Redox Reactions in Terms of Electron Transfer

The classical oxygen/hydrogen definition works well for the reactions historically used to discover redox chemistry, but it cannot describe a reaction like $\text{Zn}(s) + \text{Cu}^{2+}(aq) \to \text{Zn}^{2+}(aq) + \text{Cu}(s)$, where neither oxygen nor hydrogen appears anywhere. What is changing, unambiguously, is the distribution of electrons between zinc and copper — and this observation gives the general, modern definition of redox chemistry.

Oxidation is now defined as the loss of one or more electrons by an atom, ion, or molecule. Reduction is defined as the gain of one or more electrons. This definition subsumes the classical one exactly: when magnesium reacts with oxygen, magnesium atoms lose electrons to become $\text{Mg}^{2+}$ ions (oxidation, matching the classical picture of 'gaining oxygen'), while oxygen atoms gain those electrons to become $\text{O}^{2-}$ ions (reduction).

For the zinc–copper reaction, the process can be split cleanly into two half-reactions, each showing electrons explicitly:

Adding the two half-reactions together, and cancelling the $2e^{-}$ that appears on both sides, exactly recovers the overall observed equation $\text{Zn} + \text{Cu}^{2+} \to \text{Zn}^{2+} + \text{Cu}$. This cancellation is not a coincidence — it is a structural requirement: electrons can never appear as a net product or net reactant in a complete, balanced redox equation, because free electrons do not simply pile up in solution. Every electron released by the species undergoing oxidation must be captured by the species undergoing reduction, in the very same reaction, at the very same instant. This is why oxidation and reduction are inseparable, exactly as the classical picture already showed empirically. …