Question

A current of $3.7\, A$ is passed for $6 \,h$ between nickel electrodes in $0.5\, L$ of a $2.0 \,M$ solution of $Ni \left( NO _{3}\right)_{2}$. What will be the molarity of solution at the end of electrolysis?

Answer 1

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