Question

In the refining of silver by electrolytic method what will be the weight of 100 g Ag anode if 5 ampere current is passed for 2 hours? Purity of silver is 95% by weight. in gram is:

• A. $57$
• B. $32$
• C. $60$
• D. $13$

Answer 1

The correct option is A $57$

The quantity of electricity passed is $=\frac{I\left(A\right)\times t\left(s\right)}{96500}=\frac{5\times 2\times 3600}{96500}=0.373$ fraraday.

The mass of silver dissolved $=0.373F\times 108=40.3g$.

% purity $=95\%$

So, the mass of Ag deposited $\frac{40.3}{0.95}=42.41g$

The weight of Ag anode $=100g-42.41g=57.59g$