Question

How does `MnO_4^-` ion behave in basic solution....?

Answer 1

In a (highly) basic medium, deep-red `"permanganate ion"` is reduced to green `"manganate ion"`.....

Explanation:

And so we can write the one electron reduction.......

i.e. `Mn(VII+) rarrMn(VI+)`

`MnO_4^(-) + e^(-) rarr MnO_4^(2-)`

And so we plumb for `B`.....i.e. `158.03*g*mol^-1`