CH 11 - Chapter 11: Electrochemical Cells, Batteries, and...

Info iconThis preview shows pages 1–19. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 16
Background image of page 17

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 18
Background image of page 19
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 11: Electrochemical Cells, Batteries, and Fuel Cells Problem numbers in italics indicate that the solution is included in the Student is Solutions Manual. Questions on Concepts Qll.l) To determine standard cell potentials, measurements are carried out in very dilute solutions rather than at unit activity. Why is this the case? In order to determine the standard potential, the mean ionic activity must be known. It can be calculated in dilute solutions using the Debye-«Huckel limiting law, but there is no reliable way to calculate the activity coefficient near unit activity. Q11.2) Show that if AG‘:(H+,aq) = O for all T , the potential of the standard hydrogen electrode is zero. a 1 . 'UH" "_'uH O o _ 2 2 _ Ill/1t o __ 70 + _ (AF/H2 — ————F — -—F— and pH. _ Ac, (H ,aq) _ 0 Q11.3) How is it possible to deposit Cu on an Au electrode at a potential lower than that corresponding to the reaction Cu2+ (aq) + 2e‘ : Cu(s)? Cuzu is more tightly bound to the Au surface than to Cu. Thus a lower potential is needed to reduce Cu} onto the Au surface. Q11.4) Explain why the magnitude of the maximum work available from a battery can be greater than the magnitude of the reaction enthalpy of the overall cell reaction. = |—AG > [—AHy if AS < 0 W cleclrt‘cal Qll.5) How can one conclude from Figure 1 1.18 that Cu atoms can diffuse rapidly over a well—ordered Au electrode in an electrochemical cell? The copper atoms would be dispersed randomly over the Au surface if they could not diffuse laterally. Instead, they are seen only in islands and on edges, where they are tightly bound. They must have diffused there freely over the surface. Q11.6) The temperature dependence of the potential of a cell is vanishingly small. What does this tell you about the thermodynamics of the cell reaction? A a. 0 Because AS; 2 —[a GR} : nF ,AS; z 0 P l’ 8T 252 Chapter 11/ Electrochemical Cells, Batteries, and Fuel Cells Q11.7) Why is the capacitance of an electrolytic capacitor so high compared with conventional capacitors? The positive and negative plates of an electrolytic capacitor are separated by an extremely small distance (the thickness of the electrical double layer, essentially). And so, the electric field is very strong and the A o capacitance large. [2—, 0’ ~ 30 A] Q11.8) What is the difference in the chemical potential and the electrochemical potential for an ion and for a neutral species in solution? Under what conditions is the electrochemical potential equal to the chemical potential for an ion? [1 = ,u + z¢ . For a neutral species, the chemical potential and electrochemical potential have the same value. They differ by z¢ for an ion. The electrochemical potential is equal to the chemical potential for an ion if ¢ = 0. Q11.9) Why is it possible to achieve high—resolution electrochemical machining by applying a voltage pulse rather than a dc voltage to the electrode being machined? For a tool that is very close to a surface, the RC time constant for charging the double layer depends strongly on location. Thus, pulsed voltage charges will charge the layer in those locations where the tool is closest to the surface—allowing subsequent electrochemical reaction—but not in other locations. Therefore, the reactions are localized. This gives the high spatial selectivity needed for nanomachining. Q11.10) Can specifically adsorbed ions in the electrochemical double layer influence electrode reactions? Yes, specifically adsorbed ions can play a central role in the atomic—scale processes occurring on an electrode surface (e.g., C1“ on Cu electrode). Problems P11.1) You are given the following half-cell reactions: Pd2+(aq) + 26 : Pd(s) E" = 0.83 v PdCli‘(aq) + ZeT : Pd(s) + 4Cl‘(aq) E" = 0.64 v a. Calculate the equilibrium constant for the reaction Pd2+ (aq) + 4Cl‘(aq) 2 PdClE; (aq) b. Calculate AGO for this reaction. 253 Chapter 11/ Electrochemical Cells, Batteries, and Fuel Cells a) Pd“ (aq)+2e' :: Pd(s) E" = 0.83 V Pd(s) + 4Cl' (aq) a PdC1§'(aq)+2e' E“ = «0.64 V Pd“ (aq) + 4C1‘(aq) : PdClj‘ (aq) E z 0.83 + (-0.64) = 0.19 V E zflln K nF K : ell/“Izw/RT 2(96,485 C m01")><(0.l9 V) (8.3145 J mol“K")x(298.15 K) 2 2'65X106 K =exp b) AG” 2 ~nFE° = (—2) x (96,485 C mol" ) x (0.19 V) = —36.7 kJ mol" . P11.2) For the half—cell reaction AgBr(s) + e’ —> Ag(s) + Br“(aq), E0 = +0.0713 V. Using this result and AGof (AgBr, s) = —96.9 k] mol‘l, determine AG; (Br‘, aq). AG; = —nFE° = A6} (Br‘,aq) — AG; (AgBr,s) AG; (Br',aq) = A6} (AgBr, s) — nFE° AG; (Br',aq) = —96.9 kJ mol" —1 mol><96485 Cmol‘l x0.0713 V = —103.8 kJmor‘ P11.3) For the half—cell reaction Hg2C12(s) + 2e’ —> 2Hg(l) + 2Cl”(aq), E0 : +0.27 V. Using this result and AG} (Hg2C12,s) = “210.7 k] morl, determine AG; (Cl’, aq). AG; = —nFE° = 2m; (01mg) — AG; (Hg2C12,s) —AG;» (ngCl2 ,s) — nFE° 2 —210.7 kJ mol“1 ~ 2 mol X96485 Cmol”l >< 0.26808 V 2 AG;(C1',aq) * 131.2 kJ mol"1 || AG}. (Cl',aq) P11.4) Determine the halfocell reactions and the overall cell reaction, calculate the cell potential and determine the equilibrium constant at 298.15 K for the cell Ag(s)(AgC1(s)(C1’(aq,a(,l_ = 0.00500)||Cdz+(aq,a(fdl. = 0.100)|Cd(s) Is the cell reaction spontaneous as written? Ag(s) + cr(aq) —>AgCl(s) + e" E” = ~ 0.2223 V Cd2+(aq) + 2e' a cam E” = 7 0.4030 V Cd2+(aq) + 2Cl'(aq) + 2Ag(s) —>Cd(s) + 2AgCl(s) : — 0.6253 V E = E — 5Zln 1 , = —0.6253 V 005916102410 1 , V 2F a(y,dzr.a;,/, 2 0.100 x (000500)“ = —0.7910 V 254 Chapter 1 1/ Electrochemical Cells, Batteries, and Fuel Cells nF ., 2x96485 Cmorl x(—0.6253 V) an 2 E : I 1 RT 8.314 J K" mol‘ X298.15 K = —48.678 K = 7.23 x1042 1 #3— : 4 ><105 >> K. Another (0.00500) (0.100) The reaction is not spontaneous as written because Q = criterion is that E < 0. P115) The standard half-cell potential for the reaction 02(g) + 4H+(aq) + 46‘ ——> 2H20(l) is +1.229 V at 298.15 K. Calculate E for a 0.700-molal solution of HN03 for c102 =1 (a) assuming that the a H1 is equal to the molality and (b) using the measured mean ionic activity coefficient for this concentration, 7i = 0.717. How large is the relative error if the concentrations, rather than the activities, are used? E°=1.229V aozzl E=E°——1—€—F7:ln 1 4 ” (as) a) a“. = 0.717 (8.314 J mor‘ K")><(298.15 K) 1 E=1.229V l ln 4 =1.220v (4)><(96,485 C mol‘ ) 0.700 b) aH. = (0.757)x(0.700) (8.3145 J mor1 K'] )x(298.15 K) (4)><(96,485 C mol'l) E =1.03 1n[(0.757)x(0.700)]“ =1.211v The relative error is given by 1.211 V—1.220 V 1.211V P11.6) Determine the half—cell reactions and the overall cell reaction, calculate the cell potential, and 100%=—0.71% determine the equilibrium constant at 298.15 K for the cell . 2+ _ Cu(s)lCu (aq,a(.u.. -0.0150)| H+(aq.aH, = 0.100)|H2(g)|Pt(s) Is the cell reaction spontaneous as written? We first calculate the E0 for the reaction Cu2+(aq) + 2e' —+ Cu(s) from the tabulated values for the reactions Cu2+(aq) + e‘—>Cu+(aq) and Cu+(aq) + e’ —+ Cu(s) in the data tables. Cu2'*(aq) + e" a Cu+(aq) E” 2 + 0.153v Cu+(aq) + e‘ a Cu(s) E” = + 0.521v 255 Chapter 11/ Electrochemical Cells, Batteries, and Fuel Cells As in Example Problem 11.3, m electrons are transferred in the reaction with the potential / B , and n2 electrons are transferred in the reaction with the potential Eff/C. If m electrons are transferred in the reaction with the potential M, , then 113E"4 H, 2 n]/ B + n2EZ/(y. . Therefore _ 0.153V+0.521V Cu2+(aq) + 2e a Cu(s) E“ 2 = 0.337V Cu(s) a Cu2+(aq) + 2e E" : —0.337v mag) + e' —> 1/2 H2(g) E“ 2 0.00v 2H+(aq) + Cu(s) —» Cu2+(aq) + H2(g) °= —0.337 v a , 2+ . . tee—51111 =—0.337V—O—Q&16—V—Iogmmv2—0.342V 2F am 2 (0.100)“ 2 96485 C 1“ —0.337 v ananE": X lmo ) RT 8.314 J K' mol‘ x298.15 K = —26.235 K = 4.04x10*l2 The reaction is not spontaneous as written because Q >> K and E < 0. P11.7) Consider the Daniell cell for the indicated molalities: Zn(S)|Z1’lSO4(aq, 0.100m)llCuSO4(aq, 0.500m)lCu(s) . The activity coefficient A has the value 0.0620 for CuSO4 and 0.150 for ZnSO4 at the indicated concentrations. Calculate E by setting the activity equal to the molality and by using the above values of the yi. How large is the relative error if the concentrations, rather than the activities, are used? Oxidation: Zn(s) —> Zn2+(aq) + 2e- E0 : 0.7618 v Reduction: Cu2+(aq) + 26‘ ——> Cu(s) E0 = 0.337 V Overall: Zn(s) + Cu2+(aq) —> Zn2+(aq) + Cu(s) E0 = 1.099 v R a 2+ EZUII : E:e/l —l1n[ Z” J nF a,“ (.u (1) Activities equal to molality divided by standard molality: 83145J 14 K] 29 1 0.100 molkg4/ ESL” :1.099 ‘ m0 X 8. 5 K In m 2x96,485 C mol" 0.500 mol kg“ mo :1.ll9V 256 Chapter 11/ Electrochemical Cells, Batteries, and Fuel Cells (2) Using yi: 3145 1.,K_1 2 1 0.15x0.100molkg'l E1,,=1.099—§~____J_‘31‘1__x_?§_5_131n __________m_ 21.10“ ‘ 2 x96,485 C motl 0.062x 0.500 mol kg“ m. Relative error: 1.119 V—1.108 V 1.108V P11.8) Determine the half—cell reactions and the overall cell reaction, calculate the cell potential, and x100%=1.0% determine the equilibrium constant at 298.15 K for the cell Pt(s)an2+(aq,ai : 0.0150),Mn3+(aq,ai : O.200)|‘Zn2+(aq,ai = 0.100)|Zn(s) Is the cell reaction spontaneous as written? Mn2+(aq) a Mn3+(aq) + e' E“ = —1 .5415 V Zn2+(aq) + 2e' —+ Zn(s) E0 = ~0.7618 v 2Mn2+(aq) + Zn2+(aq) ——>2Mn3+(aq) + Zn(s) E" 2 —2.303 V a2 0.200 2 E = E" ——1£ln—é—M’L—— = —2.303 V —w—logm 4—9—— = —2.340 V 2F away. 2 (0.150) 0.100 nF _ 2x96485 C mol‘1x(—2.303 V) an = E” — l 1 RT 8.314JK‘ mol” x298.15 K =—179.3 K=1.34x10'78 The reaction is not spontaneous as written because E < 0. P11.9) Consider the half-cell reaction AgCl(s) + ew :: Ag(s) + Cl'(aq). If y° (AgCl, s) : —109.71 k] mol‘l, and if E0 = +0222 V for this half—cell, calculate the standard Gibbs energy of formation of C1“ (6161). AG; = —109.71kJ mor‘ AG; = —1><96,485 C mol'] x 0.222 V = 21.4 kJ mol" AG; = AG; (C1‘(aq))= AG1°+ AG; 2 —131.1kJ mol" P11.10) For a given overall cell reaction, AS; = 22.0 J mol"1 K" and AH; = ~315.0 kJ mol‘1 . Calculate E0 and QED/07),. Assume that n : 1. 257 Chapter 11/ Electrochemical Cells, Batteries, and Fuel Cells E, _ _ AG; _ _ AH; — ms; _ _ —225.0 k1mol" —298.1 5 K x175 JK“ mol" nF nF 1x96485 C moi" = 2.39 V [an] 109,: 17.51K"mol"‘ 1) — = _1 —1.81><10'4 V K" 6T 11F 1x96485 C mol P11.11) Consider the cell Hg(1)|Hg,so4(s)|Feso4(aq,a = 0.0100)|Fe(s) . a. Write the cell reaction. b. Calculate the cell potential, the equilibrium constant for the cell reaction, and AGO at 25°C. a) Oxidation: 2Hg(l) + 802' (aq) —> Hgst4 (s)+ 2e” EO : —0.6125 V Reduction: Fe2+ (aq) + 2e” —> Fe(s) E0 : ~0.447 V Cell reaction: 2Hg(l) + Fe” (aq) + $03 (aq) —> ngso4 (s) + Fe(s) E33,, = —0.6125 + (—0.447) = —1.0595 V Eccll 2 E26]! ——R—]:1n —1— nF a!»2+ am?“ L 8.3145 J mol’1 K'l x298.15 K = 4.0595 2 _1 ln(1/(0.0100)2) x96,485 C mol 2 —1.178 V AGjmm = —nFE° = —2><96,485 C mol‘l x(—1.0595 v) = 204.5 kJ mol'1 K = e‘AG;”L""”/RT = 1.51 x1036 P11.12) Between 0° and 90°C, the potential of the cell Pt(s)|H2(g, f =1atm)|HCl(aq,m = 0.100)|AgCl(s)|Ag(s) is described by the equation E(V) : 0.35510 —0.3422 x1041 ~3.2347 ><10"(’t2 + 6.314><10‘gt3 , Where 1 is the temperature on the Celsius scale. Write the cell reaction and calculate AG, AH, and AS for the cell reaction at 62°C. Cell reaction: 5H2 (g)+AgC1(S) —> H+(GCI)+C1" (“‘1)+Ag(5) 258 Chapter 11/ Electrochemical Cells, Batteries, and Fuel Cells At 62°C E = 0.35510 V—(0.3422><10'4 V °C" )x62°C —(3.2347x10'° v “C'2)(62)2 “C2 +(6.3l4><10‘9 V °C‘3)(62)3 °C3 =0.3420 v AG : —nFE = (—1)(96,485 C mol" )(0.3420 v) = —33.0 k] mol'l (3E , _ _( 2 [—8711 : ~0.3422x10 5 —2(3.2347x10 6)(62)+3(6.314x10 ’)(62) = —3.625x104 v K" AS = nF[§£j = —35.0 J mol'lK aT ,, (323.15 K)x(—35.0 J mol'1 K") 1000 AH = AG + TAS = —330 kJ mol‘] + = 44.7 Id 11101“ P11.13) a) Calculate AGO reaction and the equilibrium constant, K, at 298.15 K for the reaction Hg2C12(s) —9 2Hg(l) + Clg(g). b) Calculate K using Table 4.1. c) What value of AGR would make the value of K the same as calculated from the half cell potentials? a) Hg2C12(s) + 2e' —» 2Hg(l) + 2C1'(aq) E0 2 + 0.26808 V 2Cl'(aq) —+ C12(g) + 2e' E" = —1.35827 V Hg2C12(s) —> 2Hg(l) + Clz(g) E°= —1 .09019 V AGjmm = —nFE° : —2x96485 C mol" xl.09019 V = 210.4 kJ mol" m K : nF E0 2 2x96485 Clmoljx109019 V RT 8.314J K mol x298.15K = —84.8686 K =1.39x10“37 b) AGijn = —nFE° : —2x96485 C mol‘l xl.09019 V = 210.4 kJ mol“ In K 2 _ AGij 2 _ 210700 J mor' RT 8.314 J K" mol‘l X29815 K = —85.0002 K =1.22x10—37 c) The results would agree exactly if AG" 2 210374 J mol‘1 react/an P11.14) Consider the couple Ox + e‘ : Red with the oxidized and reduced species at unit activity. What must be the value of E0 for this half—cell if the reductant R is to liberate hydrogen at 1 atm from 259 Chapter 11/ Electrochemical Cells, Batteries, and Fuel Cells 21. an acid solution with a“, = l? b. water at pH = 7? C. Is hydrogen a better reducing agent in acid or basic solution? a) H+(aq) + e- —>1/2H2(g) E: 0 Red —-> Ox + e‘ E°:? Overall: H+(aq) + Red —> 1/2 H2(g) + 0x Because the activities of HJr and H2 are one, E = E ° = O V. The same is true of the Ox—Red couple. To liberate H2, E ° for the reaction must be > 0. Therefore, E ° for Ox + e‘ —> Red must be < O. b) For a pH of 7, aH, :107 a E = E —£ln “1 = ~0.059l6logl0 (1017) = ~0.414 V nF a H 1 For the overall reaction to be spontaneous, E D for Red —> Ox + e‘ must be > 0.414 V. c) It is a better reducing agent in basic solution, because the oxidation potential for the reaction l/2H2(g) —> H+(g) + e‘ becomes more positive as the pH decreases . P11.15) By finding appropriate half-cell reactions, calculate the equilibrium constant at 298.15 K for the following reactions: a. 4NiOOH(s) + 2H20(Z) —> 4Ni(OH)2(s) + 02(g) b. 4No3‘(aq) + 4H+(aq) a 4NO(g) + 2H20(1) + 302(g) a) The half cell reactions are c NiOOH(s) + H200) +e‘ —> Ni(OH)2(s) +OH‘(aq) E I +0.52 V 40H"(aq) a 02(g) + 211200) + 4e‘ E0 = - 0.401 v The overall reaction is 4NiOOH(s) + 2H20(l) —> 4Ni(OH)2(s) +02(g) E0 = + 0.1 V nF 4x96485 C mol'l x0.119 V In K = —E° = ——————T——_l——————— RT 8.314 J K 1nol x298.15 K 218.5277 K =1.11x108 b) N03'(aq) + 4H“(aq) +3e' —» NO(g) + 211200) E“ : +0957 v 2 1120(1) a 02(g) + 4H+(aq) + 4e" E” : — 1.229 v 260 Chapter 11/ Electrochemical Cells, Batteries, and Fuel Cells The overall reaction is o 4No3'(aq) + 4H+(aq) —+ 4NO(g) + 211200) + 302(g) E = 4 0.272 V nF o 12x96485 C mol“ x0272 V an 2—E =— I 1 RT 8.314 J K" mol‘ X29815 K = ~127.047 K 26.67x10’56 P11.16) The cell potential E for the cell Pt(s)|H2(g,aH2 = l)lH+(aq,aH? =1)“NaCl(aq,m = 0.300 mol L’l)'AgCl(s)lAg(s) is + 0.260 V. Determine 7m assuming that A = yW = 7“, The half cell and overall reactions are AgCl(s) + e‘ —> Ag(s) + Cl'(aq) E0 = +0.22233 V l/2H2(g) —> H+(aq) + e‘ E” = 0 AgCl(s) + 1/2H2(g) —> Ag(S) + H+(aq) + Cl'(aq) E0 = +0.22233 V a rd ,, m E: E —R—Tln H (I = E ~£lnaw = E ~£ln 7/+ ‘" nF aH2 nF " nF ‘ m° nF E0 — E m ,, lny+ =——(————)—ln “ ‘ RT m° 1x 96,485 C mol’l (0.22233 V — 0.260 V) 1n yi = ————-———_l—_l—-————-—— — In 0.30 : —0.2630 8.314 Jmol K x298 K 7: = 0.769 P11.17) The Edison storage cell is described by Fe(s)lFeO(s)|KOH(aq, a)|Ni203(s)|NiO(s)|Ni(s) and the half-cell reactions are as follows: N1203(S) + H200) + 2e‘ :: 2NiO(s) + 20H"(aq) E" z 0.40 V FeO(s) + H200) + 2e“ : Fe(s) + 20H‘(aq) E° = —0.87 V a. What is the overall cell reaction? b. How does the cell potential depend on the activity of the KOH? c. How much electrical work can be obtained per kilogram of the active materials in the cell? The mass of the reactants per mole of the overall reaction is 55.85 g mol'1 +165.38 g mol’I = 221.12 g mol". The number of moles of the reaction that includes 1 kg or reactants is 1000. g /221.12 g mol'1 = 4.52 mol. 261 Chapter 11/ Electrochemical Cells, Batteries, and Fuel Cells The maximum electrical work is given by w = AG° : 4.52 mol>< (—nFE”) : 4.52 molx2x96,485 C mol" x127 V =—1108 kJ kg" P11.18) Consider the Daniell cell, for which the overall cell reaction is Zn(s) + Cu2+(aq) —> Zn2+(aq) + Cu(s). The concentrations of CuSO4 and ZnSO4 are 9.00 and 6.50 x 10‘3 m, respectively. a. Calculate E setting the activities of the ionic species equal to their molalities. b. Calculate y: for each of the half-cell solutions using the Debyethickel limiting law. 0. Calculate E using the mean ionic activity coefficients determined in part (b). a) Calculate E setting the activities of the ionic species equal to their molalities. For this cell, E0 = 1.099 V. m» 2! "’1, 2+ mv 27 E = E" 31111 I" = E“ —flin Z" “0* "F mr'uzt HF mri'uzt my)?“ ~3 E 21099 V 0.05916 V 10g“) 6.50x104m 2 9.00x10 ’m E = 1.103 V. We next calculate the mean ionic activity coefficients m 1 7 IMO4 : 3(1)}: + [23 ) = —2—(m+z; + mng) I = 4>< 0.00650 molkg‘l + 4x0.00650 molkg") = 0.0260 molkg'I 1n 7+ : —1.173|z+zw I 1 = -1.173x4x\/0.0260 = —0.75656 ‘ molkg' yi = 0.469 1”“); = —I;Z(v+zf + vvzf ) = %(m+zi + 112]?) 1 = 4x0.00900 molkg“ + 4x0.00900 molkg'l ) = 0.0360 molkg'1 In y+ —1.173lz+z_ I l = —1.l73>< 4 xx/0.0360 = —O.89024 ‘ molkg‘ 262 Chapter 11/ Electrochemical Cells, Batteries, and Fuel Cells We next calculate E using the activities E 2 E0 _flln 612,12. 2 E0 _£]:ln aZnB aw}, HF arm“ "F arm“ asof’ 2 Z 2 2 O] EzE. erna:( nso4):E. RTlnméyiéflnS 4) 1117 ai (CuSO4) nF my: (CuSO4) 2 2 6.50x10‘3 2 0.469 2 EzE._§§m%(Z_nSO;):L099V 0.05916 vlogm( m) x( )2 "F m 7101304) 2 (9.00x10‘3m)2x(0.411) E 21.099 V + 0.0049 V = 1.104 V The relative error in this case is zero within the precision of the measurements. P11.19) The standard potential E0 for a given cell is 1.100 V at 298.15 K and (aE°/aT),, = —6.50><10“5 v K". Calculate AG“ A50 and AH? reaction 9 mac/inn ’ reaction . Assume that n = 2. AGZWM = —nFE° = —2x96485 Cmorl x1100 v = —212.3 kJmol‘] AS:CaL'll()I7 : _[ aAGreaamn ] : 67“ P N P =—2><96485 C mol" x6.50x10‘5v K'1 = —12.5 J K" AH° 2 AG" +TAS° react/(m mac/ion react I on = —212.3 M K"1 mol'1—298.15 Kx12.5 J K‘] = —216.0 kJmol'1 Pl 1.20) Determine E0 for the reaction Cr2+(aq) + 2e‘ —9 Cr(s) from the one-electron reduction potential for Cr3+ and the three-electron reduction potential for Cr3+ given in Table 11.1 (see Appendix B). Cr3+(aq) + 3e' —> Cr(s) AG° = —nFE° = —3>< 96485 C mol‘l ><(—0.744 V) = 215.4 kJ mol'1 Cr2+(aq) —» Cr3+(aq) + 6' AG" = —nFE° = —1x 96485 C mol“ x 0.407 v = —39.27 kJ mol" Cr2+(aq) + 26' —+ Cr(s) AG = 215.4 kJ mol" — 39.27 kJ mol“ = 176.1 kJ mol“ __AG° _ —176.1x103 J mol" . —— ,1 2—0.913v ” /" nF 2x96485 C mol P11.21) Harnet and Hamer [.1. American Chemical Society 57 (1935), 33] report values for the potential of the cell Pt(s)|PbSO4(s)|H2804 (aq,a)leSO4(s)|PbOZ(s)|Pt(s) over a wide range of temperature and H2804 concentrations. In 1m H2804, their results were described by E (V) = 1.91737 + 56.1 x104t +108><10‘st2 ,where t is the temperature on the Celsius scale. Calculate AG, AH, and AS for the cell reaction at 11° and 35°C. E(V)=1.91737+5.61><10‘6 t+108x10'812 263 Chapter 11/ Electrochemical Cells, Batteries, and Fuel Cells [313] =5.61x10'6+2><108x10‘8t 0T P 9 6E a 9 A6 = —nFE AS 2 111(5) AH = A6 + [AS ,1 a) 11°C E=1.91812V AG 2 —2><96,485 C mol" ><1.91812 V=—370.1kJ mol" [95-] =7.99><10'5 VK" 8T ,, AS = 2x96,485 C mol’1x7.99x10’5 V K" = 15.4 J mol’lK‘l .1 »1 AH=—370.1kJmol'1+273'15KX15'4Jm01 K e 365.8k1m61" 1000 J 101 b) 35°C E =[1.91737+(56.1x10‘6)x(25)+(108x10‘8)x(25)2]V =1.92066 V AG 2 —370.6 kJ mol‘1 [‘35] =1.317><1041 v16 6T ,1 AS = 25.4 J mol'1 K"l 298.15 KX25.4 J mol" K"‘ 1000 J kfl = ~362.8 kJ mol‘1 AH = —370.6 kJ mol‘[ + P11.22) Consider the reaction Sn(s) + Sn4+(aq) :‘ ZSn2+(aq). If metallic tin is in equilibrium with a solution of Sn2+ in which aw = 0.250, what is the activity of Sn4+ at equilibrium? 811(3) —5 sn2+(aq) + 26‘ E": 01375 V Sn4+(aq) + 26' —> Sn2+(aq) 193°: 0.151 V For 811(5) + Sn4+(aq) —> 28n2+(aq) E = 0.1375+0.151= 0.2885 v AG“ = ~nFE° = —2 x 96,845 C mor‘ x 0.2885 V = —55.7 M motl K 2 MW = 5.67 x109 (298.15 K) K 2 (61311“ )2 a Sn“ 2 2 (am) (0.250) a = =— S“ K 5.67 x109 =1.10><10’ll 264 Chapter 11/ Electrochemical Cells, Batteries, and Fuel Cells P11.23) Consider the half—cell reaction 02(g) + 4H+(aq) + 4e’ —> 2H20(l). By what factor are 11, Q, E, and E0 changed if all the stoichiometric coefficients are multiplied by the factor two? Justify your answers. 17 is proportional to the number of electrons transferred, and increases by the factor two. Q is squared if all stoichiometric factors are doubled. The factor by which it is increased depends on the activities of 02 and H+. AG° . E Q = —F 1S unchanged because both AG° and n are doubled. n , RT . . . . E = E — 7—an IS unchanged because the squarlng of Q IS offset exactly by the doubhng of n. n P11.24) Calculate AGO reaction and the equilibrium constant at 298.15 K for the reaction Cr2072T(aq) + 3H2(g) + 8H+(aq) —> 2Cr3+(aq) + 7H20(l). Cr2072'(aq) + 14H+(aq) + 6e' —» 2Cr3+(aq) + 7H20(l) E" = +1.232 V H2(g) a 2H+(aq) + 2c" E” = 0.00 V AGfWW = —nFE° = —6><96485 Cmol" ><1.232 V : — 713.2 kJ mol'1 nF , 6X96485 Cmol'1X1.232 V an : E = 1 1 RT 8.314 J K' mol' X29815 K = 287.7 K = 9.06x10‘24 P11.25) The half-cell potential for the reaction 02(g) + 4H+ (aq) + 46” —> 2H20(l) is +1.03 V at 298.15 K when c102 =1. Determine an E=E°—R—Tln 14 HF aOZaH, 1.03 V :123 V—O—DS—g-lflloglO—i— 4 aH, 1.03 V—1.23V lo a,=———-———=w3.381 g‘“ H 0.05916V a“, =4.16><10*1 P11.26) Using half-cell potentials, calculate the equilibrium constant at 298.15 K for the reaction 2 H200) —> 2H2(g) + 02(g). Compare your answer with that calculated using AG; values from Table l 1.1 (see Appendix B). What is the value of ED for the overall reaction that makes the two methods agree exactly? 265 Chapter 11/ Electrochemical Cells, Batteries, and F ue] Cells 2H20(l) +2e' a H2(g) + 20H'(aq) “= —0.8277 V 4OH'(aq) —+ 02(g) +4e‘ + 211200) E” 2 40.401 V 211200) a 2H2(g) + 02(g) °= 4.2287 v nF .. 4x96485 C mor' X12287 v an = E = I I RT 8.314JK' mol' X29815 K = —191.303 K = 8.28x10“34 ZAG". H 0,1 3 -1 an:_ .f( 2 )2 2niolx2371.1x10.111iol : 191301 RT 8.3145 Jmol‘ K' x298.15 K K = 8.30x10’84 For the two results to agree, E ° must be given by _l9l.325><8.3145 Jmol" K'I x298.15 K = —1.22869 V 4x96485 C mol'1 E°= This value lies within the error limits of the determination of E0. P11.27) The data in the following table have been obtained for the potential of the cell Pt(s)|H2(g, f :1 atm)|HCl(aq, m)|AgCl(s)|Ag(s) as a function ofm at 250C. m(m01kg“l)T E(V) m(molkg“) E(V) ‘ m(molkg“) E(V) 0.00100 0.57915 0.0200 0.43024 0.500 0.27231 0.00200 0.54425 0.0500 0.38588 1.000 4 0.23328 0.00500 0.49846 0.100 4 0.35241 1.500 0.20719 0.0100 0.46417 0.200 0.31874 2.000 0.18631 Determine Eo using a graphical method. b. Calculate 7/1 for HCl at m = 0.00100, 0.0100, and 0.100 mol kg". Cell reaction: 2AgC1(s) + H2(g) ~> 2Ag(s) + 2H+(aq) + 2Cl'(aq) 0 RT 2 0 RT E = E —2—Fln(aH.a(,l.) 2E ~—I—?—ln(aH.afl_) CIHVFCIU =0: E=E _2RT1n mi _21€T(lny+) F m° F “ 266 Chapter 11/ Electrochemical Cells, Batteries, and Fuel Cells In the low concentration limit, we can use the Debye—Hiickel result mi = 40.50926. P”: : 4.172614. m 111 Therefore, for dilute solutions E+2RT1n££j=E°+2RTX1.172614>< [E F mo F m° Using this result, a plot of E +31-1:—T—ln (y axis) vs. (x axis) will have an intercept of E ° . m m We use the data up to m = 0.100, as the Debye—Huckel model is not valid for more concentrated solutions. _m_ m Eon E+ 2” 111(4) (V) o T F m m m 0.001 0.031623 0.57915 0.224212 0.002 0.044721 0.54425 0.224909 0.005 0.070711 0.49846 0.226203 0.010 0.1 0.46417 0.227531 0.020 0.141421 0.43024 0.229218 0.050 0.223607 0.38588 0.231943 0.100 0.316228 0.35241 0.234090 The data in the table is graphed below. The best fit line gives a value for E ° of 0.223665 V. O a) L)! N ix.) N N Lu $3 B+{2R’"¥‘/F)ln (“m/m") ‘83 {\J 3.3 Chapter 11/ Electrochemical Cells, Batteries, and Fuel Cells Given E“ we can now find y: from lnyi _E)—ln(%] E = 0.223655 V m/m° I In 71 A 0.001 0.010654 0.989 0.010 00755112 0.927 0.100 0.203036 0.816 P11.28) Consider the cell Pt(s)le(g,1atm)|H+(aq,a=1)lFe3+(aq),Fe2+(aq)|Pt(s) given that Fe“ + e- .: Fe2+ and E" : 0.771 V. a. If the cell potential is 0.712 V, what is the ratio of Fe2+(aq) to Fe3+(aq)? b. What is the ratio of these concentrations if the cell potential is 0.830 V? c. Calculate the fraction of the total iron present as Fe3+(aq) at cell potentials of 0.650, 0.700, 0.750, 0.771, 0.800, and 0.900 V. Graph the result as a function of the cell potential. Overall reaction: H2(g)+ 2Fe3+(aq)—> 2H+(aq) + 2Fe2+(aq) E0 : 0.771 V RT ') a) E : E° ——2-I—7—ln(a};ez. /a14,63. a RT =E —?ln(awz. MM.) aw. 2 exp (E° — E)F aFeS. RT IfE = 0.712 V = 9.95 a. .. [(0.771v-0712 V)><96,485 c mol‘l] :exp 18 a. .1 8.314 J mol” K"l x298.15 K b e.) b) IfE= 0.830 V are. (0.771 V—O.830 V)><96,485 C mor‘ =eXp ~———fi—————— =0.100 8.314Jm01 K X298.15 K a~ 1+ 116' {363+ - - 1 c) F raction of Fe3 5 ———_ 2 W ah... + aFez, 1 + (aFez. mm ) 268 Chapter 11/ Electrochemical Cells, Batteries, and Fuel Cells We use the method described above to find aFCH / a H . We find the following results: Fe E (V) awn MR“ FraZtiin of 6 0-650 111 8.93 x 10‘3 0-700 159 0.0593 0.750 2.26 0.306 0.771 1.0 0.5 0.800 0.323 0.756 0.900 6.60 x 10'3 0993 8,8 {316 0.4 J Ffaction Fe- as £33+ 0.2 ‘ 036% (3.7 9‘8 (L9 1 E Pl 1.29) Determine Ksp for AgBr at 298.15 K using the electrochemical cell described by Ago»AgBr<s)|Br’<aq,aB,> lAg+(aq,aAg+ )lAg1s) The half cell and overall reactions are AgBr(s) + e' —> Ag(s) + Br‘(aq) E0 = + 0.07133 V Ag(s) —» Ag+(aq) + e‘ E“ = —0.7996 V AgBr(s) a Ag+(aq) + Br‘(aq) °= —0.72827 v log") K” :_0.0:9Elé V 2 09023;; = 12310 KW = 4.89x10"3 269 Chapter 11/ Electrochemical Cells, Batteries, and Fuel Cells P1130) By finding appropriate half—cell reactions, calculate the equilibrium constant at 298.15 K for the following reactions: a. 2Cd(OH)2(s) —> 2Cd(s) + 02(g) + 2H20(l) b. 2MnO42’(aq) + 2HZO(Z) —> 2Mn02(s) + 4OH”(aq) + 02(g) a) The half cell reactions are Cd(OH)2(s) + 2e" —+ Cd(s) + 2OH'(aq) E0 = —0.809V 4OH'(aq) —> 02(g) + 2H20(l) +4e‘ E0 : —0.401V The overall reaction is 2 Cd(OH)2(s) —> 2 Cd(s) + 02(g) + 2H20(l) E0 = —1.21 V nF , 4x96485C mol"><1.21V an:—-E =— I 1 RT 8.314JK‘ mol' x298.15K =—188.391 K =1.52x10‘82 b) The half cell reactions are Mno42‘(aq) + 2H20(l) + 2e' a Mn02(s) + 40H'(aq) E” = + 0.595 V 40H'(aq) —» 02(g) + 2H20(1) +4e‘ E” : — 0.401 v The overall reaction is 2Mno42‘(aq) + 211200) —+ 2Mn02(s) +40H'(aq) + 02(g) E“ = + 0.194 v nF , 4x96485C mol‘1x0.194V an= E = I 1 RT 8.314JK' mol' x298.15K =30.2049 K:1.31x10'3 270 ...
View Full Document

This note was uploaded on 04/22/2010 for the course CHEM 402 taught by Professor Sobodka during the Spring '10 term at Washington University in St. Louis.

Page1 / 19

CH 11 - Chapter 11: Electrochemical Cells, Batteries, and...

This preview shows document pages 1 - 19. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online