Electrolysis Of Aqueous Solutions (AQA GCSE Combined Science: Synergy: Physical Sciences): Revision Note

Exam code: 8465

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

The reactivity series of metals
The reactivity series of metals including hydrogen and carbon

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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