This question involves understanding redox reactions where Fe³⁺ is reduced to Fe²⁺. Let's analyze the concept step by step:

Step 1: Identify the Reduction Process
Reduction means gain of electrons. The half-reaction is:
${\mathrm{Fe}}^{3+}+e^{-}\to {\mathrm{Fe}}^{2+}$ So, we need a reducing agent that can donate electrons to Fe³⁺.

Step 2: Analyze the Options
Both Na₂O₂ and H₂O₂ can act as oxidizing or reducing agents depending on the medium. In acidic medium, H₂O₂ acts as a reducing agent. The relevant reaction is:
$H_2{O}_2\to O_2+2H^{+}+2e^{-}$ Here, H₂O₂ is oxidized to O₂, donating electrons, which can reduce Fe³⁺ to Fe²⁺.

In basic medium, H₂O₂ typically acts as an oxidizing agent. Na₂O₂ in water gives NaOH and H₂O₂, creating a basic medium, which is not suitable for reduction of Fe³⁺.

Step 3: Determine the Correct Option
Thus, H₂O₂ in acidic medium (H₂SO₄) acts as a reducing agent and can reduce Fe³⁺ to Fe²⁺. The other options either provide basic conditions or use Na₂O₂ which may not be effective.

Final Answer: H₂O₂ in presence of H₂SO₄

Related Topics

Redox Reactions: Reactions involving transfer of electrons between species. oxidizing agents accept electrons, reducing agents donate electrons.

Hydrogen Peroxide Properties: H₂O₂ can act as both oxidizing and reducing agent. In acidic medium, it is a good reducing agent for strong oxidizing agents.

Formulae

Reduction half-reaction: ${\mathrm{Fe}}^{3+}+e^{-}\to {\mathrm{Fe}}^{2+}$

Oxidation half-reaction for H₂O₂ in acid: $H_2{O}_2\to O_2+2H^{+}+2e^{-}$