Question

If 96500 coulombs of electricity liberates one gram equivalent of any substance, the time taken for a current of $0.15$ amperes to deposite $20mg$ of copper from a solution of copper sulphate is
Given:
(Equivalent weight of copper = 32 g)

• A. $\text{4 minutes 2 secconds}$
• B. $\text{3 minutes 20 secconds}$
• C. $\text{5 minutes 50 secconds}$
• D. $\text{6 minutes 42 secconds}$

Answer 1

The correct option is D $\text{6 minutes 42 secconds}$

Given :
$\text{Mass deposited , W}=20mg$
$\text{Current, I}=0.15A$

By Faraday's First law,
The mass deposited/released of any substance during electrolysis is proportional to the amount of charge passed into the electrolyte.

$W\propto Q$
$W=ZQ$
where,

W: Mass deposited or liberated
Q: Amount of charge passed
Z: Electrochemical equivalent of the substance
Since,
$Z=\frac{E}{96500}$

$W=\frac{E}{96500}\times Q$
where,
E is Equivalent weight
$\text{Charge, Q}=I\times t$

Hence,
$W=Z\times i\times t$
$\therefore$
$20\times {10}^{-3}=\left(\frac{32}{96500}\right)\times 0.15\times t$
$t=402s$
$t=\text{6 minutes 42 secconds}$