In an electrolysis of an aqueous solution of NaOH, 2.8 L of oxygen gas at STP was liberated at anode. The volume of hydrogen gas at STP liberated at cathode would be 

The reactions are  $2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2+2{\mathrm{OH}}^{-}$ and ${\mathrm{H}}_2\mathrm{O}\to \frac{1}{2}{\mathrm{O}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$

Volume of ${\mathrm{H}}_2$ will be double than that of ${\mathrm{O}}_2$.

The volume of hydrogen gas at STP liberated at cathode would be $5.6 \mathrm{L}$