3.3 - Ionic & Half Equations

Writing an ionic equation for a displacement

• When a more reactive metal reacts with a solution of a less reactive metal's compound, it displaces the less reactive metal.
• For example, an iron nail in copper(II) sulfate solution forms iron(II) sulfate solution and copper:
  • Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)

An ionic equation shows only the particles that actually change during a reaction.

The method

1. Start from the balanced symbol equation, with state symbols.

   • Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)

2. Rewrite every dissolved (aq) ionic compound as its separate ions. Solids, liquids, and gases are left whole.

   • Fe(s) + Cu²⁺(aq) + SO₄^2-(aq) → Fe²⁺(aq) + SO₄^2-(aq) + Cu(s)

3. Cross out spectator ions - Ions that appear unchanged on both sides and take no part in the reaction (here, SO₄^2-(aq)).

4. Write what's left as the ionic equation.

   • Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s)

Check the atoms and the total charge (2+ on each side).

Reading the electron transfer

• Iron loses two electrons to form Fe²⁺, so it's oxidised.
• Copper ions (Cu²⁺) gain two electrons to form copper atoms, so they're reduced.
• Every displacement is a redox reaction: the metal atom is oxidised and the metal ion in solution is reduced.