# How do you know if it is an oxidation reduction reaction?

Jun 17, 2016

If the $\text{oxidation numbers}$ of the individual elements as reactants change upon reaction.

#### Explanation:

Oxidation numbers are theoretical constructs, and have marginal physical significance. Formally, the oxidation number of an element in a compound is the charge the element would have if all the bonding electrons were broken, with the charge assigned to the most electronegative atom.

For the molecule $H - F$, we break the bond, and the 2 electrons are assigned to the fluorine centre. I reiterate that this is a theoretical exercise:

$H - F \rightarrow {H}^{+} + {F}^{-}$

Thus fluorine is conceived to have an oxidation number of $- I$, and hydrogen an oxidation number of $+ I$; Roman numerals are used.

So if oxidation numbers change upon reaction, electron transfer and redox has occurred. Typically powerful oxidants such as elemental fluorine, ${F}_{2}$, and oxygen ${O}_{2}$, after reaction (after oxidizing something) assume oxidation states of $- I$ and $- I I$ in the product compound.

Oxidation of hydrocarbons underpin our civilization. If we look at the combustion of coal, we can write the equation:

$C \left(s\right) + {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + \Delta$

The oxidation number of an element is formally $0$. The oxidation number of $C$ in $C {O}_{2}$ is $+ I V$ and of oxygen $- I I$. Electron transfer has taken place between oxygen and carbon when the reaction occurs, and thus hydrocarbon combustion is formally redox.