(A) HOCl + H2O2 H3O+ + Cl- + O2 (B) I2 + H2O2 + 2OH- 2I- + 2H2O + O2 Choose the correct option:

Step 1 — Analyse reaction (A): $\mathrm{HOCl + H_2O_2 \rightarrow H_3O^{+} + Cl^{-} + O_2}$

• In $\mathrm{H_2O_2}$: O is $-1$

• In $\mathrm{O_2}$ (product): O is $0$

• O goes from $-1 \rightarrow 0$: oxidation (loss of electrons)

• $\mathrm{H_2O_2}$ is oxidised → acts as reducing agent in (A)

• In $\mathrm{HOCl}$: Cl is $+1$; in $\mathrm{Cl^-}$: Cl is $-1$ → Cl is reduced → HOCl is the oxidizing agent

Step 2 — Analyse reaction (B): $\mathrm{I_2 + H_2O_2 + 2OH^{-} \rightarrow 2I^{-} + 2H_2O + O_2}$

• In $\mathrm{H_2O_2}$: O is $-1$

• In $\mathrm{O_2}$: O is $0$

• O goes from $-1 \rightarrow 0$: oxidation

• $\mathrm{H_2O_2}$ is oxidised → acts as reducing agent in (B)

• $\mathrm{I_2}$ (0) → $\mathrm{I^-}$ ($-1$): reduced → $\mathrm{I_2}$ is the oxidizing agent

Conclusion: $\mathrm{H_2O_2}$ acts as reducing agent in BOTH (A) and (B)

Answer: Option (4)