Chemistry - Electrochemistry Question with Solution | TestHub

Electrolysis of $50\%$ solution of sulfuric acid produces peroxydisulfuric acid (${\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8$) at the anode.

$2{\mathrm{HSO}}_4^{-}\to {\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8+2e^{-}$

Further hydrolysis of peroxydisulfuric acid yields hydrogen peroxide (${\mathrm{H}}_2{\mathrm{O}}_2$) and sulfuric acid.

${\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8+2{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{H}}_2{\mathrm{SO}}_4+{\mathrm{H}}_2{\mathrm{O}}_2$

Thus, the product obtained is ${\mathrm{H}}_2{\mathrm{O}}_2$.

Answer: ${\mathrm{H}}_2{\mathrm{O}}_2$