Question

How does permanganate ion `MnO_4^(-)` oxidize oxalic acid, `C_2O_4H_2`? What colour change would be observed in the reaction?

Answer 1

Well `"permanganate ion"` is reduced to `Mn^(2+)`......

Explanation:

`MnO_4^(-) +8H^(+) + 5e^(-) rarr underbrace(Mn^(2+))_"colourless"+4H_2O(l)` `(i)`

Permanganate ion is DEEP-PURPLE in aqueous solution...

And oxalate ion is oxidized to `CO_2`...i.e. `C(+III)rarrC(+IV)`

`C_2O_4^(2-)rarr2CO_2(g)uarr+2e^(-)` `(ii)`

And we add `2xx(i)+5xx(ii)` and cancel appropriately .....

`5C_2O_4^(2-)+2MnO_4^(-) +16H^(+) + cancel(10e^(-))rarr10CO_2(g)uarr+2Mn^(2+)+8H_2O(l)+cancel(10e^(-))`

....to give finally....

`underbrace(5C_2O_4^(2-))_"the reductant"+underbrace(2MnO_4^(-))_"the oxidant" +16H^(+) rarr`

`underbrace(10CO_2(g))_"the oxidation product"uarr+2Mn^(2+)+8H_2O(l)`

...which is balanced with respect to mass and charge, as indeed it must be if we purport to represent an actual chemical reaction....

Happy......??