In which of the following reaction ${\mathrm{H}}_2{\mathrm{O}}_2$ acts as a reducing agent?

$\text{ (P) }{\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{H}}^{+}+2e^{-}⟶2{\mathrm{H}}_2\mathrm{O}$

$\text{ (Q) }{\mathrm{H}}_2{\mathrm{O}}_2-2e^{-}⟶{\mathrm{O}}_2+2{\mathrm{H}}^{+}$

$\text{ (R) }{\mathrm{H}}_2{\mathrm{O}}_2+2e^{-}⟶2{\mathrm{OH}}^{-}$

$\text{ (S) }{\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{OH}}^{-}-2e^{-}⟶{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}$

The reaction, acts as a reducing agent as

${\mathrm{H}}_2{\mathrm{O}}_2-2e^{-}\to {\mathrm{O}}_2+2{\mathrm{H}}^{\ast }$

${\mathrm{H}}_2{\mathrm{O}}_2$ can be reduced  ${\mathrm{H}}_2\mathrm{O}$  as well as oxidized ${\mathrm{O}}_2$. As a result, it may operate as an oxidizing agent by being reduced to ${\mathrm{O}}_2$ and as a reducing agent by being oxidized to ${\mathrm{H}}_2\mathrm{O}$ . As a result, it may function as both an oxidizing and a reducing agent.

The following steps show how hydrogen peroxide acts as the reducing agent.

H₂O₂ loses one O atom and becomes decreased.

${\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{H}}^{+}+2\mathrm{e}\to 2{\mathrm{H}}_2\mathrm{O}$

 H₂O₂ molecule is oxidized after losing two electrons.

${\mathrm{H}}_2{\mathrm{O}}_2-2\mathrm{e}\to {\mathrm{O}}_2+2{\mathrm{H}}^{+}$

H₂O₂ is decreased after receiving two more electrons.

${\mathrm{H}}_2{\mathrm{O}}_2+2\mathrm{e}\to 2\left({\mathrm{OH}}^{-}\right)$

An H₂O₂ molecule is oxidized and loses hydrogen in the process.

${\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{OH}}^{-}-2\mathrm{e}\to {\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}$