# How do we represent the oxidation of ethanol to acetic acid by potassium permanganate using the oxidation number method?

Jun 1, 2017

$4 M n {O}_{4}^{-} + 12 {H}^{+} + 5 {H}_{3} C - C {H}_{2} O H \rightarrow 5 {H}_{3} C - C {O}_{2} H + 4 M {n}^{2 +} + 11 {H}_{2} O \left(l\right)$

#### Explanation:

Just to expand on the subject of oxidation of alcohols, the method of oxidation number, and gain or loss of electrons, can be utilized in the scenario, provided that we know how to assign oxidation numbers for the individual carbons in a given compound.

For the oxidation of ethanol to acetic acid, we could propose the given oxidation reaction as:

${H}_{3} \stackrel{- I I I}{C} - \stackrel{- I}{C} {H}_{2} O H + {H}_{2} O \rightarrow {H}_{3} \stackrel{- I I I}{C} - \stackrel{+ I I I}{C} \left(= O\right) O H + 4 {H}^{+} + 4 {e}^{-} \text{ (i)}$

And something, here $M n {O}_{4}^{-}$ is reduced down to $M {n}^{+ 2}$:

$\stackrel{V I I +}{\text{Mn"rarrstackrel(+II)"Mn}}$

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow \text{Mn"^(2+)+"4H"_2"O} \left(l\right)$ $\text{(ii)}$

Deep-red purple permanganate ion is reduced down to almost colourless $M {n}^{2 +}$.

And to give the final redox equation, we cross multiply $\left(i\right)$ and $\left(i i\right)$ so that electrons DO NOT appear in the final equation:

And thus $5 \times \left(i\right) + 4 \times \left(i i\right)$ gives...............

$4 M n {O}_{4}^{-} + 12 {H}^{+} + 5 {H}_{3} C - C {H}_{2} O H \rightarrow 5 {H}_{3} C - C {O}_{2} H + 4 M {n}^{2 +} + 11 {H}_{2} O \left(l\right)$

Which, if I have done my sums right, is balanced with respect to mass and charge; as indeed it must be if we purport to represent chemical reality. Note that the reaction contains a built-in indicator. $M n {O}_{4}^{-}$ is deep purple; $M {n}^{2 +}$ is almost colourless.