Explanation

(c)
$H_{2}O_{2}$ can be prepared by electrolysis of 50% $H_{2}SO_{4}.$ In this method, hydrogen is liberated at cathode.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ H_{2}SO_{4} \rightleftharpoons 2H^{+} + 2HSO_{4}^{-}$
At anode : $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2HSO_{4}\ \rightarrow H_{2}S_{2}O_{8} + 2e^{-}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ H_{2}S_{2}O_{8} + 2H_{2}O\ \rightarrow 2H_{2}SO_{4} + H_{2}O_{2}$
At cathode : $\ \ \ \ \ \ \ \ \ \ \ \ 2H^{+} + \ 2e^{-}\ \rightarrow H_{2} \uparrow$