# The importance of nature of medium while reduction of permanganate

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?

• Chem+Math MathJax formatting: Basics / Expressions/formulas/equations / Upright vs Italics / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles. Apr 1 at 4:08
• Basic medium means Mn(OH)2(s) with reduction abilities. // For MnO2, there is the equivalent equation $\ce{MnO4- + 2 H2O + 3 e- -> MnO2(s) + 4 OH-}$ Apr 1 at 6:48