# How do we find the oxidation number of an element?

Nov 16, 2016

For an element, the oxidation number is easy to assign.......

#### Explanation:

The oxidation number of any element is $0$. We assign oxidation numbers on the basis of the number of electrons donated ($\text{oxidation}$), or the number of electrons accepted ($\text{reduction}$).

Of course these designations are completely conceptual, nevertheless, these ideas are useful for writing and balancing redox reactions. We consider here the oxidation of iodine to periodate ion, $I {O}_{4}^{-}$. Now the oxidation number of oxygen in its compounds is generally $- I I$ and it is so here. Thus the oxidation number of iodine in periodate is $V I I +$, the Group oxidation number.

On the other hand, elemental iodine, has neither accepted nor donated electrons, and its oxidation state is formally $\text{zero}$.

$\frac{1}{2} {I}_{2} + 4 {H}_{2} O \rightarrow I {O}_{4}^{-} + 8 {H}^{+} + 7 {e}^{-}$

This is only the oxidation half equation, and a reduction of something (oxygen?) is necessary.

We can write a complete redox reaction for combustion reactions:

$C \left(s\right) + {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right)$

Elemental (and zerovalent) carbon is oxidized to $C \left(I V +\right)$. Elemental oxygen is oxidized to $O \left(- I I\right)$. Formally, how many electrons have been transferred in this reaction?