Eq.wt of MnO4^- in acidic, basic and neutral medium are in which ratio?A) 3:5:15B) 5:3:1C) 5:1:3D) 3:15:5(Explain in detail)

							
KMnO4 has a Molar mass of 158.04 g/mol 

There are two possibilities: 

1. KMnO4 as an oxidizer in acidic media:
MnO4- + 8 H+ + 5e- --> Mn2+ + 4H2O (gained 5 electrons from reductant) 

I.e. 5 Equivalents per mole, so equivalent mass of KMnO4 = M/5 = 158.04/5 = 31.61 gram/equivalent 


2. KMnO4 as an oxidizer in neutral or basic media:
MnO4- + 2H2O + 3e- --> MnO2(s) + 4OH- (gained 3 electrons) 

In this case: 3 equivalents per mole, so equivalent mass of KMnO4 = M/3 = 158.04/3 = 52.68 gram/equivalent
						

							
Dear student 
In Acidic medium, This Mn+7 goes to Mn+2 state and hence there is a net gain of 5 electrons. 
Now, equivalent weight = [molar mass] /[number of electrons gained or lost]
so, eq wt = 158/5 = 31.6 g.
Now, for alkaline medium, there are two possibilities
1) faintly alkaline/neutral:
 Mn+7 changes into Mn+4 therefore, gain of 3 electrons
 Hence, eq wt = 158/3 = 52.67g
2) highly alkaline:
 Mn+7 changes into Mn+6 therefore, gain of 1 electron
 Hence, eq wt = 158/1 = 158g
 
In many cases though if it's written alkaline, it mostly means the neutral one; for the highly alkaline thing, it'll especially mention it.