Question

How do we find the oxidation number of an element?

Answer 1

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 (`"oxidation"`), or the number of electrons accepted (`"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, `IO_4^-`. Now the oxidation number of oxygen in its compounds is generally `-II` and it is so here. Thus the oxidation number of iodine in periodate is `VII+`, the Group oxidation number.

On the other hand, elemental iodine, has neither accepted nor donated electrons, and its oxidation state is formally `"zero"`.

`1/2I_2 + 4H_2O rarr IO_4^(-) + 8H^(+) +7e^-`

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(s) +O_2(g) rarr CO_2(g)`

Elemental (and zerovalent) carbon is oxidized to `C(IV+)`. Elemental oxygen is oxidized to `O(-II)`. Formally, how many electrons have been transferred in this reaction?