Reduction half-reaction E (V) Ag (a) e-Ag(s) 0.80 Cu2+ (aq)+ 2e_Cu(s) | 0.34 Sn4

Question

Reduction half-reaction E (V) Ag (a) e-Ag(s) 0.80 Cu2+ (aq)+ 2e_Cu(s) | 0.34 Sn4+ (aq) + 4e-Sn(s) | 0.15 2H+ (aq) + 2e-H2 (g) Ni2+ (aq) + 2e-Ni(s) | -0.26 Fe2+ (aq)+2e-Fe(s) |-0.45 Zn2+ (aq) + 2e-Zn(s) | -0.76 Al3+ (aq) +3e--A1(s) | -1.66 Mg2+ (aq) + 2e-Mg(s)| 2.37 The equilibrium constant, K, for a redox reaction is related to the standard potential, E°,by the equation nFEo where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mole), R (the gas constant) is equal to 8.314 J/(mol K), and T is the Kelvin temperature Part A Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 °C) for the following reaction Fe(s) + Ni2+ (aq)+Fe2 (aq) + Ni(s) Express your answer numerically K= 2.68x100 Submit Hints My Answers Give Up Review Part Correct

Explanation / Answer

Solving for E° from the above equation:

lnK = nFE° / RT

RT lnK = nFE°

E° = RT lnK / nF

E° = 8.3144 * 298 ln (6.41x10-3) / 1*96485

E° = -0.1297 V

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