Question

Hydrogen peroxide acts both as an oxidising and as a reducing agent depending upon the nature of the reacting species. In which of the following cases $H_2O_2$ acts as a reducing agent in acid medium ?

• A. $MnO^{-}_{4}$
• B. $Cr_{2}O^{2-}_{7}$
• C. $SO^{2-}_{3}$
• D. $KI$

Answer 1

The Correct Option is A

To determine in which case hydrogen peroxide (H_2O_2) acts as a reducing agent in an acidic medium, we need to analyze the given options. Let's look at each option and the corresponding chemical reactions:

1. Option A: MnO^{-}_{4}
   In an acidic medium, the reduction of permanganate ion (MnO_4^-) by H_2O_2 can be represented as follows: \[ MnO_4^- + 5 H_2O_2 + 6 H^+ \rightarrow 2 Mn^{2+} + 8 H_2O + 5 O_2 \]
   Here, MnO_4^-\ is reduced to Mn^{2+}, while H_2O_2 is oxidized to O_2. Thus, H_2O_2 acts as a reducing agent.
2. Option B: Cr_{2}O^{2-}_{7}
   In an acidic medium, dichromate ion (Cr_2O_7^{2-}) is typically reduced by other agents such as iron(II) ions instead of H_2O_2. H_2O_2 usually acts as an oxidizing agent in the presence of dichromate.
3. Option C: SO^{2-}_{3}
   In the case of sulfite ions (SO_3^{2-}), H_2O_2 acts as an oxidizing agent, converting sulfites to sulfates: \[ SO_3^{2-} + H_2O_2 \rightarrow SO_4^{2-} + H_2O \]
   Here, H_2O_2 acts as an oxidizing agent, not a reducing agent.
4. Option D: KI
   When reacting with potassium iodide (KI), H_2O_2 oxidizes iodide ions (I^-\) to iodine (I_2): \[ 2 KI + H_2O_2 + H_2SO_4 \rightarrow I_2 + K_2SO_4 + 2 H_2O \]
   Again, H_2O_2 acts as an oxidizing agent in this reaction.

From this analysis, it is clear that MnO^{-}_{4} in an acidic medium is the case where H_2O_2 acts as a reducing agent. Therefore, the correct answer is:

Correct Answer: MnO^{-}_{4}