# How would we represent the oxidation of sucrose to give oxalic acid with nitric acid oxidant?

Mar 8, 2017

Sucrose (probably) is oxidized up to oxalic acid

#### Explanation:

${C}_{12} {H}_{22} {O}_{11} + 13 {H}_{2} O \rightarrow 6 H O \left(O =\right) C - C \left(= O\right) O H + 36 {H}^{+} + 36 {e}^{-}$

Carbon is formally zerovalent, i.e. a ${C}^{0}$ oxidation state, and is oxidized to ${C}^{+ I I I}$ in oxalic acid.

And nitrate is reduced to $N {O}_{2}$:

$H N {O}_{3} + {H}^{+} + {e}^{-} \rightarrow N {O}_{2} + {H}_{2} O$

So we add 36 reductions reactions to one of the oxidations:

${C}_{12} {H}_{22} {O}_{11} + 36 H N {O}_{3} \rightarrow 6 H O \left(O =\right) C - C \left(= O\right) O H + 36 N {O}_{2} + 23 {H}_{2} O$

As far as I can tell, this reaction is balanced. The reaction represents the oxidation of ${C}^{0}$ up to ${C}^{+ I I I}$.