Question

If $A$ current of $3$ amperes was passed for $3$ hour through a solution of $CuSO _{4}.3\, g$ of $Cu ^{2+}$ ions were discharged at cathode. Then current efficiency is (Atomic mass of $Cu =63.5$ )

Answer 1

$\because m _{ Cu }= E. i .\, t / 96500$

$\therefore 3=\frac{63.5 \times i \times 3 \times 60 \times}{2 \times 96500}$

$\left( E _{ Cu }= \frac{\text { Atomic mass }}{2}\right)$

Or $i =0.844$ ampere

Current efficiency $=\frac{\text { Current passed actually }}{\text { Total current passed experimentally }} \times 100$

$=\left(\frac{0.844}{3}\right) \times 100=28.14 \%$