Question

A current of 1.70 A is passed through 300mL of 0.160 M solution of $ZnS{O}_4$ for 230 sec with a current efficiency of 90%. The molarity of $Z{n}^{2+}$ after deposition of Zn assuming the volume of the solution remains constant during electrolysis:

• A. 0.154 M
• B. 0.684 M
• C. 0.334 M
• D. None of these

Answer 1

Answer: (A)

0.154 M

Using Faraday's law:
$\therefore$ Eq. of $Z{n}^{2+}$ lost $=\frac{i.t}{96500}$ = $\frac{1.70\times 90\times 230}{100\times 96500}=3.647$ $\times$ ${10}^{-3}$

$\therefore$ Meq. of $Z{n}^{2+}$ lost $=3.647$

Initial Meq. of $Z{n}^{2+}$ $=$ 300 $\times$ 0.160 $\times$ 2 $=96$ (Valence factor = 2)
Meq. of $Z{n}^{2+}$ left in solution $=96-3.647=92.353$

$\therefore$ [$ZnS{O}_4$] $=92.353/(2\times 300)=0.154$ M