What is the ratio of moles of `MnO_4^(ɵ)` used per " mol of "`C_2O_4^(2-)` in acidic medium to strong basic medium?

Explanation

(a). In strong basic medium, `MnO_4^(ɵ)` is reduced to `MnO_4^(2-)`
`[e^(-)+MnO_4^(ɵ)toMnO_4^(ɵ)(n=1)]`
`[C_2O_4^(2-)to2CO_2+2e^(-)(n=2)]`
In acidic medium `MnO_4^(ɵ)` is reduced to `MnO_4^(2-)`
`5e^(-)+MnO_4^(ɵ)toMn^(2+)(n=5)`
`MnO_4^(ɵ)` (acidic medium) `-=C_2O_4^(2-)`
`1Eq-=1Eq`
`(1)/(5)mol-=(1)/(2)mol`
1 " mol of "`C_2O_4^(2-)-=(2)/(5)` " mol of "`MnO_4^(ɵ)` in acidic medium.
(b). `MnO_4^(ɵ)` (basic medium) `-=C_2O_4^(2-)`
`1eq-=1eq`
`(1)/(1)mol-=(1)/(2)mol`
1 " mol of "`C_2O_4^(2-)=2` " mol of "`MnO_4^(ɵ)` in basic medium Ratio of `MnO_4^(ɵ)` in acidic to basic medium.
`=(2)/(5)+2=(2)/(5xx2)=(1)/(5)=15`