Electrolysis Of Aqueous Solutions (AQA GCSE Combined Science: Synergy: Physical Sciences): Revision Note
Exam code: 8465
Written by: Stewart Hird
Updated on
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Electrolysis of an aqueous solution using inert electrodes
Ions present in aqueous solutions
Aqueous solutions will always contain water molecules (H2O)
In the electrolysis of aqueous solutions, the water molecules dissociate producing H+ and OH– ions:
H2O ⇌ H+ + OH–
These ions are also involved in the electrolysis process and their chemistry must be considered
We now have an electrolyte that contains ions from the compound plus ions from the water
Which ions get discharged and at which electrode depends on the relative reactivity of the elements involved
Examiner Tips and Tricks
Note: The concentration of the solution can affect the products of electrolysis, however, this is beyond the scope of this course and you are not expected to know the specific details of this
When answering questions on this topic, it helps if you first write down all of the ions present first. Only then you should start comparing their reactivity and deducing the products formed.
Electrode reactions
What is produced at the positive electrode (anode)?
Negatively charged OH– ions and non-metal ions are attracted to the positive electrode
If halide ions (Cl-, Br-, I-) and OH- are present:
The halide ion is discharged at the anode
The halide ion loses electrons and forms a halogen (chlorine, bromine or iodine)
If no halide ions are present:
The OH- ion is discharged at the anode
The OH- loses electrons and forms oxygen
In both cases, the other negative ion remains in solution
What is produced at the negative electrode (cathode)?
Positively charged H+ and metal ions are attracted to the negative electrode but only one will gain electrons
Either hydrogen gas or the metal will be produced
If the metal is above hydrogen in the reactivity series:
Hydrogen is produced at the cathode, seen as bubbling at the cathode
This is because the more reactive ions will remain in solution, causing the least reactive ion to be discharged
If the metal is below hydrogen in the reactivity series:
The metal deposits on the cathode
The reactivity series

Worked Example
Predict the products formed at each electrode in the electrolysis of magnesium iodide solution.
Answer
1. Work out what ions are attracted to the cathode
Hydrogen (H+) and magnesium (Mg2+)
2. Decide which element will be discharged:
Magnesium is above hydrogen in the reactivity series
The less reactive element is usually formed
Therefore, hydrogen is discharged at the cathode
3. Work out what ions are attracted to the anode
Hydroxide (OH-) and iodide (I-)
4. Decide which element will be discharged:
If a halide is present, the corresponding halogen is formed, otherwise, oxygen is formed
Iodide ions are present
Therefore, iodine is discharged at the anode
Worked Example
Predict the products formed at each electrode in the electrolysis of copper sulfate solution
Answer
1. Work out what ions are attracted to the cathode
Hydrogen (H+) and copper (Cu2+)
2. Decide which element will be discharged:
Copper is below hydrogen in the reactivity series
The less reactive element is usually formed
Therefore, copper is discharged at the cathode
3. Work out what ions are attracted to the anode
Hydroxide (OH-) and sulfate (SO42-)
4. Decide which element will be discharged:
If a halide is present, the corresponding halogen is formed, otherwise, oxygen is formed
No halide ions are present
Therefore, oxygen is discharged at the anode
Determining what gas is produced
The gas produced can be tested to determine its identity
If the gas produced at the cathode burns with a ‘pop’ with a lit splint:
The gas is hydrogen
If the gas produced at the anode relights a glowing splint dipped into the gas
The gas is oxygen
If the gas produced at the anode turns damp blue litmus paper red and is then bleached white
The gas is chlorine
The halogen gases all produce their own colours (bromine is red-brown, chlorine is yellow-green)
Examiner Tips and Tricks
Once you have identified the ions, the next step is to decide towards which electrode will they be drawn and identify the product formed. It helps if you recall the reactivity series.
Half equations for electrolysis of aqueous solutions
Higher Tier Only
In aqueous electrolysis, half equations can be written for the reactions at each electrode
The half equation depends on which ion is discharged
At the cathode (reduction — gain of electrons):
If the metal is below hydrogen in the reactivity series, the metal ion is discharged:
Cu2+ + 2e- → Cu
If the metal is above hydrogen in the reactivity series, hydrogen ions are discharged instead:
2H+ + 2e- → H2
At the anode (oxidation — loss of electrons):
If halide ions are present, the halide is discharged to form a halogen:
2Cl- → Cl2 + 2e-
If no halide ions are present, hydroxide ions are discharged to form oxygen and water:
4OH- → 2H2O + O2 + 4e-
Worked Example
Write the half equations for the electrolysis of aqueous copper sulfate, CuSO4.
Answer
Ions present:
Cu2+
SO42-
H+ (from water)
OH- (from water)
At the cathode:
Cu2+ and H+ attracted
Copper is below hydrogen in the reactivity series
Therefore, Cu2+ is discharged:
Cu2+ + 2e- → Cu
At the anode:
SO42- and OH- attracted
There are no halide ions present
Therefore, OH- is discharged:
4OH- → 2H2O + O2 + 4e-
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