Question

Calculate the equilibrium constant for the reaction (in multiple of 10¹²)
$\text{Fe}^{2 +} + \text{Ce}^{4 +} \rightleftharpoons \text{Fe}^{3 +} + \text{Ce}^{3 +}$
Given, $\left(\text{E}\right)^{\text{o}} \left(\left(\text{Ce}\right)^{4 +} / \left(\text{Ce}\right)^{3 +}\right) = \left(\text{1.44 V, E}\right)^{\text{o}} \left(\left(\text{Fe}\right)^{3 +} / \left(\text{Fe}\right)^{2 +}\right) = \text{0.68 V}$ $\left\{\frac{2 . 303 RT}{F} = 0 . 0591 \text{and} antilog \left(\right. 0 . 85 \left.\right) = 7 . 07\right\}$

Answer 1

$ Fe ^{2+}+ Ce ^{4+} \rightleftharpoons Fe ^{3+}+ Ce ^{3+} $
Given, $E ^{0}\left( Ce ^{4+} / Ce ^{3+}\right)=1.44 V , \quad E ^{0}\left( Fe ^{3+} / Fe ^{2+}\right)=0.68 V$
$\left(\left(\text{E}\right)^{\text{o}} = \text{E}\right)^{\text{o}} \left(\left(\text{Ce}\right)^{4 +} / \left(\text{Ce}\right)^{3 +}\right) \left(- \text{E}\right)^{\text{o}} \left(\left(\text{Fe}\right)^{3 +} / \left(\text{Fe}\right)^{2 +}\right)$
$= \text{1.44} - \text{0.68} = \text{0.76 V}$
$\text{E}^{\text{o}}=\frac{\text{2.303RT}}{nF}\text{ log K}\because \text{E}^{\text{o}}=\text{0.059 log K}$
$\Rightarrow \text{ log K}=\frac{\text{E}^{\text{o}}}{\text{0.0592}}$
$=\frac{\text{0.76}}{\text{0.059}}=\text{12.85}$
$\Rightarrow \text{K}=\text{7.07}\times \text{10}^{1 2}$