Why KMnO4 in alkaline solution accept 3e per Mn and act as mild oxidizing agent while in acidic solution can gain 5 electron per Mn act as strong oxidizing agen???

Consider the Pourbaix diagram of #"Mn"#:

If it's easy to reduce #"MnO"_4^-#, it's a good oxidizing agent. Two of these are possible:

#stackrel(color(blue)(+7))"Mn""O"_4^(-)(aq) + 2"H"_2"O"(l) + 3e^(-) rightleftharpoons stackrel(color(blue)(+4))"Mn""O"_2(s) + 4"OH"^(-)(aq)#

#stackrel(color(blue)(+7))"Mn""O"_4^(-)(aq) + 8"H"^(+)(aq) + 5e^(-) rightleftharpoons stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4"H"_2"O"(l)#

In acidic solution, specifically below #"pH"# #4#, #stackrel(color(blue)(+7))("Mn")"O"_4^(-)# could get reduced all the way to #"Mn"^(2+)#, i.e. from #+7# to #+2#, a change in oxidation state of #-5# in the reduction half-reaction.

The diagonal #"Mn"^(2+)//"MnO"_2# equilibrium line shows a dependence on both voltage and #"pH"#. As the equilibrium between #stackrel(color(blue)(+4))"Mn""O"_2(s)# and #"Mn"^(2+)(aq)# is #bb"pH"#-dependent...

#"MnO"_2(s) + 4"H"^(+)(aq) + 2e^(-) rightleftharpoons "Mn"^(2+)(aq) + 2"H"_2"O"(l)#

The more acidic the #pH# (the more #H^+# there is), the more the equilibrium shifts to #Mn^(2+)# by Le Chatelier's principle, favoring a #5e^(-)# reduction.

In basic solution (above #"pH"# #7.5#), #"Mn"^(2+)# does not exist; as the #"pH"# increases, the #["OH"^(-)]# increases...

#"MnO"_2(s) + 2"H"_2"O"(l)(aq) + 2e^(-) rightleftharpoons "Mn"^(2+)(aq) + 4"OH"^(-)(aq)#

And that shifts the equilibrium left, towards #stackrel(color(blue)(+4))(Mn)O_2(s)#, favoring a #3e^(-)# reduction.

And so, in basic solution, we only have a change in oxidation state of #-3# in the reduction half-reaction: #+7# to #+4#.