The equivalent weights of oxidising and reducing agents can be calculated by the number of electrons gained or lost. The equivalent weight of an oxidising agent is the number of parts by weight of the substance which gains one electron. Thus, it is equal to the molecular weight of the substance divided by the number of electrons gained in the balanced chemical equation. Similarly, equivalent weight of a reducing agent is equal to the molecular weight divided by the number of electrons lost as represented in the balanced chemical equation
Equivalent weight of `MnO_(4)^(-)` in acidic, basic and neutral media are in the ratio of

Options

A. `3:5:15`
B. `5:3:1`
C. `5:1:3`
D. `3:5:5`

Answer

D

Explanation

Acidic medium :
`MnO_4^(-) + 8H^(-) +5e^(-)rarr Mn^(2+) + 4H_(2)O`
Basic medium :
`MnO_4^(-) + 2H_2O+3e^(-) rarr MnO_2+4OH^(-)`
Neutral medium :
`MnO_(4)^(-) + 2H_(2)O + 3e^(-) raarr MnO_(2) + 4OH^(-) `
If M is the molecular weight of `KMnO_4` , then its equivalent weights in acidic , basic and neutral media respectively are : `M/5:M/3:M/3`