# Question #d2fbd

Jun 5, 2016

Manganate, $M n \left(V I\right)$, is reduced to $M n \left(I I I\right)$. Oxalic acid is oxidized to $C {O}_{2}$.

#### Explanation:

Reduction half equation:

$2 M n {O}_{4}^{2 -} + 10 {H}^{+} + 6 {e}^{-} \rightarrow M {n}_{2} {O}_{3} + 5 {H}_{2} O$ $\left(i\right)$

Oxidation half equation:

$H O \left(O =\right) C - C \left(= O\right) O H \rightarrow 2 C {O}_{2} \uparrow + 2 {H}^{+} + 2 {e}^{-}$ $\left(i i\right)$

So $\left(i\right) + 3 \times \left(i i\right) =$

$2 M n {O}_{4}^{2 -} + 3 {\left\{H O \left(O =\right) C\right\}}_{2} + 4 {H}^{+} \rightarrow M {n}_{2} {O}_{3} + 5 {H}_{2} O + 6 C {O}_{2}$

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.