Question

Calculate the equilibrium constant for the reaction
S₂O₈^2-(aq) + 2 Fe²⁺(aq) --> 2 Fe³⁺(aq) + 2 SO₄^2-(aq) at 25.0°C, given that the standard cells potentials for the two half reactions at this temperature are
S₂O₈^2-(aq) + 2 e^- 2 SO₄^2-(aq) E = +2.08 V
Fe³⁺(aq) + e^- Fe²⁺(aq) E = +0.77 V

Answer 1

S₂O₈^2-(aq) + 2 Fe²⁺(aq) 2 Fe³⁺(aq) + 2 SO₄^2-(aq)

and
S₂O₈^2-(aq) + 2 e^- 2 SO₄^2-(aq)   : E^o = +2.08 V
Fe³⁺(aq) + e^- Fe²⁺(aq)     : E^o = +0.77 V

So standard potential of the cell , E^o = E^o_cathode - E^o_anode

                                                      = +2.08 - (+0.77) V

                                                      = +1.31 V

We know that G = -nFE^o    also G = -RT ln K

So nFE^o = RT lnK

Where

n = number of electrons transferred = 2

F = Faraday = 96500 C

R = gas constant = 8.314 J/(mol-K)

T = temperature = 25 oC = 25+273 = 298 K

K = Equilibrium constant = ?

Plug the values we get

lnK = (nFE^o )/ (RT)

      = 102.05

   K = e 102.05

      = 2.08x10⁴⁴

Therefore the equilibrium constant is 2.08x10⁴⁴