How do we find the oxidation number of an element?

1 Answer
Nov 16, 2016

Answer:

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?