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Unformatted text preview: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHAPTER 19 Electrochemistry INTRODUCTION ONE 19.1 REDOX REACTIONS FORM OF ENERGY THAT HAS TREMENDOUS PRACTICAL SIGNIFICANCE IS ELECTRICAL ENERGY. A 19.2 ELECTROCHEMICAL CELLS DAY WITHOUT ELECTRICITY FROM EITHER THE POWER COMPANY OR BATTERIES IS UNIMAGINABLE IN OUR TECHNOLOG- 19.3 STANDARD ELECTRODE POTENTIALS ICAL SOCIETY. 19.4 SPONTANEITY OF REDOX REACTIONS THE AREA OF CHEMISTRY THAT DEALS WITH THE INTER- CONVERSION OF ELECTRICAL ENERGY AND CHEMICAL ENERGY IS ELEC- 19.5 THE EFFECT OF CONCENTRATION ON CELL EMF TROCHEMISTRY. 19.6 BATTERIES ELECTROCHEMICAL PROCESSES ARE REDOX REACTIONS IN WHICH THE 19.7 CORROSION ENERGY RELEASED BY A SPONTANEOUS REACTION IS CONVERTED TO ELEC- 19.8 ELECTROLYSIS TRICITY OR IN WHICH ELECTRICITY IS USED TO DRIVE A NONSPONTANEOUS CHEMICAL REACTION. THIS THE LATTER TYPE IS CALLED ELECTROLYSIS. CHAPTER EXPLAINS THE FUNDAMENTAL PRINCIPLES AND AP- PLICATIONS OF ELECTROCHEMICAL CELLS, THE THERMODYNAMICS OF ELECTROCHEMICAL REACTIONS, AND THE CAUSE AND PREVENTION OF CORROSION BY ELECTROCHEMICAL MEANS. SOME SIMPLE ELECTROLYTIC PROCESSES AND THE QUANTITATIVE ASPECTS OF ELECTROLYSIS ARE ALSO DISCUSSED. 757 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 758 ELECTROCHEMISTRY 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 (oxidationreduction) 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 1 Mg(s) 2 0 2HCl(aq) 88n MgCl2(aq) H2(g) Recall that the numbers written 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 above reaction Mg metal is oxidized and H ions are reduced; the Cl ions are spectator ions. BALANCING REDOX EQUATIONS Equations for redox reactions like the one above are relatively easy to balance. However, in the laboratory we often encounter more complex redox reactions involving oxoanions such as chromate (CrO2 ), dichromate (Cr2O2 ), permanganate (MnO4 ), nitrate 4 7 (NO3 ), and sulfate (SO2 ). In principle we can balance any redox equation using the 4 procedure outlined in Section 3.7, but there are some special techniques for handling redox reactions, techniques that also give us insight into electron transfer processes. Here we will discuss one such procedure, called the ion-electron 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 Fe2 ions to Fe3 ions by dichromate ions (Cr2O2 ) in an acidic medium. As a result, the Cr2O2 7 7 ions are reduced to Cr3 ions. The following steps will help us balance the equation. Step 1. Write the unbalanced equation for the reaction in ionic form. Fe2 Cr2O2 88n Fe3 7 Cr3 Step 2. Separate the equation into two half-reactions. 2 3 Fe2 88n Fe3 Oxidation: 6 3 Reduction: Cr2O2 88n Cr3 7 Step 3. Balance the atoms other than O and H in each half-reaction separately. The oxidation half-reaction is already balanced for Fe atoms. For the reduction halfreaction we multiply the Cr3 by 2 to balance the Cr atoms. Cr2O2 88n 2Cr3 7 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.1 REDOX REACTIONS 759 Step 4. For reactions in an acidic medium, add H2O to balance the O atoms and H to balance the H atoms. Since the reaction takes place in an acidic medium, we add seven H2O molecules to the right-hand side of the reduction half-reaction to balance the O atoms: Cr2O2 88n 2Cr3 7 7H2O To balance the H atoms, we add fourteen H ions on the left-hand side: Cr2O2 88n 2Cr3 7 14H 7H2O Step 5. Add electrons to one side of each half-reaction to balance the charges. If necessary, equalize the number of electrons in the two half-reactions by multiplying one or both half-reactions by appropriate coefficients. For the oxidation half-reaction we write Fe2 88n Fe3 e We added an electron to the right-hand side so that there is a 2 charge on each side of the half-reaction. In the reduction half-reaction there are net twelve positive charges on the lefthand side and only six positive charges on the right-hand side. Therefore, we add six electrons on the left. 14H Cr2O2 7 6e 88n 2Cr3 7H2O To equalize the number of electrons in both half-reactions, we multiply the oxidation half-reaction by 6: 6Fe2 88n 6Fe3 6e Step 6. Add the two half-reactions together and balance the final equation by inspection. The electrons on both sides must cancel. The two half-reactions are added to give Cr2O2 7 14H 6Fe2 6e 88n 2Cr3 6Fe3 7H2O 6e The electrons on both sides cancel, and we are left with the balanced net ionic equation: 14H Cr2O2 7 6Fe2 88n 2Cr3 6Fe3 7H2O Step 7. Verify that the equation contains the same types and numbers of atoms and the same charges on both sides of the equation. A final check shows that the resulting equation is “atomically” and “electrically” balanced. For reaction in a basic medium, we would balance the atoms as we did in step 4 for an acidic medium. Then, for every H ion we would add an equal number of OH ions to both sides of the equation. Where H and OH appeared on the same side of the equation, we would combine the ions to give H2O. The following example illustrates the use of this procedure. EXAMPLE 19.1 Write a balanced ionic equation to represent the oxidation of iodide ion (I ) by permanganate ion (MnO4 ) in basic solution to yield molecular iodine (I2) and manganese(IV) oxide (MnO2). Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 760 ELECTROCHEMISTRY Answer Step 1: The unbalanced equation is MnO4 I 88n MnO2 I2 Step 2: The two half-reactions are 1 Oxidation: I 0 88n I2 7 4 Reduction: MnO4 88n MnO2 Step 3: To balance the I atoms in the oxidation half-reaction, we write 2I 88n I2 Step 4: In the reduction half-reaction, we add two H2O molecules on the right to bal- ance the O atoms: MnO4 88n MnO2 2H2O To balance the H atoms, we add four H ions on the left: MnO4 4H 88n MnO2 2H2O Since the reaction occurs in a basic medium and there are four H ions, we add four OH ions to both sides of the equation: MnO4 4H 4OH 88n MnO2 2H2O 4OH Combining the H and OH ions to form H2O and canceling 2H2O from both sides, we write MnO4 2H2O 88n MnO2 4OH Step 5: Next, we balance the charges of the two half-reactions as follows: 2I 88n I2 MnO4 2H2O 2e 3e 88n MnO2 4OH To equalize the number of electrons, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2: 6I 88n 3I2 2MnO4 4H2O 6e 6e 88n 2MnO2 8OH Step 6: The two half-reactions are added together to give 6I 2MnO4 4H2O 6e 88n 3I2 2MnO2 8OH 6e After canceling the electrons on both sides, we obtain 6I 2MnO4 4H2O 88n 3I2 2MnO2 8OH Step 7: 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 ionelectron method: Fe2 Back Forward Main Menu TOC MnO4 88n Fe3 Study Guide TOC Mn2 Textbook Website MHHE Website 19.2 19.2 ELECTROCHEMICAL CELLS 761 ELECTROCHEMICAL CELLS In Section 4.4 we saw that when a piece of zinc metal is placed in a CuSO4 solution, Zn is oxidized to Zn2 ions while Cu2 ions are reduced to metallic copper (see Figure 4.9): Cu2 (aq) 88n Zn2 (aq) Zn(s) Alphabetically anode precedes cathode and oxidation precedes reduction. Therefore, anode is where oxidation occurs and cathode is where reduction takes place. Half-cell reactions are similar to the half-reactions discussed above. Cu(s) The electrons are transferred directly from the reducing agent (Zn) to the oxidizing agent (Cu2 ) 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. 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). An electrochemical cell is the experimental apparatus for generating electricity through the use of a spontaneous redox reaction. (An electrochemical cell is sometimes referred to as a galvanic cell or voltaic cell, after the scientists Luigi Galvani and Allessandro Volta, who constructed early versions of this device.) Figure 19.1 shows the essential components of an electrochemical cell. A zinc bar is immersed in a ZnSO4 solution, and a copper bar is immersed in a CuSO4 solution. The cell operates on the principle that the oxidation of Zn to Zn2 and the reduction of Cu2 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 (ZnSO4 and CuSO4) is called the Daniell cell. By definition, the anode in an electrochemical cell is the electrode at which oxidation occurs and the cathode is the electrode at which reduction occurs. For the Daniell cell, the half-cell reactions, that is, the oxidation and reduction reactions at the electrodes, are Zn(s) 88n Zn2 (aq) Zn electrode (anode): 2 Cu electrode (cathode): Cu (aq) 2e 2e 88n Cu(s) Note that unless the two solutions are separated from each other, the Cu2 ions will react directly with the zinc bar: FIGURE 19.1 An electrochemical 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). Voltmeter e– Zinc anode Cl– Cu2+ Cotton plugs 2 SO4– ZnSO4 solution Forward Main Menu Copper cathode K+ Salt bridge Zn2+ Back e– TOC 2 SO4– CuSO4 solution Study Guide TOC Textbook Website MHHE Website 762 ELECTROCHEMISTRY Cu2 (aq) Zn(s) 88n Cu(s) Zn2 (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 NH4NO3, 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 (Zn2 , Cu2 , and K ) move toward the cathode, while the anions (SO2 and Cl ) move toward the anode. Without the salt 4 bridge connecting the two solutions, the buildup of positive charge in the anode compartment (due to the formation of Zn2 ions) and negative charge in the cathode compartment (created when some of the Cu2 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. Experimentally the difference in electrical potential between the anode and the cathode is measured by a voltmeter (Figure 19.2) and the reading (in volts) is called cell voltage. However, two other terms, electromotive force or emf (E) and cell potential are also used to denote cell voltage. We will see that the voltage of a cell depends not only on the nature of the 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 electrochemical cells is the cell diagram. For the Daniell cell shown in Figure 19.1, if we assume that the concentrations of Zn2 and Cu2 ions are 1 M, the cell diagram is Zn(s)Zn2 (1 M ) Cu2 (1 M )Cu(s) The single vertical line represents a phase boundary. For example, the zinc electrode is a solid and the Zn2 ions (from ZnSO4) are in solution. Thus we draw a line between Zn and Zn2 to show the phase boundary. The double vertical lines denote the FIGURE 19.2 Practical setup of the electrochemical cell described in Figure 19.1. Note the U tube (salt bridge) connecting the two beakers. When the concentrations of ZnSO4 and CuSO4 are 1 molar (1 M) at 25°C, the cell voltage is 1.10 V. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.3 763 STANDARD ELECTRODE POTENTIALS salt bridge. By convention, the anode is written first, to the left of the double lines and the other components appear in the order in which we would encounter them in moving from the anode to the cathode. 19.3 The choice of an arbitrary reference for measuring electrode potential is analogous to choosing the surface of the ocean as the reference for altitude, calling it zero meters, and then referring to any terrestrial altitude as being a certain number of meters above or below sea level. Standard states are defined in Table 18.2. STANDARD ELECTRODE POTENTIALS When the concentrations of the Cu2 and Zn2 ions are both 1.0 M, we find that the voltage or emf of the Daniell cell is 1.10 V at 25°C (see Figure 19.2). This voltage must be related directly to the redox reactions, but how? Just as the overall cell reaction 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 (from 1.10 V). It is impossible to measure the potential of just a single electrode, but if we arbitrarily set the potential value of a particular electrode at zero, we can use it to determine the relative potentials of other electrodes. The hydrogen electrode, shown in Figure 19.3, serves as the reference for this purpose. Hydrogen gas is bubbled into a hydrochloric acid solution at 25°C. The platinum electrode has two functions. First, it provides a surface on which the dissociation of hydrogen molecules can take place: H2 88n 2H 2e Second, it serves as an electrical conductor to the external circuit. Under standard-state conditions (when the pressure of H2 is 1 atm and the concentration of the HCl solution is 1 M ), the potential for the reduction of H at 25°C is taken to be exactly zero: 2H (1 M ) 2e 88n H2(1 atm) E° 0V The superscript “o” denotes standard-state conditions, and E° 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). We can use the SHE to measure the potentials of other kinds of electrodes. For example, Figure 19.4(a) shows an electrochemical 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: FIGURE 19.3 A hydrogen electrode operating under standard-state conditions. Hydrogen gas at 1 atm is bubbled through a 1 M HCl solution. The platinum electrode is part of the hydrogen electrode. H2 gas at 1 atm Pt electrode 1M HCl Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 764 ELECTROCHEMISTRY Voltmeter e– Voltmeter e– e– H2 gas at 1 atm Zn H2 gas at 1 atm Cu Salt bridge Salt bridge Pt electrode 1M ZnSO4 e– Pt electrode 1M HCl 1M CuSO4 1M HCl 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 conZn(s) 88n Zn2 (aq) 2e sisting of a copper electrode and a hydrogen electrode. Both cells The cell diagram is are operating under standardZn(s)Zn2 (1 M ) H (1 M )H2(1 atm)Pt(s) state conditions. Note that in (a) the SHE acts as the cathode, but As mentioned earlier, the Pt electrode provides the surface on which the reduction takes in (b) it is the anode. Zinc electrode place. When all the reactants are in their standard states (that is, H2 at 1 atm, H and Zn2 ions at 1 M ), the emf of the cell is 0.76 V at 25°C. We can write the half-cell reactions as follows: Zn(s) 88n Zn2 (1 M ) Anode (oxidation): Cathode (reduction): 2H (1 M ) Overall: Zn(s) 2e E° Zn/Zn2 2e 88n H2(1 atm) 2H (1 M ) 88n Zn2 (1 M ) EH ° H2(1 atm) /H2 E° cell where EZn/Zn2 is the standard oxidation potential, or voltage under standard state conditions for the oxidation half-reaction, and EH /H2 is the standard reduction potential previously defined for the SHE. The subscript Zn/Zn2 means Zn 88n Zn2 2e , and the subscript H /H2 means 2H 2e 88n H2. The standard emf of the cell, Ecell, is the sum of the standard oxidation potential and the standard reduction potential: E° cell E° ox E° red (19.1) where the subscripts “ox” and “red” indicate oxidation and reduction, respectively. For the Zn-SHE cell, E° cell E° Zn/Zn2 EH ° 0.76 V E° Zn/Zn2 0 /H2 Thus the standard oxidation potential of zinc is 0.76 V. We can evaluate the standard reduction potential of zinc, E ° 2 Zn the oxidation half-reaction: Zn2 (1 M ) 2e 88n Zn(s) /Zn, E° 2 Zn by reversing /Zn 0.76 V Remember that whenever a half-reaction is reversed, E° changes sign. The standard electrode potentials 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: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.3 Cu2 (aq) STANDARD ELECTRODE POTENTIALS 765 2e 88n Cu(s) The cell diagram is Pt(s)H2(1 atm)H (1 M ) Cu2 (1 M )Cu(s) and the half-cell reactions are Anode (oxidation): H2(1 atm) 88n 2H (1 M ) Cu2 (1 M ) Cathode (reduction): 2 Overall: H2(1 atm) 2e E ° 2/H H 2e 88n Cu(s) Cu (1 M ) 88n 2H (1 M ) E Cu2 ° Cu(s) /Cu E° cell Under standard-state conditions and at 25°C, the emf of the cell is 0.34 V, so we write E° cell E H2/H ° E Cu2 ° 0 0.34 V /Cu E Cu2 ° /Cu Thus the standard reduction potential of copper is 0.34 V and its standard oxidation potential, E ° 0.34 V. Cu/Cu2 , is For the Daniell cell shown in Figure 19.1, we can now write Zn(s) 88n Zn2 (1 M ) Anode (oxidation): 2 Cathode (reduction): Cu (1 M ) 2 Overall: Zn(s) 2e E° Zn/Zn2 2e 88n Cu(s) 2 Cu (1 M ) 88n Zn (1 M ) E Cu2 ° Cu(s) /Cu E° cell The emf of the cell is† E cell ° E° Zn/Zn2 E Cu2 ° 0.76 V 0.34 V /Cu 1.10 V The activity series in Figure 4.15 is based on data given in Table 19.1. This example illustrates how we can use the sign of the emf of a cell to predict the spontaneity of a redox reaction. Under standard-state conditions for reactants and products, the redox reaction is spontaneous in the forward direction if the standard emf of the cell is positive. If it is negative, the reaction is spontaneous in the opposite direction. It is important to keep in mind that a negative E ° does not mean that a recell action will not occur if the reactants are mixed at 1 M concentrations. It merely means that the equilibrium of the redox reaction, when reached, will lie to the left. We will examine the relationships among E ° , G°, and K later in this chapter. cell Table 19.1 lists standard reduction potentials for a number of half-cell reactions. By definition, the SHE has an E° value of 0.00 V. Above the SHE the negative standard reduction potentials increase, and below it the positive standard reduction potentials increase. It is important to know the following points about the table: The E° values apply to the half-cell reactions as read in the forward (left to right) direction. • The more positive E° is, the greater the tendency for the substance to be reduced. For example, the half-cell reaction • F2(1 atm) 2e 88n 2F (1 M ) E° 2.87 V has the highest positive E° value among all the half-cell reactions. Thus F2 is the † Some chemists prefer to calculate the standard emf of a cell as follows: E° cell E° cathode E° anode where both E° cathode and E° anode denote the standard reduction potentials, respectively. Thus for the Daniell cell, we would write E° 0.34 V ( 0.76 V) 1.10 V. cell Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ELECTROCHEMISTRY TABLE 19.1 Standard Reduction Potentials at 25°C* Increasing strength as oxidizing agent E°(V) Li (aq) e 88n Li(s) K (aq) e 88n K(s) Ba2 (aq) 2e 88n Ba(s) Sr2 (aq) 2e 88n Sr(s) Ca2 (aq) 2e 88n Ca(s) Na (aq) e 88n Na(s) Mg2 (aq) 2e 88n Mg(s) Be2 (aq) 2e 88n Be(s) Al3 (aq) 3e 88n Al(s) Mn2 (aq) 2e 88n Mn(s) 2H2O 2e 88n H2(g) 2OH (aq) Zn2 (aq) 2e 88n Zn(s) Cr3 (aq) 3e 88n Cr(s) Fe2 (aq) 2e 88n Fe(s) Cd2 (aq) 2e 88n Cd(s) PbSO4(s) 2e 88n Pb(s) SO2 (aq) 4 Co2 (aq) 2e 88n Co(s) Ni2 (aq) 2e 88n Ni(s) Sn2 (aq) 2e 88n Sn(s) Pb2 (aq) 2e 88n Pb(s) 2H (aq) 2e 88n H2(g) Sn4 (aq) 2e 88n Sn2 (aq) Cu2 (aq) e 88n Cu (aq) SO2 (aq) 4H (aq) 2e 88n SO2(g) 2H2O 4 AgCl(s) e 88n Ag(s) Cl (aq) Cu2 (aq) 2e 88n Cu(s) O2(g) 2H2O 4e 88n 4OH (aq) I2(s) 2e 88n 2I (aq) MnO4 (aq) 2H2O 3e 88n MnO2(s) 4OH (aq) O2(g) 2H (aq) 2e 88n H2O2(aq) Fe3 (aq) e 88n Fe2 (aq) Ag (aq) e 88n Ag(s) Hg2 (aq) 2e 88n 2Hg(l) 2 2Hg2 (aq) 2e 88n Hg2 (aq) 2 NO3 (aq) 4H (aq) 3e 88n NO(g) 2H2O Br2(l ) 2e 88n 2Br (aq) O2(g) 4H (aq) 4e 88n 2H2O MnO2(s) 4H (aq) 2e 88n Mn2 (aq) 2H2O Cr2O2 (aq) 14H (aq) 6e 88n 2Cr3 (aq) 7H2O 7 Cl2(g) 2e 88n 2Cl (aq) Au3 (aq) 3e 88n Au(s) MnO4 (aq) 8H (aq) 5e 88n Mn2 (aq) 4H2O Ce4 (aq) e 88n Ce3 (aq) PbO2(s) 4H (aq) SO2 (aq) 2e 88n PbSO4(s) 2H2O 4 H2O2(aq) 2H (aq) 2e 88n 2H2O Co3 (aq) e 88n Co2 (aq) O3(g) 2H (aq) 2e 88n O2(g) H2O(l) F2(g) 2e 88n 2F (aq) 3.05 2.93 2.90 2.89 2.87 2.71 2.37 1.85 1.66 1.18 0.83 0.76 0.74 0.44 0.40 0.31 0.28 0.25 0.14 0.13 0.00 0.13 0.15 0.20 0.22 0.34 0.40 0.53 0.59 0.68 0.77 0.80 0.85 0.92 0.96 1.07 1.23 1.23 1.33 1.36 1.50 1.51 1.61 1.70 1.77 1.82 2.07 2.87 m77777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777 m77777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777 HALF-REACTION Increasing strength as reducing agent 766 *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. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.3 767 STANDARD ELECTRODE POTENTIALS strongest oxidizing agent because it has the greatest tendency to be reduced. At the other extreme is the reaction Li (1 M ) e 88n Li(s) E° 3.05 V which has the most negative E° value. Thus Li is the weakest oxidizing agent because it is the most difficult species to reduce. Conversely, we say that F 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 top to bottom and the reducing agents (the species on the right-hand side of the half-reactions) increase in strength from bottom to top. • 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 cathode (H is reduced to H2) when coupled with zinc in a cell and that it becomes the anode (H2 is oxidized to H ) when used in a cell with copper. • 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 halfcell reaction located above it in Table 19.1. This principle is sometimes called the diagonal rule. In the case of the Cu/Zn electrochemical cell Zn2 (1 M ) 2 Cu (1 M ) 2e 88n Zn(s) E° 2e 88n Cu(s) E° 0.76 V 0.34 V 2 We see that the substance on the left of the second half-cell reaction is Cu and the substance on the right in the first half-cell reaction is Zn. Therefore, as we saw earlier, Zn spontaneously reduces Cu2 to form Zn2 and Cu. • Changing the stoichiometric coefficients of a half-cell reaction does not affect the value of E° because electrode potentials are intensive properties. This means that the value of E° is unaffected by the size of the electrodes or the amount of solutions present. For example, I2(s) 2e 88n 2I (1 M ) E° 0.53 V E° 0.53 V but E° does not change if we multiply the half-reaction by 2: 2I2(s) • 4e 88n 4I (1 M ) Like H, G, and S, E° is a thermodynamic quantity. Therefore, its sign changes but its magnitude remains the same when we reverse a reaction. As the following examples show, Table 19.1 enables us to predict the outcome of redox reactions under standard-state conditions, whether they take place in an electrochemical 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 (Br2) is added to a solution containing NaCl and NaI at 25°C. Assume all species are in their standard states. To predict what redox reaction(s) will take place, we need to compare the standard reduction potentials for the following half-reactions: Answer I2(s) Back Forward Main Menu TOC 2e 88n 2I (1 M ) Study Guide TOC Textbook Website E° 0.53 V MHHE Website 768 ELECTROCHEMISTRY Br2(l ) 2e 88n 2Br (1 M ) E° 1.07 V Cl2(1 atm) 2e 88n 2Cl (1 M ) E° 1.36 V Applying the diagonal rule we see that Br2 will oxidize I but will not oxidize Cl . Therefore, the only redox reaction that will occur appreciably under standard-state conditions is Oxidation: 2I (1 M ) 88n I2(s) Reduction: Br2(l ) 2e 88n 2Br (1 M ) Overall: 2I (1 M ) Comment Similar problems: 19.14, 19.17. 2e Br2(l ) 88n I2(s) 2Br (1 M ) Note that the Na ions are inert and do not enter into the redox reac- tion. PRACTICE EXERCISE Can Sn reduce Zn2 (aq) under standard-state conditions? EXAMPLE 19.3 A galvanic cell consists of a Mg electrode in a 1.0 M Mg(NO3)2 solution and a Ag electrode in a 1.0 M AgNO3 solution. Calculate the standard emf of this electrochemical cell at 25°C. Table 19.1 gives the standard reduction potentials of the two electrodes: Answer Mg2 (1 M ) 2e 88n Mg(s) E° e 88n Ag(s) E° Ag (1 M ) 2.37 V 0.80 V Applying the diagonal rule, we see that Ag will oxidize Mg: Mg(s) 88n Mg2 (1 M ) Anode (oxidation): Cathode (reduction): 2Ag (1 M ) Overall: Mg(s) 2e 2e 88n 2Ag(s) 2Ag (1 M ) 88n Mg2 (1 M ) 2Ag(s) Note that in order to balance the overall equation we multiplied the reduction of Ag by 2. We can do so because, as an intensive property, E ° is not affected by this procedure. We find the emf of the cell by using Equation (19.1) and Table 19.1: E° cell E° Mg/Mg2 2.37 V E Ag ° /Ag 0.80 V 3.17 V Similar problems: 19.11, 19.12. Comment The positive value of E ° shows that the overall reaction is spontaneous. PRACTICE EXERCISE What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution? 19.4 SPONTANEITY OF REDOX REACTIONS Our next step is to see how E ° is related to other thermodynamic quantities such as cell G° and K. In an electrochemical cell, chemical energy is converted to electrical en- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.4 SPONTANEITY OF REDOX REACTIONS 769 ergy. 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: electrical energy volts coulombs joules n is the number of moles of electrons exchanged between the reducing agent and the oxidizing agent in the overall redox equation. The total charge is determined by the number of moles of electrons (n) that pass through the circuit. By definition total charge nF † where F, the Faraday constant, is the electrical charge contained in 1 mole of electrons. Experiments have shown that 1 faraday is equivalent to 96,487 coulombs, or 96,500 coulombs, rounded off to three significant figures. Thus 1F 96,500 C/mol Since 1J 1C 1V we can also express the units of faraday as 1F A common device for measuring a cell’s emf is the potentiometer, which can match the voltage of the cell precisely without actually draining any current from it. 96,500 J/V mol The measured emf is the maximum voltage that the cell can achieve. This value is used to calculate the maximum amount of electrical energy that can be obtained from the chemical reaction. This energy is used to do electrical work (wele), so wmax wele nFEcell The sign convention for electrical work is the same as that for P-V work, discussed in Section 6.7. where wmax is the maximum amount of work that can be done. The negative sign on the right-hand side indicates that the electrical work is done by the system on the surroundings. In Chapter 18 we defined free energy as the energy available to do work. Specifically, the change in free energy ( G) represents the maximum amount of useful work that can be obtained from a reaction: G wmax Therefore we can write G nFEcell (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° nFE ° cell (19.3) Here again, E ° is positive for a spontaneous process. cell Now we can relate E ° to the equilibrium constant (K ) of a redox reaction. In cell Section 18.5 we saw that the standard free-energy change G° for a reaction is related to its equilibrium constant as follows [see Equation (18.10)]: † Michael Faraday (1791 – 1867). English chemist and physicist. Faraday is regarded by many as the greatest experimental scientist of the nineteenth century. He started as an apprentice to a bookbinder at the age of 13, but became interested in science after reading a book on chemistry. Faraday invented the electric motor and was the first person to demonstrate the principle governing electrical generators. Besides making notable contributions to the fields of electricity and magnetism, Faraday also worked on optical activity, and discovered and named benzene. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 770 ELECTROCHEMISTRY G° E° cell RT ln K E° ce ∆G °= _K RT ln __ = nF ∆G° nFE ° cell ll –n F c E°e ll Therefore, if we combine Equations (19.3) and (18.10) we obtain Solving for E ° cell RT ln K nF E° cell When T ∆ G ° = – RT ln K RT ln K (19.4) 298 K, Equation (19.4) can be simplified by substituting for R and F: K (8.314 J/K mol)(298 K) ln K n(96500 J/V mol) E° cell FIGURE 19.5 Relationships among E° , K, and ∆G°. cell 0.0257 V ln K n (19.5) Thus, if any one of the three quantities G°, K, or E ° is known, the other two can cell be calculated using Equation (18.10), Equation (19.3), or Equation (19.4) (Figure 19.5). We summarize the relationships among G°, K, and E ° and characterize the sponcell taneity of a redox reaction in Table 19.2. For simplicity, we will omit the subscript “cell” in E° and E. Examples 19.4 and 19.5 apply Equation (19.5). EXAMPLE 19.4 Calculate the equilibrium constant for the following reaction at 25°C: Sn(s) Answer 2Cu2 (aq) 34 Sn2 (aq) 2Cu (aq) The two half-reactions for the overall process are Sn(s) 88n Sn2 (aq) Oxidation: 2 Reduction: 2Cu (aq) In Table 19.1 we find that E ° 2 Sn E° 2e 88n 2Cu (aq) 0.14 V and E ° 2 Cu /Sn 2e E° 2 Sn/Sn E° 2 Cu 0.14 V /Cu 0.15 V. Thus 0.15 V /Cu 0.29 V Equation (19.5) can be written ln K In the overall reaction we find n 2. Therefore ln K K Similar problems: 19.21, 19.22. nE° 0.0257 V (2)(0.29 V) 0.0257 V e22.6 6.5 22.6 109 PRACTICE EXERCISE Calculate the equilibrium constant for the following reaction at 25°C: Fe2 (aq) Back Forward Main Menu TOC 2Ag(s) 34 Fe(s) Study Guide TOC 2Ag (aq) Textbook Website MHHE Website 19.5 TABLE 19.2 ∆G° Negative 0 Positive THE EFFECT OF CONCENTRATION ON CELL EMF 771 Relationships Among ∆G°, K, and E° cell K E° cell REACTION UNDER STANDARD-STATE CONDITIONS Positive 0 Negative 1 1 1 Spontaneous At equilibrium Nonspontaneous. Reaction is spontaneous in the reverse direction. EXAMPLE 19.5 Calculate the standard free-energy change for the following reaction at 25°C: 2Au(s) Answer 3Ca2 (1 M ) 88n 2Au3 (1 M ) 3Ca(s) First we break up the overall reaction into half-reactions: 2Au(s) 88n 2Au3 (1 M ) Oxidation: 2 Reduction: 3Ca (1 M ) In Table 19.1 we find that E° u3 A 6e 88n 3Ca(s) 1.50 V and E ° a2 C /Au E° u/Au3 A E° 6e 1.50 V E ° a2 C /Ca 2.87 V. Therefore /Ca 2.87 V 4.37 V Now we use Equation (19.3): ∆G° The overall reaction shows that n ∆G° 6, so (6 mol)(96,500 J/V mol)( 4.37 V) 2.53 106 J 2.53 Similar problem: 19.24. nFE ° 103 kJ Comment The large positive value of ∆G° tells us that the reaction is not spontaneous under standard-state conditions at 25°C. PRACTICE EXERCISE Calculate ∆G° for the following reaction at 25°C: 2Al3 (aq) 19.5 3Mg(s) 12 2Al(s) 3Mg2 (aq) 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 cell and the concentration of reactants and products in a redox reaction under non-standard-state conditions. This equation is derived below. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 772 ELECTROCHEMISTRY THE NERNST EQUATION Consider a redox reaction of the type aA bB 88n cC dD From Equation (18.9) ∆G Since ∆G nFE and ∆G° ∆G° RT ln Q nFE°, the equation can be expressed as nFE Dividing the equation through by nFE° RT ln Q nF, we get E RT ln Q nF E° (19.6) where Q is the reaction quotient (see Section 14.4). Equation (19.6) is known as the Nernst† equation. At 298 K, Equation (19.6) can be rewritten as E 0.0257 V ln Q n E° (19.7) At equilibrium, there is no net transfer of electrons, so E 0 and Q K, where K is the equilibrium constant. 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) Cu2 (aq) 88n Zn2 (aq) Cu(s) The Nernst equation for this cell at 25°C can be written as Remember that concentrations of pure solids (and pure liquids) do not appear in the expression for Q. E [Zn2 ] 0.0257 V ln [Cu2 ] 2 1.10 V If the ratio [Zn2 ]/[Cu2 ] is less than 1, ln [Zn2 ]/[Cu2 ] is a negative number, so that the second term on the right-hand side of the above equation is positive. Under this condition E is greater than the standard emf E°. If the ratio is greater than 1, E is smaller than E°. The following example illustrates the use of the Nernst equation. EXAMPLE 19.6 Predict whether the following reaction would proceed spontaneously as written at 298 K: Co(s) 2 given that [Co ] Fe2 (aq) 88n Co2 (aq) 2 0.15 M and [Fe ] Fe(s) 0.68 M. † Walter Hermann Nernst (1864 – 1941). German chemist and physicist. Nernst’s work was mainly on electrolyte solution and thermodynamics. He also invented an electric piano. Nernst was awarded the Nobel Prize in Chemistry in 1920 for his contribution to thermodynamics. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.5 Answer THE EFFECT OF CONCENTRATION ON CELL EMF 773 The half-reactions are Co(s) 88n Co2 (aq) Oxidation: 2 Reduction: Fe (aq) In Table 19.1 we find that E ° 2 Co the standard emf is 2e 88n Fe(s) 0.28 V and E ° 2 Fe /Co E° 2e E° Co/Co2 E° 2 Fe 0.28 V /Fe 0.44 V. Therefore, /Fe ( 0.44 V) 0.16 V From Equation (19.7) we write E [Co2 ] 0.0257 V ln n E° 0.16 V 0.0257 V 0.15 ln 2 0.68 0.16 V 0.019 V 0.14 V Similar problems: 19.29, 19.30. Since E is negative (or ∆G is positive), the reaction is not spontaneous in the direction written. PRACTICE EXERCISE Will the following reaction occur spontaneously at 25°C, given that [Fe2 ] and [Cd2 ] 0.010 M? Cd(s) Fe2 (aq) 88n Cd2 (aq) 0.60 M Fe(s) Now suppose we want to determine at what ratio of [Co2 ] to [Fe2 ] the reaction in Example 19.6 would become spontaneous. We can use Equation (19.7) as follows: E When E 0, Q K. 0.0257 V ln Q n E° We set E equal to zero, which corresponds to the equilibrium situation. 0 [Co2 ] 0.0257 V ln [Fe2 ] 2 0.16 V ln [Co2 ] 12.5 [Co2 ] [Fe2 ] e K 4 or 12.5 10 K 6 Thus for the reaction to be spontaneous, the ratio [Co2 ]/[Fe2 ] must be smaller than 4 10 6. As the following example shows, if gases are involved in the cell reaction, their concentrations should be expressed in atm. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 774 ELECTROCHEMISTRY EXAMPLE 19.7 Consider the electrochemical cell shown in Figure 19.4(a). In a certain experiment, the emf (E) of the cell is found to be 0.54 V at 25°C. Suppose that [Zn2 ] 1.0 M and PH2 1.0 atm. Calculate the molar concentration of H . Answer The overall cell reaction is Zn(s) 2H (? M ) 88n Zn2 (1 M ) H2(1 atm) As we saw earlier (p. 764), the standard emf for the cell is 0.76 V. From Equation (19.7), we write E 0.54 V [Zn2 ]PH2 0.0257 V ln n (1.0)(1.0) 0.0257 V 0.76 V ln [H ]2 2 E° 0.0257 V ln 2 0.22 V 17.1 2 ln [H ] 17.1 [H ] 1 ln ln [H ] Similar problems: 19.32. 1 8.6 2 10 4 M PRACTICE EXERCISE What is the emf of a cell consisting of a Cd/Cd2 half-cell and a Pt/H2/H halfcell if [Cd2 ] 0.20 M, [H ] 0.16 M, and PH2 0.80 atm? Example 19.7 shows that an electrochemical cell whose cell reaction involves H ions can be used to measure [H ] or pH. The pH meter described in Section 15.3 is based on this principle, but for practical reasons the electrodes used in a pH meter are quite different from the SHE and zinc electrode in the electrochemical cell (Figure 19.6). CONCENTRATION CELLS FIGURE 19.6 A glass electrode that is used in conjunction with a reference electrode in a pH meter. Because electrode potential depends on ion concentrations, it is possible to construct a cell from two half-cells composed of the same material but differing in ion concentrations. Such a cell is called a concentration cell. Consider a situation in which zinc electrodes are put into two aqueous solutions 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 Chatelier ’s principle, the tendency for the reduction Zn2 (aq) 2e 88n Zn(s) increases with increasing concentration of Zn2 ions. Therefore, reduction should oc- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.6 BATTERIES 775 cur in the more concentrated compartment and oxidation should take place on the more dilute side. The cell diagram is Zn(s)Zn2 (0.10 M ) Zn2 (1.0 M )Zn(s) and the half-reactions are Zn(s) 88n Zn2 (0.10 M ) Oxidation: 2 Reduction: Zn (1.0 M ) 2e 2e 88n Zn(s) Zn2 (1.0 M ) 88n Zn2 (0.10 M ) Overall: The emf of the cell is E [Zn2 ]dil 0.0257 V ln [Zn2 ] 2 conc E° where the subscripts “dil” and “conc” refer to the 0.10 M and 1.0 M concentrations, respectively. E° for this cell is zero (the same electrode and the same type of ions are involved), so E 0.10 0.0257 V ln 2 1.0 0 0.0296 V 1 mM 1 10 3 M. 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 heart beat. 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 K ions in the interior and exterior of a nerve cell are 400 mM and 15 mM, respectively. Treating the situation as a concentration cell and applying the Nernst equation, we can write E E° [K ]ex 0.0257 V ln 1 (0.0257 V) ln 15 400 0.084 V or 84 mV where “ex” and “in” denote exterior and interior. Note that we have set E° 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 K ions. 19.6 BATTERIES A battery is an electrochemical cell, or a series of combined electrochemical cells, that can be used as a source of direct electric current at a constant voltage. Although the Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 776 ELECTROCHEMISTRY + Paper spacer Moist paste of ZnCl2 and NH4Cl Layer of MnO2 Graphite cathode Zinc anode FIGURE 19.7 Interior section of a dry cell of the kind used in flashlights and transistor radios. Actually, the cell is not completely dry, as it contains a moist electrolyte paste. operation of a battery is similar in principle to that of the electrochemical cells described in Section 19.2, a battery has the advantage of being completely self-contained and requiring no auxiliary 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 cell used in flashlights and transistor radios. The anode of the cell consists of a zinc can or container that is in contact with manganese dioxide (MnO2) and an electrolyte. The electrolyte consists of ammonium chloride and zinc chloride in water, to which starch is added to thicken the solution to a pastelike consistency so that it is less likely to leak (Figure 19.7). A carbon rod serves as the cathode, which is immersed in the electrolyte in the center of the cell. The cell reactions are Zn(s) 88n Zn2 (aq) Anode: Cathode: 2NH4 (aq) 2MnO2(s) 2e 88n Mn2O3(s) 2NH3(aq) H2O(l ) 2MnO2(s) 88n Zn2 (aq) 2NH4 (aq) Overall: Zn(s) 2e 2NH3(aq) H2O(l ) Mn2O3(s) Actually, this equation is an oversimplification of a complex process. The voltage produced 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, the mercury battery consists of a zinc anode (amalgamated with mercury) in contact with a strongly alkaline electrolyte containing zinc oxide and mercury(II) oxide (Figure 19.8). The cell reactions are Anode: Zn(Hg) Cathode: HgO(s) Overall: 2OH (aq) 88n ZnO(s) H2O(l ) Zn(Hg) 2e 88n Hg(l ) HgO(s) 88n ZnO(s) H2O(l ) 2e 2OH (aq) 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. Cathode (steel) Insulation Anode (Zn can) 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 (PbO2) 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 Electrolyte solution containing KOH and paste of Zn(OH)2 and HgO FIGURE 19.8 Interior section of a mercury battery. Back Forward Main Menu Anode: Pb(s) Cathode: SO2 4 PbO2(s) Overall: Pb(s) TOC 4H (aq) PbO2(s) 4H (aq) Study Guide TOC SO2 (aq) 88n PbSO4(s) 4 (aq) 2e 88n PbSO4(s) 2SO2 (aq) 88n 2PbSO4(s) 4 2e 2H2O(l ) 2H2O(l ) Textbook Website MHHE Website 19.6 FIGURE 19.9 Interior section of a lead storage battery. Under normal operating conditions, the concentration of the sulfuric acid solution is about 38 percent by mass. BATTERIES 777 Removable cap Anode Cathode – + H2SO 4 electrolyte Negative plates (lead grills filled with spongy lead) Positive plates (lead grills filled with PbO 2 ) 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. 783.) The reactions that replenish the original materials are Anode: PbSO4(s) Cathode: PbSO4(s) Overall: 2PbSO4(s) 2e 88n Pb(s) SO2 (aq) 4 2H2O(l ) 88n PbO2(s) 2H2O(l ) 88n Pb(s) 4H (aq) PbO2(s) SO2 (aq) 4 4H (aq) 2e 2SO2 (aq) 4 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. Second, 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 electrochemical cells decreases with decreasing temperature. However, for a lead storage battery, the temperature coefficient is about 1.5 10 4 V/°C; that is, there is a decrease in voltage of 1.5 10 4 V for every degree drop in temperature. Thus, even allowing for a 40°C change in temperature, the decrease in voltage amounts to only 6 10 3 V, which is about 6 10 3 V 12 V 100% 0.05% of the operating voltage, an insignificant change. The real cause of a battery’s appar- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 778 ELECTROCHEMISTRY FIGURE 19.10 A schematic diagram of a solid-state lithium battery. Lithium metal is the anode, and TiS2 is the cathode. During operation, Li ions migrate through the solid polymer electrolyte from the anode to the cathode while electrons flow externally from the anode to the cathode to complete the circuit. e– e– Anode Cathode TiS 2 Li Li + Solid electrolyte Li Li + + e – TiS 2 + e – – TiS 2 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 resistance 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. SOLID-STATE LITHIUM BATTERY Unlike the batteries discussed so far, a solid-state battery employs a solid (rather than an aqueous solution or a water-based paste) as the electrolyte connecting the electrodes. Figure 19.10 shows a schematic solid-state lithium battery. The advantage in choosing lithium as the anode is that it has the most negative E ° value, as Table 19.1 shows. Furthermore, lithium is a lightweight metal, so that only 6.941 g of Li (its molar mass) are needed to produce 1 mole of electrons. The electrolyte is a polymer material that permits the passage of ions but not of electrons. The cathode is made of either TiS2 or V6O13. The cell voltage of a solid-state lithium battery can be as high as 3 V, and it can be recharged in the same way as a lead storage battery. Although these batteries do not yet have high reliability and long lifetime, they have been hailed as the batteries of the future. 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: CH4(g) 2O2(g) 88n CO2(g) 2H2O(l ) 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 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.6 FIGURE 19.11 A hydrogenoxygen fuel cell. The Ni and NiO embedded in the porous carbon electrodes are electrocatalysts. e– BATTERIES 779 e– Anode Cathode H2 O2 Porous carbon electrode containing Ni Porous carbon electrode containing Ni and NiO Hot KOH solution Oxidation 2H2 ( g) + 4OH– (aq) 4H2 O(l ) + 4e – Reduction O2 ( g) + 2H2O(l ) + 4e – 4OH– (aq) 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 fuel cell, an electrochemical 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: Anode: 2H2(g) Cathode: O2(g) Overall: 4OH (aq) 88n 4H2O(l ) 2H2O(l ) 2H2(g) 4e 4e 88n 4OH (aq) O2(g) 88n 2H2O(l ) The standard emf of the cell can be calculated as follows, with data from Table 19.1: E° E° ox 0.83 V E° red 0.40 V 1.23 V 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 are carried out separately at the anode and the cathode. Like platinum in the standard hydrogen electrode, the electrodes have a twofold function. They serve as electrical conductors, and they provide the necessary surfaces for the initial decomposition of the molecules into atomic species, prior to electron transfer. They are electrocatalysts. Metals such as platinum, nickel, and rhodium are good electrocatalysts. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 780 ELECTROCHEMISTRY FIGURE 19.12 A hydrogenoxygen fuel cell used in the space program. The pure water produced by the cell is consumed by the astronauts. In addition to the H2-O2 system, a number of other fuel cells have been developed. Among these is the propane-oxygen fuel cell. The half-cell reactions are Anode: C3H8(g) Cathode: 5O2(g) Overall: 20H (aq) C3H8(g) 6H2O(l ) 88n 3CO2(g) 20H (aq) 20e 20e 88n 10H2O(l ) 5O2(g) 88n 3CO2(g) 4H2O(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. However, the fuel cell does not operate like a heat engine and therefore is not subject to the same kind of thermodynamic limitations in energy conversion (see the Chemistry in Action essay on p. 736). 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 Corrosion is the term usually applied to the deterioration of metals by an electrochemical process. We see many examples of corrosion around us. Rust on iron, tarnish on silver, and the green patina formed on copper and brass are a few of them (Figure 19.13). Corrosion causes enormous damage to buildings, bridges, ships, and cars. The cost of metallic corrosion to the U.S. economy has been estimated to be over 100 bil- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.7 CORROSION 781 FIGURE 19.13 Examples of corrosion: (a) a rusted ship, (b) a half-tarnished silver dish, and (c) the Statue of Liberty coated with patina before its restoration in 1986. (a) (b) (c) lion 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: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 782 ELECTROCHEMISTRY Fe(s) 88n Fe2 (aq) 2e The electrons given up by iron reduce atmospheric oxygen to water at the cathode, which is another region of the same metal’s surface: O2(g) 4H (aq) 4e 88n 2H2O(l ) The overall redox reaction is 2Fe(s) O2(g) 4H (aq) 88n 2Fe2 (aq) 2H2O(l ) With data from Table 19.1, we find the standard emf for this process: E° The positive standard emf means that the process will occur spontaneously. Eox ° E° red 0.44 V 1.23 V 1.67 V Note that this reaction occurs in an acidic medium; the H ions are supplied in part by the reaction of atmospheric carbon dioxide with water to form H2CO3. The Fe2 ions formed at the anode are further oxidized by oxygen: 4Fe2 (aq) O2(g) (4 2x)H2O(l ) 88n 2Fe2O3 xH2O(s) 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 Fe2O3 xH2O. Figure 19.14 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 CaCl2) spread on roadways to melt ice and snow are a major cause of rust formation on automobiles. 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 (Al2O3) 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. FIGURE 19.14 The electrochemical process involved in rust formation. The H ions are supplied by H2CO3, which forms when CO2 dissolves in water. Air Water O2 Rust Fe2+ Fe3+ Iron e– Fe(s) Fe2+(aq) Back Forward Main Menu Anode Fe2+(aq) + 2e – Cathode O2(g) + 4H +(aq) + 4e – 2H2O(l) Fe3+(aq) + e – TOC Study Guide TOC Textbook Website MHHE Website 19.7 CORROSION 783 Coinage metals such as copper and silver also corrode, but much more slowly. Cu(s) 88n Cu2 (aq) Ag(s) 88n Ag (aq) 2e e In normal atmospheric exposure, copper forms a layer of copper carbonate (CuCO3), 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 (Ag2S). 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, we find that iron acts as the anode and tin as the cathode in the corrosion process: Fe(s) 88n Fe2 2 Sn (aq) 2e 2e 88n Sn(s) E° E° 0.44 V 0.14 V The protective process is different for zinc-plated, or galvanized, iron. Zinc is more easily oxidized than iron (see Table 19.1): Zn(s) 88n Zn2 (aq) 2e E° 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 an electrochemical cell. Figure 19.15 FIGURE 19.15 An iron nail that is cathodically protected by a piece of zinc strip does not rust in water, while an iron nail without such protection rusts readily. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 784 ELECTROCHEMISTRY FIGURE 19.16 Cathodic protection of an iron storage tank (cathode) by magnesium, a more electropositive metal (anode). Since only the magnesium is depleted in the electrochemical process, it is sometimes called the sacrificial anode. e– Mg Iron storage tank Mg 2+(aq) + 2e– Oxidation: Mg(s) Reduction: O2( g) + 4H +(aq) + 4e– 2H2 O(l ) shows how an iron nail can be protected from rusting by connecting the nail to a piece of zinc. Without such protection, an iron nail quickly rusts in water. Rusting of underground iron pipes and iron storage tanks can be prevented or greatly reduced by connecting them to metals such as zinc and magnesium, which oxidize more readily than iron (Figure 19.16). The Chemistry in Action essay on p. 785 shows that dental filling discomfort can result from an electrochemical phenomenon. 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 electrochemical 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 19.17(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 and Cl ions, respectively. Figure 19.17(b) is a simplified 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 Anode (oxidation): Cathode (reduction): 2Cl (l ) 88n Cl2(g) 2Na (l ) Overall: 2Na (l) 2e 2e 88n 2Na(l ) 2Cl (l ) 88n 2Na(l ) Cl2(g) This process is a major source of pure sodium metal and chlorine gas. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 19.8 785 ELECTROLYSIS Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Dental Filling Discomfort In modern dentistry the material most commonly used to fill decaying teeth is known as dental amalgam. (An amalgam is a substance made by combining mercury with another metal or metals.) Dental amalgam actually consists of three solid phases having stoichiometries approximately corresponding to Ag2Hg3, Ag3Sn, and Sn8Hg. The standard reduction potentials for these solid phases are: Hg2 /Ag2Hg3, 0.85 V; 2 Sn2 /Ag3Sn, 0.05 V; Sn2 /Sn8Hg, 0.13 V. Anyone who bites a piece of aluminum foil (such as that used for wrapping candies) in such a way that the foil presses against a dental filling will probably experience a momentary sharp pain. In effect, an electrochemical cell has been created in the mouth, with aluminum (E° 1.66 V) as the anode, the filling as the cathode, and saliva as the electrolyte. Contact between the aluminum foil and the filling short-circuits the cell, causing a weak current to flow between the electrodes. This current stimulates the sensitive nerve of the tooth, causing an unpleasant sensation. Another type of discomfort results when a less electropositive metal touches a dental filling. For example, if a filling makes contact with a gold inlay in a nearby tooth, corrosion of the filling will occur. In this case, the dental filling acts as the anode and the gold inlay as the cathode. Referring to the E° values FIGURE 19.17 (a) A practical arrangement called a Downs cell for the electrolysis of molten NaCl (m.p. 801°C). The sodium metal formed at the cathodes is in the liquid state. Since liquid sodium metal is lighter than molten NaCl, the sodium floats to the surface, as shown, and is collected. Chlorine gas forms at the anode and is collected at the top. (b) A simplified diagram showing the electrode reactions during the electrolysis of molten NaCl. The battery is needed to drive the nonspontaneous reactions. Back Forward Main Menu Gold inlay O2( g) + 4H+(aq) + 4e– 2H2O(l ) e– Sn8Hg Sn2+ Dental filling Corrosion of a dental filling brought about by contact with a gold inlay. for the three phases, we see that the Sn8Hg phase is most likely to corrode. When that happens, release of Sn(II) ions in the mouth produces an unpleasant metallic taste. Prolonged corrosion will eventually result in another visit to the dentist for a replacement filling. Cl2 gas NaCl Battery e– Anode Liquid Na e– Cathode Liquid Na Molten NaCl Iron cathode Iron cathode Carbon anode (a) TOC Study Guide TOC 2Cl– Oxidation Cl2( g) + 2e – Reduction 2Na+ + 2e – 2Na(l) (b) Textbook Website MHHE Website 786 ELECTROCHEMISTRY Theoretical estimates show that the E° value for the overall process is about 4 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°C) will not spontaneously decompose to form hydrogen and oxygen gas because the standard freeenergy change for the reaction is a large positive quantity: 2H2O(l ) 88n 2H2(g) FIGURE 19.18 Apparatus for small-scale electrolysis of water. The volume of hydrogen gas generated (left column) is twice that of oxygen gas (right column). ∆G° O2(g) 474.4 kJ However, this reaction can be induced in a cell like the one shown in Figure 19.18. This electrolytic cell consists of a pair of electrodes made of a nonreactive metal, such as platinum, immersed in water. When the electrodes are connected to the battery, nothing happens because there are not enough ions in pure water to carry much of an electric current. (Remember that at 25°C, pure water has only 1 10 7 M H ions and 1 10 7 M OH ions.) On the other hand, the reaction occurs readily in a 0.1 M H2SO4 solution because there are a sufficient number of ions to conduct electricity (Figure 19.19). Immediately, gas bubbles begin to appear at both electrodes. The process at the anode is 2H2O(l ) 88n O2(g) 4H (aq) 4e while at the cathode we have e 88n 1 H2(g) 2 H (aq) The overall reaction is given by Anode (oxidation): 2H2O(l ) 88n O2(g) Cathode (reduction): 4[H (aq) Overall: 4H (aq) 4e 2H2O(l ) 88n 2H2(g) O2(g) E° 1.23 V E° e 88n 1 H2(g)] 2 0.00 V E° 1.23 V Note that no net H2SO4 is consumed. FIGURE 19.19 A diagram showing the electrode reactions during the electrolysis of water. Battery e– Anode e– Cathode Dilute H2SO4 solution Oxidation 2H2O(l) Back Forward Main Menu O2( g) + 4H+(aq) + 4e – TOC Reduction 4H+(aq) + 4e – 2H2( g) Study Guide TOC Textbook Website MHHE Website 19.8 ELECTROLYSIS 787 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 (1) 2H2O(l ) 88n O2(g) (2) 2Cl (aq) 88n Cl2(g) 4H (aq) 4e 2e Referring to Table 19.1, we find O2(g) 4H (aq) Cl2(g) Because Cl2 is more easily reduced than O2, it follows that it would be more difficult to oxidize Cl than H2O at the anode. 4e 88n 2H2O(l ) E° 1.23 V 2e 88n 2Cl (aq) E° 1.36 V The standard reduction potentials of (1) and (2) are not very different, but the values do suggest that H2O should be preferentially oxidized at the anode. (O2 has a smaller tendency to be reduced than Cl2; therefore, it has a greater tendency to be oxidized.) However, by experiment we find that the gas liberated at the anode is Cl2, not O2! 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 O2 formation is quite high. Therefore, under normal operating conditions Cl2 gas is actually formed at the anode instead of O2. The reductions that might occur at the cathode are (3) Na (aq) e 88n Na(s) (4) 2H2O(l) 2e 88n H2(g) (5) 2H (aq) E° 0.83 V E° 2e 88n H2(g) 2.71 V E° 2OH (aq) 0.00 V Reaction (3) is ruled out because it has a very negative standard reduction potential. Reaction (5) 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 ions is too low (about 1 10 7 M ) to make (5) a reasonable choice. Thus, the half-cell reactions in the electrolysis of aqueous sodium chloride are Anode (oxidation): Cathode (reduction): 2Cl (aq) 88n Cl2(g) 2H2O(l ) Overall: 2H2O(l ) 2e 88n H2(g) 2Cl (aq) 88n H2(g) 2e 2OH (aq) Cl2(g) 2OH (aq) As the overall reaction shows, the concentration of the Cl ions decreases during electrolysis and that of the OH ions increases. Therefore, in addition to H2 and Cl2, 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. The following example deals with the electrolysis of an aqueous solution of sodium sulfate (Na2SO4). Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 788 ELECTROCHEMISTRY EXAMPLE 19.8 An aqueous Na2SO4 solution is electrolyzed, using the apparatus shown in Figure 19.19. 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. Before we look at the electrode reactions, we should consider the following facts: (1) Since Na2SO4 does not hydrolyze in water, the pH of the solution is close to 7. (2) The Na ions are not reduced at the cathode, and the SO2 ions 4 are not oxidized at the anode. These conclusions are drawn from the electrolysis of water in the presence of sulfuric acid and in aqueous sodium chloride solution. Therefore, the electrode reactions are Answer The SO2 ion is the conjugate 4 base of the weak acid HSO4 (Ka 1.3 10 2). However, the extent to which SO2 4 hydrolyzes is negligible. Anode: 2H2O(l) 88n O2(g) Cathode: 2H2O(l ) 4H (aq) 2e 88n H2(g) 4e 2OH The overall reaction, obtained by doubling the cathode reaction coefficients and adding the result to the anode reaction, is 6H2O(l ) 88n 2H2(g) O2(g) 4H (aq) 4OH (aq) If the H and OH ions are allowed to mix, then 4H (aq) 4OH (aq) 88n 4H2O(l ) and the overall reaction becomes 2H2O(l ) 88n 2H2(g) O2(g) PRACTICE EXERCISE An aqueous solution of Mg(NO3)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 Chapter 20. 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 ion accepts an electron from the electrode. To reduce 1 mole of Na ions, we must supply Avogadro’s number (6.02 1023) of electrons to the cathode. On the other hand, the stoichiometry of the anode reaction shows that oxidation of two Cl ions yields one chlorine molecule. Therefore, the formation of 1 mole of Cl2 results in the transfer of 2 moles of electrons from the Cl ions to the anode. Similarly, it takes 2 moles of electrons to reduce 1 mole of Mg2 ions and 3 moles of electrons to reduce 1 mole of Al3 ions: Mg2 Back Forward Main Menu TOC 2e 88n Mg Al3 3e 88n Al Study Guide TOC Textbook Website MHHE Website 19.8 Current (amperes) and time Charge in coulombs Number of faradays FIGURE 19.20 electrolysis. ELECTROLYSIS Moles of substance reduced or oxidized 789 Grams of substance reduced or oxidized Steps involved in calculating amounts of substances reduced or oxidized in Therefore 2F 1 mol Mg2 3F 1 mol Al3 where F is the faraday. 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 1C 1A 1s that is, a coulomb is the quantity of electrical charge passing any point in the circuit in 1 second when the current is 1 ampere. Figure 19.20 shows the steps involved in calculating the quantities of substances produced in electrolysis. Let us illustrate the approach by considering molten CaCl2 in an electrolytic cell. Suppose a current of 0.452 amperes is passed through the cell for 1.50 hours. 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 oxidized at the anode and which species will be reduced at the cathode. Here the choice is straightforward since we only have Ca2 and Cl ions in molten CaCl2. Thus we write the half- and overall reactions as Anode (oxidation): Cathode (reduction): 2Cl (l ) 88n Cl2(g) Ca2 (l ) Overall: Ca2 (l ) 2e 2e 88n Ca(l ) 2Cl (l ) 88n Ca(l ) Cl2(g) 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 time, or charge: ?C 0.452 A 1.50 h 3600 s 1h 1C 1A s 2.44 103 C Since 1 F 96,500 C and 2 F are required to reduce 1 mole of Ca2 ions, the mass of Ca metal formed at the cathode is calculated as follows: ? g Ca 2.44 103 C 1F 96,500 C 1 mol Ca 2F 40.08 g Ca 1 mol Ca 0.507 g Ca The anode reaction indicates that 1 mole of chlorine is produced per 2 F of electricity. Hence the mass of chlorine gas formed is ? g Cl2 2.44 103 C 1F 96,500 C 1 mol Cl2 2F 70.90 g Cl2 1 mol Cl2 0.896 g Cl2 The following example applies this approach to the electrolysis in an aqueous solution. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 790 ELECTROCHEMISTRY EXAMPLE 19.9 A current of 1.26 A is passed through an electrolytic cell containing a dilute sulfuric acid solution for 7.44 hours. Write the half-cell reactions and calculate the volume of gases generated at STP. Earlier in this chapter (see p. 786) we saw that the half-cell reactions for this process are Answer Anode (oxidation): 2H2O(l ) 88n O2(g) Cathode (reduction): 4[H (aq) 4H (aq) 2H2O(l ) 88n 2H2(g) Overall: 4e e 88n 1 H2(g)] 2 O2(g) First we calculate the number of coulombs of electricity that pass through the cell: ?C 1.26 A 7.44 h 3600 s 1h 1C 1A s 3.37 104 C We see that for every mole of O2 formed at the anode, 4 moles of electrons are generated so that ? g O2 3.37 104 C 1F 96,500 C 1 mol O2 4F 32.00 g O2 1 mol O2 2.79 g O2 The volume of 2.79 g O2 at STP is given by V nRT P (2.79 g/32.00 g mol 1)(0.0821 L atm/K mol)(273 K) 1 atm 1.95 L Similarly, for hydrogen we write ? g H2 3.37 104 C 1F 96,500 C 1 mol H2 2F 2.016 g H2 1 mol H2 0.352 g H2 The volume of 0.352 g H2 at STP is given by V nRT P (0.352 g/2.016 g mol 1)(0.0821 L atm/K mol)(273 K) 1 atm 3.91 L Note that the volume of H2 is twice that of O2, 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). Comment Similar problem: 19.46. PRACTICE EXERCISE A constant current is passed through an electrolytic cell containing molten MgCl2 for 18 hours. If 4.8 105 g of Cl2 are obtained, what is the current in amperes? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website KEY WORDS SUMMARY OF KEY EQUATIONS • E° cell E° ox E° red (19.1) 791 Calculating the standard emf of an electrochemical cell. Relating free energy change to the emf of the cell. Relating the standard free energy change to the standard emf. • ∆G • ∆G° nFEcell nFE° cell (19.2) (19.3) • E° cell RT ln K nF (19.4) Relating the standard emf of the cell to the equilibrium constant. (19.6) The Nernst equation. For calculating the emf of a cell under non-standard-state conditions. •E SUMMARY OF FACTS AND CONCEPTS E° RT ln Q nF 1. Redox reactions involve the transfer of electrons. Equations representing redox processes can be balanced using the ion-electron method. 2. All electrochemical reactions involve the transfer of electrons and therefore are redox reactions. 3. In an electrochemical cell, electricity is produced by a spontaneous chemical reaction. Oxidation and reduction take place separately at the anode and cathode, respectively, and the electrons flow through an external circuit. 4. The two parts of an electrochemical cell are the half-cells, and the reactions at the electrodes are the half-cell reactions. A salt bridge allows ions to flow between the half-cells. 5. The electromotive force (emf) of a cell is the voltage difference between the two electrodes. In the external circuit, electrons flow from the anode to the cathode in an electrochemical cell. In solution, the anions move toward the anode and the cations move toward the cathode. 6. The quantity of electricity carried by 1 mole of electrons is called a faraday, which is equal to 96,500 coulombs. 7. Standard reduction potentials show the relative likelihood of half-cell reduction reactions and can be used to predict the products, direction, and spontaneity of redox reactions between various substances. 8. The decrease in free energy of the system in a spontaneous redox reaction is equal to the electrical work done by the system on the surroundings, or ∆G nFE. 9. The equilibrium constant for a redox reaction can be found from the standard electromotive force of a cell. 10. The Nernst equation gives the relationship between the cell emf and the concentrations of the reactants and products under non-standard-state conditions. 11. Batteries, which consist of one or more electrochemical cells, are used widely as self-contained power sources. Some of the better-known batteries are the dry cell, such as the Leclanché cell, the mercury battery, and the lead storage battery used in automobiles. Fuel cells produce electrical energy from a continuous supply of reactants. 12. The corrosion of metals, such as the rusting of iron, is an electrochemical phenomenon. 13. Electric current from an external source is used to drive a nonspontaneous chemical reaction in an electrolytic cell. The amount of product formed or reactant consumed depends on the quantity of electricity transferred at the electrode. KEY WORDS Anode, p. 761 Battery, p. 775 Back Forward Cathode, p. 761 Cell voltage, p. 762 Main Menu TOC Corrosion, p. 780 Electrolysis, p. 783 Study Guide TOC Electrolytic cell, p. 784 Electrochemistry, p. 758 Textbook Website MHHE Website 792 ELECTROCHEMISTRY Electrochemical cell, p. 761 Electromotive force (emf), p. 762 Faraday, p. 769 Standard emf (E°), p. 764 Standard oxidation potential, p. 764 Fuel cell, p. 779 Half-cell reaction, p. 761 Nernst equation, p. 772 Overvoltage, p. 787 Standard reduction potential, p. 763 QUESTIONS AND PROBLEMS BALANCING REDOX EQUATIONS Problems 19.1 Balance the following redox equations by the ionelectron method: H2O (in acidic solu(a) H2O2 Fe2 88n Fe3 tion) NO H2O (in acidic (b) Cu HNO3 88n Cu2 solution) MnO4 88n CNO MnO2 (in basic (c) CN solution) Br (in basic solution) (d) Br2 88n BrO3 I2 88n I S4O2 (in acidic solu(e) S2O2 3 6 tion) 19.2 Balance the following redox equations by the ionelectron method: H2O2 88n MnO2 H2O (in basic solu(a) Mn2 tion) Bi (in basic (b) Bi(OH)3 SnO2 88n SnO2 2 3 solution) C2O2 88n Cr3 CO2 (in acidic (c) Cr2O2 7 4 solution) Cl 88n Cl2 ClO2 (in acidic solu(d) ClO3 tion) ELECTROCHEMICAL CELLS AND STANDARD EMFs Review Questions 19.3 Define the following terms: anode, cathode, electromotive force, standard oxidation potential, standard reduction potential. 19.4 Describe the basic features of an electrochemical cell. Why are the two components of the cell separated from each other? 19.5 What is the function of a salt bridge? What kind of electrolyte should be used in a salt bridge? 19.6 What is a cell diagram? Write the cell diagram for an electrochemical cell consisting of an Al electrode placed in a 1 M Al(NO3)3 solution and a Ag electrode placed in a 1 M AgNO3 solution. 19.7 What is the difference between the half-reactions discussed in redox processes in Chapter 4 and the half-cell reactions discussed in Section 19.2? Back Forward Main Menu TOC 19.8 After operating a Daniell cell (see Figure 19.1) for a few minutes, a student notices that the cell emf begins to drop. Why? 19.9 Use the information in Table 2.1, and calculate the Faraday constant. 19.10 Discuss the spontaneity of an electrochemical reaction in terms of its standard emf (E° ). cell Problems 19.11 Calculate the standard emf of a cell that uses the Mg/Mg2 and Cu/Cu2 half-cell reactions at 25°C. Write the equation for the cell reaction that occurs under standard-state conditions. 19.12 Calculate the standard emf of a cell that uses Ag/Ag and Al/Al3 half-cell reactions. Write the cell reaction that occurs under standard-state conditions. 19.13 Predict whether Fe3 can oxidize I to I2 under standard-state conditions. 19.14 Which of the following reagents can oxidize H2O to O2(g) under standard-state conditions? H (aq), Cl (aq), Cl2(g), Cu2 (aq), Pb2 (aq), MnO4 (aq) (in acid). 19.15 Consider the following half-reactions: MnO4 (aq) NO3 (aq) 8H (aq) 4H (aq) 5e 88n Mn2 (aq) 3e 88n NO(g) 4H2O(l ) 2H2O(l ) Predict whether NO3 ions will oxidize Mn2 to MnO4 under standard-state conditions. 19.16 Predict whether the following reactions would occur spontaneously in aqueous solution at 25°C. Assume that the initial concentrations of dissolved species are all 1.0 M. (a) Ca(s) Cd2 (aq) 88n Ca2 (aq) Cd(s) (b) 2Br (aq) Sn2 (aq) 88n Br2(l ) Sn(s) (c) 2Ag(s) Ni2 (aq) 88n 2Ag (aq) Ni(s) (d) Cu (aq) Fe3 (aq) 88n Cu2 (aq) Fe2 (aq) Study Guide TOC Textbook Website MHHE Website 793 QUESTIONS AND PROBLEMS 19.17 Which species in each pair is a better oxidizing agent under standard-state conditions? (a) Br2 or Au3 , (b) H2 or Ag , (c) Cd2 or Cr3 , (d) O2 in acidic media or O2 in basic media 19.18 Which species in each pair is a better reducing agent under standard-state conditions? (a) Na or Li, (b) H2 or I2, (c) Fe2 or Ag, (d) Br or Co2 SPONTANEITY OF REDOX REACTIONS 19.19 Write the equations relating ∆G° and K to the standard emf of a cell. Define all the terms. 19.20 Compare the ease of measuring the equilibrium constant electrochemically with that by chemical means [see Equation (18.10)]. Problems What is the equilibrium constant for the following reaction at 25°C? Mg(s) 19.22 19.23 19.24 19.25 19.26 Zn2 (aq) 34 Mg2 (aq) Mg2 (aq) 34 Sr2 (aq) Mg(s) is 2.69 1012 at 25°C. Calculate E° for a cell made up of Sr/Sr2 and Mg/Mg2 half-cells. Use the standard reduction potentials to find the equilibrium constant for each of the following reactions at 25°C: (a) Br2(l ) 2I (aq) 34 2Br (aq) I2(s) (b) 2Ce4 (aq) 2Cl (aq) 34 Cl2(g) 2Ce3 (aq) 2 (c) 5Fe (aq) MnO4 (aq) 8H (aq) 34 Mn2 (aq) 4H2O 5Fe3 (aq) Calculate G° and Kc for the following reactions at 25°C: (a) Mg(s) Pb2 (aq) 34 Mg2 (aq) Pb(s) (b) Br2(l ) 2I (aq) 34 2Br (aq) I2(s) (c) O2(g) 4H (aq) 4Fe2 (aq) 34 2H2O(l ) 4Fe3 (aq) 3 (d) 2Al(s) 3I2(s) 34 2Al (aq) 6I (aq) Under standard-state conditions, what spontaneous reaction will occur in aqueous solution among the ions Ce4 , Ce3 , Fe3 , and Fe2 ? Calculate ∆G° and Kc for the reaction. Given that E° 0.52 V for the reduction Cu (aq) e 88n Cu(s), calculate E°, ∆G°, and K for the following reaction at 25°C: 2Cu (aq) 88n Cu2 (aq) Back Zn(s) The equilibrium constant for the reaction Sr(s) Forward Main Menu Review Questions 19.27 Write the Nernst equation and explain all the terms. 19.28 Write the Nernst equation for the following processes at some temperature T: (a) Mg(s) Sn2 (aq) 88n Mg2 (aq) Sn(s) (b) 2Cr(s) 3Pb2 (aq) 88n 2Cr3 (aq) 3Pb(s) Problems Review Questions 19.21 THE EFFECT OF CONCENTRATION ON CELL EMF Cu(s). TOC 19.29 What is the potential of a cell made up of Zn/Zn2 and Cu/Cu2 half-cells at 25°C if [Zn2 ] 0.25 M and [Cu2 ] 0.15 M? 19.30 Calculate E°, E, and ∆G for the following cell reactions. (a) Mg(s) Sn2 (aq) 88n Mg2 (aq) Sn(s) [Mg2 ] 0.045 M, [Sn2 ] 0.035 M (b) 3Zn(s) 2Cr3 (aq) 88n 3Zn2 (aq) 2Cr(s) [Cr3 ] 0.010 M, [Zn2 ] 0.0085 M 19.31 Calculate the standard potential of the cell consisting of the Zn/Zn2 half-cell and the SHE. What will the emf of the cell be if [Zn2 ] 0.45 M, PH2 2.0 atm, and [H ] 1.8 M? 19.32 What is the emf of a cell consisting of a Pb/Pb2 half-cell and a Pt/H2/H half-cell if [Pb2 ] 0.10 M, [H ] 0.050 M, and PH2 1.0 atm? 19.33 Referring to the arrangement in Figure 19.1, calculate the [Cu2 ]/[Zn2 ] ratio at which the following reaction is spontaneous at 25°C: Cu(s) Zn2 (aq) 88n Cu2 (aq) Zn(s) 19.34 Calculate the emf of the following concentration cell: Mg(s)Mg2 (0.24 M ) Mg2 (0.53 M )Mg(s) BATTERIES AND FUEL CELLS Review Questions 19.35 Explain the differences between a primary electrochemical cell—one that is not rechargeable—and a storage cell (for example, the lead storage battery), which is rechargeable. 19.36 Discuss the advantages and disadvantages of fuel cells over conventional power plants in producing electricity. Problems 19.37 The hydrogen-oxygen fuel cell is described in Section 19.6. (a) What volume of H2(g), stored at 25°C at a pressure of 155 atm, would be needed to Study Guide TOC Textbook Website MHHE Website 794 ELECTROCHEMISTRY run an electric motor drawing a current of 8.5 A for 3.0 h? (b) What volume (liters) of air at 25°C and 1.00 atm will have to pass into the cell per minute to run the motor? Assume that air is 20 percent O2 by volume and that all the O2 is consumed in the cell. The other components of air do not affect the fuel-cell reactions. Assume ideal gas behavior. 19.38 Calculate the standard emf of the propane fuel cell discussed on p. 780 at 25°C, given that ∆G° for f propane is 23.5 kJ/mol. 19.47 Considering only the cost of electricity, would it be cheaper to produce a ton of sodium or a ton of aluminum by electrolysis? 19.48 If the cost of electricity to produce magnesium by the electrolysis of molten magnesium chloride is $155 per ton of metal, what is the cost (in dollars) of the electricity necessary to produce (a) 10.0 tons of aluminum, (b) 30.0 tons of sodium, (c) 50.0 tons of calcium? 19.49 One of the half-reactions for the electrolysis of water is CORROSION 2H2O(l ) 88n O2(g) Review Questions 19.39 Steel hardware, including nuts and bolts, is often coated with a thin plating of cadmium. Explain the function of the cadmium layer. 19.40 “Galvanized iron” is steel sheet that has been coated with zinc; “tin” cans are made of steel sheet coated with tin. Discuss the functions of these coatings and the electrochemistry of the corrosion reactions that occur if an electrolyte contacts the scratched surface of a galvanized iron sheet or a tin can. 19.41 Tarnished silver contains Ag2S. The tarnish can be removed by placing silverware in an aluminum pan containing an inert electrolyte solution, such as NaCl. Explain the electrochemical principle for this procedure. [The standard reduction potential for the half-cell reaction Ag2S(s) 2e 88n 2Ag(s) S2 (aq) is 0.71 V.] 19.42 How does the tendency of iron to rust depend on the pH of solution? ELECTROLYSIS 19.50 19.51 19.52 19.53 19.54 Review Questions 19.43 What is the difference between an electrochemical cell (such as a Daniell cell) and an electrolytic cell? 19.44 What is Faraday’s contribution to quantitative electrolysis? Problems 19.55 19.45 The half-reaction at an electrode is Mg2 (molten) 2e 88n Mg(s) Calculate the number of grams of magnesium that can be produced by supplying 1.00 F to the electrode. 19.46 Consider the electrolysis of molten barium chloride, BaCl2. (a) Write the half-reactions. (b) How many grams of barium metal can be produced by supplying 0.50 A for 30 min? Back Forward Main Menu TOC 19.56 4H (aq) 4e If 0.076 L of O2 is collected at 25°C and 755 mmHg, how many faradays of electricity had to pass through the solution? How many faradays of electricity are required to produce (a) 0.84 L of O2 at exactly 1 atm and 25°C from aqueous H2SO4 solution; (b) 1.50 L of Cl2 at 750 mmHg and 20°C from molten NaCl; (c) 6.0 g of Sn from molten SnCl2? Calculate the amounts of Cu and Br2 produced in 1.0 h at inert electrodes in a solution of CuBr2 by a current of 4.50 A. In the electrolysis of an aqueous AgNO3 solution, 0.67 g of Ag is deposited after a certain period of time. (a) Write the half-reaction for the reduction of Ag . (b) What is the probable oxidation half-reaction? (c) Calculate the quantity of electricity used, in coulombs. A steady current was passed through molten CoSO4 until 2.35 g of metallic cobalt was produced. Calculate the number of coulombs of electricity used. A constant electric current flows for 3.75 h through two electrolytic cells connected in series. One contains a solution of AgNO3 and the second a solution of CuCl2. During this time 2.00 g of silver are deposited in the first cell. (a) How many grams of copper are deposited in the second cell? (b) What is the current flowing, in amperes? What is the hourly production rate of chlorine gas (in kg) from an electrolytic cell using aqueous NaCl electrolyte and carrying a current of 1.500 103 A? The anode efficiency for the oxidation of Cl is 93.0 percent. Chromium plating is applied by electrolysis to objects suspended in a dichromate solution, according to the following (unbalanced) half-reaction: Cr2O2 (aq) 7 e H (aq) 88n Cr(s) H2O(l ) How long (in hours) would it take to apply a Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 19.57 19.58 19.59 19.60 chromium plating 1.0 10 2 mm thick to a car bumper with a surface area of 0.25 m2 in an electrolytic cell carrying a current of 25.0 A? (The density of chromium is 7.19 g/cm3.) The passage of a current of 0.750 A for 25.0 min deposited 0.369 g of copper from a CuSO4 solution. From this information, calculate the molar mass of copper. A quantity of 0.300 g of copper was deposited from a CuSO4 solution by passing a current of 3.00 A through the solution for 304 s. Calculate the value of the faraday constant. In a certain electrolysis experiment, 1.44 g of Ag were deposited in one cell (containing an aqueous AgNO3 solution), while 0.120 g of an unknown metal X was deposited in another cell (containing an aqueous XCl3 solution) in series with the AgNO3 cell. Calculate the molar mass of X. One of the half-reactions for the electrolysis of water is 2H (aq) ADDITIONAL PROBLEMS 19.61 For each of the following redox reactions, (i) write the half-reactions; (ii) write a balanced equation for the whole reaction, (iii) determine in which direction the reaction will proceed spontaneously under standard-state conditions: (a) H2(g) Ni2 (aq) 88n H (aq) Ni(s) (b) MnO4 (aq) Cl (aq) 88n Mn2 (aq) Cl2(g) (in acid solution) (c) Cr(s) Zn2 (aq) 88n Cr3 (aq) Zn(s) 19.62 The oxidation of 25.0 mL of a solution containing Fe2 requires 26.0 mL of 0.0250 M K2Cr2O7 in acidic solution. Balance the following equation and calculate the molar concentration of Fe2 : Cr2O2 7 Fe2 H 88n Cr3 Fe3 19.63 The SO2 present in air is mainly responsible for the phenomenon of acid rain. The concentration of SO2 can be determined by titrating against a standard permanganate solution as follows: 5SO2 2MnO4 2H2O 88n 5SO2 4 2Mn2 4H Calculate the number of grams of SO2 in a sample of air if 7.37 mL of 0.00800 M KMnO4 solution are required for the titration. Back Forward 19.64 A sample of iron ore weighing 0.2792 g was dissolved in an excess of a dilute acid solution. All the Fe(II) was converted to Fe(III) ions. The solution required 23.30 mL of 0.0194 M KMnO4 for titration. Calculate the percent by mass of iron in the ore. 19.65 The concentration of a hydrogen peroxide solution can be conveniently determined by titration against a standardized potassium permanganate solution in an acidic medium according to the following unbalanced equation: MnO4 Main Menu TOC H2O2 88n O2 Mn2 (a) Balance the above equation. (b) If 36.44 mL of a 0.01652 M KMnO4 solution are required to completely oxidize 25.00 mL of a H2O2 solution, calculate the molarity of the H2O2 solution. 19.66 Oxalic acid (H2C2O4) is present in many plants and vegetables. (a) Balance the following equation in acid solution: 2e 88n H2(g) If 0.845 L of H2 is collected at 25°C and 782 mmHg, how many faradays of electricity had to pass through the solution? 795 MnO4 C2O2 88n Mn2 4 CO2 (b) If a 1.00-g sample of H2C2O4 requires 24.0 mL of 0.0100 M KMnO4 solution to reach the equivalence point, what is the percent by mass of H2C2O4 in the sample? 19.67 Complete the following table. State whether the cell reaction is spontaneous, nonspontaneous, or at equilibrium. ∆G E CELL REACTION 0 0 0 19.68 Calcium oxalate (CaC2O4) is insoluble in water. This property has been used to determine the amount of Ca2 ions in blood. The calcium oxalate isolated from blood is dissolved in acid and titrated against a standardized KMnO4 solution as described in Problem 19.66. In one test it is found that the calcium oxalate isolated from a 10.0-mL sample of blood requires 24.2 mL of 9.56 10 4 M KMnO4 for titration. Calculate the number of milligrams of calcium per milliliter of blood. 19.69 From the following information, calculate the solubility product of AgBr: AgBr(s) e 88n Ag(s) Ag (aq) Br (aq) e 88n Ag(s) E° 0.07 V E° 0.80 V 19.70 Consider an electrochemical cell composed of the Study Guide TOC Textbook Website MHHE Website 796 ELECTROCHEMISTRY SHE and a half-cell using the reaction Ag (aq) e 88n Ag(s). (a) Calculate the standard cell potential. (b) What is the spontaneous cell reaction under standard-state conditions? (c) Calculate the cell potential when [H ] in the hydrogen electrode is changed to (i) 1.0 10 2 M and (ii) 1.0 10 5 M, all other reagents being held at standard-state conditions. (d) Based on this cell arrangement, suggest a design for a pH meter. 19.71 An electrochemical cell consists of a silver electrode in contact with 346 mL of 0.100 M AgNO3 solution and a magnesium electrode in contact with 288 mL of 0.100 M Mg(NO3)2 solution. (a) Calculate E for the cell at 25°C. (b) A current is drawn from the cell until 1.20 g of silver have been deposited at the silver electrode. Calculate E for the cell at this stage of operation. 19.72 Explain why chlorine gas can be prepared by electrolyzing an aqueous solution of NaCl but fluorine gas cannot be prepared by electrolyzing an aqueous solution of NaF. 19.73 Calculate the emf of the following concentration cell at 25°C: phenolphthalein have been added is electrolyzed using an apparatus like the one shown below: 19.78 Cu(s)Cu2 (0.080 M) Cu2 (1.2 M)Cu(s) 19.74 The cathode reaction in the Leclanché cell is given by 2MnO2(s) Zn2 (aq) 2e 88n ZnMn2O4(s) If a Leclanché cell produces a current of 0.0050 A, calculate how many hours this current supply will last if there are initially 4.0 g of MnO2 present in the cell. Assume that there is an excess of Zn2 ions. 19.75 Suppose you are asked to verify experimentally the electrode reactions shown in Example 19.8. In addition to the apparatus and the solution, you are also given two pieces of litmus paper, one blue and the other red. Describe what steps you would take in this experiment. 19.76 For a number of years it was not clear whether mercury(I) ions existed in solution as Hg or as Hg2 . 2 To distinguish between these two possibilities, we could set up the following system: Hg(l )soln A soln BHg(l ) where soln A contained 0.263 g mercury(I) nitrate per liter and soln B contained 2.63 g mercury(I) nitrate per liter. If the measured emf of such a cell is 0.0289 V at 18°C, what can you deduce about the nature of the mercury(I) ions? 19.77 An aqueous KI solution to which a few drops of Back Forward 19.79 Main Menu TOC 19.80 19.81 19.82 Describe what you would observe at the anode and the cathode. (Hint: Molecular iodine is only slightly soluble in water, but in the presence of I ions, it forms the brown color of I3 ions. See Problem 12.102.) A piece of magnesium metal weighing 1.56 g is placed in 100.0 mL of 0.100 M AgNO3 at 25°C. Calculate [Mg2 ] and [Ag ] in solution at equilibrium. What is the mass of the magnesium left? The volume remains constant. Describe an experiment that would enable you to determine which is the cathode and which is the anode in an electrochemical cell using copper and zinc electrodes. An acidified solution was electrolyzed using copper electrodes. A constant current of 1.18 A caused the anode to lose 0.584 g after 1.52 103 s. (a) What is the gas produced at the cathode and what is its volume at STP? (b) Given that the charge of an electron is 1.6022 10 19 C, calculate Avogadro’s number. Assume that copper is oxidized to Cu2 ions. In a certain electrolysis experiment involving Al3 ions, 60.2 g of Al is recovered when a current of 0.352 A is used. How many minutes did the electrolysis last? Consider the oxidation of ammonia: 4NH3(g) 3O2(g) 88n 2N2(g) 6H2O(l ) (a) Calculate the ∆G° for the reaction. (b) If this reaction were used in a fuel cell, what would the standard cell potential be? 19.83 An electrochemical cell is constructed by immersing a piece of copper wire in 25.0 mL of a 0.20 M CuSO4 solution and a zinc strip in 25.0 mL of a 0.20 M ZnSO4 solution. (a) Calculate the emf of the cell at 25°C and predict what would happen if a small Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS amount of concentrated NH3 solution were added to (i) the CuSO4 solution and (ii) the ZnSO4 solution. Assume that the volume in each compartment remains constant at 25.00 mL. (b) In a separate experiment, 25.0 mL of 3.00 M NH3 are added to the CuSO4 solution. If the emf of the cell is 0.68 V, calculate the formation constant (Kf) of Cu(NH3)2 . 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 2Hg2 (aq) Hg2 2 (aq) 2e 88n Hg2 (aq) 2 E° 0.92 V 2e 88n 2Hg(l ) E° 0.85 V calculate ∆G° and K for the following process at 25°C: Hg2 (aq) 88n Hg2 (aq) 2 Hg(l ) (The above reaction is an example of a disproportionation reaction in which an element in one oxidation state is both oxidized and reduced.) 19.87 Fluorine (F2) is obtained by the electrolysis of liquid hydrogen fluoride (HF) containing potassium fluoride (KF). (a) Write the half-cell reactions and the overall reaction for the process. (b) What is the purpose of KF? (c) Calculate the volume of F2 (in liters) collected at 24.0°C and 1.2 atm after electrolyzing the solution for 15 h at a current of 502 A. 19.88 A 300-mL solution of NaCl was electrolyzed for 6.00 min. If the pH of the final solution was 12.24, calculate the average current used. 19.89 Industrially, copper is purified by electrolysis. The impure copper acts as the anode, and the cathode is made of pure copper. The electrodes are immersed in a CuSO4 solution. During electrolysis, copper at the anode enters the solution as Cu2 while Cu2 ions are reduced at the cathode. (a) Write half-cell reactions and the overall reaction for the electrolytic process. (b) Suppose the anode was contaminated with Zn and Ag. Explain what happens to these impurities during electrolysis. (c) How many hours will it take to obtain 1.00 kg of Cu at a current of 18.9 A? Back Forward Main Menu TOC 797 19.90 An aqueous solution of a platinum salt is electrolyzed 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. Calculate the charge on the Pt ions in this solution. 19.91 Consider an electrochemical cell consisting of a magnesium electrode in contact with 1.0 M Mg(NO3)2 and a cadmium electrode in contact with 1.0 M Cd(NO3)2. Calculate E° for the cell, and draw a diagram showing the cathode, anode, and direction of electron flow. 19.92 A current of 6.00 A passes through an electrolytic cell containing dilute sulfuric acid for 3.40 h. If the volume of O2 gas generated at the anode is 4.26 L (at STP), calculate the charge (in coulombs) on an electron. 19.93 Gold will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. However, the metal does dissolve in a mixture of the acids (one part HNO3 and three parts HCl by volume), called aqua regia. (a) Write a balanced equation for this reaction. (Hint: Among the products are HAuCl4 and NO2.) (b) What is the function of HCl? 19.94 Explain why most useful electrochemical cells give voltages of no more than 1.5 to 2.5 V. What are the prospects for developing practical electrochemical 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, Ag2C2O4, at 25°C. The measured potential difference between the rod and the SHE is 0.589 V, the rod being positive. Calculate 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 oxidation potential is 1.36 V for the reaction Zn(s) 4OH (aq) 88n Zn(OH)2 (aq) 4 2e Calculate the formation constant (Kf) for the reaction Zn2 (aq) 4OH (aq) 34 Zn(OH)2 (aq) 4 19.97 Use the data in Table 19.1 to determine whether or not hydrogen peroxide will undergo disproportionation in an acid medium: 2H2O2 88n 2H2O O2. 19.98 The magnitudes (but not the signs) of the standard electrode potentials of two metals X and Y are X2 Y2 2e 88n X 2e 88n Y E° E° 0.25 V 0.34 V where the notation denotes that only the magnitude (but not the sign) of the E° value is shown. When the half-cells of X and Y are connected, elec- Study Guide TOC Textbook Website MHHE Website 798 ELECTROCHEMISTRY trons flow from X to Y. When X is connected to a SHE, electrons flow from X to SHE. (a) Are the E° values of the half-reactions positive or negative? (b) What is the standard emf of a cell made up of X and Y? 19.99 An electrochemical cell is constructed as follows. One half-cell consists of a platinum wire immersed in a solution containing 1.0 M Sn2 and 1.0 M Sn4 ; the other half-cell has a thallium rod immersed in a solution of 1.0 M Tl . (a) Write the half-cell reactions and the overall reaction. (b) What is the equilibrium constant at 25°C? (c) What is the cell voltage if the Tl concentration is increased tenfold? 0.34 V.) (E° /Tl Tl 19.100 Given the standard reduction potential for Au3 in Table 19.1 and Au (aq) e 88n Au(s) E° 19.105 19.106 1.69 V answer the following questions. (a) Why does gold not tarnish in air? (b) Will the following disproportionation occur spontaneously? 3Au (aq) 88n Au3 (aq) 19.101 19.102 19.103 19.104 Back 19.107 2Au(s) (c) Predict the reaction between gold and fluorine gas. The ingestion of a very small quantity of mercury is not considered too harmful. Would this statement still hold if the gastric juice in your stomach were mostly nitric acid instead of hydrochloric acid? When 25.0 mL of a solution containing both Fe2 and Fe3 ions is titrated with 23.0 mL of 0.0200 M KMnO4 (in dilute sulfuric acid), all of the Fe2 ions are oxidized to Fe3 ions. Next, the solution is treated with Zn metal to convert all of the Fe3 ions to Fe2 ions. Finally, 40.0 mL of the same KMnO4 solution are added to the solution in order to oxidize the Fe2 ions to Fe3 . Calculate the molar concentrations of Fe2 and Fe3 in the original solution. Consider the Daniell cell in Figure 19.1. When viewed externally, the anode appears negative and the cathode positive (electrons are flowing from the anode to the cathode). Yet in solution anions are moving toward the anode, which means that it must appear positive to the anions. Since the anode cannot simultaneously be negative and positive, give an explanation for this apparently contradictory situation. Lead storage batteries are rated by ampere hours, that is, the number of amperes they can deliver in an hour. (a) Show that 1 A h 3600 C. (b) The lead anodes of a certain lead-storage battery have a total mass of 406 g. Calculate the maximum theo- Forward Main Menu TOC 19.108 19.109 19.110 retical capacity of the battery in ampere hours. Explain why in practice we can never extract this much energy from the battery. (Hint: Assume all of the lead will be used up in the electrochemical reaction and refer to the electrode reactions on p. 776.) (c) Calculate E° and ∆G° for the battery. cell The concentration of sulfuric acid in the leadstorage battery of an automobile over a period of time has decreased from 38.0 percent by mass (density 1.29 g/mL) to 26.0 percent by mass (1.19 g/mL). Assume the volume of the acid remains constant at 724 mL. (a) Calculate the total charge in coulombs supplied by the battery. (b) How long (in hours) will it take to recharge the battery back to the original sulfuric acid concentration using a current of 22.4 amperes. Consider a Daniell cell operating under non-standard-state conditions. Suppose that the cell’s reaction is multiplied by 2. What effect does this have on each of the following quantities in the Nernst equation? (a) E, (b) E°, (c) Q, (d) ln Q, and (e) n? A spoon was silver-plated electrolytically in a AgNO3 solution. (a) Sketch a diagram for the process. (b) If 0.884 g of Ag was deposited on the spoon at a constant current of 18.5 mA, how long (in minutes) did the electrolysis take? Comment on whether F2 will become a stronger oxidizing agent with increasing H concentration. In recent years there has been much interest in electric cars. List some advantages and disadvantages of electric cars compared to automobiles with internal combustion engines. Calculate the pressure of H2 (in atm) required to maintain equilibrium with respect to the following reaction at 25°C: Pb(s) 2H (aq) 34 Pb2 (aq) H2(g) Given that [Pb2 ] 0.035 M and the solution is buffered at pH 1.60. 19.111 A piece of magnesium ribbon and a copper wire are partially immersed in a 0.1 M HCl solution in a beaker. The metals are joined externally by another piece of metal wire. Bubbles are seen to evolve at both the Mg and Cu surfaces. (a) Write equations representing the reactions occurring at the metals. (b) What visual evidence would you seek to show that Cu is not oxidized to Cu2 ? (c) At some stage, NaOH solution is added to the beaker to neutralize the HCl acid. Upon further addition of NaOH, a white precipitate forms. What is it? 19.112 The zinc-air battery shows much promise for electric cars because it is lightweight and rechargeable: Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 799 battery every second? Assume that the temperature is 25°C and the partial pressure of oxygen is 0.21 atm. 19.113 Calculate E° for the reactions of mercury with (a) 1 M HCl and (b) 1 M HNO3. Which acid will oxidize Hg to Hg2 under standard-state conditions? 2 Can you identify which test tube below contains HNO3 and Hg and which contains HCl and Hg? The net transformation is Zn(s) 1 O2(g) 88n 2 ZnO(s). (a) Write the half-reactions at the zinc-air electrodes and calculate the standard emf of the battery at 25°C. (b) Calculate the emf under actual operating conditions when the partial pressure of oxygen is 0.21 atm. (c) What is the energy density (measured as the energy in kilojoules that can be obtained from one kilogram of the metal) of the zinc electrode? (d) If a current of 2.1 105 A is to be drawn from a zinc-air battery system, what volume of air (in liters) would need to be supplied to the Back Forward Main Menu TOC Answers to Practice Exercises: 19.1 5Fe2 MnO4 8H 88n 5Fe3 Mn2 4H2O. 19.2 No. 19.3 0.34 V. 19.4 1.1 10 42. 19.5 ∆G° 4.1 102 kJ. 19.6 Yes. 19.7 0.38 V. 19.8 Anode, O2; cathode, H2. 19.9 2.0 104 A. Study Guide TOC Textbook Website MHHE Website C HEMICAL M YSTERY Tainted Water † T he salesman was persuasive and persistent. “Do you realize what’s in your drinking water?” he asked Tom. Before Tom could answer, he continued: “Let me demonstrate.” First he filled a glass of water from the kitchen faucet, then he produced an electrical device that had a pair of probes and a light bulb. It resembled a standard conductivity tester. He inserted the probes into the water and the bulb immediately beamed brightly. Next the salesman poured some water from a jar labeled “distilled water” into another glass. This time when he inserted the probes into the water, the bulb did not light. “Okay, can you explain the difference?” the salesman looked at Tom with a triumphant smile. “Sure,” Tom began to recall an experiment he did in high school long ago, “The tap water contains minerals that caused. . . .” “Right on!” the salesman interrupted. “But I’m not sure if you realize how harmful the nation’s drinking water has become.” He handed Tom a booklet entitled The Miracle of Distilled Water. “Read the section called ‘Heart Conditions Can Result from Mineral Deposits’,” he told Tom. “The tap water may look clear, although we know it contains dissolved minerals. What most people don’t realize is that it also contains other invisible substances that are harmful to our health. Let me show you.” The salesman proceeded to do another demonstration. This time he produced a device that he called a “precipitator,” which had two large electrodes attached to a black box. “Just look what’s in our tap water,” he said, while filling another large glassful from the faucet. The tap water appeared clean and pure. The salesman plugged the precipitator into the ac (alternating current) outlet. Within seconds, bubbles rose from both electrodes. The tap water took on a yellow hue. In a few minutes a brownish scum covered the surface of the water. After 15 minutes the glass of water was filled with a black-brown precipitate. When he repeated the experiment with distilled water, nothing happened. Tom was incredulous. “You mean all this gunk came from the water I drink?” “Where else?” beamed the salesman. “What the precipitator did was to bring out all the heavy metals and other undesirable substances. Now don’t worry. There is a remedy for this problem. My company makes a distiller that will convert tap water to distilled water, which is the only safe water to drink. For a price of $600 you will be able to produce distilled water for pennies with the distiller instead of paying 80 cents for a gallon of water from the supermarket.” Tom was tempted but decided to wait. After all, $600 is a lot to pay for a gadget that he only saw briefly. He decided to consult his friend Sarah, the chemistry teacher † Adapted with permission from “Tainted Water,” by Joseph J. Hesse, CHEM MATTERS, February, 1988, p. 13. Copyright 1988 American Chemical Society. 800 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website The precipitator with its electrodes immersed in tap water. Left: Before electrolysis has started. Right: 15 minutes after electrolysis commenced. at the local high school, before making the investment. The salesman promised to return in a few days and left the precipitator with Tom so that he could do further testing. CHEMICAL CLUES 1. 2. 3. 4. 5. After Sarah examined the precipitator, she concluded that it was an electrolytic device consisting of what seemed like an aluminum electrode and an iron electrode. Since electrolysis cannot take place with ac (why not?), the precipitator must contain a rectifier, a device that converts ac to dc (direct current). Why does the water heat up so quickly during electrolysis? From the brown color of the electrolysis product(s), deduce which metal acts as the cathode and which electrode acts as the anode. Write all possible reactions at the anode and the cathode. Explain why there might be more than one type of reaction occurring at an electrode. In analyzing the solution, Sarah detected aluminum. Suggest a plausible structure for the aluminum-containing ion. What property of aluminum causes it to dissolve in the solution? Suggest two tests that would confirm Sarah’s conclusion that the precipitate originated from the electrodes and not from the tap water. 801 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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This document was uploaded on 07/27/2009.

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