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, ${\left(\text{E}\right)}^{\text{o}}\left({\left(\text{Ce}\right)}^{4+}/{\left(\text{Ce}\right)}^{3+}\right)={\left(\text{1.44 V, E}\right)}^{\text{o}}\left({\left(\text{Fe}\right)}^{3+}/{\left(\text{Fe}\right)}^{2+}\right)=\text{0.68 V}$ $\left\{\frac{2.303RT}{F}=0.0591\text{and}antilog(0.85)=7.07\right\}$

Answer 1

Answer: 7.07

$F{e}^{2+}+C{e}^{4+}\rightleftharpoons F{e}^{3+}+C{e}^{3+}$
Given, $E^0\left(C{e}^{4+}/C{e}^{3+}\right)=1.44V,E^0\left(F{e}^{3+}/F{e}^{2+}\right)=0.68V$
${\left({\left(\text{E}\right)}^{\text{o}}=\text{E}\right)}^{\text{o}}\left({\left(\text{Ce}\right)}^{4+}/{\left(\text{Ce}\right)}^{3+}\right){\left(-\text{E}\right)}^{\text{o}}\left({\left(\text{Fe}\right)}^{3+}/{\left(\text{Fe}\right)}^{2+}\right)$
$=\text{1.44}-\text{0.68}=\text{0.76 V}$
${\text{E}}^{\text{o}}=\frac{\text{2.303RT}}{nF}\text{ log K}\because {\text{E}}^{\text{o}}=\text{0.059 log K}$
$\Rightarrow \text{ log K}=\frac{{\text{E}}^{\text{o}}}{\text{0.0592}}$
$=\frac{\text{0.76}}{\text{0.059}}=\text{12.85}$
$\Rightarrow \text{K}=\text{7.07}\times {\text{10}}^{12}$