Q.Show that $2\text{Cu}^{+} \to \text{Cu} + \text{Cu}^{2+}$ is a disproportionation reaction by tracking the oxidation number of copper on both sides.

Oxidation is the loss of an electron and reduction is the gain of an electron (the general definition, superseding the older oxygen/hydrogen-based one); the two always occur together in a redox reaction. Using oxidation-number change, every redox reaction can be classified into one of five types:

1. Combination reaction -- two substances combine into a single compound. Example: C + O₂ → CO₂ (C: 0→+4, oxidised; O: 0→−2, reduced).
2. Decomposition reaction -- the reverse: one compound breaks into two or more products, with different elements within it changing oxidation number in opposite directions. Example: 2KClO₃ → 2KCl + 3O₂ (Cl: +5→−1, reduced; O: −2→0, oxidised).
3. Displacement reaction -- an ion/atom in a compound is replaced by an ion/atom of another element. Metal displacement example: CuSO₄ + Zn → Cu + ZnSO₄ (Zn: 0→+2, oxidised; Cu: +2→0, reduced). Non-metal displacement example: Zn + 2HCl → ZnCl₂ + H₂ (Zn: 0→+2, oxidised; H: +1→0, reduced).
4. Disproportionation reaction (auto-redox) -- the same element in the same starting compound is simultaneously oxidised and reduced. Example: 2H₂O₂ → 2H₂O + O₂ (O: −1 splits into −2 in H₂O, reduced, and 0 in O₂, oxidised). …