Chemistry - Electrochemistry Question with Solution | TestHub

$2{\mathrm{Cl}}^{-}\to \underset{1\text{ mole }}{{\mathrm{Cl}}_2}+\underset{2\times 96500\text{ coulomb }}{2e^{-}\text{(At anode) }}$ $\begin{array}{rl}Q& =it=1\times 30\times 60\\ & =1800\text{ coulomb }\end{array}$ The amount of chlorine liberated by passing 1800 coulomb of electric charge $=\frac{1\times 1800\times 71}{2\times 96500}=0.66\mathrm{g}$