Final answer:
The process of writing out all ions present in a reaction is called writing ionic equations, which include oxidation and reduction half-reactions. Zinc undergoes oxidation, losing electrons, and copper ions undergo reduction, gaining electrons, leading to the formation of zinc ions and solid copper without spectator ions.
Step-by-step explanation:
The process of splitting species in a reaction into all the ions present is referred to as writing the ionic equations for the reaction. In the case of a single displacement reaction like that of zinc metal with copper(II) sulfate, the process involves two half-reactions: the oxidation half-reaction and the reduction half-reaction. The oxidation half-reaction for zinc is Zn(s) → Zn²⁺ (aq) + 2e⁻, where zinc loses electrons (is oxidized). The reduction half-reaction for copper is Cu²⁺ (aq) + 2e⁻ → Cu(s), where copper ions gain electrons (are reduced). When combined, these half-reactions give the overall reaction: Zn(s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu(s). Ions that appear on both sides of the complete ionic equation, such as NO3⁻ in this case, are called spectator ions and are not included in the net ionic equation.