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?

1 Answer
anor277
Aug 21, 2016

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.

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