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I was just going through some reactions related to $\ce{KMnO4}$ when I came across this.

In acidic medium,
$\ce{MnO4- +5e- + 8H+ -> Mn^2+ +4H2O}$
Here, the $\text{n-factor}$ is 5.

In neutral (slightly acidic) medium,
$\ce{MnO4- +3e- +4H+ -> MnO2 +2H2O}$
Here, the $\text{n-factor}$ is 3.

In basic medium,
$\ce{MnO4- +e- -> MnO4^2-}$
Here, the $\text{n-factor}$ is 1.

My Question:
Though I can see that the presence of protons is necessary for the reaction to proceed, what exactly, is the importance of the medium in such a reaction. Is there no other way in which we can reduce the permanganate ion to.... say $\ce{MnO2}$ or $\ce{Mn^2+}$ in basic medium?

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