Question

The oxidation state of manganese in the product obtained in a reaction of potassium permanganate and hydrogen peroxide in basic medium is____.

Answer 1

Correct Answer: 4

To determine the oxidation state of manganese in the product obtained from the reaction of potassium permanganate (KMnO₄) with hydrogen peroxide (H₂O₂) in a basic medium, we need to analyze the redox reaction. In basic medium, the half-reactions can be represented as follows:

• Reduction half-reaction (Mn):
  \(MnO_4^- + e^- \rightarrow MnO_2(s)\)
  This represents a change from +7 to an oxidation state of +4 for manganese, as \(MnO_4^-\) is reduced to \(MnO_2\).
• Oxidation half-reaction (H₂O₂):
  \(H_2O_2 \rightarrow O_2 + 2e^- + 2H^+\)
  In basic medium, this simplifies to:
  \(2OH^- \rightarrow O_2 + 2e^- + 2H_2O\)

By combining and balancing these reactions, we ensure that the electrons lost in the oxidation half-reaction match those gained in the reduction half-reaction, achieving a balanced equation. Ultimately, the manganese from KMnO₄ is reduced to MnO₂, where manganese has an oxidation state of +4 in \(MnO_2\).

Therefore, the oxidation state of manganese in the product is +4. This value is confirmed to fall within the given range of 4.