Equivalent weight of `MnO_(4)^(-)` in acidic, weakly basic and neutral medium 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

Acdic medium
` MnO_(4)^(- ) +8H^(+) +5E^(- ) to Mn^(2+) +4H_ 2O`
Weakly basic medium :
`MnO_(4)^(- ) + 2H_2 O +3e^(- ) to MnO_2 +4 OH^(-) `
Neutral medium :
`MnO_(4)^(- ) +2H_2 O+ 3e^(-) to MnO_2 +4OH^(-)`
if M is the molecular weight of `KMnO_4 ` then its equivalent weight in acidic . acidic basic and neutral media respectively are :
`M/5 :M/3 :M/3 or 3 :5:5`