Chemistry

4.3 - Ionic Equations

3 min

note 1 of 2

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

Start from the balanced symbol equation, with state symbols.
- Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)
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)
Cross out spectator ions - Ions that appear unchanged on both sides and take no part in the reaction (here, SO₄^2-(aq)).
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).

Three stacked equations: iron plus copper sulfate, then as separate ions with both SO4 2- ions struck through and labelled spectator ions, then the boxed ionic equation Fe(s) + Cu2+(aq) gives Fe2+(aq) + Cu(s).

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.

note 2 of 2

Worked example - Zinc and copper(II) sulfate

Zinc is added to copper(II) sulfate solution. Write the ionic equation for the reaction.

Step 1 - Balanced symbol equation with state symbols

Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

One zinc, one copper and one sulfate on each side.

Step 2 - Split every dissolved ionic compound into its ions

CuSO₄(aq) becomes Cu²⁺(aq) and SO₄^2-(aq); ZnSO₄(aq) becomes Zn²⁺(aq) and SO₄^2-(aq). The zinc and copper metal are solids, so they stay whole.

Zn(s) + Cu²⁺(aq) + SO₄^2-(aq) → Zn²⁺(aq) + SO₄^2-(aq) + Cu(s)

Step 3 - Cross out spectator ions

SO₄^2-(aq) is unchanged on both sides, so it's deleted.

Step 4 - Write the ionic equation

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Check the atoms and the charge: one zinc and one copper on each side, and a total charge of 2+ on each side.

key terms

TermDefinition

Ionic equationAn equation that shows only the particles that actually take part in a reaction, leaving out any spectator ions.

Spectator ionAn ion that is unchanged by a reaction, so it appears on both sides of the full equation and is left out of the ionic equation.

end of notes

Key points to remember

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

Start from the balanced symbol equation, with state symbols.

Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)

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)

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

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).

Worked example - Zinc and copper(II) sulfate

Zinc is added to copper(II) sulfate solution. Write the ionic equation for the reaction.

Step 1 - Balanced symbol equation with state symbols

Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

One zinc, one copper and one sulfate on each side.

Step 2 - Split every dissolved ionic compound into its ions

CuSO₄(aq) becomes Cu²⁺(aq) and SO₄^2-(aq); ZnSO₄(aq) becomes Zn²⁺(aq) and SO₄^2-(aq). The zinc and copper metal are solids, so they stay whole.

Zn(s) + Cu²⁺(aq) + SO₄^2-(aq) → Zn²⁺(aq) + SO₄^2-(aq) + Cu(s)

Step 3 - Cross out spectator ions

SO₄^2-(aq) is unchanged on both sides, so it's deleted.

Step 4 - Write the ionic equation

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Check the atoms and the charge: one zinc and one copper on each side, and a total charge of 2+ on each side.