What is meant by #"oxidation"#?

1 Answer
Aug 27, 2017

Answer:

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

Explanation:

See this old answer.

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"#.