Question

The number of electrons that are involved in the reduction of permanganate to manganese(II) salt, manganate and manganese dioxide respectively are

• A. 5, 1, 3
• B. 5, 3, 1
• C. 2, 7, 1
• D. 5, 2, 3
• E. 2, 3, 1

Hint:

Memorize the mnemonic "Ban 153": - Basic medium: Change is 1. - Acidic medium: Change is 5. - Neutral medium: Change is 3. This refers to the electron count for $MnO_4^-$ reduction!

Answer 1

The Correct Option is A

Concept: The number of electrons involved in a redox process corresponds to the change in the oxidation state of the atom being reduced.
• Initial State: In Permanganate ($MnO_4^-$), the oxidation state of Mn is +7.
• Reduction: $e^- \text{ involved} = (\text{Initial Oxidation State}) - (\text{Final Oxidation State})$.

Step 1: Calculate the change for each Manganese species. 1. To Manganese(II) salt ($Mn^{2+}$): Change = $(+7) - (+2) = 5$ electrons. (Occurs in acidic medium). 2. To Manganate ($MnO_4^{2-}$): Change = $(+7) - (+6) = 1$ electron. (Occurs in strongly alkaline medium). 3. To Manganese dioxide ($MnO_2$): Change = $(+7) - (+4) = 3$ electrons. (Occurs in neutral or faintly alkaline medium).

Step 2: Identify the sequence. The calculated electron counts are 5, 1, and 3 respectively. This matches the sequence in Option (A).