# The number of electrons involved in the manufacture of 1mole of $H_2O_2$ from 50% of $H_2SO_4$

$\begin{array}{1 1}(a)\;2moles\\(b)\;3moles\\(c)\;1mole\\(d)\;4moles\end{array}$

By electrolysis of $50\% \; H_2SO_4$
Hydrogen peroxide is manufactured by the electrolysis of a cold $50\%$ solutions of $H_2SO_4$ in an electrolytic cell using platinum as anode and graphite as cathode.
$2H_2SO_4 \Rightarrow 2H^+ + 2HSO_4^-$
$2HSO_4^- \Rightarrow H_2S_2O_8 + 2e^-$ ( at anode)
Peroxy- disulphuric acid $H_2S_2O_8$ reacts with water during distillation to form hydrogen peroxide .
$H_2S_2O_8 (aq)+2H_2O(l) \Rightarrow 2H_2SO_4(aq) + H_2O_2 (aq)$
$2H^+ + 2e^- \Rightarrow H_2 (g)$ (at cathode)
The number of electrons involved in the manufacture of 1mole of $H_2O_2$ from $50\%$ of $H_2SO_4$ is 2 moles
Hence (a) is the correct option.
edited Mar 24