Chemistry part 31, Julia Burdge,2e (2009) doc

760 CHAPTER 19 Electrochemistry . ~ ! . - _. Multimedia Electrochemistry~xidation - reduction reactions . Now is a go od t im e to r evi ew how oxi d at i on n umbers are ass i gne d [ ~ Section 4.4) . • Balancing Redox Reactions In Chapter 4 we briefly discussed oxidation-reduction or "redox" reactions, those in which elec- trons are transferred from one species to another. In this section we review how to identify a reac- tion as a redox reaction and look more closely at how such reactions are balanced. A redox reaction is one in which there are changes in oxidation states, which we identify . . . . . . . using the rules introduced in Chapter 4. The following are examples of redox reactions: <+ L yt: S-'» c;:- 1);- 1 o , +4 '-2 + 1-2 Sn(s) + Cu?+(aq) • Cu(s) + Sn2+(aq) r- + 2 Equations for redox reactions, such as those shown here, can be balanced by inspection, the method of balancing introduced in Chapter 3 [ ~~ Section 3.3], but remember that redox equations must be balanced for mass (number of atoms) and for charge (number of electrons) [ ~~ Section 4.4] . In this section we introduce the half-reaction method to balance equations that cannot be balanced simply by inspection. Consider the aqueous reaction of the iron (II) ion (Fe 2+ ) with the dichromate ion (Cr20~ - ): Because there is no species containing oxygen on the product side of the equation, it would not be possible to balance this equation simply by adjusting the coefficients of reactants and products. However, there are two things about the reaction that make it possible to add species to the equa- tion to balance it without changing the chemical reaction it represents: • The reaction takes place in aqueous solution, so we can add H 2 0 as needed to balance the equation . • This particular reaction takes place in acidic solution, so we can add H+ as needed to bal- ance the equation. (Some reactions take place in basic solution, enabling us to add OH - as needed for balancing. We will learn more about this shortly.) After writing the unbalanced equation, we balance it stepwise as follows: 1. Separate the unbalanced reaction into half-reactions. A half-reaction is an oxidation or a reduction that occurs as part of the overall redox reaction. Oxidation: Reduction: Fe 2 + -_. Fe H Cr 2 0 ~ - -_. Cr 3 + 2. Balance each of the half-reactions with regard to atoms other than 0 and H. In this case, no change is required for the oxidation half-reaction. We adjust the coefficient of the chromium(III) ion to balance the reduction half-reaction. Oxidation: Reduction: Fe 2 + -_. Fe 3 + Cr 2 0 ~ - _ . 2 Cr 3 + 3. Balance both half-reactions for 0 by adding H 2 0. Again, the oxidation in this case requires no change, but we must add seven water molecules to the product side of the reduction. Oxidation: Fe 2 + +. Fe H Reduction: Cr 2 0 ~ - -_. 2Cr H + 7H 2 0 4. Balance both half-reactions for H by adding H+. Once again, the oxidation in this case requires no change, but we must add 14 hydrogen ions to the reactant side of the reduction. Oxidation: Fe 2+ -_ . Fe 3 + Reduction: 14H+ + Cr 2 0 ~ - +. 2Cr 3 + + 7H 2 0 SECTION 19.1 Balancing Redox Reactions 76 5. Balance both half-reactions for charge by adding electrons. To do this, we determine the total charge on each side and add electrons to make the total charges equal. In the case of the oxidation, there is a charge of +2 on the reactant side and a charge of +3 on the product side. Adding one electron to the product side makes the charges equal. Oxidation: Fe 2 + Fe 3+ + e- ~ ; Total charge: +2 +2 In the case of the reduction, there is a total charge of [(14)( + 1) + (1)(-2)] = + 12 on the reactant side and a total charge of [(2)( +3)] = + 6 on the product side. Adding six electrons to the reactant side makes the charges equal. Reduction: 6e - + 14H+ + Cr 2 0~ - • 2Cr 3+ + 7H ?O ., Total charge: +6 +6 6. If the number of electrons in the balanced oxidation half-reaction is not the same as the number of electrons in the balanced reduction half -reaction, multiply one or both of the half- reactions by the number(s) required to make the number of electrons the same in both. In this case, with one electron in the oxidation and six in the reduction, multiplying the oxida- tion by six accomplishes this. Oxidation: 6(Fe 2+ Fe 3+ + e- ) 6Fe 2+ • 6Fe 3+ + ~ e ) Reduction: ~ e ) + l4H + + Cr20~ - • 2Cr 3+ + 7H 2 0 7. Finally, add the balanced half-reactions back together and cancel the electrons, in addition to any other identical terms that appear on both sides. 6Fe 2+ • 6Fe 3+ + W A final check shows that the resulting equation is balanced both for mass and for charge. Some redox reactions occur in basic solution. When this is the case, balancing by the half- reaction method is done exactly as described for reactions in acidic solution, but it requires two additional steps: 8. For each H+ ion in the final equation, add one OH- ion to each side of the equation, combin- ing the H+ and OH- ions to produce H 2 0. 9. Make any additional cancellations made necessary by the new H 2 0 molecules. Sample Problem 19.1 shows how to use the half-reaction method to balance a reaction that takes place in basic solution. Pennanganate ion and iodide ion react in basic solution to produce manganese(IV) oxide and molecular iodine. Use the half-reaction method to balance the equation: Mn0 4 + r Mn0 2 + 12 Strategy The reaction takes place in basic solution, so apply steps 1 through 9 to balance for mass and for charge. Setup Identify the oxidation and reduction half-reactions by assigning oxidation numbers. Solution Step 1. Separate the unbalanced reaction into half-reactions. Oxidation: 1- 1 2 Reduction: Mn0 4 Mn0 2 (Continued) • 762 CHAPTER 19 Electrochemistry I Think About It Verify that the final equation is balanced for mass and for charge. Remember that electrons cannot appear in the overall balanced equation. Step 2. Balance each half-reaction for mass, excluding 0 and H. Mn04 +. Mn02 Step 3. Balance both half-reactions for 0 by adding H 2 0. Mn04 -_. Mn02 + 2H 2 0 Step 4. Balance both half-reactions for H by adding H+. 4H+ + Mn04 +. Mn0 2 + 2H 2 0 Step 5. Balance the total charge of both half-reactions by adding electrons. 3e - + 4H + + Mn0 4 +. Mn02 + 2H 2 0 Step 6. Multiply the half-reactions to make the number s of electrons the same in both. 3( 21 - + . 12 + 2e- ) 2(3e- + 4H + + Mn04 +. Mn02 + 2H 2 0) Step 7. Add the half-reactions back together, cancelling electrons. 61 - +. 31 2 + V V + 8H+ + 2Mn04 • 2Mn0 2 + 4H 2 0 8H + + 2Mn04 + 61 - • 2Mn0 2 + 31 2 + 4H 2 0 Step 8. For each H+ ion in the final equation, add one OH - ion to each s id e of the equation, combining the H+ and OH - ions to produce H 2 0. 8H+ + 2Mn0 4 + 61 - -_ . 2Mn02 + 31 2 + 4H 2 0 + 8 0W + 80W Step 9. Carry out any cancellations made necessary by the additional H 2 0 molecules. 4H 2 0 + 2Mn04 + 61- -_. 2Mn0 2 + 31 2 + 80H - Practice Problem A Use the half-reaction method to balance the following equation in acidic solution: Practice Problem B Use the half-reaction method to balance the following equation in basic solution: CN - + Mn04 -_. CNO - + Mn0 2' Checkpoint 19.1 Balancing Redox Reactions 19.1.1 Which of the following equations does not represent a redox reaction? (Select all that apply.) 19.1.2 Mn04 and C 2 0 ~ - react in basic solution to form Mn0 2 and CO ~- . What are the coefficients of MnO 4 and C20~ - in the balanced equation? a) NH 3 + HCI • NH 4 Cl b) 2H 2 0 2 • 2H 2 0 + O 2 a) I and 1 c) 20 3 • 30 2 b) 2 and 1 d) 3N0 2 + H 2 0 • NO + 2HN0 3 c) 2 and 3 e) LiCl Li+ + Cl- d) 2 and 6 e) 2 and 2 SECTION 19.2 Galvanic Cells 1 53 Galvanic Cells When zinc metal is Qlaced in a solution containing copper(II) ions, Zn is oxidized to Zn 2 + ions whereas Cu 2+ ions are reduced to Cu [ ~~ Section 4.4]: Media Player/ MPEG Content Electrochemistry-Operatle- :; voltaic ce ll. . . Zn(s) + Cu 2 +(aq) +. Zn2+(aq) + Cu(s) The electrons are transferred directly from the reducing agent, Zn, to the oxidizing agent, Cu 2 + , in solution. However, if we physically separate two half-reactions from each other, we can arrange it such that the electrons must travel through a wire in order to pass from the Zn atoms to the Cu 2+ ions. As the reaction progresses, it generates a flow of electrons through the wire and thereby gen- erates electricity. The experimental apparatus for generating electricity through the use of a spontaneous reac- This reaction is shown in Figure 4.5 . . . . . . . . . . . . . . . . . . tion is called a galvanic cell. Figure 19.1 shows the essential components of a galvanic cell. A zinc bar is immersed in an aqueous ZnS0 4 solution in one container, and a copper bar is immersed in an aqueous CUS04 solution in another container. The cell operates on the principle that the oxi- dation of Zn to Zn 2 + and the reduction of Cu 2+ 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. By definition, the anode in a galvanic cell is the electrode at which oxidation occurs and the cathode is the electrode at which reduction occurs. (Each combination of container, electrode, and solution is called a half-cell.) This particu- lar arrangement of electrodes and electrolytes is called a Daniell cell. The half-reactions for the galvanic cell shown in Figure 19.1 are Oxidation: Zn(s). Zn 2+ (aq) + 2e- Reduction: Cu ?+ (aq) + 2e- +. Cu(s) To complete the electric circuit, and allow electrons to flow through the external wire, the solutions must be connected by a conducting medium through which the cations and anions can move from one half-cell 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 KCI or NH 4 N0 3 . The ions o 2e- lost perZn atom oxidized Anode (-) Zn o Zn 2+ (l M) . Salt bridge 2e- gained per Cu 2 + ion reduced Cathode (+) Cu • Na + == :~ __ ' ''' _ v I In 1.1 U Cu2+ (l M) o e , A galvanic cell can also be called a voltaic C? Both t erms refer to a cell in which a spoma"!?_ , s chem i cal reaction generates a flow of elee O"S. Figure 19.1 A galvanic cell. The salt bridge (an inverted U tube) containing an Na 2 S04 solution provides an electrically conducting medium between two solutions. Th e openings of the U tube are loosely plugged with cotton balls to prevent me Na 2 S04 solution from flowing into me containers while allowing the anions and cations to move across. Electrons flow externally from the Zn electr ode (anode) to the eu electrode (cathode). 764 CHAPTER 19 Electrochemistry This is the origin of the terms cathode and anode: • Cations move toward the ca t hode. • Anions move toward the anode. The volt is a derived SI unit: 1 V = 1 J/1 C. , The terms cell potential, cell voltage, cell electromotive force, and cell emf are used interchangeably and are all symbolized the same way with f eell . · . == _ "'" Multimedia Electrochemistry - an interactive voltaic cell . A cell in which different half- reactions take place wi ll have a different emf. Figure 19.2 The galvanic cell de sc ribed in Figure 19.1. Note th e U tube (salt bridg e) connecting the two beakers. When the concentrations of Zn 2 + a nd Cu 2+ are 1 molar (1 M) at 25 ° C, the ce ll vo ltage is 1.10 V. • • . . in the sa lt bridge must not react with the other ions in solution or with the electrodes (see Figure 19.1). During the course of the redox reaction, electrons flow through the external wire from the anode (Zn electrode) to the cathode (Cu electrode). In the solution, the cations (Zn 2 +, Cu 2 +, and . + ~ . "' ?' ., .• • • • • K ) move toward the cathode, while the anions (S04 - and Cl- ) move toward the anode. Without the salt bridge connecting the two solutions, the buildup of po sitive charge in the anode compart- ment (due to the departure of electrons and the resulting formation of Zn 2 +) and the buildup of negative charge in the cathode compartment (created by the arrival of electrons and the reduction of Cu 2 + ions to Cu) would quickly prevent the cell from operating. An electric current fl ows from the anode to the cathode becau se there is a difference in elec- trical potential energy between the electrodes. This flow of electric current is analogous to the flow of water down a waterfall, which occurs becau se there is a difference in the 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 the cell potential (Eceu). The potential of a cell depends not only on the nature of the electrodes and the ions in solution,: but also on the concentra- • tions of the ions and the temperature at which the cell is operated. : . . . . . . . . . . . . . The conventional notation for representing galvanic cells is the cell diagram. For the cell shown in Figure 19.1, if we assume that the concentrations of Zn 2+ and Cu 2 + ions are 1 M, the cell diagram is Zn (s) I Zn ?+ (1 M) II Cu 2 + (1 M) I Cu(s) The sing le vertical line re pr esents a phase boundar y. For example, the zinc electrode is a solid and the Zn 2+ ions are in solution. Thus, we draw a line between Zn and Zn 2 + to show the pha se bound- ary. The double vertical lines denote the sa lt bridge. By convention, the anode is written first, to the left of the double line s, and the other components appear in the order in which we would encounter them in mo ving from the anode to the cathode (from left to right in the cell diagram). Standard Reduction Potentials When the concentrations of the Cu 2 + and Zn 2 + ions are both 1.0 M, the cell potential of the cell . . . . " . described in Section 19.2 is 1.10 V at 25°C (see Figure 19.2). This measured potential is related to the half-reactions that take place in the anode and cathode compartments. The overall cell poten- tial is the difference between the electric potentials at the Zn and Cu electrodes-known as the half-cell potentials. Just as it is impossible for only one of the half-reactions to occur indepen- dently, it is impossible to measure the potential of ju st a single half-cell. However, if we arbitrarily define the potential value of a particular half-cell as zero, we can use it to determine the relative potentials of other half-cells. Then we can use these relative potentials to determine overall cell potentials. 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 ha s two functions. First, it provides a surface on which the dissociation (and oxidation) of hydrogen molecules can take place: Second, it serves as an electrical conductor to the external circuit. Salt bridge + , SECTION 19.3 Standard Reduction Potentials 76: • 1 MHCI -H 2 gas at I atm + Pt electrode Under standard-state conditions [ ~~ Section 18.4], when the pressure of H2 is 1 atm and the concentration of HCl solution is 1 M, the potential for the reduction of H+ at 25°C is defined as exactly zero: 2H+ (l M) + 2e- -_. H2 (1 atm) As before, the ° superscript denotes standard-state conditions, and EO is the standard reduction potential; that is, EO is the potential associated with a reduction half-reaction at an electrode when the ion concentration is 1 M and the gas pressure is 1 atm. Because the hydrogen electrode is used to determine all other electrode potentials, it is called the standard hydrogen electrode (SHE). A half-cell potential is measured by constructing a galvanic cell in which one of the electrodes is the SHE. The measured voltage is then used to determine the potential for the other half-cell. Figure 19.4(a) shows a galvanic cell with a zinc electrode and a SHE. When the circuit is completed in this cell, electrons flow from the zinc electrode to the SHE, thereby oxidizing Zn to Zn 2+ and + . . . reducing H ions to H 2 . In this case, therefore, the zinc electrode is the anode (where oxidation takes place) and the SHE is the cathode (where reduction takes place). The cell diagram is Zn(s) I Zn 2 + (l M) II H+ (1 M) I HzCl atm) I Pt(s) The measured potential for this cell is 0.76 V at 25°C. We can write the half-cell reactions as follows: Anode (oxidation): Zn(s) -_. Zn2+(aq)(1 M) + 2e - Cathode (reduction): 2H +(aq)(1 M) + 2e - • H2(g)(1 atrn) Overall: Zn(s) + 2H + (aq)(l M) - Zn2+(aq)(l M) + H 2 (g) (l atm) Anode (-) Zn Zn 2 + (1 M) Salt bridge n 5 u. I (a) Pt elec:l[o(ie 1 MHCI -H 2 gas at 1 atm Hz gas at • 1 aIm 1 MHCI Figure 19.3 A hydrogen electrode operating under standard-state conditions. Hydrogen gas at 1 atrn bubbled through aIM Hel solution. The electrode itself is made of platinum. One indication of the direction of electro n fla is the fact that as the reaction proceeds , the mass of the z in c electrode decreases as a result of thE oxidation half-reactio n: Salt bridge I v O.3LJ (b) Zn(s) • Zn2+(aq) + 2e - Cathode (+) Cu Figure 19.4 (a) A cell consisting of a zinc electrode and a hydrogen electrode. (b) A cell consisting of a copper electrode and a hydrogen electrode_ Both cells are operating under standard-state conditions. Note that in (a) the SHE is the cathode, but in (b) it is the anode. 766 CHAPTER 19 Electrochemistry I Again, the direction of electron flow is determined in part by the mass of the electrode. Reduction of Cu '+ ions causes the mas s of the Cu electrode to increase. The half-reaction with the greater reduction potential has the greater potential to occur as a reduction. By convention, the standard cell potential, E~elj, which is composed of a contribution from the anode and a contribution from the cathode, is given by Equation 19.1 E o - EO EO cell - cathode - anode where E ~athod e and E~ode are the standard reduction potentials of the cathode and anode, respec- tively. For the Zn -S HE cell, we write 0.76 V = 0 - Ezn 2+/ zn where the subscript "H + 1H 2 " means "2H + + 2e - • H 2 " and the subscript "Zn2+ IZn" means "Z n 2+ + 2e - • Zn". Thus, the standard reduction potential of zinc, Ezn 2+/z n is - 0.76 V. The standard reduction potential of copper can be determined in a similar fashion, by using a ce ll with a copper electrode and a SHE [Figure 19.4(b)]. In this ca se, electrons flow from the SHE to the copper electrode when the circuit is completed; that is, the co pper electrode is the cathode . ,. . . . . . and the SHE is the anode. The cell diagram is Pt(s) H 2 (1 atm) I H+ (l M) I Cu 2 + (l M) I Cu(s) and the half-cell reactions are Anode (ox idati on): H 2 (g)(1 atm) . 2H+(aq)(l M) + 2e - Cathode (reduction): Cu 2+ (aq)(1 M) + 2e - • Cu(s) O verall : H2(g)(1 atm) + Cu 2 +( aq)(l M) • 2H +(aq)(1 M) + Cu(s) Under standard-state conditions and at 25°C, the measured potential for the cell is 0.34 V, so we write EO - EO E }" Sf cell - cathode - anode = ECu 2+/ Cu - E H + ~ 0.34 V = ECu 2+/ Cu - 0 Thus, the standard reduction potential of copper, ECu 2+/ c u, is 0.34 V, where the subscript "Cu 2 + ICu" means "Cu 2 + + 2e - • Cu." , Having determined the standard reduction potentials of Zn and Cu, we can use Equation 19.1 to calculate the cell potential for the Daniell cell described in ection 19.2: EO - EO EO ce ll - cathode - anode = 0.34 V - (-0 .76 V) = 1.10 V As in the case of /1G o [ ~~ Section 18.4], we can use the sign of EO to predict whether a reaction lies to the right or to the left. A positive EO means that the redox reaction will favor the formation of products at equilibrium. Conversely, a negative EO indicates that reactants will be favored at equilibrium. We wi ll examine how E ~ elj, /1G o , and K are related in Section 19.4. ~ . Table 19.1 lists standard reduction potentials, in order of decreasing reduction potential, for a number of half-cell reactions. To avoid ambiguity, all half-cell reactions are shown as. redud ions. A galvanic cell is composed of two half-cells, and therefore two half-cell reactions. Whether not a particular half-cell reaction occurs as a reduction in a galvanic cell depends on how its red ction potential compares to that of the other half-cell reaction. If it has the greater (or more positive educ- . . . . . . . . . . . . . . tion potential of the two, it will occur as a reduction. If it has the smaller (or more negative) reduction , potential of the two, it will occur in the reverse direction, as an oxidation. Consider the example of the cell described in Section 19.2. The two half-cell reactions and their standard reduction potentials are Cu 2 + + 2e - • Cu EO = 0.34 V Zn 2 + + 2e- Zn EO = -0.76V - Ha lf -Reaction Co3+(aq) + e- • Co 2+ (aq) HzOz(aq) + 2H+(aq) + 2e - • 2H z 0(l) PbOz(s) + 4H+(aq) + SO~ - (aq) + 2e- • PbS0 4 (s) + 2H 2 0(l) Ce 4 +(aq) + e- • Ce 3 +(aq) Mn04(aq) + 8H +(aq) + 5e - • Mn 2+ (aq) + 4H z O(l) Au 3 +(aq) + 3e - • Au(s) Clz(g) + 2e- • 2Cqaq) CrzO~ - (aq) + 14H+(aq) + 6e - - _. 2Cr 3+ (aq) + 7H z 0(l) MnOz(s) + 4H+(aq) + 2e- - _. Mn 2+ (aq) + 2H z 0(l) 02(g) + 4H+(aq) + 4e - • 2H z 0(l) N0 :J (aq) + 4H+(aq) + 3e - - _. NO(g) + 2H z O(I) Hg~ + (aq) + 2e - • 2Hg(l) • Ag +(aq) + e- • Ag(s) Fe3+(aq) + e- • Fe 2+ (aq) OzCg) + 2H+(aq) + 2e - • HzOz(aq) Mn04(aq) + 2H z 0(l) + 3e - • MnO z(s) + 40H - (aq) Iz(s) + 2e - • 2C (aq) 0z(g) + 2H z 0(l) + 4e - • 40W(aq) Cu2+(aq) + 2e - • Cu(s) AgCl(s) + e - • Ag(s) + cqaq) SO~-(aq) + 4H +(aq) + 2e - • SO z(g) + 2H 2 0(l) Cu 2 +(aq) + e- • Cu+(aq) Sn4 +(aq) + 2e- • Sn 2+ (aq) 2H +(aq) + 2e - • Hz(g) Pb2+(aq) + 2e- • Pb(s) Sn2+(aq) + 2e- • Sn(s) Ni 2+ (aq) + 2e - • Ni(s) Co 2 +(aq) + 2e- • Co(s) PbS0 4 (s) + 2e - • Pb(s) + so t (aq) Cd 2+ (aq) + 2e- • Cd(s) Fe 2 +(aq) + 2e- • Fe(s) Cr 3+ (aq) + 3e- • Cr(s) Zn2+(aq) + 2e - • Zn(s) 2H z 0(l) + 2e- • Hig) + 20W(aq) Mn2+(aq) + 2e- • Mn(s) Al 3+ (aq) + 3e- • Al(s) Be 2+ (aq) + 2e- • Be(s) Mg z+ (aq) + 2e- • Mg(s) Na +(aq) + e- • Na(s) Ca 2 +(aq) + 2e - • Ca(s) S? +(aq) + 2e- • Sr(s) Ba2+(aq) + 2e- • Ba(s) K+(aq) + e- • K(s) Li +(aq) + e- • Li(s) ' F or a ll half-reactions the concentrat ion is 1 M for dissolved species and the pressure is 1 atm for gases. These are the stan dard-state va lu es. 768 CHAPTER 19 Electrochemistry The electrode where reduct io n o ccur s is the cath o de . There fore, E ~a thode = 0. 34 V The el e ctro de w her e o xidat ion o cc urs is th e anode. Therefore , E ~node = - 0 .76 V • • , The Cu half-reaction, having a greater (more positive) reduction potential, is the half-reaction that . . . . . . . . . , . . . . . . . . . . . . • • • • • will occur as a reduction: Cu 2 + + 2e - -_I Cu The Zn half-reaction has a smaller (less positive) reduction potential and will occur, instead, as an oxidation: Zn -_I Zn 2 + + 2e- Adding the two half-reactions gives the overall cell reaction: Zn + Cu ?+ -_I Zn 2+ + Cu Consider, however, what would happen if we were to construct a galvanic eell combining t 'C half-cell with an Mn half-cell. The reduction potential ofMn is - 1.18 V. Zn 2+ + 2e- +. Zn Mn ?+ + 2e- -_I Mn EO = -0.76V EO= - U8V In this ca se, the greater (less negative) reduction potential is that of Zn, so the Zn half-reaction will occur as a reduction and the Zn electrode will be the cathode. The Mn electrode will be the anode. The cell potential is therefore The overall cell reaction is EO - EO EO ce ll - cathode - anode = (-0.76V) - (-U8V) = 0.42 V Mn + Zn 2+ -_I Mn 2+ + Zn By using Equation 19.1, we can predict the direction of an overall cell reaction. . It is important to understand that the standard reduction potential is an intensive property (like temperature and density), not an extensive property (like mass and volume) [I ~~ Section 1.4]. This means that the value of the standard reduction potential does not depend on the amount of a substance involved. Therefore, when it is necessary to multiply one of the half-reactions by a coef- ficient in order to balance the overall equation, the value of EO for the half-reaction remains the same. Con sider a galvanic cell made up of a Zn half-cell and an Ag half-cell: The half-cell reactions are Zn(s) I Zn 2 + (1 M) Ag + (1 M) I Ag(s) Ag+ + e- -_. Ag Zn 2+ + 2e- +. Zn EO = 0.80 V EO = -0.76V The Ag half-reaction, with the more positive standard reduction potential, will occur as a reduc- tion, and the Zn half-reaction will occur as an oxidation. Balancing the equation for the overall cell reaction requires mUltiplying the reduction (the Ag half-reaction) by 2: 2(Ag + + e- • Ag) We can then add the two half-reactions and cancel the electrons to get the overall, balanced equatior:: 2Ag + + X' -_I 2Ag + Zn • Zn 2 + + X' 2Ag + + Zn • 2Ag + Zn 2+ The standard cell potential can be calculated using Equation 19,1: EO - EO EO cell - cathode - an ode = 0.80 V - (-0.76 V) = 1.56 V , SECTION 19.3 Standard Reduction Potentials 769 Although we !:l:~ . 1 . t!p.~~~4 . ~~~ ~g . ~~~~ ~ r . c: a :-; ~~,?~ . ? y' . ?" . ~~ . ~i . d I! ?t. !:l:~ . l . t~p~ y ~~~ . ~ . t . ~~~ . ~~ ~~ . ~~~~! ?~ . potential by 2. Table 19.1 is really a more extensive and quantitative version of the activity series [ ~~ Section 4.4]. Standard reduction potentials can be used to determine what, if any, redox reac- tion will take place in solution. Sample Problems 19.2 and 19.3 illustrate how to use standard reduction potentials to predict the direction of the overall reaction in a galvanic cell and to predict what, if any, redox reaction will take place in solution. A galvanic cell consists of an Mg electrode in a 1.0 M Mg(N0 3 )2 solution and a Cd electrode in a 1.0 M Cd(N0 3 )2 solution. Determine the overall cell reaction, and calculate the standard cell potential at 25°C. Strategy Use the tabulated values of EO to determine which electrode is the cathode and which is the anode, combine cathode and anode half-cell reactions to get the overall cell reaction, and use Equatio'n 19.1 to calculate E ~e ll' Setup The half-cell reactions and their standard reduction potentials are Mg 2+ + 2e - +. Mg Cd 2+ + 2e- • Cd EO = -2 .37V EO = -OAOV Because the Cd half-cell reaction has the greater (less negative) standard reduction potential, it will occur as the reduction. The Mg half-cell reaction will occur as the oxidation. Therefore, E ~a ' hode = -OAO V and E~nod e = -2.37 V. Solution Adding the two half-cell reactions together gives the overall cell reaction: The standard cell potential is Overall: Mg + Cd 2 + -_. Mg 2+ + Cd EO - EO EO cell - cathode - anode = (- OAO V) - ( -2.3 7 V) = 1.97 V Practice Problem A Determine the overall cell reaction and E ~ell (at 25 °C) of a galvanic cell made of a Cd electrode in a 1.0 M Cd(N0 3 )2 solution and a Pb electrode in a 1.0 M Pb (N0 3 )2 solution. Practice Problem B Determine the overall cell reaction and E ~ell (at 25 °C) of a galvanic cell made of an Al electrode in a 1.0 M AI(N0 3 )3 solution and a Cu electrode in a 1.0 M CU (N0 3 )2 solution. Predict what redox reaction will take place, if any, when molecular bromine ( Br 2) is added to (a) a 1 M solution of NaI and (b) aIM solution of NaCI. (Assume a temperature of 25°C.) Strategy In each case, write the equation for the redox reaction that might take place and use EO values to determine whether or not the propos ed reaction will actually occur. Setup From Table 19.1: BriZ) + 2e- -_. 2Br-(aq) I 2 (s) + 2e- • 2r(aq) CI 2 (g) + 2e - • 2CI-(aq) EO = 1.07 V EO = 0.53 V EO = 1.36 V ( Continued) It is a common error to multi ply E" values by the same number as the half-cell equ atio n. Th ink of the reduction potential as the he ight of a waterfall. Just as water fal ls'from a hig her elevat i on to a lower elevat i on, electrons mo e from an electrode with a hig h er poten tia l to :.3::: one with a lower potential. The amoun t of w ater falling does not affect the net cha nge II' its elevation. Likewise, the nu mbe r of elect rons moving from higher potential to lower potentic does not aff ect the si ze of the change in potential. Think About It If you ever calculate a negative voltage for a galvanic cell potential, you have done something wrong-check your work. Under standard-state condition s, the overall cell reaction will proceed in the direction that gives a positive E ~ell ' ! , [...]... concentration cell is typically 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 equal, E becomes zero and no further change occurs Bringing Chemistry to Life Biological Concentration Cells A biological cell can be compared to a concentration cell for the purpose of... but differing in ion concentrations Such a cell is called a concentration cell Consider a galvanic cell consisting of a zinc electrode in 0.10 M zinc sulfate in one compartment and a zinc electrode in l.0 M zinc sulfate in the other compartment (Figure 19.5) According to Le Chatelier's principle, the tendency for the reduction Zn2+(aq) + 2e- • Zn(s) Figure 19.5 A concentration cell Oxidation occurs... bridge Cathode (+ ) N;:a~::;=.~N~a~-======-::s-::o~= """'" _ + + ~ Zn • o o Zn2+ (0.10 M) 02 o Zn2 + (1.0 M) o 776 CHAPTER 19 Electrochemistry to occur increases with increasing concentration of Zn2+ ions Therefore, reduction should occur in the more concentrated compartment and oxidati on should take place on the more dilute side The cell diagram is Zn(s) I Zn 2+(aq)(O lO M) II Zn2+ (aq)(1.0 M) I... interested in science after reading a book on chemistry Faraday invented the electric motor and was the first person to demonstrate the principle governing electric generators Besides making notable contributions to the fields of electricity and magnetism, Faraday also worked on optical activity and di scovered and named benzene • 77 772 CHAPTER 19 Electrochemistry • And, by converting to the base-lO... notable application of fuel cells is their use in space vehicles Hydrogen-oxygen fuel cells provide electric power (and drinking water!) for space flight • 780 _ [ ~ CHAPTER 19 Electrochemistry ~ Electrolysis Multimedia Electrochemistry-the electrolysis of potassium iodide Cell Type Chemical Reaction Electric Energy Galvanic Electrolytic Spontaneous Nonspontaneous Producea Consumed In Section 19.6, we... l.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 the weak acid, carbonic acid (H 2C0 3) The Fe2+ ions formed at the anode are further oxidized by oxygen as follows': o M edia Player/ M PEG Content Eledrochemistry- The produdion of aluminum From Table 19 1: + 2e- - _ Fe(s) EO = - 0.44 V • AI(s)... Walter Hermann Nernst (1864-1941) German chem ist and physicist Nernst's work was mainly on electrolyte solutions and thermodynamics He also invented an electric piano Nernst was awarded the Nobel Prize in Chemistry in 1920 for hi s contribution to thermo ~ SECTION 19.5 Spontaneity of Redox Reactions Under Conditions Other Than Standard State 77':> Think About It For this reaction to be spontaneous as written,... cap + 2e - + 4H+(aq) + SO~ - (aq) + 2e- - _ I PbSOis) + 2H20(l) + Pb02(s) + 4H+(aq) + 2S0~-(aq) barrier • 2PbS0 4 (s) + 2H20(l) • Figure 19.7 alkaline battery Interior view of an 778 CHAPTER 19 Electrochemistry Figure 19.8 Interior view of a lead storage battery Under normal operating conditions, the concentration of the sulfuric acid solution is about 38 percent by mass Anode I - Removable cap / Cathode... carbon electrode -~""'- containing Ni and NiO o o Hot KOH solution -\-• Motor Oxidation 2H2(g) + 40H-(aq) Reduction OzCg) + 2H20(l) + 4e- • 4H20(l) + 4e - • 40H - (aq) through the anode and cathode compartments (Figure 19.9), where the following reactions take place: Anode: 2H2(g) Cathode: 0 2(g) + 40H-(aq) + 4H20(l) + 4e- + 2H20(l) + 4e- Overall: 2H 2(g) • 40H - (aq) + 0 2(g) • 2H20(l) Using EOvalues... conductors, and they provide the necessary surfaces for the initial decomposition of the molecules into atomic species, which must take place before electrons can be transferred Electrodes that serve this particular purpose are called "electrocatalysts." Metals such as platinum, nickel, and rhodium also make good electrocatalysts In addition to the Hz-0 2 system, a number of other fuel cells have been developed . charge in the anode compart- ment (due to the departure of electrons and the resulting formation of Zn 2 +) and the buildup of negative charge in the cathode compartment (created by the. concentrations in the two compartments approach each other. When the concentrations of the ions in the two compartments are equal, E becomes zero and no further change occurs. Bringing Chemistry to Life. cathode, but in (b) it is the anode. 766 CHAPTER 19 Electrochemistry I Again, the direction of electron flow is determined in part by the mass of the electrode. Reduction of Cu '+