Question

Balance the following reaction? `(MnO_4^-) + C_2O_4^(2-) + H^(+) -> Mn^(2+) + CO_2 + H_2O`

Answer 1

You gots a redox reaction....the which we solve by the method of half equations....

Explanation:

Permanganate ion, `MnO_4^(-)`, the which is a POTENT oxidant, is REDUCED to colourless `Mn^(2+)`:

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

The number of electrons accounts for the `Mn(VII+)rarrMn(+II)` reduction....

And oxalate ion is oxidized to carbon dioxide....

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

i.e. `C(+III)rarrC(+IV)`

And so we takes `2xx(i)+5xx(ii)` to eliminate the electrons as particles of convenience....

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

...to give after cancellation of common reagents.....

`2underbrace(MnO_4^(-))_"purple"+5C_2O_4^(2-)+16H^+rarrunderbrace(2Mn^(2+))_"colourless" + 10CO_2(g)+8H_2O(l)`

The which (I think) is balanced with respect to mass and charge, as indeed it must be if we purport to represent chemical reality. And what would we see in this reaction? Well, the deep purple colour of permanganate ion would dissipate to give (almost) colourless `Mn^(2+)`.