Question

A current of 9.65 ampere is passed through the aqueous solution $NaCl$ using suitable electrodes for $1000s$ . The amount of $NaOH$ formed during electrolysis is :

• A. $8.0g$
• B. $6.0g$
• C. $4.0g$
• D. $2.0g$

Answer 1

The correct option is C $4.0g$

Given :
$\text{Current, I}=9.65A\text{Time, t}=1000s$

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 mass
$E=\frac{\text{Molecular Mass}}{n-factor}$

$N{a}^{+}+O{H}^{-}\to NaOH$
Molecular mass of $NaOH$ is $40gmo{l}^{-1}$
n-factor the reaction is 1

$E=\frac{\text{40}}{1}$

$\text{Charge, Q}=I\times t$
$Q=9.65\times 1000$
Hence,
$W=Z\times i\times t$
$W=\frac{E}{96500}\times 9.65\times 1000$

$W=\frac{\left(\frac{40}{1}\right)}{96500}\times 9.65\times 1000$
$W=4g$