Consider the following redox reaction: MnO _ 4 ^ - + H ^ + + H _ 2...

For the reaction at equilibrium,

$2{MnO}_{4}^{-} + 6{H}^{+} + 5{H}_{2}{C}_{2}{O}_{4} ⇌ 2{Mn}^{2 +} + 8{H}_{2}O + 10{CO}_{2} .$

Given : ${E}_{{MnO}_{4}^{-} / {Mn}^{2 +}}^{0} = 1 . 51Vand{E}^{0}{H}_{2}{C}_{2}{O}_{4}^{} / {CO}_{2} = 0 . 49V .$

So, ${E}_{cell}^{0} = 1 . 51 + 0 . 49 = 2V .$

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

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

$n = 10$

Now, ${E}_{cell}^{0} = \dfrac{0 . 0591}{n}logK$

$logK = \dfrac{10 \times 2}{0 . 0591}$

$logK = 338 . 4$