Calculate the equilibrium constant for the reaction:
$F e + {C u S O}_{4} ⇌ {F e S O}_{4} + C u$
Given, $E_{F e / F e}^{o} = 0.44 V; E_{C u / {C u}^{2 +}}^{o} = - 0.337 V$

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Solution

The cell potential of the reaction is
$Δ E_{0} = E_{0} (F e | F e^{+ 2}) + E_{0} (C u^{+ 2} | C u)$
= 0.44V + 0.337 V
= 0.777 V

Gibbs free energy change due to reaction and cell potential are related as follows:
$Δ G = n ∙ F ∙ Δ E_{0}$
with
n number of electrons exchanged, here n=2
F Faraday constant

Equilibrium constant and Gibbs free energy of reaction are related as:
$Δ G = - R ∙ T ∙ l n (K)$
with
R universal gas constant
T thermodynamic temperature

Hence,
$- n ∙ F ∙ Δ E_{0} = - R ∙ T ∙ l n (K)$
$K = e^{n ∙ F ∙ Δ E_{0} / (R ∙ T)}$

For this reaction:
$K = e^{2 ∙ 96485 C ∙ m o l ⁻ ¹ \times 0.777 V / (8.3145 J ∙ K ⁻ ¹ ∙ m o l ⁻ ¹ \times 298 K)}$
$= e^{60.5144}$
$= 1.91 \times 10^{26}$

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F e + C u S O_{4} ⟶ F e S O_{4} + C u

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