Half Equations in Electrolysis (AQA GCSE Chemistry: Combined Science)

Higher tier only

• In electrochemistry we are mostly concerned with the transfer of electrons, hence the definitions of oxidation and reduction are applied in terms of electron loss or gain rather than the addition or removal of oxygen
• Oxidation is when a substance loses electrons and reduction is when a substance gains electrons
• As the ions come into contact with the electrode, electrons are either lost or gained and they form neutral substances
• These are then discharged as products at the electrodes
• At the anode, negatively charged ions lose electrons and are thus oxidised
• At the cathode, the positively charged ions gain electrons and are thus reduced
• This can be illustrated using half equations which describe the movement of electrons at each electrode
• It is important to make sure that the charges as well as the number of atoms/ions on each side of the equation are balanced 

Electrolysis of molten lead(II)bromide

• In the electrolysis of molten lead(II) bromide the half equation at the negative electrode (cathode) is:

Pb²⁺ + 2e^– ⟶ Pb

• At the positive electrode (anode) bromine gas is produced by the discharge of bromide ions:

2Br^- – 2e^– ⟶ Br₂

OR

2Br^- ⟶ Br₂ + 2e^–

Electrolysis of molten aluminium oxide

• Aluminium ions are discharged at the negative electrode (cathode) and the aluminium is collected at the bottom of the cell:

Al³⁺ + 3e^– ⟶ Al

• At the positive electrode (anode) oxygen gas is produced:

2O^2- – 4e^– ⟶ O₂

OR

2O^2- ⟶ O₂  + 4e^–

Electrolysis of aqueous solutions

• At the negative electrode (cathode), when the metal is more reactive than hydrogen, hydrogen is discharged and the half equation is:

2H⁺ + 2e^– ⟶ H₂

• When the metal is less reactive than hydrogen, the metal is discharged, e.g.:

Cu²⁺ +  2e^– ⟶ Cu   

• At the positive electrode (anode), if a halide ion is present, the corresponding halogen is formed e.g.:

2Cl^– – 2e^– ⟶ Cl₂

OR

2Cl^– ⟶ Cl₂ + 2e^–

• When a halide ion is not present, oxygen is formed as hydroxide ions are discharged, e.g.:

4OH^– ⟶ O₂ + 2H₂O + 4e^–

OR

4OH^– - 4e^– ⟶ O₂ + 2H₂O 

Summary of the half equations at the anode and cathode

Half equations at electrodes
Electrolyte	Anode reaction	Cathode reaction
Copper(II) chloride (CuCl2)	2Cl– → Cl2 + 2e–	Cu2+ + 2e–→ Cu
Sodium chloride (NaCl)	2Cl– → Cl2 + 2e–	2H+ + 2e– → H2
Copper(II) sulftate (CuSO4)	4OH– → O2 + 2H2O + 4e–	Cu2+ + 2e–→ Cu
Sodium sulfate (Na2SO4)	4OH– → O2 + 2H2O + 4e–	2H+ + 2e– → H2
Concentrated hydrochloric acid (HCl)	2Cl– → Cl2 + 2e–	2H+ + 2e– → H2
Water acidified with sulfuric acid (H2SO4)	4OH– → O2 + 2H2O + 4e–	2H+ + 2e– → H2