Hydrogen

Chlorine water on standing loses yellow colour due to formation of HCl and HOCl which in turns provide nascent oxygen.

$\mathrm{Cl}_2+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HCl}+\mathrm{HOCl}$

$\mathrm{HClO} \longrightarrow \mathrm{HCl}+$ $\mathrm{\mathop {[O]}\limits_{Nascent\,oxygen}}$

Thus, production of $\mathrm{HCl}$ and $\mathrm{HOCl}$ are responsible for fading of yellow colour of chlorine water.

$\mathrm{H}_2 \mathrm{O}_2$ can be obtained by $\mathrm{BaO}_2 . 8 \mathrm{H}_2 \mathrm{O}$ as follows:

$\mathop {Ba{O_2}.8{H_2}O(s)}\limits_{Barium\,peroxide} + {H_2}S{O_4}(aq)\buildrel {} \over \longrightarrow BaS{O_4}(s) + \mathop {{H_2}{O_2}(aq)}\limits_{Hydrogen\,peroxide} + 8{H_2}O(l)$

Anhydrous barium oxide is not used here because the precipitate $\mathrm{BaSO}_4$ forms a protective layer on the unreacted barium peroxide and this prevents its further participation in the reaction. Hence, option (a) is the correct.