# How do you determine which element oxidized in a reaction?

Dec 10, 2016

The species whose $\text{oxidation number}$ increased during reaction is said to have been $\text{oxidized}$.

#### Explanation:

Oxidation number is an imaginary charge possessed by atoms within molecules, that help us to conceive of electron transfer. The element whose $\text{formal oxidation number}$ increases during a reaction is said to have been $\text{oxidized}$. The element whose $\text{formal oxidation number}$ decreases during a reaction is said to have been $\text{reduced}$.

For a simple combustion reaction:

$C {H}_{4} + 2 {O}_{2} \rightarrow C {O}_{2} + 2 {H}_{2} O$

The oxidation number of (reactants) $C = - I V$, $H = + I$, and $O = 0$ (because oxygen is an element that has neither donated nor accepted electrons). The carbon is oxidized up to $C \left(+ I V\right)$ in carbon dioxide gas, and the oxygen is reduced to $O \left(- I I\right)$ in water.

These types of combustion reactions are fairly easy to balance. However, suppose we oxidized $C {r}^{I I I +}$ to $C {r}^{V I +}$?

$C {r}^{3 +} + 4 {H}_{2} O \left(l\right) \rightarrow C r {O}_{4}^{2 -} + 8 {H}^{+} + 3 {e}^{-}$

Here, first off, I accounted for electron transfer: the difference in oxidation number between the chromium species, i.e. $3$, were represented by electrons as actual particles. And then I balanced the oxygens and then the hydrogens. When we invoke these electrons in the equation, we have to have somewhere for them to go (i.e. for every oxidation, there must be a reduction).

Confused yet? Keep trying; remember that this may be pitched at a higher level than which you are at currently.