Question

An aqueous solution of $NaCl$ on electrolysis gives $H_2(g),C{l}_2(g)$ and $NaOH$ according to the reaction.
$2C{l}^{-}(aq)+2H_{2}O\rightleftharpoons 2OH(aq)+H_2(g)+C{l}_2(g)$
A direct current of $25A$ with a current efficiency of $62\%$ is passed through $20L$ of $NaCl$ solution ( $20\%$ by weight). Write down the reactions taking place at the anode and cathode. How long will it take to produce $1kg$ of $C{l}_2$ ? What will be the molarity of the solution with respect to hydroxide ion? (Assume no loss due to evaporation)

Answer 1

At anode $2C{l}^{-}⟶C{l}_2+2e^{-}$
At cathode $2H_{2}O+2e^{-}⟶H_2+2O{H}^{-}$
$1kgC{l}_2=\frac{1000}{35.5}$ equivalent of $C{l}_2=28.17$ equivalent
$\Rightarrow$ Theoretical electricity requirement $=28.17F$
$\because$ Efficiency is only $62\%$
$\therefore$ Electricity requirement (experimental)
$=\frac{28.17\times 100}{62}F=45.44F$
$\Rightarrow 45.44\times 96500=25t$ (in second)
$\Rightarrow t=48.72h$
Also, gram equivalent of $H{O}^{-}$produced $=28.17$
$\Rightarrow$ Molarity of $H{O}^{-}=\frac{28.17}{20}=1.4085M$