# How do you find the oxidizing agent in a reaction?

Aug 19, 2017

The oxidizing agent is FORMALLY the species that is reduced.........

#### Explanation:

See this old answer for the formal rules of oxidation numbers. We conceive that a species, an atom LOSES electron upon oxidation.

And thus for the oxidation of methane.....

$C {H}_{4} \left(g\right) + {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right)$

We can put in oxidation numbers if we like.....

$C {H}_{4} \left(g\right) + {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right)$

$C \left(- I V\right) + O \left(0\right) \rightarrow C \left(+ I V\right) + 4 \times O \left(- I I\right)$

A formal EIGHT electron oxidation.

We could formalize this process even further by writing the separate redox reactions:

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

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

$4 \times \left(i i\right) + \left(i\right)$ gives the balanced combustion reaction.