The oxidising agent accepts electrons and is itself reduced. To identify it in an ionic equation, look for the reactant whose oxidation state decreases. The reducing agent is the reactant that loses electrons and is itself oxidised.
The 2027 SEC G3 Chemistry K324 syllabus covers electron transfer and changes in oxidation state in redox reactions. This example follows the electrons in a metal displacement reaction.
- Write the oxidation states before and after the reaction.
- Find the reactant whose oxidation state decreases.
- Name that reactant as the oxidising agent.
- Use half-equations to verify the electron transfer.
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Read the reactants first
Consider Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Zinc starts in elemental form with oxidation state 0 and ends as Zn²⁺ with oxidation state +2. Copper starts as Cu²⁺ with oxidation state +2 and ends in elemental form with oxidation state 0.
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Worked example | Follow both electrons
The zinc half-equation is Zn(s) → Zn²⁺(aq) + 2e⁻. Zinc loses electrons, so it is oxidised. The copper half-equation is Cu²⁺(aq) + 2e⁻ → Cu(s). Copper(II) ions gain electrons, so they are reduced.
The oxidising agent is Cu²⁺, because it accepts electrons from zinc and causes zinc to be oxidised. The reducing agent is Zn, because it supplies the electrons that reduce Cu²⁺.
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CHAPTER 3 OF 4 · Back to contents
Distinguish the reacting ion from its product
The equation shows copper metal on the product side. If asked for the oxidising agent, name the starting Cu²⁺ ion. If the original question gives copper(II) sulfate solution, the net ionic equation shows that Cu²⁺ is the ion taking part in the electron transfer. Sulfate stays in solution in this example.
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CHAPTER 4 OF 4 · Back to contents
Frequently asked questions
Why is the oxidising agent reduced?
It causes another species to lose electrons by accepting those electrons itself.
What check can I do before finishing?
Make sure the number of electrons lost in the oxidation half-equation equals the number gained in the reduction half-equation.