Chemistry - Electrochemistry Question with Solution | TestHub

For the reaction at equilibrium,

$2{\mathrm{MnO}}_4^{-}+6{\mathrm{H}}^{+}+5{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\rightleftharpoons 2{\mathrm{Mn}}^{2+}+8{\mathrm{H}}_2\mathrm{O}+10{\mathrm{CO}}_2.$

Given : ${\mathrm{E}}_{{\mathrm{MnO}}_4^{-}/ {\mathrm{Mn}}^{2+}}^0=1.51 \mathrm{V} \mathrm{and} {{\mathrm{E}}^0}_{{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4^{}/{\mathrm{CO}}_2}=0.49 \mathrm{V}.$

So,  ${\mathrm{E}}_{\mathrm{cell}}^0 =1.51 + 0.49=2 \mathrm{V}.$

Number of electrons involved in reaction can be calculated as follows:

${\mathrm{Mn}}^{+7 } \mathrm{changes} {\mathrm{Mn}}^{2+}$ so, each  $\mathrm{Mn}$ gain five electrons.

$\mathrm{n} = 10$

Now, ${\mathrm{E}}_{\mathrm{cell}}^0 =\frac{0.0591}{\mathrm{n}}\log \mathrm{K}$

$\log \mathrm{K} =\frac{10\times 2}{0.0591}$

$\log \mathrm{K} =338.4$