Question

In acid medium MnO\(_4^-\) is reduced to Mn\(^{2+}\) by a reducing agent. Then the equivalent mass of KMnO\(_4\) is given by (M = molecular mass)

• A. M/2
• B. M
• C. M/5
• D. M/3

Hint:

In basic medium, MnO\(_4^-\) reduces to MnO\(_2\) (n=3).

Answer 1

The Correct Option is C

To determine the equivalent mass of KMnO₄ in an acidic medium, we need to consider the reaction and the concept of equivalent mass.

In acidic conditions, the permanganate ion (MnO₄^-) is reduced to manganese ion (Mn²⁺). The balanced redox reaction in acidic medium is:

MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

Explanation:

1. The change in oxidation state for manganese goes from +7 in MnO₄^- to +2 in Mn²⁺, resulting in a change of 5 units.
2. Equivalent mass is calculated using the formula: \(\text{Equivalent Mass} = \frac{\text{Molar Mass}}{\text{n-factor}}\), where the n-factor is the number of electrons exchanged per molecule.
3. Here, the n-factor for KMnO₄ is 5, as 5 electrons are exchanged per MnO₄^- ion.

Therefore, the equivalent mass of KMnO₄ is:

\(\frac{M}{5}\)

Conclusion: The equivalent mass for KMnO₄ in acid medium is \(\frac{M}{5}\), which aligns with option M/5.