Question

In the oxidation of oxalate ion, `C_2O_4^(2-)` to give carbon dioxide by potassium permanganate, `K^(+)MnO_4^(-)`, how do we vizualize the endpoint?

Answer 1

Disappearance of the deep purple/violet colour.

Explanation:

Permanganate ion, which gives intense purple solutions, is reduced to `Mn^(2+)`, which as a `d^5` metal is almost colourless in aqueous solution:

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

The first drop of permanganate titrant delivered after the end point, delivers a very noticable purple colour. And thus the endpoint can be easily perceived.

Oxalate ion is oxidized up to carbon dioxide:

`C_2O_4^(2-) rarr 2CO_2 +2e^(-)` `(ii)`

So to balance the redox equations, we cross mulitply to remove the electrons:

`2xx(i)+5xx(ii)=`

`2MnO_4^(-) + 16H^(+) +5C_2O_4^(2-) rarr 2Mn^(2+) + 8H_2O(l) +10CO_2`

Very concentrated solutions of `Mn^(2+)` salts give a pale rose-coloured solution, but as a `d^5` system, `d-d` transitions are spin-forbidden, and as such very weak. The given equation (I think) is stoichiometrically balanced with respect to mass and charge, as it must be if it reflects chemical reality.