# What is meant by "oxidation"?

Aug 27, 2017

Oxidation describes the combination of an element with oxygen, or AN increase in the oxidation state of the element.......

#### Explanation:

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 $\text{oxidized}$. 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 $\text{reduced}$.

In these types of reactions, we introduce an electron as a virtual particle to represent the loss $\text{(oxidation)}$ or gain $\text{(reduction)}$ of electrons.

So let us represent the reduction of a metal ion, $C r \left(V I +\right)$, in $C {r}_{2} {O}_{7}^{2 -}$(typically we use Roman numerals for oxidation numbers.) And we represent the difference in oxidation numbers by exchange of electrons......

$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 {e}^{-} \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O$ $\left(i\right)$

Now methanol $C \left(- I I\right)$ is oxidized up to $C \left(+ I V\right)$ in $\left(O =\right) C {\left(O H\right)}_{2}$.

""^(-II)CH_3OH +2H_2O rarr(O=)stackrel(+IV)C(OH)_2+6H^+ +6e^- $\left(i i\right)$

And we add these together to eliminate the electrons, i.e. $\left(i\right) + \left(i i\right) :$

$C {r}_{2} {O}_{7}^{2 -} + \cancel{14} 8 {H}^{+} + \cancel{6 {e}^{-}} + C {H}_{3} O H + \cancel{2 {H}_{2} O} \rightarrow 2 C {r}^{3 +} + \cancel{7} 5 {H}_{2} O + \left(O =\right) C {\left(O H\right)}_{2} + \cancel{6 {H}^{+} + 6 {e}^{-}}$

$C {r}_{2} {O}_{7}^{2 -} + 8 {H}^{+} + C {H}_{3} O H \rightarrow 2 C {r}^{3 +} + 5 {H}_{2} O + \left(O =\right) C {\left(O H\right)}_{2}$

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

This is more than 2 paragraphs. But oxidation is $\text{FORMAL LOSS OF ELECTRONS}$.