In the reaction $\mathrm{H_2O_2}(aq) + \mathrm{Cl}_2(g) \rightarrow 2\,\mathrm{HCl}(aq) + \mathrm{O}_2(g)$, what role does $\mathrm{H_2O_2}$ play?

1. Track the species' oxidation states. 2. In $\mathrm{H_2O_2}$, oxygen is in the peroxide state, $-1$. In product $\mathrm{O}_2(g)$, oxygen is $0$. So O goes UP by 1 — $\mathrm{H_2O_2}$ is OXIDISED. 3. In $\mathrm{Cl}_2$, Cl is $0$. In product $\mathrm{HCl}$, Cl is $-1$. So Cl goes DOWN — $\mathrm{Cl}_2$ is reduced. 4. The species that is OXIDISED is the REDUCING AGENT (because it donated electrons to whatever was reduced). So $\mathrm{H_2O_2}$ here is the reducing agent. 5. Note: $\mathrm{H_2O_2}$ can play EITHER role depending on the other reactant. With strong oxidisers like $\mathrm{Cl}_2$ or $\mathrm{KMnO}_4$, it acts as a reducing agent (O: $-1 \rightarrow 0$). With reducing species like KI, it acts as an oxidising agent (O: $-1 \rightarrow -2$). This is a recurring NEET trap. 6. Option A and C describe the WRONG direction. Option D is wrong because oxidation states change → this IS a redox reaction. _Source: NCERT Class 11 Chemistry Part 2, Ch 7, §7.2 + §7.3 (H₂O₂ in oxidising vs reducing roles), p. 3 + p. 10._