How does permanganate ion behave in redox reactions?

1 Answer
Apr 25, 2017

Answer:

Well, permanganate anion is generally reduced to (colourless) #Mn^(2+)# in acidic solution..........

Explanation:

#MnO_4^(-) +8H^+ +5e^(-) rarr Mn^(2+) + 4H_2O(l)#

Note that #Mn^(2+)# is almost colourless in aqueous solution; it is a #d^5# system, for which electronic transition is spin forbidden. Very concentrated solutions are pale rose.

In basic solution, generally, permanganate is reduced to #MnO_2(s)#, a 3 electron reduction:

#MnO_4^(-) + 4H^+ +3e^(-) rarr MnO_2(s) + 2H_2O(l)# (but BASIC conditions were specified, so we add #4xxHO^-# to both sides.....

#MnO_4^(-) + 2H_2O +3e^(-) rarr MnO_2(s) + 4HO^(-)#

Of course you need the corresponding oxidation equation. But there is formal #5e^-# transfer in acid, and #3e^-# transfer in base.