(A) $$HOCl + H_2O_2 \to H_3O^+ + Cl^- + O_2$$
(B) $$I_2 + H_2O_2 + 2OH^- \to 2I^- + 2H_2O + O_2$$
Choose the correct option.

We need to determine the role of $$H_2O_2$$ in both reactions.

In equation (A): $$HOCl + H_2O_2 \to H_3O^+ + Cl^- + O_2$$

The oxidation state of oxygen in $$H_2O_2$$ is $$-1$$. In the product $$O_2$$, the oxidation state of oxygen is $$0$$. Since the oxidation state increases from $$-1$$ to $$0$$, oxygen is oxidised. A species that gets oxidised acts as a reducing agent. So $$H_2O_2$$ is a reducing agent in reaction (A).

In equation (B): $$I_2 + H_2O_2 + 2OH^- \to 2I^- + 2H_2O + O_2$$

Again, the oxidation state of oxygen in $$H_2O_2$$ is $$-1$$. In the product $$O_2$$, the oxidation state of oxygen is $$0$$. The oxidation state increases from $$-1$$ to $$0$$, so oxygen is oxidised. Therefore, $$H_2O_2$$ acts as a reducing agent in reaction (B) as well.

Hence, $$H_2O_2$$ acts as a reducing agent in both equations (A) and (B).

Hence, the correct answer is Option 4.