the standard reduction potentials are
Fe3+ (aq) + e- ----->Fe2+(aq) E0 = +0.77V
Ag+(aq) + e- -------> Ag(s) E0 = +0.80V
the two half cell reactions can be written as
anode: Fe2+(aq) ------->Fe3+ (aq) + e-
cathode: Ag+(aq) + e- -------> Ag(s)
-------------------------------------------------
overall: Fe2+(aq) +Ag+(aq) ----->Fe3+ (aq) + Ag(s)
E
0cell =E
0cathode - E
0anode
E0cell = (0.80)-(0.77)
E0cell = + 0.03V
the gibbs free energy is related to the E0cell as
ΔG0 = - nFE0cell
where,n=no. of moles of electrons transfered through the wire= 1
F= 1 Faraday= 96500 coulombs
∴ΔG0 = - nFE0cell = - (1*96500 J/V.mol*( - 0.03V))
∴ΔG0 =-2895J= -2.895KJ
we have a relation between gibbs energy and equilibrium constant K
ΔG0 = - RTln K
where R= 8.314 J/mol.K, T=298 K
∴K = exp(-ΔG0 /RT) =exp(-(-2895J)/(8.314 J/mol.K*298 K))
∴K= 3.217