick to a car grams of barium metal can be produced by supply- bumper with a surface area of 0.25 m 2 in an elec- ing 0.50 A for 30 min? trolytic cell carrying a current of 25.0 A? (The den- 19.47 Considering only the cost of electricity, would it be sity of chromium is 7.19 g ͞cm 3.) cheaper to produce a ton of sodium or a ton of alu- 19.57 The passage of a current of 0.750 A for 25.0 min de- minum by electrolysis? posited 0.369 g of copper from a CuSO 4 solution. 19.48 If the cost of electricity to produce magnesium by From this information, calculate the molar mass of the electrolysis of molten magnesium chloride is copper. $155 per ton of metal, what is the cost (in dollars) 19.58 A quantity of 0.300 g of copper was deposited from of the electricity necessary to produce (a) 10.0 tons of a CuSO 4 solution by passing a current of 3.00 A aluminum, (b) 30.0 tons of sodium, (c) 50.0 tons of through the solution for 304 s. Calculate the value of calcium? the faraday constant. 19.49 One of the half-reactions for the electrolysis of 19.59 In a certain electrolysis experiment, 1.44 g of Ag water is were deposited in one cell (containing an aqueous 2H O( l) ¡ O (g) ϩ 4H ϩ(aq ) ϩ 4eϪ AgNO 3 solution), while 0.120 g of an unknown 2 2 metal X was deposited in another cell (containing an If 0.076 L of O 2 is collected at 25 ЊC and 755 mmHg, aqueous XCl 3 solution) in series with the AgNO 3 how many faradays of electricity had to pass through cell. Calculate the molar mass of X. the solution? 19.60 One of the half-reactions for the electrolysis of 19.50 How many faradays of electricity are required to water is produce (a) 0.84 L of O 2 at exactly 1 atm and 25 ЊC ϩ Ϫ 2H (aq ) ϩ 2e 88n H2(g) from aqueous H 2SO 4 solution; (b) 1.50 L of Cl 2 at 750 mmHg and 20 ЊC from molten NaCl; (c) 6.0 g of If 0.845 L of H 2 is collected at 25 ЊC and 782 mmHg, Sn from molten SnCl 2? how many faradays of electricity had to pass through 19.51 Calculate the amounts of Cu and Br 2 produced in the solution? 1.0 h at inert electrodes in a solution of CuBr 2 by a current of 4.50 A. Additional Problems 19.52 In the electrolysis of an aqueous AgNO 3 solution, 0.67 g of Ag is deposited after a certain period of 19.61 For each of the following redox reactions, (i) write time. (a) Write the half-reaction for the reduction the half-reactions; (ii) write a balanced equation for the of Ag ϩ. (b) What is the probable oxidation half- whole reaction, (iii) determine in which direction the reaction? (c) Calculate the quantity of electricity reaction will proceed spontaneously under standard- used, in coulombs. state conditions: 2ϩ ϩ (a) H2(g) ϩ Ni (aq ) ¡ H (aq ) ϩ Ni( s) 19.53 A steady current was passed through molten CoSO 4 Ϫ Ϫ until 2.35 g of metallic cobalt was produced. Calcu- (b) MnO 4 (aq ) ϩ Cl (aq ) ¡ 2ϩ late the number of coulombs of electricity used. Mn (aq ) ϩ Cl 2(g) (in acid solution) ϩ ϩ 19.54 A constant electric current flows for 3.75 h through (c) Cr( s) ϩ Zn 2 (aq ) ¡ Cr 3 (aq ) ϩ Zn( s) two electrolytic cells connected in series. One con- 19.62 The oxidation of 25.0 mL of a solution containing tains a solution of AgNO and the second a solution 2ϩ 3 Fe requires 26.0 mL of 0.0250 M K2Cr 2O7 in of CuCl 2. During this time 2.00 g of silver are de- acidic solution. Balance the following equation and posited in the first cell. (a) How many grams of cop- calculate the molar concentration of Fe 2ϩ: per are deposited in the second cell? (b) What is the 2Ϫ 2ϩ ϩ 3ϩ 3ϩ current flowing, in amperes? Cr 2O7 ϩ Fe ϩ H ¡ Cr ϩ Fe cha48518_ch19_642-683.qxd 12/21/06 12:57 PM Page 679 CONFIRMING PAGES

Questions and Problems 679

19.63 The SO 2 present in air is mainly responsible for the 19.69 From the following information, calculate the solu- phenomenon of acid rain. The concentration of SO 2 bility product of AgBr: can be determined by titrating against a standard ϩ Ϫ Ag (aq ) ϩ e 88n Ag( s) E° ϭ 0.80 V permanganate solution as follows: Ϫ Ϫ AgBr( s) ϩ e 88n Ag( s) ϩ Br (aq ) E° ϭ 0.07 V Ϫ 5SO 2 ϩ 2MnO 4 ϩ 2H 2O ¡ 2Ϫ 2ϩ ϩ 19.70 Consider a galvanic cell composed of the SHE and a 5SO ϩ 2Mn ϩ 4H ϩ Ϫ 4 half-cell using the reaction Ag (aq ) ϩ e → Ag( s). Calculate the number of grams of SO 2 in a sample of (a) Calculate the standard cell potential. (b) What is air if 7.37 mL of 0.00800 M KMnO 4 solution are re- the spontaneous cell reaction under standard-state quired for the titration. conditions? (c) Calculate the cell potential when [H ϩ] ϫ 19.64 A sample of iron ore weighing 0.2792 g was dis- in the hydrogen electrode is changed to (i) 1.0 Ϫ2 ϫ Ϫ5 solved in an excess of a dilute acid solution. All the 10 M and (ii) 1.0 10 M, all other reagents iron was ﬁrst converted to Fe(II) ions. The solution being held at standard-state conditions. (d) Based on this cell arrangement, suggest a design for a pH meter. then required 23.30 mL of 0.0194 M KMnO 4 for oxidation to Fe(III) ions. Calculate the percent by mass 19.71 A galvanic cell consists of a silver electrode in con- of iron in the ore. tact with 346 mL of 0.100 M AgNO 3 solution and a 19.65 The concentration of a hydrogen peroxide solution magnesium electrode in contact with 288 mL of can be conveniently determined by titration against a 0.100 M Mg(NO 3)2 solution. (a) Calculate E for the Њ standardized potassium permanganate solution in an cell at 25 C. (b) A current is drawn from the cell acidic medium according to the following unbal- until 1.20 g of silver have been deposited at the sil- anced equation: ver electrode. Calculate E for the cell at this stage of operation. MnO Ϫ ϩ H O ¡ O ϩ Mn 2ϩ 4 2 2 2 19.72 Explain why chlorine gas can be prepared by elec- (a) Balance the above equation. (b) If 36.44 mL of trolyzing an aqueous solution of NaCl but ﬂuorine a 0.01652 M KMnO 4 solution are required to com- gas cannot be prepared by electrolyzing an aqueous pletely oxidize 25.00 mL of a H 2O2 solution, calculate solution of NaF. the molarity of the H 2O2 solution. 19.73 Calculate the emf of the following concentration cell Њ 19.66 Oxalic acid (H 2C2O4) is present in many plants and at 25 C: vegetables. (a) Balance the following equation in ϩ ϩ Cu( s)|Cu 2 (0.080 M)|| Cu 2 (1.2 M)|Cu( s) acid solution: Ϫ Ϫ ϩ 19.74 The cathode reaction in the Leclanché cell is MnO ϩ C O2 ¡ Mn 2 ϩ CO 4 2 4 2 given by

(b) If a 1.00-g sample of H 2C2O4 requires 24.0 mL 2ϩ Ϫ 2MnO 2(s) ϩ Zn (aq ) ϩ 2e 88n ZnMn 2O4(s) of 0.0100 M KMnO 4 solution to reach the equiva- lence point, what is the percent by mass of H 2C2O4 If a Leclanché cell produces a current of 0.0050 A, in the sample? calculate how many hours this current supply will last if there are initially 4.0 g of MnO 2 present in the 19.67 Complete the following table. State whether the 2ϩ cell reaction is spontaneous, nonspontaneous, or at cell. Assume that there is an excess of Zn ions. equilibrium. 19.75 Suppose you are asked to verify experimentally the electrode reactions shown in Example 19.8. In addi- ⌬ E G Cell Reaction tion to the apparatus and the solution, you are also Ͼ 0 given two pieces of litmus paper, one blue and the Ͼ0 other red. Describe what steps you would take in this ϭ 0 experiment. 19.76 For a number of years it was not clear whether mer- ϩ 2ϩ 19.68 Calcium oxalate (CaC 2O4) is insoluble in water. cury(I) ions existed in solution as Hg or as Hg 2 . This property has been used to determine the amount To distinguish between these two possibilities, we of Ca 2ϩ ions in blood. The calcium oxalate isolated could set up the following system: from blood is dissolved in acid and titrated against a Hg( l)|soln A||soln B|Hg( l) standardized KMnO 4 solution, as described in Prob- lem 19.66. In one test it is found that the calcium ox- where soln A contained 0.263 g mercury(I) nitrate alate isolated from a 10.0-mL sample of blood per liter and soln B contained 2.63 g mercury(I) ni- Ϫ4 requires 24.2 mL of 9.56 ϫ 10 M KMnO 4 for trate per liter. If the measured emf of such a cell is titration. Calculate the number of milligrams of cal- 0.0289 V at 18 ЊC, what can you deduce about the na- cium per milliliter of blood. ture of the mercury(I) ions? cha48518_ch19_642-683.qxd 12/21/06 12:57 PM Page 680 CONFIRMING PAGES

680 CHAPTER 19 Redox Reactions and Electrochemistry

19.77 An aqueous KI solution to which a few drops of phe- 25.0 mL. (b) In a separate experiment, 25.0 mL of nolphthalein have been added is electrolyzed using 3.00 M NH 3 are added to the CuSO 4 solution. If the an apparatus like the one shown here: emf of the cell is 0.68 V, calculate the formation con- 2ϩ stant ( Kf) of Cu(NH 3)4 . 19.84 In an electrolysis experiment, a student passes the same quantity of electricity through two electrolytic cells, one containing a silver salt and the other a gold salt. Over a certain period of time, she ﬁnds that 2.64 g of Ag and 1.61 g of Au are deposited at the cathodes. What is the oxidation state of gold in the gold salt? 19.85 People living in cold-climate countries where there is plenty of snow are advised not to heat their garages in the winter. What is the electrochemical basis for this recommendation? 19.86 Given that Describe what you would observe at the anode and 2ϩ Ϫ 2ϩ 2Hg (aq ) ϩ 2e 88n Hg 2 (aq ) E° ϭ 0.92 V the cathode. ( Hint: Molecular iodine is only slightly 2ϩ Ϫ Ϫ Hg (aq ) ϩ 2e 88n 2Hg( l) E° ϭ 0.85 V soluble in water, but in the presence of I ions, it 2 Ϫ forms the brown color of I 3 ions.) calculate ⌬GЊ and K for the following process at 25 ЊC:

19.78 A piece of magnesium metal weighing 1.56 g is 2ϩ 2ϩ Hg 2 (aq ) ¡ Hg (aq ) ϩ Hg( l) placed in 100.0 mL of 0.100 M AgNO 3 at 25 ЊC. Cal- ϩ ϩ culate [Mg 2 ] and [Ag ] in solution at equilibrium. (The preceding reaction is an example of a What is the mass of the magnesium left? The volume disproportionation reaction in which an element in remains constant. one oxidation state is both oxidized and reduced.)

19.79 Describe an experiment that would enable you to 19.87 Fluorine (F 2) is obtained by the electrolysis of liquid determine which is the cathode and which is the hydrogen fluoride (HF) containing potassium fluo- anode in a galvanic cell using copper and zinc ride (KF). (a) Write the half-cell reactions and the electrodes. overall reaction for the process. (b) What is the pur- 19.80 An acidified solution was electrolyzed using cop- pose of KF? (c) Calculate the volume of F 2 (in liters) per electrodes. A constant current of 1.18 A caused collected at 24.0 ЊC and 1.2 atm after electrolyzing the anode to lose 0.584 g after 1.52 ϫ 10 3 s. (a) the solution for 15 h at a current of 502 A. What is the gas produced at the cathode and what 19.88 A 300-mL solution of NaCl was electrolyzed for is its volume at STP? (b) Given that the charge of 6.00 min. If the pH of the final solution was 12.24, Ϫ an electron is 1.6022 ϫ 10 19 C, calculate Avo- calculate the average current used. gadro’s number. Assume that copper is oxidized to 2ϩ 19.89 Industrially, copper is purified by electrolysis. The Cu ions. impure copper acts as the anode, and the cathode is ϩ 19.81 In a certain electrolysis experiment involving Al 3 made of pure copper. The electrodes are immersed in ions, 60.2 g of Al is recovered when a current of a CuSO 4 solution. During electrolysis, copper at the 0.352 A is used. How many minutes did the elec- anode enters the solution as Cu 2ϩ while Cu 2ϩ ions trolysis last? are reduced at the cathode. (a) Write half-cell reac- 19.82 Consider the oxidation of ammonia: tions and the overall reaction for the electrolytic process. (b) Suppose the anode was contaminated ϩ ϩ 4NH 3(g) 3O 2(g) ¡ 2N 2(g) 6H 2O( l) with Zn and Ag. Explain what happens to these im- (a) Calculate the ⌬GЊ for the reaction. (b) If this purities during electrolysis. (c) How many hours will reaction were used in a fuel cell, what would the it take to obtain 1.00 kg of Cu at a current of 18.9 A? standard cell potential be? 19.90 An aqueous solution of a platinum salt is elec- 19.83 A galvanic cell is constructed by immersing a piece trolyzed at a current of 2.50 A for 2.00 h. As a result, 9.09 g of metallic Pt are formed at the cathode. Cal- of copper wire in 25.0 mL of a 0.20 M CuSO 4 solu- culate the charge on the Pt ions in this solution. tion and a zinc strip in 25.0 mL of a 0.20 M ZnSO 4 solution. (a) Calculate the emf of the cell at 25 ЊC and 19.91 Consider a galvanic cell consisting of a magnesium predict what would happen if a small amount of con- electrode in contact with 1.0 M Mg(NO 3)2 and a cad- centrated NH 3 solution were added to (i) the CuSO 4 mium electrode in contact with 1.0 M Cd(NO 3)2. Њ solution and (ii) the ZnSO 4 solution. Assume that the Calculate E for the cell, and draw a diagram show- volume in each compartment remains constant at ing the cathode, anode, and direction of electron flow. cha48518_ch19_642-683.qxd 12/22/06 7:51 AM Page 681 CONFIRMING PAGES

Questions and Problems 681

19.92 A current of 6.00 A passes through an electrolytic cell answer the following questions. (a) Why does gold containing dilute sulfuric acid for 3.40 h. If the volume not tarnish in air? (b) Will the following dispropor- of O 2 gas generated at the anode is 4.26 L (at STP), tionation occur spontaneously? calculate the charge (in coulombs) on an electron. ϩ ϩ 3Au (aq ) ¡ Au 3 (aq ) ϩ 2Au( s) 19.93 Gold will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. However, (c) Predict the reaction between gold and fluorine the metal does dissolve in a mixture of the acids (one gas. part HNO 3 and three parts HCl by volume), called 19.101 Calculate EЊ for the reactions of mercury with (a) 1 M aqua regia. (a) Write a balanced equation for this re- HCl and (b) 1 M HNO 3. Which acid will oxidize Hg 2ϩ action. ( Hint: Among the products are HAuCl 4 and to Hg 2 under standard-state conditions? Can you NO 2.) (b) What is the function of HCl? identify which test tube below contains HNO 3 and 19.94 Explain why most useful galvanic cells give voltages Hg and which contains HCl and Hg? of no more than 1.5 to 2.5 V. What are the prospects for developing practical galvanic cells with voltages of 5 V or more? 19.95 A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag 2C2O4, at 25 ЊC. The measured potential difference between the rod and the SHE is 0.589 V, the rod being positive. Calcu- late the solubility product constant for silver oxalate. 19.96 Zinc is an amphoteric metal; that is, it reacts with both acids and bases. The standard reduction poten- Based on your answer, explain why ingestion of a tial is Ϫ1.36 V for the reaction very small quantity of mercury is not considered too 2Ϫ Ϫ Ϫ Zn(OH) 4 (aq ) ϩ 2e ¡ Zn( s) ϩ 4OH (aq ) harmful. 2ϩ Calculate the formation constant ( K ) for the reaction 19.102 When 25.0 mL of a solution containing both Fe f and Fe 3ϩ ions is titrated with 23.0 mL of 0.0200 M 2ϩ ϩ Ϫ 2Ϫ 2ϩ Zn (aq ) 4OH (aq ) ∆ Zn(OH) 4 (aq ) KMnO 4 (in dilute sulfuric acid), all of the Fe ions 3ϩ 19.97 Use the data in Table 19.1 to determine whether or are oxidized to Fe ions. Next, the solution is treated with Zn metal to convert all of the Fe 3ϩ ions not hydrogen peroxide will undergo disproportiona- 2ϩ tion in an acid medium: 2H O → 2H O ϩ O . to Fe ions. Finally, 40.0 mL of the same KMnO 4 2 2 2 2 solution are added to the solution in order to oxidize 19.98 The magnitudes (but not the signs) of the standard the Fe 2ϩ ions to Fe 3ϩ. Calculate the molar concen- reduction potentials of two metals X and Y are trations of Fe 2ϩ and Fe 3ϩ in the original solution. ϩ Ϫ Y2 ϩ 2e 88n Y ͉E°͉ ϭ 0.34 V 19.103 Consider the Daniell cell in Figure 19.1. When ϩ Ϫ X2 ϩ 2e 88n X ͉E°͉ ϭ 0.25 V viewed externally, the anode appears negative and the cathode positive (electrons are flowing from the an- where the || notation denotes that only the magnitude ode to the cathode). Yet in solution anions are moving (but not the sign) of the EЊ value is shown. When the toward the anode, which means that it must appear half-cells of X and Y are connected, electrons flow positive to the anions. Because the anode cannot si- from X to Y. When X is connected to a SHE, elec- multaneously be negative and positive, give an expla- trons flow from X to SHE. (a) Are the EЊ values of nation for this apparently contradictory situation. the half-reactions positive or negative? (b) What is the standard emf of a cell made up of X and Y? 19.104 Lead storage batteries are rated by ampere hours, that is, the number of amperes they can deliver in an 19.99 A galvanic cell is constructed as follows. One half- hour. (a) Show that 1 A и h ϭ 3600 C. (b) The lead cell consists of a platinum wire immersed in a solu- ϩ ϩ anodes of a certain lead storage battery have a total tion containing 1.0 M Sn 2 and 1.0 M Sn 4 ; the other mass of 406 g. Calculate the maximum theoretical half-cell has a thallium rod immersed in a solution of ϩ capacity of the battery in ampere hours. Explain why 1.0 M TI . (a) Write the half-cell reactions and the in practice we can never extract this much energy overall reaction. (b) What is the equilibrium constant ϩ from the battery. ( Hint: Assume all of the lead will at 25 ЊC? (c) What is the cell voltage if the TI concen- ϩ ϭ Ϫ be used up in the electrochemical reaction and refer tration is increased tenfold? (E°T1 /T1 0.34 V.) to the electrode reactions on p. 662.) (c) Calculate 19.100 Given the standard reduction potential for Au 3ϩ in EЊcell and ⌬GЊ for the battery. Table 19.1 and 19.105 The concentration of sulfuric acid in the lead storage Au ϩ(aq ) ϩ eϪ 88n Au( s) E° ϭ 1.69 V battery of an automobile over a period of time has cha48518_ch19_642-683.qxd 12/21/06 12:57 PM Page 682 CONFIRMING PAGES

682 CHAPTER 19 Redox Reactions and Electrochemistry

decreased from 38.0 percent by mass (density ϭ ϩ 2ϩ Pb( s) ϩ 2H (aq ) ∆ Pb (aq ) ϩ H2(g) 1.29 g͞mL) to 26.0 percent by mass (1.19 g ͞mL). 2ϩ ϭ Assume the volume of the acid remains constant at Given that [Pb ] 0.035 M and the solution is 724 mL. (a) Calculate the total charge in coulombs buffered at pH 1.60. supplied by the battery. (b) How long (in hours) will 19.111 Because all alkali metals react with water, it is not it take to recharge the battery back to the original possible to measure the standard reduction poten- sulfuric acid concentration using a current of 22.4 A. tials of these metals directly as in the case of, say, 19.106 Consider a Daniell cell operating under nonstandard- zinc. An indirect method is to consider the following state conditions. Suppose that the cell’s reaction is hypothetical reaction multiplied by 2. What effect does this have on each ϩ ϩ 1 ϩ ϩ Li (aq ) 2H2(g) ¡ Li( s) H (aq ) of the following quantities in the Nernst equation? (a) E, (b) EЊ, (c) Q, (d) ln Q, and (e) n? Use the appropriate equation presented in this chapter and the thermodynamic data in Appendix 2, cal- 19.107 A spoon was silver-plated electrolytically in a culate EЊ for Li ϩ(aq ) ϩ eϪ → Li( s) at 298 K. AgNO 3 solution. (a) Sketch a diagram for the Compare your result with that listed in Table 19.1. process. (b) If 0.884 g of Ag was deposited on the (See back endpaper for the Faraday constant.) spoon at a constant current of 18.5 mA, how long (in 2ϩ 2ϩ minutes) did the electrolysis take? 19.112 A galvanic cell using Mg ͞Mg and Cu ͞Cu half- cells operates under standard-state conditions at 25 ЊC 19.108 Comment on whether F 2 will become a stronger ox- ϩ and each compartment has a volume of 218 mL. The idizing agent with increasing H concentration. cell delivers 0.22 A for 31.6 h. (a) How many grams of 19.109 In recent years there has been much interest in elec- Cu are deposited? (b) What is the [Cu 2ϩ] remaining? tric cars. List some advantages and disadvantages of 19.113 Given the following standard reduction potentials, electric cars compared to automobiles with internal calculate the ion-product, K , for water at 25 ЊC: combustion engines. w ϩ ϩ Ϫ ϭ 19.110 Calculate the pressure of H (in atm) required to 2H (aq ) 2e ¡ H2(g) E° 0.00 V 2 ϩ Ϫ ϩ Ϫ maintain equilibrium with respect to the following 2H 2O( l) 2e ¡ H2(g) 2OH (aq ) ϭ Ϫ reaction at 25 ЊC: E° 0.83 V

SPECIAL PROBLEMS

ϩ 1 n 19.114 A piece of magnesium ribbon and a copper wire are The net transformation is Zn( s) 2O2(g) ZnO( s) partially immersed in a 0.1 M HCl solution in a (a) Write the half-reactions at the zinc-air electrodes beaker. The metals are joined externally by another and calculate the standard emf of the battery at 25ЊC. piece of metal wire. Bubbles are seen to evolve at (b) Calculate the emf under actual operating conditions both the Mg and Cu surfaces. (a) Write equations when the partial pressure of oxygen is 0.21 atm. (c) representing the reactions occurring at the metals. What is the energy density (measured as the energy in (b) What visual evidence would you seek to show kilojoules that can be obtained from 1 kg of the metal) that Cu is not oxidized to Cu 2ϩ? (c) At some stage, of the zinc electrode? (d) If a current of 2.1 ϫ 10 5 A is NaOH solution is added to the beaker to neutralize to be drawn from a zinc-air battery system, what vol- the HCl acid. Upon further addition of NaOH, a ume of air (in liters) would need to be supplied to the white precipitate forms. What is it? battery every second? Assume that the temperature is Њ 19.115 The zinc-air battery shows much promise for electric 25 C and the partial pressure of oxygen is 0.21 atm. cars because it is lightweight and rechargeable: 19.116 Calculate the equilibrium constant for the following reaction at 298 K:

ϩ ϩ Zn( s) ϩ Cu 2 (aq ) ∆ Zn 2 (aq ) ϩ Cu( s) 19.117 A construction company is installing an iron culvert (a long cylindrical tube) that is 40.0 m long with a radius of 0.900 m. To prevent corrosion, the culvert must be galvanized. This process is carried out by ﬁrst passing an iron sheet of appropriate dimensions through an electrolytic cell containing Zn 2ϩ ions, using graphite as the anode and the iron sheet as the cha48518_ch19_642-683.qxd 12/21/06 12:57 PM Page 683 CONFIRMING PAGES

Answers to Practice Exercises 683

cathode. If the voltage is 3.26 V, what is the cost of was totally immersed. Next, she added a few table- electricity for depositing a layer 0.200 mm thick if spoonfuls of baking soda (sodium bicarbonate), the efficiency of the process is 95 percent? The elec- which readily dissolved. Finally, she placed some tricity rate is $0.12 per kilowatt hour (kWh), where aluminum foil at the bottom of the pan in contact 1 W ϭ 1 J͞s and the density of Zn is 7.14 g ͞cm 3. with the spoon and then heated the solution to about 2 80 ЊC. After a few minutes, the spoon was removed 19.118 A 9.00 ϫ 10 -mL 0.200 M MgI 2 was electrolyzed. As a result, hydrogen gas was generated at the cath- and rinsed with cold water. The tarnish was gone and ode and iodine was formed at the anode. The volume the spoon regained its original shiny appearance. (a) of hydrogen collected at 26 ЊC and 779 mmHg was Describe with equations the electrochemical basis for 1.22 ϫ 10 3 mL. (a) Calculate the charge in coulombs the procedure. (b) Adding NaCl instead of NaHCO 3 consumed in the process. (b) How long (in min) did would also work because both compounds are the electrolysis last if a current of 7.55 A was used? strong electrolytes. What is the added advantage of (c) A white precipitate was formed in the process. using NaHCO 3? ( Hint: Consider the pH of the solu- What was it and what was its mass in grams? As- tion.) (c) What is the purpose of heating the solu- sume the volume of the solution was constant. tion? (d) Some commercial tarnish removers containing a fluid (or paste) that is a dilute HCl 19.119 To remove the tarnish (Ag S) on a silver spoon, a stu- 2 solution. Rubbing the spoon with the fluid will also dent carried out the following steps. First, she placed remove the tarnish. Name two disadvantages of using the spoon in a large pan filled with water so the spoon this procedure compared to the one described here.

ANSWERS TO PRACTICE EXERCISES

2ϩ ϩ Ϫ ϩ ϩ n 3ϩ ϩ 2ϩ ϩ 19.1 5Fe MnO 4 8H 5Fe Mn 4H 2O. 19.7 0.38 V. 19.8 Anode, O 2; cathode, H 2. Ϫ 19.2 No. 19.3 0.34 V. 19.4 1 ϫ 10 42 . 19.9 2.0 ϫ 10 4 A. 19.5 ⌬GЊ ϭ Ϫ 4.1 ϫ 10 2 kJ ͞mol. 19.6 Yes, E ϭ ϩ 0.01 V.