Question

If $0.50\text{L}$ of a $0.60\text{M}$ $SnS{O}_4$ solution is electrolyzed for a period of $30\text{min}$ using a current of $4.60\text{A}$. If inert electrodes are used, what is the molarity of remaining solution?
[Atomic wt. of $Sn=119\text{u}$]
(Give your answer upto to decimals)

Answer 1

At cathode,
$S{n}^{2+}+2e^{-}\to Sn\left(s\right)$
Given,
$0.5\text{L}$ of $0.6\text{M}$ $SnS{O}_4$
So, moles of $SnS{O}_4$ $=(0.5\times 0.6)\text{mol}=0.3\text{moles}$
and
$\text{Time (t)}=30\text{min}=30\times 60\text{s}=1800\text{s}$;
$\text{Current applied}I=4.6\text{A}$
$\therefore$ Moles of $Sn$ deposited
$=\frac{I\times t}{n\times F}=\frac{4.6\times 1800}{2\times 96500}$
$=0.0429\text{mol}$
So, moles of $Sn$ remaining
$=0.3–0.0429\text{mol}$
$=0.2571\text{mol}$
Therefore,
$\text{Molarity of the final solution}=\frac{0.257}{0.5}\text{M}=0.514\text{M}$