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Calculate the equilibrium constant for the following reaction carried out in 1 M perchloric acid: 2Fe³+ + 2I- = 2FE²+ +I_(aq) See appendix H in the textbook for standard reduction potentials.

Calculate the equilibrium constant for the following reaction carried out in 1 M perchloric acid: 2Fe³+ + 2I- = 2FE²+ +I_(aq) See appendix H in the textbook for standard reduction potentials.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Reaction Dysprosium Dy + 3e Dy E (volts) EMT K) Re Erbium Fr + 3e" Er) Europium Eu +e Eu Eu + 3e"Eula) Eu + 2e Eats) -2.295 -2.331 0.38 -0.35 -1.991 1.53 0.338 -0.26 Fluorine -2.812 FAR) + 2e 2 FOlg) + 2H* + 4e 2F+ H0 2.800 2.168 Ciadolinium Gd + 3e" Gdr) -1.20 Gallium Ga+ 3e=Gala) GaOOH(s) + H,0 + 3e Gals) + 301 Germanium Ge* + 2e = Ge(s) HGeO, + 4H + 4c"Ge() + 4H0 -2.279 0.315 -0.549 0.61 -1320 0.1 -0.039 -0429 Gold Au +e Auls) Au" + 2e Au' AuCl, +e - Au(s) + 2CI AuCl, + 2e= AuCl + 201 149 -1.1 1.41 1.154 0.926 Hafnium HI* + 4e - Hs) HIO:(s) + 4H" + 4e -HRs) + 2H,) -1.55 -1.591 068 -0.355 Holmium Ho+ 3e =Ho(s) -2.33 0.371 Hydrogen 2H* + 2e H(g) H0 +e=IH,(g) + OII" Indium In* + 3e + Hg In(in lg) In* + 3e In(s) In* + 2e In' In(OH),() + 3e In(s) + 30H Iodine 1O, + 2H + 2e 10, + H0 HIO, + 2H + 2e= HIO, + 3H,0 HOI + H + e=1(4) + H,O ICI,(s) + 3c= 1,() + 3CI" ICI() +e=L(s) + CI 10, + 6H + Se s) + 3H-O I0, + 51" + 4e= HOI + 2H,0 Ilag) + 2e21 I(s) + 2e= 21 5 + 2e 31 10, + 3H,0 + 6e1 + 60H Iridium IC +e- IrCl IrBr +e= IrBr IrC + 4e Irts) + 6CI IrO,(r) + 4H* + 4e Irls) + 2H,0 0.000 0 -0.828 0 -0.8360 -0.313 -0.338 -0.444 -0.99 042 -095 1589 1.567 1.430 1.28 --0.85 -0.12 -039 1.22 1.210 1.154 0.620 0535 0.535 -0.367 -0.374 -0.234 -0.125 -0.186 -1.163 0.269 1.026 IF HCI 0.947 2 FNaBr 0.835 -036 0.73 0.485 IFKI 1.147 Iron Felphenanthroline) +e Felphenanthroline Fe(bipyridyl), + e Fe(bipyridyl)" FEOH" + H* +e Fe + H;0 Feo-+ 3H-0 + 3e = FE0OH(s) + SOH 1.120 0,096 0.900 -159 1.175 0.80 (0.771 0.732 1F HCI 0.767 1F HCI0, 0.746 IF HNO, Fe" +e Fe 0.68 1F H,SO, APPENDIX H Standard Reduction Polentials
Transcribed Image Text:Reaction Dysprosium Dy + 3e Dy E (volts) EMT K) Re Erbium Fr + 3e" Er) Europium Eu +e Eu Eu + 3e"Eula) Eu + 2e Eats) -2.295 -2.331 0.38 -0.35 -1.991 1.53 0.338 -0.26 Fluorine -2.812 FAR) + 2e 2 FOlg) + 2H* + 4e 2F+ H0 2.800 2.168 Ciadolinium Gd + 3e" Gdr) -1.20 Gallium Ga+ 3e=Gala) GaOOH(s) + H,0 + 3e Gals) + 301 Germanium Ge* + 2e = Ge(s) HGeO, + 4H + 4c"Ge() + 4H0 -2.279 0.315 -0.549 0.61 -1320 0.1 -0.039 -0429 Gold Au +e Auls) Au" + 2e Au' AuCl, +e - Au(s) + 2CI AuCl, + 2e= AuCl + 201 149 -1.1 1.41 1.154 0.926 Hafnium HI* + 4e - Hs) HIO:(s) + 4H" + 4e -HRs) + 2H,) -1.55 -1.591 068 -0.355 Holmium Ho+ 3e =Ho(s) -2.33 0.371 Hydrogen 2H* + 2e H(g) H0 +e=IH,(g) + OII" Indium In* + 3e + Hg In(in lg) In* + 3e In(s) In* + 2e In' In(OH),() + 3e In(s) + 30H Iodine 1O, + 2H + 2e 10, + H0 HIO, + 2H + 2e= HIO, + 3H,0 HOI + H + e=1(4) + H,O ICI,(s) + 3c= 1,() + 3CI" ICI() +e=L(s) + CI 10, + 6H + Se s) + 3H-O I0, + 51" + 4e= HOI + 2H,0 Ilag) + 2e21 I(s) + 2e= 21 5 + 2e 31 10, + 3H,0 + 6e1 + 60H Iridium IC +e- IrCl IrBr +e= IrBr IrC + 4e Irts) + 6CI IrO,(r) + 4H* + 4e Irls) + 2H,0 0.000 0 -0.828 0 -0.8360 -0.313 -0.338 -0.444 -0.99 042 -095 1589 1.567 1.430 1.28 --0.85 -0.12 -039 1.22 1.210 1.154 0.620 0535 0.535 -0.367 -0.374 -0.234 -0.125 -0.186 -1.163 0.269 1.026 IF HCI 0.947 2 FNaBr 0.835 -036 0.73 0.485 IFKI 1.147 Iron Felphenanthroline) +e Felphenanthroline Fe(bipyridyl), + e Fe(bipyridyl)" FEOH" + H* +e Fe + H;0 Feo-+ 3H-0 + 3e = FE0OH(s) + SOH 1.120 0,096 0.900 -159 1.175 0.80 (0.771 0.732 1F HCI 0.767 1F HCI0, 0.746 IF HNO, Fe" +e Fe 0.68 1F H,SO, APPENDIX H Standard Reduction Polentials
Calculate the equilibrium constant for the following reaction carried out in 1 M perchloric acid: 2Fe3+ + 21- = 2FE²+ + I½(aq) See appendix H in the textbook for standard reduction potentials.
Transcribed Image Text:Calculate the equilibrium constant for the following reaction carried out in 1 M perchloric acid: 2Fe3+ + 21- = 2FE²+ + I½(aq) See appendix H in the textbook for standard reduction potentials.
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