$Ca^{2+} + 2e^{-} → Ca$
$2 F = 2 × 96500$ કુલમ્બ વિદ્યુતજથ્થો $= 1$ મોલ $(6.022 × 10^{23})$ પરમાણુ $Ca$
$\therefore {\text{1}}{\text{.5 }}$ કુલમ્બ વિદ્યુતજથ્થો $ = \frac{{1.5\, \times 6.022 \times {{10}^{23}}}}{{2 \times 96500}} = 4.68 \times {10^{18}}$
આપેલ: $E _{ x ^{2+} \mid x }^0=-2.36\,V$
$E _{ Y ^{3+} \mid Y }^0=+0.36\,V$
$\frac{2.303\,RT }{ F }=0.06\,V$