In which of the following reactions, hydrogen peroxide acts as an oxidizing agent?

We start by recalling the basic idea of redox behaviour for hydrogen peroxide. In $$H_2O_2$$ the oxidation number of each oxygen atom is $$-1$$. Whenever $$H_2O_2$$ itself is converted to $$O_2$$, the oxidation number of its oxygen rises from $$-1$$ to $$0$$, so $$H_2O_2$$ is getting oxidised; therefore it must be acting as a reducing agent in that case. Conversely, whenever $$H_2O_2$$ changes to $$H_2O$$ or $$OH^-$$, the oxidation number of its oxygen falls from $$-1$$ to $$-2$$; hence $$H_2O_2$$ is being reduced and therefore behaves as an oxidising agent.

Now we examine each option one by one, writing the oxidation numbers step by step.

Option A is

$$HOCl + H_2O_2 \rightarrow H_3O^+ + Cl^- + O_2$$

• In $$HOCl$$ the chlorine is in the $$+1$$ state.
• It becomes $$Cl^-$$ where chlorine is $$-1$$, so chlorine is reduced.
• Simultaneously $$H_2O_2$$ turns into $$O_2$$, changing oxygen from $$-1$$ to $$0$$, i.e. oxygen is oxidised.
Because $$H_2O_2$$ is oxidised, it must be the reducing agent here. So $$H_2O_2$$ is not an oxidising agent in reaction A.

Option B is

$$I_2 + H_2O_2 + 2OH^- \rightarrow 2I^- + 2H_2O + O_2$$

• Iodine goes from $$0$$ in $$I_2$$ to $$-1$$ in $$I^-$$, so iodine is reduced.
• Again $$H_2O_2$$ becomes $$O_2$$ (oxygen $$-1 \rightarrow 0$$), meaning $$H_2O_2$$ is oxidised and acts as the reducing agent.
Therefore in reaction B, $$H_2O_2$$ is not an oxidising agent.

Option C is

$$PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O$$

Let us assign oxidation numbers carefully:

• In $$PbS$$, sulphur is $$-2$$.
• In $$PbSO_4$$, sulphur is $$+6$$ (because the overall charge is zero and $$Pb$$ is $$+2$$).

So sulphur changes from $$-2$$ to $$+6$$, a rise of $$8$$ units; sulphur is oxidised.

Next we track $$H_2O_2$$:

• Oxygen in $$H_2O_2$$ is $$-1$$.
• Oxygen in the product $$H_2O$$ is $$-2$$.

Thus the oxygen of peroxide is reduced. Since $$H_2O_2$$ itself is reduced while the other species (sulphur in $$PbS$$) is oxidised, $$H_2O_2$$ must be supplying the oxygen and causing oxidation of $$PbS$$. That means $$H_2O_2$$ acts as the oxidising agent in reaction C.

Option D is

$$2MnO_4^- + 3H_2O_2 \rightarrow 2MnO_2 + 3O_2 + 2H_2O + 2OH^-$$

• Manganese goes from $$+7$$ in $$MnO_4^-$$ to $$+4$$ in $$MnO_2$$, so manganese is reduced.
• Yet again $$H_2O_2$$ produces $$O_2$$, oxygen moving from $$-1$$ to $$0$$, so $$H_2O_2$$ is oxidised and therefore a reducing agent here.

Summarising all four cases:

• A - $$H_2O_2$$ is reducing agent
• B - $$H_2O_2$$ is reducing agent
• C - $$H_2O_2$$ is oxidising agent
• D - $$H_2O_2$$ is reducing agent

Hence only reaction C (which corresponds to Option 3) features hydrogen peroxide acting as an oxidising agent.

Hence, the correct answer is Option C.