4.5 - 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
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Start from the balanced symbol equation, with state symbols.
- Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)
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Rewrite every dissolved (aq) ionic compound as its separate ions. Solids, liquids, and gases are left whole.
- Fe(s) + Cu2+(aq) + SO42-(aq) → Fe2+(aq) + SO42-(aq) + Cu(s)
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Cross out spectator ions - Ions that appear unchanged on both sides and take no part in the reaction (here, SO42-(aq)).
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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.
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 SO42-(aq); ZnSO4(aq) becomes Zn2+(aq) and SO42-(aq). The zinc and copper metal are solids, so they stay whole.
Zn(s) + Cu2+(aq) + SO42-(aq) → Zn2+(aq) + SO42-(aq) + Cu(s)
Step 3 - Cross out spectator ions
SO42-(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.
Writing a half equation
A half equation shows what happens to just one substance in a redox reaction, with electrons written explicitly as e-.
Losing electrons (oxidation)
- Electrons are lost, so they're written on the right-hand side.
- A sodium atom loses its outer electron to form a sodium ion:
- Na → Na+ + e-
- A magnesium atom loses two electrons:
- Mg → Mg2+ + 2e-
Gaining electrons (reduction)
- Electrons are gained, so they're written on the left-hand side.
- A chlorine molecule gains two electrons to form two chloride ions:
- Cl2 + 2e- → 2Cl-
Balancing a half equation
- Balance the atoms first, then balance the charge.
- Add electrons (each counting as 1−) to whichever side is more positive until the total charge on both sides matches.
Worked example - Half equations for calcium and oxygen
Write the half equation for a calcium atom forming a calcium ion, and for an oxygen atom forming an oxide ion.
Step 1 - Find the charge on each ion from its group
- Calcium is in Group 2, so a calcium atom loses two electrons to form a Ca2+ ion.
- Oxygen is in Group 6, so an oxygen atom gains two electrons to form an O2- ion.
Step 2 - Put the electrons on the side they move to
Calcium loses its two, so they go on the right with the ion:
Ca → Ca2+ + 2e-
Oxygen gains its two, so they go on the left with the atom:
O + 2e- → O2-
Step 3 - Check the charge on both sides
- Calcium: left is 0 (neutral atom); right is 2+ with two electrons at 1− each = 0. Balanced.
- Oxygen: left is 0 + 2− = 2−; right is 2−. Balanced.