Question

KMnO₄ acts as an oxidizing agent in an acidic medium.

Hint:

In acidic medium, $KMnO_4$ is reduced to $Mn^{2+}$ with an electron exchange ratio of 2:5. The reaction follows: $$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$

Answer 1

Step 1: Understanding the Oxidizing Action of KMnO₄ in Acidic Medium 
- Potassium permanganate (KMnO₄) is a strong oxidizing agent.
- In acidic medium, it undergoes reduction, converting Mn in ⁺⁷ oxidation state to Mn in ⁺² oxidation state. 
Step 2: Balanced Redox Reaction in Acidic Medium 
The half-reaction for the reduction of permanganate ion in acidic solution is: $$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$ This means that one mole of MnO₄^- accepts 5 electrons to form Mn²⁺. 
Step 3: Ratio of Electrons Gained to MnO₄^- Ions Reduced 
- The equivalent weight concept states that the oxidizing action in acidic medium follows a 2:5 ratio.
- This means that for every 2 moles of MnO₄^-, 5 moles of electrons are transferred. 
Step 4: Conclusion 
Thus, the oxidizing action of KMnO₄ in acidic solution follows the ratio $\frac{2}{5}$.