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?

Answer

2

Explanation

The reaction is `MnO_(4)^(-) +8H^(+) + 5e ^(-) to Mn^(2+) + 4H_2 O`
` A^(n+) + 3 H_2 O to AO_(3)^(-) + 6H^(+) +(5 -n) e^(-)`
Amount of electrons involved in the given amount of `MnO_(4)^(-) =5 xx 1.61 xx 10^(-3)` mol. Equating these two we get `5 xx 1.61 xx 10^(-3)=(5-n)2.68 xx 10^(-3) " " therefore n=2`(appox.)