Why are oxidation-reduction reactions always coupled?

1 Answer
Aug 9, 2017

Consider the formalism of a redox reaction..........

Explanation:

Redox reactions proceed on the basis of #"(i) reduction"#, i.e. #"a formal GAIN"# in the number of electrons, e.g.

#1/2stackrel(0)N_2 + 3e^(-) +3H^(+) rarr stackrel(-III)" NH"_3# #(i)#

.....and on #"(ii) oxidation"#, #"a formal LOSS"# in the number of electrons.....

#stackrel(0)Fe rarr Fe^(3+) + 3e^(-)# #(ii)#

I write #"formal"# because these reactions are very much a #"formalism"#, i.e. a description of something in very abstract terms for reasons of pedagogy or as a model. The appearance of electrons in the stoichiometric equation allows us to balance redox equations fairly easily by balancing mass and charge.......i.e. for the above reaction we take....#1xx(i) +1xx(ii)# to give......

#1/2N_2 + Fe +3H^(+) rarr NH_3 +Fe^(3+)#

And since charge as well as mass are always conserved in every chemical reaction, every oxidation must be formally accompanied by the appropriate reduction...