Which one of the following reactions indicates the reducing ability of hydrogen peroxide in basic medium?

We need to identify the reaction where H$$_2$$O$$_2$$ acts as a reducing agent in basic medium. When H$$_2$$O$$_2$$ acts as a reducing agent, it gets oxidized (oxygen goes from -1 oxidation state in H$$_2$$O$$_2$$ to 0 in O$$_2$$).

Analysis of each option:

Option A: HOCl + H$$_2$$O$$_2$$ $$\rightarrow$$ H$$_3$$O$$^+$$ + Cl$$^-$$ + O$$_2$$

H$$_2$$O$$_2$$ is oxidized to O$$_2$$ (reducing agent), but this reaction produces H$$_3$$O$$^+$$, indicating an acidic medium, not basic.

Option B: PbS + 4H$$_2$$O$$_2$$ $$\rightarrow$$ PbSO$$_4$$ + 4H$$_2$$O

Here, oxygen in H$$_2$$O$$_2$$ (-1) goes to O in H$$_2$$O (-2). So H$$_2$$O$$_2$$ is reduced and acts as an oxidizing agent, not a reducing agent.

Option C: 2MnO$$_4^-$$ + 3H$$_2$$O$$_2$$ $$\rightarrow$$ 2MnO$$_2$$ + 3O$$_2$$ + 2H$$_2$$O + 2OH$$^-$$

H$$_2$$O$$_2$$ is oxidized to O$$_2$$ (oxygen: -1 $$\rightarrow$$ 0), so it acts as a reducing agent. The presence of OH$$^-$$ in the products confirms basic medium. Also, Mn goes from +7 in MnO$$_4^-$$ to +4 in MnO$$_2$$, confirming MnO$$_4^-$$ is reduced.

Option D: Mn$$^{2+}$$ + H$$_2$$O$$_2$$ $$\rightarrow$$ Mn$$^{4+}$$ + 2OH$$^-$$

Here, Mn is oxidized from +2 to +4, and oxygen in H$$_2$$O$$_2$$ (-1) goes to OH$$^-$$ (-2). So H$$_2$$O$$_2$$ acts as an oxidizing agent.

Hence, the correct answer is Option C, where H$$_2$$O$$_2$$ acts as a reducing agent in basic medium.