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Q.
Electrolysis of $H _{2} SO _{4}$ (conc.) gives the following at anode
Electrochemistry
Solution:
Electrolysis of concentration $H _{2} SO _{4} \rightarrow H ^{+}+ HSO _{4}^{-}$
At Cathode
$2 H ^{+}+2 e ^{-} \rightarrow H _{2}$
At Anode
$2 HSO _{4}^{-} \rightarrow \underset{\text { Marshall acid }}{\stackrel{ H _{2} S _{2} O _{8}}{\downarrow}}+2 e ^{-}$