Question

Question #d2fbd

Answer 1

Manganate, `Mn(VI)`, is reduced to `Mn(III)`. Oxalic acid is oxidized to `CO_2`.

Explanation:

Reduction half equation:

`2MnO_4^(2-) + 10H^(+) + 6e^(-) rarr Mn_2O_3 +5H_2O` `(i)`

Oxidation half equation:

`HO(O=)C-C(=O)OH rarr 2CO_2uarr +2H^(+) +2e^(-)` `(ii)`

So `(i)+3xx(ii)=`

`2MnO_4^(2-) + 3{HO(O=)C}_2 +4H^(+)rarr Mn_2O_3 +5H_2O +6CO_2`

Which (I think) is balanced with respect to mass and charge as required.

You've got potassium oxide in your equation, for which I have not accounted. The redox couple is manganese and carbon.