Reduction of Ferric Ion to Ferrous Ion using Hydrogen Peroxide

To determine which reagents can reduce $\text{Fe}^{3+}$ to $\text{Fe}^{2+}$, we analyze the redox behavior of hydrogen peroxide ($\text{H}_2\text{O}_2$) and sodium peroxide ($\text{Na}_2\text{O}_2$) in different media (alkaline vs. acidic).

1. Reduction of $\text{Fe}^{3+}$ in Alkaline Medium

In an alkaline (basic) medium, $\text{H}_2\text{O}_2$ acts as a strong reducing agent. The oxidation half-reaction for $\text{H}_2\text{O}_2$ in the presence of base ($\text{OH}^-$) is: $\text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 2e^-$

The reduction half-reaction for ferric ion ($\text{Fe}^{3+}$) is: $\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$

Combining these two half-reactions gives the overall balanced redox reaction: $2\text{Fe}^{3+} + \text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow 2\text{Fe}^{2+} + \text{O}_2 + 2\text{H}_2\text{O}$

In terms of hydroxides, ferric hydroxide ($\text{Fe(OH)}_3$) is reduced to ferrous hydroxide ($\text{Fe(OH)}_2$): $2\text{Fe(OH)}_3 + \text{H}_2\text{O}_2 \rightarrow 2\text{Fe(OH)}_2 + 2\text{H}_2\text{O} + \text{O}_2 \uparrow$

• Option (A): $\text{H}_2\text{O}_2$ in the presence of $\text{NaOH}$ provides an alkaline medium, which successfully reduces $\text{Fe}^{3+}$ to $\text{Fe}^{2+}$. (Correct)

• Option (B): Sodium peroxide ($\text{Na}_2\text{O}_2$) reacts with water to produce $\text{NaOH}$ and $\text{H}_2\text{O}_2$: $\text{Na}_2\text{O}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\text{O}_2$ Since this reaction generates both $\text{H}_2\text{O}_2$ and an alkaline medium ($\text{NaOH}$), it reduces $\text{Fe}^{3+}$ to $\text{Fe}^{2+}$. (Correct)

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2. Behavior in Acidic Medium

In an acidic medium (in the presence of $\text{H}_2\text{SO}_4$), $\text{H}_2\text{O}_2$ acts as an oxidizing agent, not a reducing agent. Instead of reducing $\text{Fe}^{3+}$, it oxidizes $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ according to the reaction: $2\text{Fe}^{2+} + \text{H}_2\text{O}_2 + 2\text{H}^+ \rightarrow 2\text{Fe}^{3+} + 2\text{H}_2\text{O}$

• Option (C): $\text{H}_2\text{O}_2$ in the presence of $\text{H}_2\text{SO}_4$ acts as an oxidizing agent, so reduction of $\text{Fe}^{3+}$ does not occur. (Incorrect)

• Option (D): $\text{Na}_2\text{O}_2$ in the presence of $\text{H}_2\text{SO}_4$ reacts to form $\text{H}_2\text{O}_2$ and $\text{Na}_2\text{SO}_4$ in an acidic environment, which also acts as an oxidizing agent. (Incorrect)

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Conclusion

The correct options are (A) and (B).