Question

How do you write the full redox equation for the titration of sodium oxalate with potassium permanganate?

Answer 1

We use the method of half-equations to get....

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

Explanation:

Permanganate ion, `Mn(VII+)`. is reduced to colourless `Mn^(2+)`...

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

And oxalate ion, `C_2O_4^(2-)`, `C(+III)`, is oxidized to carbon dioxide...`CO_2`, `C(+IV)`.

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

...And we add the half-equations in such a way as to retire the electrons...`2xx(i)+5xx(ii)`

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

...after cancellation....

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

The which I think is balanced with respect to mass and charge. The reaction has a built-in indicator...with respect to the distinct metal species...the endpoints are easily vizualized...

`underbrace(MnO_4^(-))_"deep purple"`

`underbrace(Mn^(2+))_"almost colourless"`

Of course oxalate ion and carbon dioxide are colourless...