One mole of `AO_(2)^(-)` is oxidised to `A^(n+)` in acidic solutions by 0.4 mole of permanganate. Calculate the value of n in `A^(n+)`. 

Explanation

In a redox reaction, the product of number of moles and change in oxidation number is same for oxidant and reductant.
The change in oxidation number of Mn is 5.
The change in oxidation number of A say x.
`"(0.4 mol) (5)= (1 mol) (x), x = 2 "`
Oxidation number of A in `AO_(2)^(-)` is `+3`.
It is oxidised to `A^(n+1)` where the increases in the oxidation number is 2.
The value of n is, `n = +3 + 2 = +5`.