Question

A current of 3 ampere was passed for 2 hours through a solution of $CuSO_{4}.3g$of$Cu^{2 +}$ ions were discharged at cathode. The current efficiency is (At. wt. of Cu=63.5)

• A. 20%
• B. 42.2%
• C. 60%
• D. 80%

Answer 1

Answer: (B)

42.2%

Answer 2

$W_{Cu^{2 +}} = \frac{E \times I \times t}{\text{96500}}$

$\mathbf{3 =}\frac{\mathbf{63.5}\mathbf{\times}\mathbf{I}\mathbf{\times}\mathbf{2}\mathbf{\times}\mathbf{60}\mathbf{\times}\mathbf{60}}{\mathbf{2}\mathbf{\times}\mathbf{96500}}$

$\mathbf{I = 1.266}$Ampere

Current efficiency

$= \frac{\text{Current passed actually}}{\text{Total current passed experimentally }} \times 100$ $= \frac{1.266}{3} \times 100 = 42.2\%$