Question

Through a solution of $CuS{O}_4$, a current of $3$ amperes was passed for $2$ hours. At cathode, $3g$ of $C{u}^{2+}ions$ were discharged. The current efficiency is : [At. wt. of $Cu=63.5$]

• A. $33.3$ %
• B. $42.2$ %
• C. $48.7$ %
• D. $54.4$ %

Answer 1

Answer: (B)

$42.2$ %

According to law of electrolysis,
Mass deposited $\left(m\right)=Zit$

or $i=\frac{m\times 96500}{t\times Eq.wt}$

Here, $m=3g,t=2\times 60\times 60=7200sec$

$z=\frac{Eq.wt}{96500};$

$Eq.wt.=\frac{At.wt}{\text{Oxidation number}}$

$\therefore i=\frac{3\times 96500\times 2}{63.5\times 7200}$

$=1.266A$

Efficiency of current is given by,

$=\frac{\text{Current used}}{\text{Total current passed}}\times 100$

$=\frac{1.266}{3}\times 100=42.22$ %

Hence, option B is correct.