How does oxidizing work?

Aug 25, 2017

Oxidation is the formal loss of electrons.........

Explanation:

And this can be rationalized by invoking the electron as a FORMAL particle that is LOST on oxidation, and GAINED in a corresponding reduction.

For the simple oxidation of methane, we could balance the reaction by means of oxidation numbers:

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

But zerovalent dioxygen gas is REDUCED to $\stackrel{- I I}{O}$ in water....

$\frac{1}{2} {O}_{2} + 2 {H}^{+} + 2 {e}^{-} \rightarrow {H}_{2} O$ $\left(i i\right)$

We add $4 \times \left(i i\right) + \left(i\right)$ to eliminate the electrons.....

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

We are making a meal of this oxidation reaction. If we are given an hydrocarbon, and asked to represent the combustion reaction, we would balance the carbons as carbon dioxide, balance the hydrogens as water, and then balance the oxygens appropriately.