How to Write Half-Equations (Edexcel IGCSE Chemistry (Modular))
Revision Note
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–
Examiner Tip
In electrode half equations the charges on each side of the equation should always balance.It may seem odd that water molecules are discharged and not hydroxide ions, but remember that acidic solutions will not contain any hydroxide ions. Even copper(II)sulfate is slightly acidic in water, so will not contain hydroxide ions.
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