Writing Electrolysis Half Equations
This topic covers how to represent the reactions at the electrodes during electrolysis using half-equations, which show the specific gain or loss of electrons for each substance.
Part of the ESAT Chemistry syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- A half-equation represents either oxidation (electron loss) or reduction (electron gain) at one electrode.
- At the anode (positive electrode), negative ions (anions) are oxidised, so electrons (e-) appear on the product (right) side of the equation.
- At the cathode (negative electrode), positive ions (cations) are reduced, so electrons (e-) appear on the reactant (left) side of the equation.
- Half-equations must be balanced for both atoms and total charge. The number of electrons is chosen to make the total charge on both sides equal.
- For aqueous solutions, remember that water can be oxidised to produce O2 and H+ ions, or reduced to produce H2 and OH- ions.
Formulae
Anion → Product(s) + n e- General form for an oxidation half-equation at the anode.
Cation + n e- → Product General form for a reduction half-equation at the cathode.
Definitions
- Half-equation
- An equation showing either the oxidation or reduction part of a redox reaction, explicitly including the electrons (e-) that are lost or gained.
- Oxidation
- The loss of electrons by a species, which occurs at the anode during electrolysis. Remember OIL: Oxidation Is Loss.
- Reduction
- The gain of electrons by a species, which occurs at the cathode during electrolysis. Remember RIG: Reduction Is Gain.
Worked example
Write the half-equations for the reactions occurring at the anode and the cathode during the electrolysis of molten lead(II) bromide (PbBr2) using inert graphite electrodes.
- 1
Identify the substance and its state.
'Molten' means there is no water, only Pb2+ and Br^- ions are present and mobile.
- 2
Identify the electrodes and the ions attracted to them.
The cathode is negative and attracts the positive Pb2+ cations.
The anode is positive and attracts the negative Br^- anions.
- 3
Write the half-equation for the cathode.
Pb2+ ions gain electrons (reduction) to form lead metal.
To balance the 2+ charge, two electrons are needed:
Pb2+ + 2e- → Pb.
- 4
Write the half-equation for the anode.
Br^- ions lose electrons (oxidation) to form bromine.
Bromine is diatomic (Br2), so two bromide ions are needed.
This releases two electrons:
2Br- → Br2 + 2e-.
- 5
Check that both equations are balanced for atoms and charge.
Answer: Cathode: Pb2+ + 2e- → Pb(l); Anode: 2Br- → Br2(g) + 2e-
Common mistakes
- ×Forgetting to balance equations for diatomic elements (like H2, O2, Cl2, Br2). For example, writing Cl- → Cl + e- instead of the correct 2Cl- → Cl2 + 2e-. This 'factor of 2' error is very common.
- ×Incorrectly balancing the charge. The number of electrons must make the total charge on both sides of the arrow equal. For Al3+, three electrons are needed for reduction (Al3+ + 3e- → Al), not one or two.
- ×Mixing up oxidation and reduction equations. Remember electrons are a product (lost) in oxidation at the anode, and a reactant (gained) in reduction at the cathode.
No-calculator tips
- ✓Always do a final 'charge check' on your half-equation. Sum the charges of all species on the left and check it equals the sum on the right. For example, in 2Br- → Br2 + 2e-, the left side is 2*(-1) = -2, and the right side is 0 + 2*(-1) = -2. They match.
- ✓Use the mnemonic 'PANIC': Positive Anode, Negative Is Cathode to remember the electrode polarities in electrolysis.
- ✓For diatomic products like chlorine (Cl2), start by writing the product, then work backwards to balance the atoms (2Cl-) and finally the charge (2e-).