What is the rate of equivalent weights of `MNO_(4)^(-)` in acidic ,basic & neutral medium?

Explanation

The reaction that `MnO_(4)^(-)` ion undergoes in acidic ,basic and neutral medium are as follows.
Acidic Medium: `MnO_(4)^(-)(aq)+8H^(+)(aq)+5etoMn^(2-)(aq)+4H_(2)O(l)`
Equivalent weight `E_(1) =(M)/(5)`,where M= molecular mass of `KMnO_(4)`.
Basic Medium `MnO_(4)^(-)(aq)`
Equivalent weight `E_(2)=(M)/(1)`
Netural Medium
`MnO_(4)^(-)(aq)+2H_(2)O(l)+3etoMnO_(2)(s)+4OH^(-)(aq)`
Equivalent weight ,`E_(3)=(M)/(3)`
Therefore`E_(1):E_(2):E_(3)=(1)/(5):1(1)/(3)=:3:15:5`