Question

The equivalent weight of Potassium permanganate in acidic medium is:

• A. 31.6
• B. 51.6
• C. 41.6
• D. 21.6

Hint:

To calculate the equivalent weight of a substance, divide its molar mass by the number of electrons involved in the redox reaction. For (\( \text{KMnO}_4 \)), it is 158 g/mol divided by 5 electrons, resulting in 31.6 g/equiv in acidic medium.

Answer 1

The Correct Option is A

In acidic medium, the equivalent weight of Potassium permanganate (\( \text{KMnO}_4 \)) is calculated based on its change in oxidation state. Potassium permanganate undergoes a reduction from +7 to +2 oxidation state, and since 5 electrons are involved, the equivalent weight of \( \text{KMnO}_4 \) is 31.6 g/equiv.

Answer 2

The equivalent weight of an oxidising or reducing agent equals its molar mass divided by the number of electrons it gains or loses per molecule (the n-factor), and that n-factor changes with the reaction medium. Let's see which medium each option corresponds to.

1. 31.6: In an acidic medium, permanganate ion is reduced all the way from manganese in the +7 state down to \( \text{Mn}^{2+} \), a gain of 5 electrons per \( \text{MnO}_4^- \) ion. Dividing the molar mass of \( \text{KMnO}_4 \), which is 158 g/mol, by this n-factor of 5 gives \( 158/5 = 31.6 \) g/equiv, matching the acidic-medium condition asked about in the question.
2. 51.6: This value is close to what would be obtained in a neutral or faintly alkaline medium, where manganese is reduced only to the +4 state as \( \text{MnO}_2 \), a 3-electron change, giving \( 158/3 \approx 52.7 \) g/equiv. It reflects a different reaction medium than the acidic one specified here.
3. 41.6: This number does not correspond to any whole-number electron change for \( \text{KMnO}_4 \); dividing 158 by 41.6 gives a non-integer n-factor of about 3.8, which has no real chemical basis in any of permanganate's standard reduction pathways.
4. 21.6: Like the previous option, this does not match a genuine n-factor either; 158 divided by 21.6 gives roughly 7.3 electrons transferred, which is more electrons than manganese's entire oxidation-state range from +7 down to 0 could account for.

Since the question specifically asks for the acidic medium, where the 5-electron reduction to \( \text{Mn}^{2+} \) applies, only the first value fits.

Therefore, the correct answer is 31.6.