# Why are oxidation and reduction reactions linked?

Sep 20, 2017

Why? Because we introduce electrons as conceptual particles in order to help us balance redox equations......

#### Explanation:

$\text{Oxidation}$ is conceived to involve the formal loss of electrons, and oxidation state is introduced to allow us to keep track of electron transfer.

We could write the oxidation of ammonia to give dinitrogen gas, i.e. $N \left(- I I I\right) \rightarrow N \left(O\right)$:

$N {H}_{3} \left(a q\right) \rightarrow \frac{1}{2} {N}_{2} + 3 {H}^{+} + 3 {e}^{-}$ $\left(i\right)$

The difference between oxidation numbers is accounted for by the three electrons....these are conceptual particles, i.e. particles of convenience.

And her mass and charge are balanced, as they must be if we purport to represent chemical reality. But the electrons are conceived to GO somewhere, i.e. they cause a corresponding REDUCTION reaction. And we could invoke $\text{cupric oxide}$ as a potential reducing agent.....

$C u O \left(s\right) + 2 {H}^{+} + 2 {e}^{-} + \Delta \rightarrow C u \left(s\right) + {H}_{2} O$ $\left(i i\right)$

And so we add $\left(i\right)$ and $\left(i i\right)$ together in such a way that the electrons are eliminated. And so....$2 \times \left(i\right) + 3 \times \left(i i\right)$ gives.....

$3 C u O \left(s\right) + \cancel{6 {H}^{+}} + \cancel{6 {e}^{-}} + 2 N {H}_{3} \left(a q\right) + \Delta \rightarrow 3 C u \left(s\right) + 3 {H}_{2} O + {N}_{2} + \cancel{6 {H}^{+}} + \cancel{6 {e}^{-}}$

And after cancelling we gets.....

$3 C u O \left(s\right) + 2 N {H}_{3} \left(a q\right) + \Delta \rightarrow 3 C u \left(s\right) + 3 {H}_{2} O \left(l\right) + {N}_{2} \left(g\right) \uparrow$

Which is balanced with respect to mass and charge, as it must be if we represent a valid chemical reaction. Oxidation states, and electrons, are here used as particles of convenience to help us represent the redox reaction. As written, the given reduction (with respect to ammonia) represents an actual chemical transformation.

And if this seems a bit abstract, all I have done is to make sure both mass and charge are conserved in the individual redox equations.....the electrons are introduced as a means to this end.