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Q.
What amount of $ C{{l}_{2}} $ gas liberated at anode, if $1\, A$ current is passed for $30\, min$ from $ NaCl $ solution?
BHUBHU 2005
Solution:
The problem will be solved by Faraday's law of electrolysis.
$ Q=it $
Given, $ i=1\text{ }A,\text{ }t=30\times 60=1800\text{ }s $
$ \therefore $ $ Q=1\times 1800 $
$ =1800C. $ $ \underset{\begin{smallmatrix} 1\,mol \\ 71\,g. \end{smallmatrix}}{\mathop{2C{{l}^{-}}}}\,\xrightarrow{{}}\underset{2\times 96500\,C}{\mathop{C{{l}_{2}}\text{ }+\text{ }2{{e}^{-}}}}\, $ (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 $