Question

In which of the following reactions of H$_2$O$_2$ acts as an oxidizing agent (either in acidic, alkaline, or neutral medium)? 
(i) \( 2Fe^{2+} + H_2O_2 \rightarrow \) 
(ii) \( 2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow \) 
(iii) \( I_2 + H_2O_2 + 2OH^- \rightarrow \) 
(iv) \( Mn^{2+} + H_2O_2 \rightarrow \) 
 

• A. (ii), (iii)
• B. (i), (iv)
• C. (i), (iii)
• D. (ii), (iv)

Hint:

Hydrogen peroxide (\( H_2O_2 \)) is a versatile redox agent that can act as either an oxidizing or reducing agent, depending on the reaction medium.

Answer 1

The Correct Option is B

Step 1: Understanding the Role of H$_2$O$_2$ as an Oxidizing Agent 
Hydrogen peroxide (\( H_2O_2 \)) can act as both an oxidizing and a reducing agent, depending on the reaction conditions. 
In acidic and neutral conditions, it tends to act as an oxidizing agent. 
In alkaline medium, it can act as either an oxidizing or reducing agent. 
Step 2: Completing the Given Reactions 
\[ (i) \quad 2Fe^{2+} + H_2O_2 \rightarrow 2Fe^{3+} + 2OH^- \] Here, \( H_2O_2 \) acts as an oxidizing agent by converting \( Fe^{2+} \) to \( Fe^{3+} \). \[ (ii) \quad 2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O \] Here, \( H_2O_2 \) acts as a reducing agent because it is oxidized to \( O_2 \). \[ (iii) \quad I_2 + H_2O_2 + 2OH^- \rightarrow 2I^- + O_2 + 2H_2O \] Here, \( H_2O_2 \) acts as a reducing agent by reducing iodine (\( I_2 \)) to iodide (\( I^- \)). \[ (iv) \quad Mn^{2+} + H_2O_2 \rightarrow MnO_2 + 2H^+ \] Here, \( H_2O_2 \) oxidizes \( Mn^{2+} \) to \( MnO_2 \), acting as an oxidizing agent. 
Step 3: Identifying the Correct Answer 
\( H_2O_2 \) acts as an oxidizing agent in reactions (i) and (iv). 
 

Answer 2

Step 1: Understand the role of H₂O₂
Hydrogen peroxide (H₂O₂) is an amphoteric substance – it can act as both an oxidizing agent and a reducing agent depending on the reaction medium and the species involved.
To determine whether it acts as an oxidizing agent, we must see if it is causing other species to increase their oxidation number (i.e., lose electrons), while H₂O₂ itself gets reduced.

Step 2: Analyze each reaction
(i) \( 2Fe^{2+} + H_2O_2 \rightarrow 2Fe^{3+} + 2OH^- \)
Fe²⁺ is oxidized to Fe³⁺, so H₂O₂ is accepting electrons → acts as an oxidizing agent ✔️

(ii) \( 2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O \)
Here, MnO₄⁻ is reduced from Mn⁺⁷ to Mn²⁺.
H₂O₂ is oxidized to O₂, so it acts as a reducing agent ❌

(iii) \( I_2 + H_2O_2 + 2OH^- \rightarrow 2I^- + 2H_2O + O_2 \)
I₂ is reduced to I⁻, and H₂O₂ is oxidized to O₂.
So H₂O₂ is acting as a reducing agent ❌

(iv) \( Mn^{2+} + H_2O_2 \rightarrow Mn^{4+} + 2OH^- \)
Mn²⁺ is oxidized to Mn⁴⁺, hence H₂O₂ accepts electrons and acts as an oxidizing agent ✔️

Step 3: Final selection
H₂O₂ acts as an oxidizing agent in:
(i) \( 2Fe^{2+} + H_2O_2 \rightarrow \)
(iv) \( Mn^{2+} + H_2O_2 \rightarrow \)

Final Answer: (i), (iv)