How much current is required to produce $C l_{2} (g)$ at the rate of $10^{- 3} L s^{- 1}$ electrolysis of molten KCl? (in A)

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Solution

Rate of deposition = $10^{- 3} L s^{- 1} = \frac{10^{- 3}}{22.4} m o l s^{- 1}$
$2 C l^{⊖} \to C l_{2} + 2 e^{-}$
$2 F = 1 m o l C l_{2}$
$\Rightarrow \frac{10^{- 13}}{22.4} m o l s^{- 1} C l_{2} = \frac{2 \times 10^{- 3}}{22.4} F s^{- 1}$
$\Rightarrow \frac{I \times 1 s}{96500} = \frac{2 \times 10^{- 3}}{22.4} \Rightarrow I = 8.616 A$
$m E q o f K d e p o s i t e d = N u m b e r o f F a r a d a y s \times 10^{3}$

$= \frac{2 \times 10^{- 3}}{22.4} \times 10^{3} = 0.0892$

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