Question

Calculate the equilibrium constant for the reaction
$ Fe^{2+} + Ce^{4+} \rightleftharpoons Fe^{3+} + Ce^{3+}$
Given $ E^\circ (Ce^{4+}/Ce^{3+})=1.44\, V, E^\circ ( Fe^{3+}/ Fe^{2+})=0.68\, V$

Answer 1

$Fe^{2+} + Ce^{4+} \rightleftharpoons Fe^{3+} + Ce^{3+}$
$ E^\circ =E^\circ (Ce^{4+}/Ce^{3+})- E^\circ(Fe^{3+}/Fe^{2+})$
$ = 1.44 - 0.68 = 0.76\, V$
$\because E^\circ = 0.0592\, \log\, K $
$\Rightarrow \log\, K = \frac {E^\circ} {0.0592} =\frac{0.76}{0.0592}=12.83 $
$\Rightarrow k = 6.88 \times 10^{12}$