Question

If $A$ current of $3$ amperes was passed for $3$ hour through a solution of $CuS{O}_4.3g$ of $C{u}^{2+}$ ions were discharged at cathode. Then current efficiency is (Atomic mass of $Cu=63.5$ )

Answer 1

Answer: 28.14

$\because m_{Cu}=E.i.t/96500$

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

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

Or $i=0.844$ ampere

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

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