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For example, consider:

$$\ce{NH3 + H2O -> NH4+ + OH-}$$

Here, the $\ce{NH3}$ is acting as a base (By Lewis theory, it is an electron pair donor).

Why don't we consider this to be a redox (there is a transfer of electrons taking place as $\ce{NH3}$ is acting as an electron donor and $\ce{H2O}$ as an electron acceptor)?

I'm aware that calculation of oxidation states is the best way to figure out if a reaction is redox or not. Then I would like to ask: why exactly do we assign oxidation states according to those rules? Is there an intuitive way to explain why a certain reaction is acid-base and not redox?

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In contrary to acid-base reactions(1), a redox reaction is such a reaction, which can be formally divided into two half-reactions. At least one reactant in such a half-reaction explicitly formally acts as a donor or acceptor of one or more electrons.

$$\ce{Cl2(aq) + 2 Fe^2+(aq) -> 2 Cl-(aq) + 2 Fe^3+(aq)}$$

can be formally divided to:

$$\ce{Cl2(aq) + 2 e- -> 2 Cl-(aq)}$$

$$\ce{2 Fe^2+(aq) -> 2 e- + 2 Fe^3+(aq)}$$

In the example, there was no electron transfer between $\ce{NH3}$ and $\ce{H2O}$.

Oxidation states are nothing more but valence electron inventory with defined formal rules of electron distribution. Their sum does not change during a reaction, due the charge conservation law.


(1) A very generalized Usanovich theory of acids and basis does consider redox reactions as a special case of acid-base reactions. But this theory is not much known nor applied within chemistry community.

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