Half Equations & Ionic Equations (AQA GCSE Combined Science: Trilogy): Revision Note
Exam code: 8464
Half equations & ionic equations
Higher tier only
What is a half equation?
A half equation shows what happens to one species during a redox reaction
Each half equation shows:
Atoms balanced
Charges balanced
The gain or loss of electrons
You need two half equations to make a full redox equation
How to write a half equation
Steps to write a half equation:
Identify the species involved, e.g. a metal ion being reduced or a non-metal ion being oxidised
Write the correct formula and charge for the species before and after the change
Balance the atoms on both sides of the equation
Add electrons (e⁻) to one side to balance the charges
Check that both atoms and charges are balanced
Examples of half equations
Copper(II) ions gains electrons to form copper metal
The species involved are:
Copper(II) ions as a reactant
Copper as a product
The formulas and charges are:
Copper(II) ions = Cu2+
Copper = Cu
Balancing the atoms on both sides of the equation gives:
Cu2+ → Cu
Two electrons are required, on the reactant side, to balance the equation:
Cu2+ + 2e- → Cu
Check that the atoms and charges are balanced:
Atoms
Each side has one Cu
Charges
A 2+ charge on the copper(II) ion
A 1- charge on each electron = 2- charge overall
The 2+ and 2- charges cancel
Bromide ions lose electrons to form bromine
The species involved are:
Bromide ions as a reactant
Bromine as a product
The formulas and charges are:
Bromide ions = Br-
Bromine = Br2
Balancing the atoms on both sides of the equation gives:
2Br- → Br2
Two electrons are required to balance the equation:
2Br- - 2e- → Br2
This is more commonly written as:
2Br- → Br2 + 2e-
Check that the atoms and charges are balanced:
Atoms
Each side has two Br
Charges
Bromide ions each have a 1− charge
So, two bromide ions give a total charge of 2−
Two electrons (1− each) are released to balance this
Examiner Tips and Tricks
For half equations:
Electrons on the right
The species is losing electrons (oxidation)
Electrons on the left
The species is gaining electrons (reduction)
This links to the mnemonic OIL RIG
Oxidation Is Loss
Reduction Is Gain
What is an ionic equation?
An ionic equation shows only the particles that change during a reaction
Two half equations can be combined to form an ionic equation
They are useful for reactions in solution such as:
Neutralisation
Precipitation
How to write an ionic equation
Steps to write an ionic equation:
Write the full balanced chemical equation
Split aqueous compounds into ions
Only aqueous substances are split into ions
Solids, liquids, and gases stay in their molecular form
Write the ionic form of the equation
Cancel the spectator ions
Spectator ions do not take part in the chemical change
Write the final ionic equation
Ionic equations must be balanced for both atoms and charge
Examples of ionic equations
The neutralisation reaction of hydrochloric acid and sodium hydroxide
The full balanced chemical equation is:
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
The aqueous ions are:
HCl → H+ and Cl-
NaOH → Na+ and OH-
NaCl → Na+ and Cl-
H2O stays as a molecule
The ionic form of the equation is:
H+ + Cl- + Na+ + OH- → Na+ + Cl- + H2O
Cancelling out the spectator ions:
Na+ and Cl- are spectator ions because they appear unchanged on both sides
H+
+ Cl-+ Na++ OH- →Na++ Cl-+ H2O
The final ionic equation is:
H+ (aq) + OH- (aq) → H2O (l)
Worked Example
The precipitation reaction of lead(II) nitrate and potassium iodide.
A student mixes aqueous solutions of lead(II) nitrate and potassium iodide. A yellow precipitate of lead(II) iodide forms.
Write the ionic equation for the formation of the yellow precipitate.
Answer:
The full balanced chemical equation is:
Pb(NO3)2 (aq) + 2KI (aq) → PbI2 (s) + 2KNO3 (aq)
The aqueous ions are:
Pb(NO3)2 → Pb2+ and 2NO3-
KI → K+ and I-
KNO3 → K+ and NO3-
PbI2 is a solid, so it remains as PbI2 (s)
The ionic form of the equation is:
Pb2+ + 2NO3- + 2K+ + 2I- → PbI2 + 2K+ + 2NO3-
Cancelling out the spectator ions:
2NO3- and 2K+ are unchanged on both sides
The final ionic equation is:
Pb2+ + 2I- → PbI2 (s)
Half equations and ionic equations feature in:
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