Calculate equilibrium constant for the equilibrium
${2 m n O}_{4}^{-} + 6 H^{+} + {5 H}_{2} C_{2} O_{4} ⇌ {2 M N}^{2 +} + 8 H_{2} O + {10 C O}_{2}$
Given that, $E_{{M n O}_{4} M n^{2 +},}^{0} = 1.51 V$ and $E_{{C O}_{3} C_{2} O_{4}^{2 -}}^{0} = - 0.49 V$

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Solution

$E_{c e l l}^{o} = E_{c a t h o d e} - E_{a n o d e}$
$= 1.51 + 0.49$
$= 2$ Volts
At equilibrium $E_{c e l l} = 0$
So $E_{c e l l}^{o} = - 0.059 / n l o g K_{c}$
$\Rightarrow 2 = \frac{- 0.059}{5} log K_{c}$
$\Rightarrow log K_{c} = \frac{10}{- 0.059}$
$\Rightarrow K_{C} = 2.51$

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