# How are oxidation numbers determined and used?

Aug 12, 2017

They are a convenient fiction to help balance chemical reactions and processes, and to rationalize reactivity.

#### Explanation:

For general rules on the assignment of oxidation numbers see this old answer.

And so how do you use it? Well, let's examine the oxidation of ethanol to acetic acid by potassium permanganate, $K M n {O}_{4}$.

Carbon undergoes a 4 electron oxidation........

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

But permanganate ion undergoes a 5 electron reduction.....

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

We cross multiply to remove the electrons.....$3 \times \left(i\right) + 4 \times \left(i i\right)$

$\text{3H"_3"CCH"_2"OH" +4MnO_4^(-) +4H^+ rarr 3"H"_3"CC(=O)OH} + 4 M n {O}_{2} \left(s\right) + 5 {H}_{2} O$

And this (I think) is balanced with respect to mass and charge...And what would you observe in this reaction? The disappearance of the deep purple colour of permanganate to give almost colourless $M {n}^{2 +}$ ion.

Note that we could also represent the oxidation of ethanol ALL the way up to give quadrivalent $C {O}_{2}$....

${H}_{3} \stackrel{- I I I}{C} - \stackrel{- I}{C} {H}_{2} O H + 3 {H}_{2} O \rightarrow 2 \stackrel{+ I V}{C} {O}_{2} + 12 {H}^{+} + 12 {e}^{-}$

Here, I treated $H$ as $\stackrel{+ I}{H}$. And again I accounted for the difference in oxidation numbers of CARBON, by individual electrons; this is a 12 electron oxidation.

Now oxygen is the standard oxidant, and it is reduced to oxide....

${O}_{2} \left(g\right) + 4 {e}^{-} \rightarrow 2 {O}^{2 -}$

We add the equations in such a way as to eliminate the electrons; and we thus take one of the former, and THREE of the latter.......

${H}_{3} \stackrel{- I I I}{C} - \stackrel{- I}{C} {H}_{2} O H + 3 {H}_{2} O + 3 {O}_{2} \left(g\right) + 12 {e}^{-} \rightarrow 2 \stackrel{+ I V}{C} {O}_{2} + {\underbrace{12 {H}^{+} + 6 {O}^{2 -}}}_{6 {H}_{2} O} + 12 {e}^{-}$

And from this equation we cancel out common reactants and products....

${H}_{3} C - C {H}_{2} O H + \cancel{3 {H}_{2} O} + 3 {O}_{2} \left(g\right) \rightarrow 2 C {O}_{2} + \cancel{6} 3 {H}_{2} O$

To give finally......

${H}_{3} C - C {H}_{2} O H + 3 {O}_{2} \left(g\right) \rightarrow 2 C {O}_{2} + 3 {H}_{2} O$

Please note that I would NEVER advocate using this method of oxidation numbers to approach a combustion reaction. The normal rigmarole is to balance the carbons as $C {O}_{2}$, then the hydrogens as water, and then add in stoichiometric dioxygen.