Question

The time required to remove electrolytically one-fourth of Ag from 0.2 L of 0.1 M $\text{AgN}{\text{O}}_3$ solution by a current of 0.1 amp is approximately:

• A. 320 min
• B. 160 min
• C. 80 min
• D. 40 min

Answer 1

The correct option is A 320 min

$\omega =\text{Z}\times \text{Q}$

$\text{moles}=\frac{\text{I}\times t}{n\times 96500}$

$\text{Moles of}AgN{O}_3=0.1\times 0.2=0.02$

$A{g}^{+}+e^{-}⟶Ag$ $\therefore n=1$

$0.02=\frac{0.1\times t}{96500}$

$t=\frac{0.02\times 96500}{0.1}\sec$

t = 320 minutes.

Option (A) is correct.