Question

How does oxidizing work?

Answer 1

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:

`CH_4 +2H_2O rarr CO_2 +8H^+ + 8e^(-)` `(i)` (i.e. `stackrel(-VI)Crarrstackrel(+IV)C`)

But zerovalent dioxygen gas is REDUCED to `stackrel(-II)O` in water....

`1/2O_2 + 2H^(+) + 2e^(-) rarr H_2O` `(ii)`

We add `4xx(ii) + (i)` to eliminate the electrons.....

`CH_4 +2O_2 rarr CO_2 +2H_2O`

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.