Question

A current of $0.1A$ was passed for $2$ hr through a solution of cupro cyanide and $0.3745$ g of copper was deposited on the cathode.

Calculate the current efficiency for the copper deposition.

[$Cu=63.5$ g/mole]

• A. $39.5$ %
• B. $79.6$ %
• C. $63.25$ %
• D. $63.5$ %

Answer 1

The correct option is B $79.6$ %

From faraday's law of electrolysis, $m=\frac{It}{F}\times \frac{M}{z}$

To calculate the current effeciency , we need to calculate the amount of current used from $0.1$ ampere. let it be $x$.

$m$ = mass of substance liberated at electrode $=0.3745$ gm

$t$ = time = $2$ hours $=2\times 3600=7200$ second

$M$ = molecular weight $=63.5$

$F$ - faraday connstant $=96500$

$z$ = valency number of ions of the substance (electrons transferred per ion).

$x=m\times \frac{F}{t}\times \frac{z}{M}$ $=0.3745\times \frac{96500}{7200}\times \frac{1}{63}$ $=0.0796$ amp

Current effeciency is $\frac{x}{I}\times 100=\frac{0.0796}{0.1}\times 100=79.6$%

Hence, the correct option is $\text{B}$