Chapter 11 Solutions - 22. 23. CHAPTER 11 ELECTROCHEMISTRY...

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Unformatted text preview: 22. 23. CHAPTER 11 ELECTROCHEMISTRY . 35] 1%. Pt I Mn2+ (1.0 M), M110; [1.0 M), H+ (1.0 M) IO; (1.0M), 103' (1.0M), H" (1.0M) 1 Pt 190. Pt I 1130100 11/1), I-I+(1.0 M) | 02 (1.0 atm) If H202 (1.0 M), H*(1.0 M)|Pt 19d. MnIMn2* (1.0M)||Fe3+(1.0M)[Fe The reduction half-reaction for the SCE is: Hg2C12 + 2 e‘ -> 2 Hg + 2 Cl‘ ESCE = 0.242 V For a spontaneous reaction to occur, Emll must be positive. Using the standard reduction potentials in Table 1 I .1 and the given SCE potential, deduce which combination will produce a positive overall cell potential. 21. Cu3++2e‘->Cu E°=0.34V Been = 0.34 - 0.242 = 0.10 V; SCE is the anode. b. Fe3+ + e' —> Fe2+ E" = 0.77 V E9,“ = 0.77 - 0.242 = 0.53 V; SCE is the anode. c. AgCI + e' -> Ag + Cl‘ E° = 0.22 V Ewu= 0.242 - 0.22 = 0.02 V; SCE is the cathode. d. A13++3e'—*Al E°=-1.66V Emu= 0.242 + 1.66 = 1.90 V; SCE is the cathode. e. NF? 2 e‘ -> Ni ° = -0.23 V Em" = 0.242 + 0.23 = 0.47 V; SCE is the cathode. a. 2H”+2e'*>I-I2 °=0.00V; Cu—*Cu1++2e‘ —B°=-0.34V Esau = —0.34 V; No, H+ cannot oxidize Cu to Cu” at standard conditions (Egg11 < 0). b. Fe3+ + e' -> Fe1+ ED = 0.77 V; 21' 412+ 2 e' -E° = -0.54 V Egg" = 0.77 - 0.54 = 0.23 V; Yes, Fe3+ can oxidize I' to 12. c. H2-+2H++2e' -E°=0.00V; Ag*+e'—+Ag E” =0.80V E25" = 0.80 V; Yes, H2 can reduce Ag” to Ag at stande conditions (BEBE > 0). CHAPTER 1 l ELECTROCHEMISTRY 352 24. 25. 26. 27. d. Fe2+ -P Fe“ + e‘ -E° = -0.77 V; C13“ + e" -+ Cr1+ ° = ~0.50 V EEC“ = -0.50 - 0.77 = -l .27 V; No, Fe1+ cannot reduce (113+ to Cr2+ at standard conditions. Good oxidizing agents are easily reduced. Oxidizing agents are on the left side of the reduction half- reactions listed in Table l 1.1. We look for the largest, most positive standard reduction potentials to correspond to the best oxidizing agents. The ordering from worst to best oxidizing agents is: K+ < H20 < Cd2+ < I2 < AuCL,’ < 103' E°(V) -2.92 -0.83 -0.40 0.54 0.99 1.20 Good reducing agents are easily oxidized. The reducing agents are on the right side of the reduction half-reactions listed in Table 11.1. The best reducing agents have the most negative standard reduction potentials (E°), i.e., the best reducing agents have the most positive ~E° value. F‘< C13+<Fe2* < H2 < Zn < Li —E°(V) -2.37 -1.33 —0.77 0.00 0.76 3.05 C12+2e‘-VZC1' E" =1.36V Ag*+e'-+Ag E°=0.80V Pb2*+2e'—>Pb E°=-0.13V Zn1++2e‘->Zn E°=-0.76V Na"+e'->Na E°=-2.71V a. Oxidizing agents (species reduced) are on the hit side of the above reduction half-reactions. Of the species available, Ag+ would be the best oxidizing agent since it has the most positive E° value. b. Reducing agents (species oxidized) are on the right side of the reduction half-reactions. 0f the species available, Zn would be the best reducing agent since it has the most positive —E° value. c. 5042' + 4 H’ + 2 e‘ -+ H2803 + H20 E: = 0.20 V; 8042‘ can oxidize Pb and Zn at standard conditions. When 803' is coupled with these reagents, Egg“ is positive. (1. Al -> Al3+ + 3 c' — E: = 1.66 V; A1 can oxidize Ag” and Zn“ at standard conditions since E° > 0. cell a. 2 Br" —> Br1+2 c" - = -l.09 V; 2 Cl'—* 011-126 rE: =—1.36 V; E: 3- 1.09V to oxidize Br"; E: < 1.36 V to not oxidize Cl‘; C1307”, 02, M1102, and 103' are all possible since when all of these oxidizing agents are coupled with Br' give EEC“ > 0 and when coupled with C1" give BEE" < 0 (assuming standard conditions). b. Mn —> an" + 2 e" — E: =1.18; Ni —* Ni2+ + 2 e‘ — B: = 0.23 V; Any oxidizing agent with -0.23 V >E: > «1.18 V will work. PbSO4, Cd“, Fe”, C132 Zn2+ and H30 will be able to oxidize Mn but not oxidize Ni (assuming standard conditions). CHAPTER 1 1 ELECTROCHEMISTRY 353 ______—___________._._——-——-—-—-——— 28. 29. 30. 31. 32. a. 0112+ + 2 e‘ —> Cu B: = 0.34 V; Cu2+ + e‘ -+ Cu+ E: = 0.16 V; To reduce Cu2+ to Cu but not reduce Cu“ to Cu”, the reducing agent must have a — value between —0.34 V and —0. 16 V (50 ESE" is positive only for the Cu2+ to Cu reduction). The reducing agents (species oxidized) are on the right side of the half-reactions in Table 11.1. The reagents at standard conditions which have a — 13; value between -0.34 V and —0.16 V are Ag (in 1.0 MCl') and HQSO, b. Br2 + 2 e' -+ 2 Br‘ E: =1.09 V; I2 + 2 e' -> 21‘ = 0.54 V; From Table 11.1, V0”, Au (in 1.0 M Cl‘), N0, C102; Hgf”, Ag, Hg, Fe”, H202 and MnO,‘ are all capable at standard conditions of reducing Br2 to Br“ but not reducing I; to I‘. When these reagents are coupled with Brz, E92”)- 0, and when coupled with 12, E“ < 0. cell ClO'+HZO+2e'->20H'+Cl‘ 2NH3+20H'->N2H,+2H20+2e‘ E" = 0.90 V —E° = 0.10 V (3101an + 2 NH3(aq) —> Cl‘(aq) + N,H,(aq) + H100) E = 1.00 v cell Since Egan is positive for this reaction, then at standard conditions C10" can spontaneously oxidize NH3 to the somewhat toxic Nil-I4. T13*+2 e'->T1+ B" =1.25v 31-—>1,-+2e -E°=-O.SSV T1“ + 3 I“ ~+ T1+ +1; E36" = 0.70 v In solution, T131can oxidize I' to 13‘. Thus, we expect 'l‘lIll to be thalliumfl) triiodidc. 1-1202 + 2 I—F + 2 e' -> 2 H20 E: = 1.218 V; H101 is the oxidizing agent. H,o,—» 02 + 2 H* + 2 e' — E: = -0 .68 V; H202 is the reducing agent. H202+2H*+2e'—>2H20 E:=1.78V H,O,—+O,+2H*+2e' —E:=-0.68V 2 H,o,(aq) a 2 1120(1) + 01(3) E :8“ = 1.10 v Consider the strongest oxidizing agent combined with the strongest reducing agent from Table 11.1: FZ+2e‘—P2F' 133:2.37v (Li-+Li++e‘)><2 —E;=3.05V mg) + 2 Li(s) 4 2 Li+(aq) + 2 F'(aq) 13;, = 5.92 v The claim is impossible. The strongest oxidizing agent: and reducing agent when combined only give E" of about 6 V. CE“ 358 CHAPTER 1 1 ELECTROCHEMISTRY 43c. 2 Hc102(aq) -> 0103-016) + HTaq) + HClO(aq) E3811 = 0.44 v, n = 2 mol c‘ 013° = —nFE;;“ = —(2 11101 e')(96,485 (216161 6')(0.44 J/C) = -s4,900 J = —35 kJ logK= “ED = 2(0’44) =14.s9, K= 7.3 x 10” 0.0591 0.0591 45. A13++3e'—>A1 E: =-1.66V Al+6F'—PA1F63‘+3e'_ E§=207v Al3+(aq) + 6 F'(aq) «r 4119,31an 15;;11 = 0.41 v K = 2 log K= “E0 = 3‘0“) = 20.31, K = 101°31= 6.5 >< 1020 0.0591 0.0591 46. Ag*+e'-+Ag E: =0.80V Ag + 2 82032. _+ Ag(3103)23' + e— — E: = “0.017 V A3139) '1‘ 2 320.5%“) A 1435209231”) For this overall reaction, E3311 = 00591 log K n logK = “Bo = m = 13.20, K =1013-1°=1.6 ><1013 0.0591 0.0591 3;” = 0.78 V K= ? 47. CdS+2c‘->Cd+Sz' Cd -> Cd2+ + 2 e” g = —1.21 v — Eg= 0.402 v CdS(s) —» Cd=+(aq) + sl-(aq) Egan = —0.31 v ' K = Ks? = 7 ,' _ 0.0591 For this overall reaction, Emu " log Kw n _ 11E." 2( 43.81) H _ .2741 .28 lo -— ’=————- —-27.41, — 10 =39 X10 g Kg}! 0.0591 0.0591 KW 43. CuI+e-—»Cu+1- 9311:? C11 "‘1' 611* + e' -E°= -0.52 V ._—._.___.._..._.—_———.——a—-———-——--- Cu1(s) + cu+(aq) + I'(aq) 15° = Ea” - 0.52 V ceil For this overali reaction, K = K3p = 1.1 X 10'”: o _ 0.0591 0.059l coll _ 10g Kw = log (1.1 >< 10'”)=-0.71 V n 1 15° =-0.71V= E3“. .052, Ear-0.191! cell _ _ “FW— _ ___ . CHAPTER 11 ELECTROCHEMISTRY 359 Galvanic Cells: Concentration Dependence 49. a. Au3+ + 3 e' -+ An E: = 1.50 V (Tl-+Tl++e')x3_ ~E;=0.34v Au3+(aq) + 3 T1(s) —> Au(s) + 3 T1*(aq) E;" = 1.84 V b. AG° = -nFE;" = -(3 mol e‘)(96,485 Clmol e')(l.84 Ja’C) = -5.33 X 105 J = -533 kJ 11E° _ 3(1.84) log K = _. 0.0591 0.0591 =93.401, K = 1091"”1 = 2.52 x1093 ' + 3 0.0591 10g Q where Q : [Tl ] c. At 25°C, Ecell = E“ 11 [Au 3*] cell - [T113 =134_ 0.0591 10g (I.0><10“)3 [M31 3 1.0 x 10'2 Egg“ = 1.84 - (-0.20) = 2.04 v 13cull = 1.34v - 00:91 log 50. (Cr2+ —> (313* + e')X2 C02+ + 2c" -—+ Co 2 Cr2+(aq) + 002120;) + 2 Ct3+(aq) + 00(5) :en = 00:91 logK= 0‘02591 log (2.79 x 10"): 0.220 v 3+ 2 2 E =E° - 0'0”! log ———[C‘ 1 = 0.220 v - 99L” 10g —(2‘O) = 0.151 v n [Cr2"}2[C02’] 2 (030)2(020) AG = -nFE = -(2 Inol e')(96,485 C/mcl c”)(0.151 JI'C) = -2.91 X 10“ J = -29.1 k] 51. (Pb2*+Ze'-r Pb)><3 E: =-0.13v ' (A1 4' A13++3e-)x2 --E; =1.66V 3 sz’Taq) + 2 A1(s) 4* 3 Pb(s) + 2 Alwaq) 15;" = 1.53 V From the balanced reaction, when the [AP‘] has increased by 0.60 mol/L (Al3+ is a product in the spontaneous reaction), then the Pbfi concentration has decreased by 312 (0.60 Incl/L) = 0.90 M. 3v 2 2 0.059110gm1 1 =15} 0.0591.10g (1.60) 6 [Pb2'13 6 (0.10)3 Ecfl= 1.53 V -0.034V=1.50V Euu=1.53V- CHAPTER 1 1 58. 59. ELEC TROCHEMI STRY 3 63 d. Emu = EEC“ - (0.05902) log (1f[Cu2*]) = Egan + 0.0296 log [Cufi]; This equation is in the form of a straight line equation, y = mx + b. A graph of Ecell vs. log [CuP’] will yield a straight line with slope equal to 0.0296 V or 29.6 mV. 3 Ni2+(aq) + 2 Al(s) a 2 A13+(aq)+ 3 Ni(s) Eje" = -023 v + 1.66 v = 1.43 v; n = 5 3+ 2 3* 2 0.0591 10g [A] ] a 132‘}: 143 V _ 0.0591 10g [Al ] n [Nizr13 6 (1.0)3 __ 0 E6811 _ cell _ log [20.13"]2 = 69.59, [Al3+ 2 = 1039'”, [AIM] = 1.6 X 10'20 M Al{OH)3(s) ‘——‘ Al’Taq) + 3 ‘OH'(aq) K5,, = [A131 [OH‘]3; From the problem, [OH'] = 1.0 x 10“ M. K“, = (1.6 X 10'2") (1.0 X104? =16 X1032 From Exercise 11.1?a: 3 C12(g) + 2 Cr3+(aq) + 7 H200) re 14 H“(aq) + Cr2072'(aq) + 6 Cl'(aq) E° = 0.03 V cell : E. 0.0591 [Crzof'] [H '*]”[c1 '16 cell- 6 _"—H_ [Cr 3‘]2 P512 When KZCrZO1r and Cl' are added to concentrated H2304, Q becomes a large number due to [H"]“ term. The log of a large number is positive. EDell becomes negative, which means the reverse reaction becomes spontaneous. The pungent finnes were Cl1(g). E log cell Electrolysis 60. 61. a. Al3+ + 3 e' -* A1; 3 mo] e‘ are needed to produce 1 mol Al from AP". 111101111X3mole'x96,485C>< ls 27.0 g mol A! 100.0 C 1J3><103g>< =1.07X105s=3.0><10’hours mole" 1 mol 2 mole ' 96,485 C 1 s __ x x —-— x _— — 33 s 58.7 g mol Ni b. 1.0 gNi x 100.0(: mole" lmole' X 96,485C x 15 mol Ag 100.0 C =4.8 ><103 s= 1.3 hours c. 5.0 mol Ag X mo] e ' 15 C x 605 x 60mm 5 min h 15A = = 5.4 x 10“ C of charge passed in 1 hour a S4xlo4Cx1mole"xlmolCox58.9g ' ‘ 96.485C 2m013- mo] =16gCo lmole' X lmoleX 178.5g =2Sng C 4 [110} e _ mol b. 5.4x10‘C x 364 62. 63. CHAPTER 1 1 ELECTROCHEMISTRY - lmol 253.8 I 1 mole x [1X g 2 c. 21'w12+26";5.4 X104C X ~— C 2 m0](:_ 1110112 =71g12 d. Cr is in the +6 oxidation state in CrO3. Six mo]. of e' are needed to produce 1 mol Cr from molten Cr03. lmole‘ X lmolCr X 52.0gCr 96,485 C 6 mole ' mol Cr 5.4X10‘CX =4.9gCr 2.30min x 6’05 =13s 5,133 5 x 200‘: x “11°16'— xlflfli‘ig =2.86 x10'3molAg min S 96,485 C me] e ' [Ag] = 2.86 x 10-3 mol Aguazso L = 1.14 x 10-2 M First determine the species present, then reference Table 11.1 to help you identify each species as a possible oxidizing agent (species reduced) or as a possible reducing agent (species oxidized). Of all the possible oxidizing agents, the species that will be reduced at the cathode will have the most positive EC” value; the species that will be oxidized at the anode will be the reducing agent with the most positive - E: value. :1. Species present: Ni2+ and Br”; Ni2+ can be reduced to Ni and Br' can be oxidized to Br1 (from Table 11.1). The reactions are: Cathode: Ni“ + 2e' -r Ni E: = e 0.23 V Anode: 2 Br" -> Br, + 2 e" -E: = —1.09 V b. Species present: All3+ and F“; Al3+ can be reduced and F‘ can be oxidized. The reactions are: E;=—1.66v —E;=-2.srv Cathode: A13+ + 3 e' ~+ A1 Anode: 2F‘-rF2+2e‘ c. Species present: Mn2+ and I‘; Mn2+ can be reduced and I‘ can be oxidized. The reactions are: Cathode: Mn2++2e‘->Mn E: =—1.18V Anode: 21' —> I2 + 2 e' — E: = --0.54 V d. For aqueous solutions, we must now consider H20 as a possible oxidizing agent and a possible reducing agent. Species present: Ni“, Br' and H20. Possible cathode reactions are: Ni2++2e‘*+Ni 2H20+2e'—>H2+20H' E3: «0.23 v 13:: —o.33 V Since it is easier to reduce Niz’ than H20 (assuming standard conditions), then Ni1+ will be reduced by the above cathode reaction. CHAPTER 1 1 ELECTROCHEMISTRY 367 69. Alkaline earth metals form +2 ions, so 2 mol of e' are transferred to form the metal, M. 5.00Cx lrnole' X lmolM S C 2 m0} 3' molM=74Ss>< =].94XI0'2moIM 0.471 g M molar mass of M = m 1.94 x 10—2 11001 M = 24.3 g/mol; MgCl2 was electrolyzed. 150. X 103 g CGHgN2 1 h 1 mm x 1 mo] CfiHgN2 X 2 m0] 3- x 96,435 C 70. W X X _ h 60 min ()0 S [4 g CE'HSN2 mol CfiHgl"%l2 m0] 3' é 7.44 x 10*1 Cfs oracurrent of7.44 x lO‘A 71. F1 is produced at the anode: 2 F' -> F1 + 2 e' 60 min 605 10.0C lmole‘ x ——- x x #H— 2.00 h x = 0.746 mol c' min 5 96,485 c 1 mol P 0.746 mol e' x 2 = 0.373 mol F1; PV = nRT, v = “RT 2 me] e‘ P V = (0.373 11101) (0.08206 L atm K " mol '1) (298 K_) = 9.12 L F2 1.00 atrn I K is produced at the cathode: K” + e‘ —> K lmolK X 39.10gK mole’ mOIK 0.746 mol e‘ X = 29.2 gK 15000Jh X§_s_x60min S min h 72. 15 kWh = = 5.4 x 10" I or 5.4 X 10“ kJ (Hall process) “1231141 X 10.1r R] 26.98 g mol Al It is feasible to recycle AI by melting the metal because in theory, it takes less than 1% of the energy required to produce the same amount of Al by the Hall process. To melt 1.0 kg A1 requires: 1.0 X 103 g Al x = 4.0 x 102 kJ 73. In the electrolysis of aqueous sodium chloride, H20 is reduced in preference to Na+ and Cl' is oxidized in preference to H20. The anode reaction is 2 Cl' -* C1; + 2 e‘ and the cathode reaction is 2 H20 + 2 e‘ *+ H2 + 2 OH“. The overall reaction is 2 H100) + 2 Cl'(aq) -> Clz(g) + H2(g) + 2 OH‘(aq). From the 1:1 moi ratio between C12 and H2 in the overall balanced reaction, if 6.00 L of H2(g) are produced, then 6.00 L of C12(g) will also be produced since moles and volume of gas are directly proportional at constant T and P (see Chapter 5 of text). ...
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This note was uploaded on 04/16/2009 for the course CHEM 2150 taught by Professor Chirik,p during the Fall '08 term at Cornell University (Engineering School).

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Chapter 11 Solutions - 22. 23. CHAPTER 11 ELECTROCHEMISTRY...

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