# 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?

Aug 21, 2016

Disappearance of the deep purple/violet colour.

#### Explanation:

Permanganate ion, which gives intense purple solutions, is reduced to $M {n}^{2 +}$, which as a ${d}^{5}$ metal is almost colourless in aqueous solution:

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O \left(l\right)$ $\left(i\right)$

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:

${C}_{2} {O}_{4}^{2 -} \rightarrow 2 C {O}_{2} + 2 {e}^{-}$ $\left(i i\right)$

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

$2 \times \left(i\right) + 5 \times \left(i i\right) =$

$2 M n {O}_{4}^{-} + 16 {H}^{+} + 5 {C}_{2} {O}_{4}^{2 -} \rightarrow 2 M {n}^{2 +} + 8 {H}_{2} O \left(l\right) + 10 C {O}_{2}$

Very concentrated solutions of $M {n}^{2 +}$ 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.