Question

What amount of $C{l}_2$ gas liberated at anode, if $1A$ current is passed for $30min$ from $NaCl$ solution?

• A. 0.66 mol
• B. 0.33 mol
• C. 0.66 g
• D. 0.33 g

Answer 1

Answer: (C)

The problem will be solved by Faraday's law of electrolysis.
$Q=it$
Given, $i=1A,t=30\times 60=1800s$
$\therefore$ $Q=1\times 1800$
$=1800C.$ $\underset{\begin{array}{c}1mol\\ 71g.\end{array}}{2C{l}^{-}}\to \underset{2\times 96500C}{C{l}_2+2e^{-}}$ (at anode)
$\therefore$ the amount of $C{l}_2$ gas liberated by passing 1800 C of electric charge
$=\frac{1\times 1800\times 71}{2\times 96500}=0.66g$