Question

When electrolysis of $KCl$ is done in alkaline medium, 10 g of $KCl{O}_3$ is produced as follows :
$C{l}^{\ominus }+6\stackrel{\ominus }{O}H\to Cl{O}_3^{\ominus }+3H_{2}O+6e^{-}$
A current of 2 A is passed for 10.941 hours. The current efficiency (in %) used in the process is $(MwtofKCl{O}_3=122.5)$ :

Answer 1

Let x% is the current efficiency of $KCl{O}_3$ = Number of Faradays.
$\frac{10g}{122.5/6}=\frac{2\times x\times 10.941\times 3600}{100\times 96500}$
$\therefore$ x = 60%