1.1.4Atomic Structure and the Periodic Table
Half Equations and Ionic Equations
5 min read 8 parts Lesson 5 of 19 1 quiz
Section 1 of 8Half equations
What they show
- What they show A half equation shows what happens to one species, with electrons written as e−.
- Two half equations can be added together, with the electrons cancelling, to give the full ionic equation for a redox reaction.
Losing electrons
- Losing electrons Metal atoms form positive ions: Na → Na+ + e−. Electrons appear on the product side.
Gaining electrons
- Gaining electrons Non-metal atoms form negative ions: Cl2 + 2e− → 2Cl−.
- Electrons appear on the reactant side.
Balanced twice
- Balanced twice The number of atoms and the total charge must be equal on both sides.
Section 2 of 8Writing a half equation
1
Write the species before and after the change, such as Mg and Mg2+.
2
Balance the atoms, adding a coefficient where needed, as in Cl2 → 2Cl−.
3
Work out the total charge on each side.
4
Add electrons to the side with the more positive charge so that the charges match.
5
Check that atoms and charge balance: Mg → Mg2+ + 2e− is neutral on the left and 2+ plus 2− on the right.
Section 3 of 8Writing an ionic equation
1
Write the balanced symbol equation with state symbols.
2
Split every aqueous ionic compound into its ions. Leave solids, liquids and gases whole.
3
Cross out the ions that appear unchanged on both sides, the spectator ions.
4
Write what is left as the ionic equation.
5
Check that atoms and charges balance.
Section 4 of 8Worked example: silver nitrate and sodium chloride
| Stage | Equation |
|---|---|
| Full equation | AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) |
| Split into ions | Ag+ + NO3− + Na+ + Cl− → AgCl(s) + Na+ + NO3− |
| Remove spectators | Na+ and NO3− |
| Ionic equation | Ag+(aq) + Cl−(aq) → AgCl(s) |
Section 5 of 8More ionic equations
| Reaction | Ionic equation | Observation |
|---|---|---|
| Neutralisation | H+(aq) + OH−(aq) → H2O(l) | Solution becomes neutral |
| Barium sulfate precipitate | Ba2+(aq) + SO42−(aq) → BaSO4(s) | White precipitate forms |
| Copper hydroxide precipitate | Cu2+(aq) + 2OH−(aq) → Cu(OH)2(s) | Blue precipitate forms |
Section 6 of 8Electrolysis half equations
Electrodes
Cathode
- Cathode Reduction: Cu2+ + 2e− → Cu, or 2H+ + 2e− → H2.
- Positive ions are attracted to the negative cathode, where they gain electrons.
Anode
- Anode Oxidation: 2Cl− → Cl2 + 2e−, or 4OH− → O2 + 2H2O + 4e−.
- Negative ions are attracted to the positive anode, where they lose electrons.
Electron balance
- Electron balance The electrons gained at the cathode equal those lost at the anode, which is why the overall equation has no electrons in it.
Section 7 of 8Key terms
| Term | Meaning |
|---|---|
| Half equation | An equation showing the loss or gain of electrons by one species. |
| Ionic equation | An equation showing only the ions or particles that take part in a reaction. |
| Spectator ion | An ion that is present but unchanged during the reaction. |
| Precipitate | An insoluble solid formed in a reaction between solutions. |
| Electron | A negatively charged particle, written e−. |
| Aqueous | Dissolved in water, shown by the state symbol (aq). |
Section 8 of 8Exam technique
1
Splitting a solid or gas into ions.
2
Not balancing the charge, only the atoms.
3
Putting electrons on the wrong sideelectrons are lost on the product side for positive ions.
4
Leaving spectator ions in the final ionic equation.
5
Forgetting that diatomic elements like Cl2 take 2 electrons.