What is meant by `"oxidation"`?

See this old [answer.](https://socratic.org/questions/how-do-you-write-oxidation-reduction-half-reactions#278689)

Now normally redox reactions require the assignment of oxidation states. When the oxidation number of an element or an element in a compound INCREASES, the element is said to have been `"oxidized"`. On the other hand, when the oxidation number of an element or an element in a compound DECREASES, the element is said to have been `"reduced"`.

In these types of reactions, we introduce an electron as a virtual particle to represent the loss `"(oxidation)"` or gain `"(reduction)"` of electrons.

So let us represent the reduction of a metal ion, `Cr(VI+)`, in `Cr_2O_7^(2-)`(typically we use Roman numerals for oxidation numbers.) And we represent the difference in oxidation numbers by exchange of electrons......

`Cr_2O_7^(2-) +14H^(+) +6e^(-) rarr 2Cr^(3+) +7H_2O` `(i)`

Now methanol `C(-II)` is oxidized up to `C(+IV)` in `(O=)C(OH)_2`.

`""^(-II)CH_3OH +2H_2O rarr(O=)stackrel(+IV)C(OH)_2+6H^+ +6e^-` `(ii)`

And we add these together to eliminate the electrons, i.e. `(i) + (ii):`

`Cr_2O_7^(2-) +cancel(14)8H^(+) +cancel(6e^(-)) +CH_3OH +cancel(2H_2O)rarr 2Cr^(3+) +cancel(7)5H_2O+(O=)C(OH)_2+cancel(6H^+ +6e^-)`

`Cr_2O_7^(2-) +8H^(+) +CH_3OH rarr 2Cr^(3+) +5H_2O+(O=)C(OH)_2`

The which, I think is balanced with respect mass and charge.

This is more than 2 paragraphs. But oxidation is `"FORMAL LOSS OF ELECTRONS"`.