Question

A current of $2A$ was passed for $1h$ through a solution of $CuS{O}_{4}0.237g\text{of}C{u}^{2+}$ ions were discharged at cathode, The current efficiency is:

• A. $26.1$%
• B. $10$%
• C. $42.2$%
• D. $40.01$%

Answer 1

The correct option is B $10$%

Given $I=2A$
$t=1hour=60\times 60sec$
According to faraday's $I{I}^{nd}$ law :
$\frac{1F}{Q}=\frac{\text{Atomic weight/valency factor}}{\text{weight}}$
$Q=I\times t=2\times 60\times 60$
$\text{Weight (w)}=\frac{63.5\times 2\times 60\times 60}{2\times 96500}=2.37g$
% efficiency $=\frac{\text{Mass obtained}}{\text{Maximum Mass}}\times 100$
$=\frac{0.237}{2.37}\times 100=10$%.