Which of the following equation depicts the oxidizing nature of $$H_2O_2$$?

When $$H_2O_2$$ acts as an oxidizing agent, it gets reduced (its oxygen goes from -1 oxidation state to -2). We need to identify the reaction where $$H_2O_2$$ oxidizes another species.

In option (1): $$2I^- + H_2O_2 + 2H^+ \to I_2 + 2H_2O$$. Here, iodide ($$I^-$$, oxidation state -1) is oxidized to $$I_2$$ (oxidation state 0), and $$H_2O_2$$ (oxygen in -1 state) is reduced to $$H_2O$$ (oxygen in -2 state). So $$H_2O_2$$ acts as an oxidizing agent.

In option (2): $$I_2 + H_2O_2 + 2OH^- \to 2I^- + 2H_2O + O_2$$. Here, $$H_2O_2$$ is oxidized to $$O_2$$ (oxygen goes from -1 to 0), so $$H_2O_2$$ acts as a reducing agent.

In option (3): $$Cl_2 + H_2O_2 \to 2HCl + O_2$$. Here, $$H_2O_2$$ is oxidized to $$O_2$$, so $$H_2O_2$$ acts as a reducing agent.

In option (4): $$KIO_4 + H_2O_2 \to KIO_3 + H_2O + O_2$$. Here too, $$H_2O_2$$ is oxidized to $$O_2$$, acting as a reducing agent.

Therefore, the equation depicting the oxidizing nature of $$H_2O_2$$ is option (1): $$2I^- + H_2O_2 + 2H^+ \to I_2 + 2H_2O$$.