8. ELECTROCHEMICAL CELLS. n Electrode Reactions and Electrode Potentials a. H 2 2H + + 2e. Cl 2 + 2e 2Cl. H 2 + Cl 2 2H + + 2Cl ; z = 2

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1 8. ELECTROCHEMICAL CELLS n Electrode Reactions and Electrode Potentials 8.1. a. H H + + e Cl + e Cl H + Cl H + + Cl ; z = E = E RT F ln ( a H +a Cl ) b. Hg(l)+ Cl Hg Cl + e H + + e H Hg + H + + Cl Hg Cl + H ; z = E = E + RT F ln ( a H +a Cl ) c. Ag + Cl AgCl(s)+ e e + Hg Cl (s) Hg + Cl Ag(s) + Hg Cl (s) AgCl(s) + Hg(s) u u d. E = E (no concentration dependence) 1 H (g) H+ + e AuI(s) + e Au(s) + I AuI(s) + 1 H (g) Au(s) + H+ + I ; z = 1 E = E RT F ln (a H +a I )u

2 158 n CHAPTER 8 e. Ag(s) + Cl (a 1 ) AgCl(s) + e AgCl(s) + e Ag(s) + Cl (a ) Cl (a 1 ) Cl (a ); z = 1 E = RT F ln a 1 a 8.. E a. A + H + + e AH 0.60 V B + H + + e BH 0.16 V AH + B A + BH AH is oxidized by B; the half-cell A + H + + e AH is written as a reduction. In the overall reaction this equation is reversed and therefore represents an oxidation. Because the net reaction has a positive emf (exergonic), the reaction is spontaneous in the forward direction. b V ( 0.60 V) = 0.44 V c. [H 3 O + ] does not appear in the equilibrium expression, and the hydrogen-containing entities cancel in the numerator and denominator. Therefore, there is no effect of ph on the equilibrium ratio The G values for the two reactions are Cr e Cr G 1 = ze 1 F = J mol 1 (1) Cr 3+ + e Cr + G = ze F = J mol 1 () The reaction Cr + + e Cr is obtained by subtracting reaction () from reaction (1), and the G value for Cr + + e Cr is obtained by subtracting G from G 1 : G = ( ) J mol 1 = J mol 1 Since Cr + + e Cr involves two electrons, and since G = ze F, it follows that 1.81 F = (E /V)F or E = 0.90 V 8.4. Left-hand electrode H H + (1 m) + e Right-hand electrode e + H + (aq) + F S Overall reaction H + (aq) + F + H H + (1 m) + S The expression for the emf of the cell is E = E RT F ln [S ] [1 m] [F ] c

3 ELECTROCHEMICAL CELLS n a. In writing the representation of the cell, the oxidation reaction occurs at the anode, which is placed at the left-hand position of the cell. In this case Fe + is losing electrons, and the oxidation process is Fe + Fe 3+ + e The cathode reaction is written on the right-hand side of the cell and is Ce 4+ + e Ce 3+ where reduction occurs. The overall reaction is the sum of these two reactions. The cell representation is Fe 3+ (aq) Fe + (aq) MMCe 4+ (aq) Ce 3+ (aq) The symbols for both sides are always written as though they were a reduction process, i.e., oxidized form reduced form. The voltage of this cell is then the reduction potential of the right-hand electrode minus the reduction potential of the left-hand electrode. Thus E = E Ce 4+ Ce 3+ E Fe 3+ Fe+ = 1.44 V V = 0.67 V b. Upon examining the standard reduction potentials in Table 8.1, we see that the following halfcell reactions can be combined to give the cited reaction. Ag + (aq) + e Ag V (1) AgCl(s) + e Ag + Cl 0.4 V () Reversal of the second equation and addition gives the desired equation: Ag + (aq) + Cl (aq) AgCl(s) Equation (1) is the reduction reaction and is placed on the right-hand side. The anode reaction is placed on the left-hand side of the cell representation. Ag AgCl(s) Cl (aq) MM Ag + (aq) Ag(s) The voltage of this cell is the right-hand reduction potential minus the left-hand reduction potential. E = E Ag + Ag E AgCl Ag = V 0.4 V = V c. HgO undergoes reduction to Hg and is the cathode. H is oxidized and is the anode. The electrode potentials are found in the CRC Handbook. H O + e H + OH V (3) HgO + H O + e Hg + OH V (4) Reversing the sense of equation (3) and adding to (4) gives HgO(s) + H (g) Hg(l) + H O(l) The cell is represented by Pt, H O(l) H (g), OH (aq) MM OH (aq), H O HgO(s) Hg(l) The cell potential is E = E HgO Hg E H O H = V ( V) = 0.954

4 160 n CHAPTER 8 n Thermodynamics of Electrochemical Cells 8.6. Fe 3+ /Fe + = V; I /I = V E = = V ln K u c = = K c = dm 6 mol 8.7. The E for the process Sn + Fe + Sn + + Fe is ( 0.136) = V Since z = we have = ln K u c ln K = K c = G = z J; z = 1 = J mol 1 = 4.1 kj mol From Table 8.1 the relevant values are O + 4H + + 4e H O E = 1.3 V (1) H + + e H E = 0 () Subtraction of () from 1 (1) gives the required equation H + 1 O H O with E = 1.3 V and z =. Then, G = J mol 1 = J mol 1 = 37.4 kj mol From Table 8.1, Cu + + e Cu E 1 = V (1) Cu + + e Cu + E = V ()

5 ELECTROCHEMICAL CELLS n 161 If we subtract () from (1) we obtain Cu + Cu + + Cu E = E 1 E = V; z = ln K u F = (8.3145)(98.15) = K = dm 3 mol 1 = 14.3 If Cu O is dissolved in dilute H SO 4, half will form Cu + and half Cu Note that the G given is for the reaction of 3 moles of H to form moles of Sb. The electrode reactions may be written as 3H 6H + + 6e, Sb O 3 + 6H + + 6e Sb + 3H O. Using Eq. (8.), we get E = J/( ) = V. Since this reaction is spontaneous, the electron flow is from the hydrogen electrode (negative) to the antimony electrode (positive). n Nernst Equation and Nernst Potentials 8.1. E = RT zf ln m 1 m = ln = V The process is Pyruvate + H + + e lactate and the Nernst equation is E = E RT F ln [lactate ] [pyruvate ] Then E /V = E = V a. From Table 8.1, ln Fe 3+ + e Fe + E 1 Fe + + e Fe E = V (1) = V () The corresponding G values are:

6 16 n CHAPTER 8 G 1 = = J mol 1 (1) G = ( 0.440) = J mol 1 () The required reaction is obtained by adding (1) and (): Fe e Fe G = kj mol 1 Then, since z = 3, E = = V b. The electrode reactions are Sn + Sn 4+ + e E 1 3e + Fe 3+ Fe E 1 = 0.15 V (1) = V () The cell reaction is 3Sn + + Fe 3+ 3Sn 4+ + Fe E = V; z = 6 From the Nernst equation, Eq. 8.13, E = ln = = 0.16 V From K a = [H+ ][CH 3 COO ] [CH 3 COOH] (0.01) 3 (0.1) 3 (0.5), we obtain [H + ] = K a [CH 3 COOH] [CH 3 COO ]. The concentration of acetate ions formed from the dissociation of acetic acid will be negligible compared to the acetate concentration from the fully dissociated sodium salt. Therefore, [H + ] = = M. The cell reactions are (see Table 8.1): H (g) H + + e ; Hg Cl + e Hg + Cl ; E = V (by definition) E = V However, since the cathode reaction (reduction) is accounted for by the standard reduction potential of the electrode, the Nernst equation contains only the hydrogen ion concentration. For this concentration, the Nernst equation gives E = ( ) ln [ ] = V The reactions taking place during the electrolysis are

7 ELECTROCHEMICAL CELLS n 163 Cu + + e Cu; E = V Br Br + e ; E = V If the reverse reaction were taking place in a galvanic cell, the initial cell voltage is given by E = E ln [Cu + ][Br ]. E = ( ) ln ( ) = V. Therefore, a minimum voltage of V will have to be applied at the beginning in order for the electrolysis reaction to occur. At the end of the electrolysis, the concentrations are: [Cu + ] = M (.87/63.456) = M. [Br ] = M (.87/63.456) = M. Therefore, the voltage required will be ( ) ln ( ) = V For the reaction I + I I 3 E = V and z = K c = [I 3 ] [I ] = [I3 ] 0.5 mol dm = [I 3 ln ] 0.5 mol dm 3 ln [I 3 ]u 0.5 = Φ/V = Φ [I 3 ]u 0.5 = 0.95 [I 3 ] = M = V = 10.4 mv ln From Table 8.1, we see that Au + has a much higher reduction potential than Pb +. Therefore, the gold will be deposited first. As the Au + concentration falls, the lead begins to be deposited, i.e., we have

8 164 n CHAPTER 8 Au(s) + Pb + Pb(s) + Au + for which we write (see Equation 8.7) E = ( ) V ln [Au+ ] or [Au + ] = e ( /0.057) = Therefore, [Au + ] = M. The conclusion is that only an infinitesimal amount of gold will be left in the solution by the time the lead starts to deposit at the electrode. Therefore, this is an acceptable way of separating the two metals Reaction at the right-hand electrode: e + H + (0. m) 1 H (10 bar) At the left-hand electrode: 1 H (1 bar) e + H + (0.1 m) Overall reaction: H + (0. m) + 1 H (1 bar) 1 H (10 bar) + H+ (0.1 m); z = 1 Cell emf: E = RT F E/V E 0. (1 bar)1/ ln 0.1 (10 bar) 1/ = ln = 11.8 mv 10 = V 8.1. The reactions at the two electrodes are 1 H (1 bar) H+ (0.1 m) + e H + (0. m) + e 1 H (10 bar) Every H + ion produced in the left-hand solution will have to pass through the membrane to preserve electrical neutrality. H + (0.1 m) H + (0. m) The net reaction is 1 H (1 bar) 1 H (10 bar) E = RT F ln 1 10 = ln 1 10 = V = 9.6 mv

9 ELECTROCHEMICAL CELLS n The capacitance of the cell membrane (see Eq. 8.0) is C = F = F a. With a potential difference of V, the net charge on either side of the wall is Q = CV = = C b. The number of K + ions required to produce this charge is Q e = = The number of ions inside the cell is = The fraction of ions at the surface is therefore = At the left-hand electrode: 1 H H+ + e At the right-hand electrode: CrSO 4 (s) + e SO 4 + Cr(s) Overall reaction: CrSO 4 (s) + H H + + SO 4 + Cr(s) E = 0.40 V and z = a. E = E RT F ln ([H+ ] [SO 4 ]) u E/V = E = 0.15 V b. The ionic strength is ln (0.00) (0.001) I = 1 { ( ) } = M Historically, common logs have been used in activity coefficient calculations where the value of B is easily remembered as log 10 γ ± = = γ ± = 0.879

10 166 n CHAPTER 8 E = E RT F {ln ([H + ] [SO 4 ]) u + ln γ + γ } = E RT F {ln ([H + ] [SO 4 ]) u + ln γ 3 ± } E/V = At the left-hand electrode: Cu(s) Cu + + e ln (0.879) 3 E = = At the right-hand electrode: AgCl(s) + e Ag(s) + Cl Overall reaction: AgCl(s) + Cu(s) Ag(s) + Cu + + Cl, z = To a good approximation, it can be assumed that the activity coefficients are unity at 10 4 M (DHLL gives γ ± = 0.988). Then E = E RT zf ln ([Cu+ ][Cl ] ) u = E /V = E /V E = V ln [ 10 4 ( 10 4 ) ] u Suppose that at 0.0 M the activity coefficients are γ + and γ. Then = ln [0.0(0.40) ] ln γ + γ = ln γ 3 ± ln γ 3 ± =.179 γ 3 ± = γ ± = a. The anode reaction: Tl(s) + Cl (0.0 m) TlCl(s) + e The cathode reaction: Cd + (0.01 m) + e Cd(s) The overall cell reaction:

11 ELECTROCHEMICAL CELLS n 167 Tl(s) + Cd + (0.01 m) + Cl (0.0 m) TlCl(s) + Cd(s); z = b. This reaction can alternatively be written as () Tl(s) + Cd + (0.01 m) Tl + (in 0.01 m CdCl ) + Cd(s); z = for which the Nernst equation is E = E Cd + Cd E Tl + Tl V E = 0.40 ( 0.34) = 0.06 V ln ([Tl + ] /[Cd + ]) u When a CdCl = 1, E 1 is the value of E, and substituting [Tl+ ] = K sp /[Cl ] we have E 1 /V = ln (K sp /[Cd+ ][Cl ] ) u = ln ( ) E 1 = 0.06 ( 0.165) = V When m = 0.01 m, E/V = ln ( ) /(0.01)(.0) E = 0.06 ( ) = V 8.6. The diffusible K + ions are at a higher potential on the right-hand side of the membrane; there is thus a tendency for a few of them to cross to the left-hand side and create a positive potential there. (The same conclusion is reached by considering the diffusible Cl ions; they are at a higher concentration on the left-hand side, and a few tend to cross to the right-hand side and create a negative potential there.) The Nernst potential is given by Eq. 8.19: Φ = RT zf ln c 1 c = = V = 37 mv ln The overall reaction is Lactate + cytochrome c (Fe 3+ ) pyruvate + cytochrome c (Fe + ) + H + with z = and E = = V. If K is the ratio at ph 7, E = = RT F ln K = ln K ln K = K = If K is the ratio at ph 6,

12 168 n CHAPTER 8 so that K true = K (10 7 ) = K (10 6 ) K = K 10 = The processes are AgCl(s) + e Ag(s) + Cl (0.1 m) Ag(s) + Cl (0.01 m) AgCl(s) + e The electrical neutrality is maintained by the passage of H + ions from right to left: H + (0.01 m) H + (0.10 m) The overall process is H + (0.01 m) + Cl (0.01 m) H + (0.10 m) + Cl (0.10 m) The emf is E = RT F ln E/V = ln 100 E = V 8.9. a. The processes at the electrodes are 1 H H+ (m 1 ) + e H + (m ) + e 1 H To maintain electrical neutrality of the solutions, for every mole of H + produced in the lefthand solution, t + mol of H + ions will cross the membrane from left to right, and t mol of Cl ions will pass from right to left. In the left-hand solution there is therefore a net gain of (l t + ) mol = t mol of H + and of t mol of Cl. In the right-hand solution, the net loss is (l t + ) mol = t mol of H + and t mol of Cl. The overall process is thus The emf is t H + (m ) + t Cl (m ) t H + (m 1 ) + t Cl (m 1 )

13 ELECTROCHEMICAL CELLS n 169 E = RT F = t RT F t m ln 1 mt 1 m t mt ln m m 1 b. For m 1 = 0.01 m and m = 0.10 m, = t ln 10 t = t + = M M + (0.1 m) + e Cell reaction: H + (0. m) + e 1 H (1 bar) H + (0. m) + M M + (0.1 m) + 1 H (1 bar) E = E H + H E M + M The Nernst equation is (0.1) 0.4 = E M + /V ln M (0.) = E M + /V M E M + M = V Upon addition of KCl, almost all of the M + precipitates, and 0.10 m Cl is in excess. The value of M + in solution is found from the K sp, namely K sp = [M + ][Cl ] = (M + ) (0.10 M) and from the Nernst equation 0.1 = E M + M ln K sp /(0.10) = ln K u sp ln = ln K u sp ln K u sp = = 7.08 K sp = mol kg The E values for the half reactions (Table 8.) are pyruvate + H + + e lactate E = 0.19 V (1) NAD + + H + + e NADH E = 0.34 V ()

14 170 n CHAPTER 8 The required reaction is obtained by subtracting () from (1). Hence E = 0.19 ( 0.34) = 0.15 V Since n =, the corresponding G is G = J = J a. The equilibrium ratio at ph 7 can be calculated from the relationship G = RT ln K = ln K ln K = 8 946/( ) = K = dm 3 mol 1 Alternatively, K could have been calculated directly from E, using Eq. (8.7): 0.15 = ln K ln K = K = dm 3 mol 1 This K at ph 7.0 is related to the true (ph-independent) K by the equation K = [lactate ][NAD + ] [pyruvate ][NADH][H + ] = K [H + ] = K 10 7 b. Similarly the K at ph 8.0 is related to K by K = K 10 8 Thus and K K 10 8 = 10 7 K = =

15 ELECTROCHEMICAL CELLS n 171 n Temperature Dependence of Cell emfs 8.3. a. We are given that fumarate + H + + e succinate E = V (1) pyruvate + H + + e lactate E = V () Subtraction of () from (1) gives fumarate + lactate succinate + pyruvate E = 0.16 V; z = (Note that this is also E, the hydrogen ions having canceled out.) The Gibbs energy change is G = zfe = = 41.7 kj mol 1 The entropy change is obtained by use of Eq. 8.3: S = J K 1 mol 1 = 4.1 J K 1 mol 1 b. The enthalpy change can be calculated by use of Eq. 8.5, or more easily from the G and S values: H = G + T S = ( ) J mol 1 = J mol 1 = 40.4 kj mol a. The individual electrode processes are Cd(Hg) Cd + + e Hg + + e Hg and the overall reaction is Cd(Hg) + Hg + Cd+ + Hg; z = Since the solution is saturated with Hg SO H O the overall reaction can be written as Cd(Hg) + Hg SO 4 (s) H O(l) CdSO H O(s) + Hg(l) b. G = = J mol 1 = kj mol 1 S = ( ) = 9.65 J K 1 mol 1 H = ( ) = J mol 1 = kj mol 1

16 17 n CHAPTER Since we are only interested in the slope (i.e., not the intercept) of the straight line that best fits the E vs. t data, we need not bother to convert the temperature to Kelvin. We perform a linear regression analysis using t as the independent variable and E as the dependent variable. The result is: E = t. Differentiation with respect to t gives E t P = V ( C) The entropy change for the conversion of one mole of (½)Br to Br (z = 1) is C mol V K 1 = 7.80 J K 1 mol 1. 1 E = T = The cell reaction is Mg(s) + Cl (g) Mg + (aq) + Cl (aq); i.e., z =. From Eq. (8.3), we get P 4 V K 1 E T P S = zf JK mol 3 1 = = V K C mol E = E = = 0.8 V G = = J mol 1 a. If x kj mol 1 = G f (fumarate), x = x = G f = kj mol 1 b. S = ( ) = 8.0 J K 1 mol 1 H = G + T S = ( ) = J mol 1 If y kj mol 1 = H f (fumarate), y = 5.35 y = H f = kj mol a. The cell reaction can be written Tl + H + (a = 1) Tl + (in HBr; a = 1) + 1 H (1 bar) E = E H + H E Tl + T ln ([T1 + ]/[H + ]) u

17 ELECTROCHEMICAL CELLS n 173 Since Tl + = K sp /[Br ] E/V = ln (K sp /[H + ][Br ]) u = ln K u sp = E = 0.58 V b. H for the reaction is H for the half cell Tl Tl + + e H = zf(e TdE/dT) = [( ( 0.003)] = (1.34) J mol 1 = rounded to 119 kj mol 1 n Applications of emf Measurements Subtraction of the second reaction from the first gives AgBr(s) Ag + + Br E = V; z = 1 Then = ln ([Ag + ][Br ]) u ln ([Ag + ][Br ]) u = K sp = [Ag + ][Br ] = mol kg Solubility = K sp = mol kg The standard emf of the AgCl Ag electrode is 0.4 V and the cell reaction is AgCl(s) + 1 H H+ + Cl + Ag(s); z = 1 E = E RT F E RT F ln (a H +)u ln (a H +)u = log 10 (a H +) u log 10 (a H +) u = ph = = The cell is represented as Ag AgI(s) I (aq) MM Ag + (aq) Ag(s).

18 174 n CHAPTER 8 At the two electrodes: e + AgI(s) Ag(s) + I Ag(s) Ag + + e Overall reaction: AgI(s) Ag + + I ; z = 1 E = RT F ln[m Ag+ m Cl ]u = RT F ln K sp = ln K u sp K sp = mol kg Solubility = K sp = mol kg The standard cell potential for the cell shown is E = ( ) = V. Since the molality is exactly 1.0, Eq. (8.43) simplifies to = ln γ ±. Solving for the mean activity coefficient, we calculate γ ± = exp[ ( )/( 0.057)] = a. The electrical work is G. G = H T S G /J mol 1 = ( 43.7) = J mol 1 = 750 kj mol 1 Electrical work available = 750 kj mol 1 b. The total work done is A. G = A ΣνRT Σν = = 3.5 ΣνRT = = 8676 J mol 1 A = = 739 kj mol 1 Total available work = 740 kj mol a. The overall process is Cd + AgCl Ag + CdCl and is a two-electron process. G = G + RT F ln a Cd +a Cl where a is the activity of the respective ions.

19 ELECTROCHEMICAL CELLS n 175 E = E RT F ln a Cd +a Cl = RT F ln (m Cd +γ Cd +m Cl γ Cl ) = RT F ln[(0.01γ Cd +)(0.0) γ Cl )] ln γ 3 ± = γ ± = = (98.15) (96 485) ln ( )γ 3 ± b. The ionic strength of a m solution of CdCl is I = 1 Σm i z i = 1 [(0.010 ) + ( )] = mol kg 1 From the Debye-Hückel limiting law, log 10 γ ± γ ± = Az + z I = = = = 0.67 (to the degree of accuracy allowed by the problem) The two values are in good agreement The LiOH is required for the hydrogen electrode and the LiCl salt is used to complete the AgCl electrode. Both the Cl ion and the H + ion will behave according to their activities in solution. Begin by determining the emf of the cell. E cell = E AgCl E H = E AgCl RT F ln a Cl RT F ln a H + This can be rewritten using K w = a H + a OH, and so the desired relationship is established. E cell = E AgCl RT F ln a Cl RT F Combining the two activity terms gives E cell = E AgCl RT a F ln Cl a RT OH F ln K w + RT F ln a OH ln K w

20 176 n CHAPTER 8 Rewriting this in terms of activity coefficients and molalities gives, after rearrangement, (E cell E AgCl )F RT + ln (m Cl /m OH ) = ln K w ln γ Cl γ OH The value of m OH is 0.01 mol kg 1. With that substitution, plot the left-hand side of the equation with E AgCl = 0.4 V at K against the ionic strength, which varies with the concentration, and extrapolate to zero ionic strength. At zero ionic strength the γ Cl and γ OH approach unity. Then the value of the curve is ln K w. In the following data, I is based on m m OH. The latter is constant. For example, I = 1/[ ( 1) ( 1) ]. m/mol kg I/mol kg (E cell E AgCl )F RT m ln Sum of last terms From the indicated plot shown, the value of ln K w = 3.6. The value of K w is

21 ( ( ELECTROCHEMICAL CELLS n 177 m 0.01 mol kg 1 (E cell E AgCl )F + ln RT ( 3.4. Plot of (E cell E AgCl )F m + ln RT ( 0.01 mol kg 1 against I I/mol kg 1

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