Question

Electrolysis of a solution of $Mn{SO}_4$ in aqueous sulphuric acid is a method for the preparation of $Mn{O}_2$. Passing a current of $27A$ for $24$ hours gives $1kg$ of $Mn{O}_2$. The current efficiency in this process is:

• A. $100$%
• B. $95.185$%
• C. $80$%
• D. $82.951$%

Answer 1

Answer: (B)

$95.185$%

The reaction taking place on the given reaction are,

$M{n}^{2+}+2H_{2}O\to Mn{O}_2+2H^{+}+H_2$

Amount of current is given by,
$m=Z\times I\times t$

$I=m\frac{F\times x}{t\times M}$

$=1000g\times \frac{96500\times 2}{24\times 60\times 60\times 86.9}$

$⟹I=25.70A$

Now current efficiency$=\frac{25.7}{27}\times 100=95.185$ %

Hence,option B is correct answer.