Cell Potential and Equilibrium Standard reduction potentials The equilibrium con

Question

Cell Potential and Equilibrium Standard reduction potentials The equilibrium constant, K, for a redox reaction is related to the standard cell potential, Ecel, by the equation Reduction half-reaction (V) Ag+ (aq) + e-Ag(s) Cu2+ (aq) + 2e-Cu(s) 0.34 Sn (a) 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-Al(s) -1.66 Mg2+ (aq) + 2e-Mg(s) -2.37 0.80 n FEcell where n is the number of moles of electrons transferred, F (the Faraday's constant) is equal to 96 500 C mol R (the gas constant) is equal to 8.314 JK-1 mol-1 , 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 K2.68x10 Submit Hints My Answers Give Up Review Part Correct When Ecell0 and K > 1 the reaction favours the products

Explanation / Answer

Part A :

Eo = 0.45-0.26 = 0.19 V

lnk = nFE / RT

n = 2

F = 98500

R = 8.314

T =298K

K = 2.74*10^-7

part B :

lnK = nFEo / RT

we know all values except Eo

Eo = RTlnK/nF = - 0.12V

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