Calculate the cell potential and the equilibrium constant for the following reaction at 298 K: Fe2+(aq) + H2(g) Fe(s) + 2H+(aq)

Question

Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm.

Answer 1

Fe²⁺(aq) + H₂(g) ==> Fe(s) + 2H⁺(aq)

oxidation @ the anode:

H₂(g) ==> 2H⁺(aq) + 2e- ... Eº = 0

reduction @ the cathode:

Fe²⁺(aq) + 2e- ==> Fe(s) ... Eº = -0.44 V

Eºcell = Eºcathode - Eºanode = -0.44 V (non spontaneous)

-nFEº = -RT ln K

n = 2 mols electrons

F = 96,500 coulombs / mol electrons

Eº = -0.44 V = 0.44 Joules / coulomb = -0.44 J/C

R = gas constant = 8.314 J/Kmol

T = temp in K = 298K

K = ?

Since ln K is a negative number, K will be very small, again suggesting a non spontaneous reaction.