Question

A current of $3.7A$ is passed for $6h$ between nickel electrodes in $0.5L$ of a $2.0M$ solution of $Ni{\left(N{O}_3\right)}_2$. What will be the molarity of solution at the end of electrolysis?

Answer 1

During electrolysis, $N{i}^{2+}$ will be reduced at cathode and $H_{2}O$ will be oxidised at anode.
Number of Faraday's passed $=\frac{3.7\times 6\times 60\times 60}{96500}=0.828$
$\Rightarrow 0.828g$ equivalent of $N{i}^{2+}$ will be deposited at cathode. Initial moles of $N{i}^{2+}$ ion $=2\times 0.5=1.0$
Moles of $N{i}^{2+}$ ion remaining after electrolysis $=1.0-\frac{0.828}{2}$
$=0.586$
$\Rightarrow$ Molarity of $N{i}^{2+}$ in final solution $=\frac{0.586}{0.50}=1.172M$