How does $M n O_{4}^{-}$ $MnO_4^-$ ion behave in basic solution....?

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Answer 1 · 2017-09-16T17:22:49

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

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

i.e. $M n (V I I +) \to M n (V I +)$ $Mn(VII+) rarrMn(VI+)$

$M n O_{4}^{-} + e^{-} \to M n O_{4}^{2 -}$ $MnO_4^(-) + e^(-) rarr MnO_4^(2-)$

And so we plumb for $B$ $B$ .....i.e. $158.03 \cdot g \cdot m o l^{- 1}$ $158.03*g*mol^-1$

How does MnO−4MnO_4^- ion behave in basic solution....?