Question

A current of 3.7 A is passed for 6 hours between two platinum electrodes dipped in 500 mL of a 2.0 M solution of $Ni(N{O}_3{)}_2$. The concentration (in $M$) of the solution after the electrolysis is:

Answer 1

Number of Faradays $=\frac{3.7\times 6\times 3600}{96500}=0.828F$

$N{i}^{2+}+2e^{-}\to Ni$

$2F=1$ mol of Ni

$\Rightarrow$ Moles of Ni deposited $=\frac{0.828}{2}=0.414$
Initial moles $=2\times 0.5=1$

$\Rightarrow$ Moles of $N{i}^{2+}$ left in solution $=1-0.414=0.586$.

Final concentration $=\frac{0.586}{0.5}=1.172M$

If instead of Pt electrodes, Ni electrodes are used, then the final concentration of solution will remain same as Ni deposited from the solution will be filled up by the amount of Ni obtained from electrodes.

Hence, final concentration $\approx 1M$.