8. Using the standard reduction potentials provided below, determine the equilibrium constant at 298 Kfor the following reaction:4 H(aq) + MnO:(s) + 2 Fe²*(aq) Mn"(aq) + 2 Fe"(aq) + 2 H:O(1)IronE° (V)Fe2 + 2e= Fe(s)-0.44Fe + 3e" Fe(s)-0.037Fe* +e Fe2+0.771Fe(CN), +e Fe(CN),+0.356CopperE° (V)Cu* +e Cu(s)0.520Cu* +e Cu0.159Cu* + 2e= Cu(s)0.3419Cu* +r +e Cul(s)0.86Cu* + CI" +e = CuCl(s)0.559MangaE° (V)Mn + 2e" Mn(s)-1.17Mn +e" Mn21.5MnO,(s) + 4H* + 2e" Mn* + 2H½O()1.23Mno,(s) + 4H* + e° Mn (aq) + 2H,O(1)0.95MnO, + 4H* +3e= MnO2(s) + 2H,0()1.70MnO,+ 8H* + 5e¯ = Mn* + 4H,O()1.51Mno, + 2H,0(1) + 3e" MnO2(s) + 40H"0.60HydrogenE° (V)2H +2e" = H2(g)0.00000H20 +e= VH2(g) + OH-0.8282

Answered: Image /qna-images/answer/6383781a-1488-4fd5-9916-a04fbaaa37fa.jpg

8. Using the standard reduction potentials provided below, determine the equilibrium constant at 298 K for the following reaction: 4 H(aq) + MnO:(s) + 2 Fe²*(aq) – Mn (aq) + 2 Fe (aq) + 2 H:O(1) Iron E° (V) Fe + 2e" Fe(s) -0.44 Fe + 3e" Fe(s) -0.037 Fe+e Fe2+ 0.771 Fe(CN) +e= Fe(CN), 0.356 Сopper E° (V) Cu +e= Cu(s) 0.520 Cu* +e= Cu 0.159 Cu* + 2e= Cu(s) 0.3419 Cu* +r +e= Cul(s) 0.86 Cu* + CI" +e - CuCl(s) 0.559 Mang E° (V) Mn + 2e" Mn(s) -1.17 Mn +e" Mn2 1.5 MnO,(s) + 4H* + 2e Mn* + 2H,O() 1.23 MnO,(s) + 4H* + e= Mn" (aq) + 2H;O(1) 0.95 MnO, + 4H* +3e" MnO(s) + 2H,0() 1.70 Mno,+ 8H* + 5e" Mn* + 4H,O() 1.51 MnO, + 2H,0() + 3e"= MnO2(s) + 4 +40H 0.60 Hydrogen E° (V) 2H +2e= H2(9) 0.00000 H20 +e= VH2(g) + OH --0.828

8. Using the standard reduction potentials provided below, determine the equilibrium constant at 298 K for the following reaction: 4 H(aq) + MnO:(s) + 2 Fe²(aq) – Mn (aq) + 2 Fe (aq) + 2 H:O(1) Iron E° (V) Fe + 2e" Fe(s) -0.44 Fe + 3e" Fe(s) -0.037 Fe+e Fe2+ 0.771 Fe(CN) +e= Fe(CN), 0.356 Сopper E° (V) Cu +e= Cu(s) 0.520 Cu +e= Cu 0.159 Cu* + 2e= Cu(s) 0.3419 Cu* +r +e= Cul(s) 0.86 Cu* + CI" +e - CuCl(s) 0.559 Mang E° (V) Mn + 2e" Mn(s) -1.17 Mn +e" Mn2 1.5 MnO,(s) + 4H* + 2e Mn* + 2H,O() 1.23 MnO,(s) + 4H* + e= Mn" (aq) + 2H;O(1) 0.95 MnO, + 4H* +3e" MnO(s) + 2H,0() 1.70 Mno,+ 8H* + 5e" Mn* + 4H,O() 1.51 MnO, + 2H,0() + 3e"= MnO2(s) + 4 +40H 0.60 Hydrogen E° (V) 2H +2e= H2(9) 0.00000 H20 +e= VH2(g) + OH --0.828

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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