$x + 2\,( - 2) = 0$; $x - 4 = 0$; $x = + 4$.
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Given $: \frac{2.303 RT }{ F }=0.06 V$
$Pd _{( aq )}^{2+}+2 e ^{-} \rightleftharpoons Pd ( s ) \quad E ^{\circ}=0.83\,V$
$PdCl _4^{2-}( aq )+2 e ^{-} \rightleftharpoons Pd ( s )+4 Cl ^{-}( aq )$
$E ^{\circ}=0.65\,V$
Given : $1 \,F =96500\, C \,mol ^{-1}$
Atomic mass of $Fe =56\, g \,mol ^{-1}$
$6 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \rightarrow \mathrm{ClO}_{3}^{-}+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{e}^{-}$
A current of $x A$ has to be passed for $10 h$ to produce $10.0 \mathrm{~g}$ of potassium chlorate. The value of $\mathrm{x}$ is $.......$ (Nearest integer)
(Molar mass of $\left.\mathrm{KClO}_{3}=122.6 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{~F}=96500 \mathrm{C}\right)$