If `2.68 xx 10^(-3) `mol of a solution containing an ion `A^(n+)` requires `1.6 xx 10^(-3)` mol of `MnO_4^(-)` for the oxidation of `A^(n+)` to `AO_3^(-)` in acid medium, what is the value of n?

Options

A. 3
B. 5
C. 2
D. 4

Answer

C

Explanation

Since in acidic medium, `A^(n+)`, is oxidized to `AO_3^(-)`, the change in oxidation state from (+n) to (+5)=5-n. Total number of electrons that have been given out during oxidation of `2.68 xx 10^(-3)`mol of `A^(n+) = 2:68 xx 10^(-3) xx (5-n)`
Thus, the number of electrons added to reduce `1.61 xx 10^(-3) `mol of `MnO_4^(-)` to `Mn^(2+)`, that is `(+7)` to`(+2) = 1.61 xx 10^(-3) xx 5,` Therefore, `1.61 xx 10^(-3) xx 5 = 2.68 xx 10^(-3) xx (5 - n)`, Solving, we get n=2