Question

`2.68xx10^(-3)` moles of solution containing anion `A^(n+)` require `1.61xx10^(-3)` moles of `MnO_(4)^(-)` for oxidation of `A^(n+)` to `AO_(3)^(-)` in acidic medium. What is the value of `n`?

Answer 1

`A^(n+)` is oxidised to `AO_(3)^(-)`
Change in oxidation number `=5(" in" AO_(3)^(-)) -n (" in" A^(n+))`
=5-n
`2.68xx10^(-3) ` mol of `A^(n+)` ion react with `1.6xx10^(-3)` mol of `MnO_(4)^(-)` ions.
`therefore 1 " mol of " A^(n+)` ion will react with `(1.6xx10^(-3))/(2.68xx10^(-3))` mol of `MnO_(4)^(-)` ions
=0.579 mol of `MnO_(4)^(-)` ions
`2K overset(+7)Mn O_(4)+3H_(2)SO_(4) to K_(2)SO_(4)+2 overset(+2)Mn SO_(4) +3H_(2)O+5[O]`
Number of equivalents of `MnO_(4)^(-)` used in oxidation of `A^(n+)` to `AO_(3)^(-) =0.597xx5=2.985=3`
Thus, from equation (i), 5-n=3 n=2