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 Question:
We must determine the total electrical charge (in Faradays, F) required to completely reduce 2 moles of potassium permanganate into manganese sulfate.

Step 2: Key Formula or Approach:
The charge required in Faradays is directly equal to the total moles of electrons transferred in the balanced half-reaction.
$$\text{Charge (F)} = \text{Moles of reactant} \times \text{Change in Oxidation State per molecule}$$

Step 3: Detailed Explanation:
1. Determine the oxidation state of Manganese (Mn) in the reactant, $KMnO_4$:
Potassium (K) is +1, Oxygen (O) is -2. Let Mn be $x$.
$$1 + x + 4(-2) = 0 \implies x - 7 = 0 \implies x = +7$$
2. Determine the oxidation state of Mn in the product, $MnSO_4$:
Sulfate ($SO_4$) is a polyatomic ion with a -2 charge. Let Mn be $y$.
$$y + (-2) = 0 \implies y = +2$$
3. Calculate the number of electrons required for one mole:
To reduce $Mn^{7+}$ to $Mn^{2+}$, the atom must accept exactly 5 electrons.
Therefore, 1 mole of $KMnO_4$ requires 5 moles of electrons, which equals 5 Faradays (5 F) of charge.
4. Calculate the charge for the required amount (2 moles):
Total Charge = $2 \text{ moles} \times 5 \text{ F/mole} = 10 \text{ F}$.

Step 4: Final Answer:
The required charge is 10 F, matching option (d).