Question

How do you turn KMnO4 (aq) + H2C2O4 (aq) into a molecular equation, then break it down through decomposition?

I don't understand how you get MnO3, isn't it suppose to be a double displacement, and then decomposition? Also, is the equation considered an acid base equation? Thank you.

Answer 1

I think you are OXIDIZING oxalic acid to carbon dioxide...

Explanation:

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

And of course, every oxidation requires a corresponding reduction...

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

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

We take `5xx(i)+2xx(ii)` to eliminate the electrons, virtual particles of convenience:

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

to give...

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

The which I think is balanced with respect to mass and charge, as indeed it must be if we purport to represent a valid chemical reaction. And this is an example of an `"oxidation-reduction reaction"`. The deep purple colour of permanganate would dissipate to give colourless `Mn^(2+)`..