In which of the following reactions $$H_2O_2$$ acts as a reducing agent?
(a) $$H_2O_2 + 2H^+ + 2e^- \to 2H_2O$$
(b) $$H_2O_2 - 2e^- \to O_2 + 2H^+$$
(c) $$H_2O_2 + 2e^- \to 2OH^-$$
(d) $$H_2O_2 + 2OH^- - 2e^- \to O_2 + 2H_2O$$

We begin by recalling that a species behaves as a reducing agent when it itself gets oxidised, that is, when it loses electrons and the oxidation number of (at least one of) its atoms increases.

In hydrogen peroxide, $$H_2O_2$$, the oxidation number of each oxygen atom is $$-1$$. If $$H_2O_2$$ is oxidised, the oxidation number of oxygen must rise from $$-1$$ to $$0$$ (in $$O_2$$) or to $$-2$$ (in $$OH^-$$ or $$H_2O$$ it actually goes down, so that would represent reduction). Oxidation will therefore be accompanied by the loss of electrons, and in a half-reaction those lost electrons will appear on the right-hand side.

Now we analyse each given half-reaction one by one, carefully watching the position of electrons and the change in oxidation number.

(a) $$H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O$$

Here the electrons are written on the left. So $$H_2O_2$$ is gaining electrons. This means that $$H_2O_2$$ is being reduced, acting as an oxidising agent, not as a reducing agent.

(b) $$H_2O_2 \;-\; 2e^- \rightarrow O_2 + 2H^+$$

We can rewrite this more explicitly as $$H_2O_2 \rightarrow O_2 + 2H^+ + 2e^-$$. Now the electrons are on the right; $$H_2O_2$$ is losing electrons. Its oxygen atoms go from $$-1$$ in $$H_2O_2$$ to $$0$$ in $$O_2$$, an increase in oxidation number. Thus $$H_2O_2$$ is oxidised and therefore acts as a reducing agent.

(c) $$H_2O_2 + 2e^- \rightarrow 2OH^-$$

Electrons again appear on the left; $$H_2O_2$$ is gaining electrons, so it is being reduced. Its oxygen atoms change from $$-1$$ in $$H_2O_2$$ to $$-2$$ in $$OH^-$$ (a decrease in oxidation number). Hence $$H_2O_2$$ functions as an oxidising agent, not as a reducing agent.

(d) $$H_2O_2 + 2OH^- \;-\; 2e^- \rightarrow O_2 + 2H_2O$$

Again, writing the electrons on the right yields $$H_2O_2 + 2OH^- \rightarrow O_2 + 2H_2O + 2e^-$$. The electrons are produced, so $$H_2O_2$$ is losing electrons. The oxidation number of oxygen in $$H_2O_2$$ increases from $$-1$$ to $$0$$ in $$O_2$$, confirming that $$H_2O_2$$ is oxidised and therefore acts as a reducing agent.

Summarising our findings:

  • Reaction (a): oxidising agent.
  • Reaction (b): reducing agent.
  • Reaction (c): oxidising agent.
  • Reaction (d): reducing agent.

So the reactions in which $$H_2O_2$$ acts as a reducing agent are (b) and (d).

Looking at the options provided, option D corresponds to (b) and (d).

Hence, the correct answer is Option D.