Question

Which one of the following is reduced by H$_2$O$_2$ in alkaline medium?

• A. Fe$^{2+}$
• B. HOCl
• C. KMnO$_4$
• D. PbS
• E. Mn$^{2+}$

Hint:

In alkaline medium, always remember: \(H_2O_2\) behaves as a reducing agent and reduces strong oxidizers like \(KMnO_4\).

Answer 1

The Correct Option is C

Concept: Hydrogen peroxide (\(H_2O_2\)) is an amphoteric redox reagent, meaning it can act both as an oxidizing agent and a reducing agent depending on the medium:
• In acidic medium → acts mainly as an oxidizing agent
• In alkaline medium → acts mainly as a reducing agent

Step 1: Understand what the question asks.
We need to find which substance is reduced by \(H_2O_2\). Since \(H_2O_2\) is acting as a reducing agent (in alkaline medium), it will: \[ \text{donate electrons} \Rightarrow \text{itself gets oxidized} \] and hence it will reduce the other species.

Step 2: Check oxidation states of given species.
• Fe$^{2+}$ → already in lower oxidation state, tends to get oxidized to Fe$^{3+}$ → not reduced
• HOCl → Cl is in +1 state → can be reduced but weak oxidizing behavior here
• KMnO$_4$ → Mn is in +7 oxidation state (very high) → strong oxidizing agent → easily gets reduced
• PbS → sulphide form (S$^{2-}$), tends to be oxidized, not reduced
• Mn$^{2+}$ → already in low oxidation state → cannot be reduced further easily

Step 3: Focus on KMnO$_4$.
In alkaline medium, \(KMnO_4\) is reduced typically to \(MnO_2\) or \(MnO_4^{2-}\). Example reaction: \[ 2MnO_4^- + H_2O_2 + 2OH^- \rightarrow 2MnO_4^{2-} + 2H_2O + O_2 \] Here:
• Mn goes from +7 to +6 → reduction
• \(H_2O_2\) gets oxidized → acts as reducing agent

Step 4: Conclusion. Only \(KMnO_4\) undergoes reduction by \(H_2O_2\) in alkaline medium.

Step 5: Final answer. \[ \boxed{KMnO_4} \]