# Why are oxidation-reduction reactions always coupled?

Aug 9, 2017

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

#### Explanation:

Redox reactions proceed on the basis of $\text{(i) reduction}$, i.e. $\text{a formal GAIN}$ in the number of electrons, e.g.

$\frac{1}{2} {\stackrel{0}{N}}_{2} + 3 {e}^{-} + 3 {H}^{+} \rightarrow {\stackrel{- I I I}{\text{ NH}}}_{3}$ $\left(i\right)$

.....and on $\text{(ii) oxidation}$, $\text{a formal LOSS}$ in the number of electrons.....

$\stackrel{0}{F} e \rightarrow F {e}^{3 +} + 3 {e}^{-}$ $\left(i i\right)$

I write $\text{formal}$ because these reactions are very much a $\text{formalism}$, 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....$1 \times \left(i\right) + 1 \times \left(i i\right)$ to give......

$\frac{1}{2} {N}_{2} + F e + 3 {H}^{+} \rightarrow N {H}_{3} + F {e}^{3 +}$

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