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(D) The electrolysis of $50 \%$ sulfuric acid $(H_{2}SO_{4})$ at high current density yields peroxydisulfuric acid $(H_{2}S_{2}O_{8})$,also known as M

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$H_{2}O_{2}$

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(D) The electrolysis of $50 \%$ sulfuric acid $(H_{2}SO_{4})$ at high current density yields peroxydisulfuric acid $(H_{2}S_{2}O_{8})$,also known as Marshall's acid,at the anode.$2H_{2}SO_{4} \rightarrow H_{2}S_{2}O_{8} + H_{2(g)}$Upon further hydrolysis of peroxydisulfuric acid,it reacts with water to produce hydrogen peroxide $(H_{2}O_{2})$ and sulfuric acid.$H_{2}S_{2}O_{8} + 2H_{2}O \rightarrow 2H_{2}SO_{4} + H_{2}O_{2}$Thus,the final product obtained after hydrolysis is hydrogen peroxide $(H_{2}O_{2})$.

What product is obtained by electrolysis of $50 \%$ solution of sulfuric acid on further hydrolysis?

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