How to Write Half-Equations (Edexcel IGCSE Chemistry (Modular))

Electrolysis half equations

• 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

• 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

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         Reduction

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

2Br^– – 2e^– ⟶ Br₂          Oxidation

OR

2Br^– ⟶ Br₂  + 2e^–

Electrolysis of aqueous sodium chloride

• In the electrolysis of aqueous sodium chloride the half equation at the negative electrode (cathode) is:

2H⁺ + 2e^– ⟶ H₂           Reduction

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

2Cl^– – 2e^– ⟶ Cl₂          Oxidation

OR

2Cl^– ⟶ Cl₂  + 2e^–

Electrolysis of dilute sulfuric acid

• In the electrolysis of dilute sulfuric acid the half equation at the negative electrode (cathode) is:

2H⁺ + 2e^– ⟶ H₂                     Reduction

• At the positive electrode (anode) oxygen gas is produced by the discharge of water molecules:

2H₂O – 4e^– ⟶ O₂  +  4H⁺    Oxidation

OR

2H₂O  ⟶ O₂  +  4H⁺ + 4e^– 

Electrolysis of aqueous copper(II) sulfate

• In the electrolysis of aqueous copper(II) sulfate the half equation at the negative electrode (cathode) is:

Cu²⁺ + 2e^– ⟶ Cu                Reduction

• At the positive electrode (anode) oxygen gas is produced by the discharge of water molecules:

2H₂O – 4e^– ⟶ O₂  +  4H⁺    Oxidation

OR

2H₂O  ⟶ O₂  +  4H⁺ + 4e^–