Question

What is the charge required to convert 2 mol $KMnO_4$ to $MnSO_4$?

• A. 2 F
• B. 4 F
• C. 5 F
• D. 10 F

Hint:

Remember this classic stoichiometry fact for titration and electrolysis problems: Permanganate ($MnO_4^-$) always undergoes a 5-electron change in strongly acidic medium (like forming $MnSO_4$).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The charge required ($Q$) for a redox reaction is related to the change in oxidation state ($n$) by Faraday's Law: $Q = nF$ per mole.

Step 2: Formula Application:
Determine the oxidation state of Mn in both compounds. $KMnO_4$: Mn is $+7$. $MnSO_4$: Mn is $+2$.

Step 3: Explanation:
The reduction half-reaction is: $MnO_4^- + 5e^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O$. Since 5 moles of electrons ($5F$) are required to convert 1 mole of $KMnO_4$ to $Mn^{2+}$, for 2 moles, the charge required is: $2 \times 5F = 10F$.

Step 4: Final Answer:
The total charge required is 10 F.