Question:medium

The reaction of aqueous $KMnO_4$ with $ H_2O_2 $ in acidic conditions gives

Updated On: May 22, 2026
  • $Mn^{4+} $ and $ O_2 $
  • $Mn^{2+}$ and $O_2 $
  • $Mn^{2+}$ and $ O_3 $
  • $Mn^{4+}$ and $ MnO_2 $
Show Solution

The Correct Option is B

Solution and Explanation

The reaction between potassium permanganate \(KMnO_4\) and hydrogen peroxide \(H_2O_2\) in acidic conditions is a classic redox reaction. Let's analyze this step-by-step to determine the products:

  1. In acidic conditions, \(KMnO_4\) acts as an oxidizing agent and gets reduced. The manganese in \(KMnO_4\) has an oxidation state of +7.
  2. During the reaction, the Mn in \(KMnO_4\) is reduced to Mn2+ (oxidation state +2) under acidic conditions. The half-reaction for this reduction is: \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}
  3. Hydrogen peroxide (\(H_2O_2\)) acts as a reducing agent and gets oxidized to oxygen gas (\(O_2\)). The half-reaction for this oxidation is: \text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2\text{e}^-
  4. Balancing the overall redox reaction involves ensuring the electrons lost in oxidation match those gained in reduction. To do this, the half-reactions must be multiplied accordingly:
    • Oxidation: \(2\text{H}_2\text{O}_2 \rightarrow 2\text{O}_2 + 4\text{H}^+ + 4\text{e}^-\)
    • Reduction: \(2(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O})\)
  5. Combining and balancing the electrons, the reactions give: 2\text{MnO}_4^- + 5\text{H}_2\text{O}_2 + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{O}_2 + 8\text{H}_2\text{O}
  6. The products of this reaction are therefore \(Mn^{2+}\) and \(O_2\), confirming that the correct answer is “\(Mn^{2+}\) and \(O_2\).”
  7. This explanation eliminates other options because:
    • \(Mn^{4+}\) is not a common reduction product in acidic conditions.
    • \(O_3\) (ozone) and \(MnO_2\) are not typical products in this reaction under acidic conditions.

Thus, the reaction correctly results in \(Mn^{2+}\) and \(O_2\) as products under acidic conditions.

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