1. Key idea: A reducing agent reduces another species and itself gets oxidized.

2. So we must identify in which given half-reactions $H_2O_2$ is undergoing oxidation.

3. Oxidation state of oxygen in $H_2O_2$: In peroxide, oxygen has oxidation number $-1$.

4. Check each reaction:

Reaction 1

$H_2O_2 + 2H^+ + 2e^- \to 2H_2O$

• Oxygen goes from $-1$ in $H_2O_2$ to $-2$ in $H_2O$.
• This is reduction (gain of electrons).
• Hence, here $H_2O_2$ acts as an oxidizing agent, not reducing agent.

Reaction 2

$H_2O_2 - 2e^- \to O_2 + 2H^+$

• Oxygen goes from $-1$ in $H_2O_2$ to $0$ in $O_2$.
• This is oxidation (loss of electrons).
• Hence, here $H_2O_2$ acts as a reducing agent.

Reaction 3

$H_2O_2 + 2e^- \to 2OH^-$

• Oxygen goes from $-1$ in $H_2O_2$ to $-2$ in $OH^-$.
• This is reduction.
• Hence, here $H_2O_2$ acts as an oxidizing agent.

Reaction 4

$H_2O_2 + 2OH^- - 2e^- \to O_2 + 2H_2O$

• Oxygen in $H_2O_2$ goes from $-1$ to $0$ in $O_2$.
• This is oxidation.
• Hence, here $H_2O_2$ acts as a reducing agent.

5. Therefore, $H_2O_2$ acts as a reducing agent in reactions 2 and 4.

6. So the correct option is: $\boxed{\text{D: } 2,4}$