From a consideration of the following two half reactions at 298 K. Pb2+(aq) + 2e- → Pb(s) -0.126 V Fe2+(aq) + 2e- → Fe(s) -0.447 V What is the equilibrium constant for the following equation? Pb2+(aq) + Fe(s) → Pb(s) + Fe2+(aq) { 1F = 96485 C/mol, R = 8.314 J/K mol} a. 3.7 x 10-6 b. 2.7 x 105 c. 7.2 x 1010 d. 2.4 x 1019

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