Question

If $m$ is the mass of silver deposited at the cathode by $2A$ current flowing for $25min$ through a silver voltameter, then the mass deposited by $1.5A$ current flowing for $600s$ is

• A. $m$
• B. $5m$
• C. $0.3m$
• D. $1.02m$

Answer 1

The correct option is C $0.3m$

By Faradays first law of electrolysis,
$W\propto Q$
$W=ZQ$
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,
$W=Z\times I\times t$
When,
$I=2At=25\times 60$
$W=Z\times I\times t$
$m=Z\times I\times t......\left(1\right)$
When,
$I^{\prime }=1.5A{t}^{\prime }=600$
$W^{\prime }=Z\times I^{\prime }\times t^{\prime }.......\left(2\right)$

Divide equation (2) by (1)

$\frac{W^{\prime }}{m}=\frac{I^{\prime }{t}^{\prime }}{It}=\frac{1.5\times 600}{2\times 25\times 60}=\frac{900}{3000}=\frac{3}{10}=0.3$
$\Rightarrow W^{\prime }=0.3m$