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In an electrolysis of an aqueous solution of sodium sulphate,2.4L of oxygen at STP was liberated at anode.The volume of hydrogen at STP liberated at cathode would be

$\begin{array}{1 1}1.2L\\2.4L\\2.6L\\4.8L\end{array} $

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Answer : 4.8L
The reactions are $2H^++2e^-\rightarrow H_2$ and $2H_2O\rightarrow 4H^++O_2+4e^-$
The volume of $H_2$ liberated will be twice that of $O_2$
answered Jul 23, 2014 by sreemathi.v

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