A half equation shows the reaction at one electrode during electrolysis — electrons gained at the cathode (reduction) or lost at the anode (oxidation). Every electrolysis can be split into two half equations. Writing and balancing them links electrode chemistry to the wider concept of oxidation and reduction.
What is a half equation?
A half equation represents only one half of an overall redox (reduction–oxidation) reaction. In electrolysis, the two halves are:
- Cathode half equation — shows reduction: ions gain electrons from the cathode and become neutral atoms (or discharge as a gas).
- Anode half equation — shows oxidation: ions or atoms lose electrons to the anode.
Electrons appear explicitly as e⁻ in a half equation. The number of electrons must balance the charge change of the species involved.
Memory aid: OIL RIG — Oxidation Is Loss (of electrons); Reduction Is Gain (of electrons). The cathode attracts positive ions and gives them electrons (reduction). The anode attracts negative ions or oxidises materials at its surface.
How do you write half equations step by step?
Step 1 — Identify what is discharged at each electrode
Different ions are discharged depending on the electrolyte and the electrode material. For a simple solution of a metal salt:
- Cathode: the positive ion (cation) is reduced — e.g. Cu²⁺, Na⁺, H⁺ (from water)
- Anode: the negative ion (anion) or the electrode material is oxidised — e.g. Cl⁻, OH⁻, O²⁻, or the anode itself (if copper)
Step 2 — Write the species on each side
Write the ion on the left and the product on the right (for cathode: ions → atoms; for anode: atoms/ions → ions/molecules).
Step 3 — Balance atoms
Ensure the same number of each atom appears on both sides.
Step 4 — Balance charge by adding electrons (e⁻)
Add electrons to the side that is more positive (for cathode: left side; for anode: right side) until the total charge on both sides is equal.
Step 5 — Check
Count atoms and charges on both sides — they must match.
Worked examples at the cathode
Example 1 — copper ions being reduced:
Cu²⁺ is a 2+ ion. To become neutral copper, it must gain 2 electrons:
Cu²⁺ (aq) + 2e⁻ → Cu (s)
Check: left side charge = 2+ + 2− = 0; right side charge = 0. ✓
Example 2 — hydrogen ions being reduced (when dilute acid or water is involved):
2H⁺ (aq) + 2e⁻ → H₂ (g)
Check: left 2+ + 2− = 0; right 0. Atoms: 2H each side. ✓
Example 3 — sodium ions (from molten sodium chloride):
Na⁺ (l) + e⁻ → Na (l)
Note: molten sodium chloride produces liquid sodium (not aqueous — the conditions matter).
Worked examples at the anode
Example 1 — chloride ions oxidised (concentrated hydrochloric acid or molten NaCl):
2Cl⁻ (aq) → Cl₂ (g) + 2e⁻
Check: left 2−; right 0 + 2− = 2−. ✓ Atoms: 2Cl each side. ✓
Example 2 — oxygen produced from water (dilute solution, inert anode):
2H₂O (l) → O₂ (g) + 4H⁺ (aq) + 4e⁻
Check: left 0; right 4+ + 4− = 0. ✓ Atoms: 4H and 2O each side. ✓
Example 3 — copper anode dissolving (reactive copper electrode):
Cu (s) → Cu²⁺ (aq) + 2e⁻
Summary table of common half equations
| Electrode | Ion/species | Half equation | State symbol |
|---|---|---|---|
| Cathode | Cu²⁺ | Cu²⁺ + 2e⁻ → Cu | aq → s |
| Cathode | 2H⁺ | 2H⁺ + 2e⁻ → H₂ | aq → g |
| Cathode | Na⁺ (molten) | Na⁺ + e⁻ → Na | l → l |
| Cathode | Al³⁺ (molten) | Al³⁺ + 3e⁻ → Al | l → l |
| Anode | 2Cl⁻ | 2Cl⁻ → Cl₂ + 2e⁻ | aq → g |
| Anode | 2H₂O | 2H₂O → O₂ + 4H⁺ + 4e⁻ | l → g |
| Anode | Cu (reactive) | Cu → Cu²⁺ + 2e⁻ | s → aq |
| Anode | 2O²⁻ (molten) | 2O²⁻ → O₂ + 4e⁻ | l → g |
Which ion is discharged when there is a mixture?
When the electrolyte contains more than one type of ion, the ion actually discharged depends on:
At the cathode (reduction):
- Metal ions are preferred over hydrogen ions, UNLESS the metal is very reactive (above zinc in the reactivity series, e.g. Na, Ca, Mg, Al) — in aqueous solution, H⁺ from water is reduced instead.
- Rule: hydrogen is produced at the cathode when the metal is very reactive; the metal is deposited when it is copper, silver, nickel, or less reactive.
At the anode (oxidation):
- Halide ions (Cl⁻, Br⁻, I⁻) are preferentially oxidised over hydroxide ions/water.
- If no halide is present, water is oxidised and oxygen is produced.
- Rule: if the electrolyte is concentrated enough in Cl⁻, chlorine gas forms; otherwise oxygen forms.
Frequently asked questions
How do I know how many electrons to include in a half equation?
The number of electrons equals the charge on the ion being discharged. Cu²⁺ needs 2e⁻ to become neutral; Al³⁺ needs 3e⁻; Na⁺ needs 1e⁻; Cl⁻ gives up 1e⁻; O²⁻ gives up 2e⁻. After writing the ions and products, count the overall charge on each side, then add e⁻ to the more positive side to balance charges. This always gives the correct number of electrons.
Is the cathode positive or negative?
The cathode is the negative electrode. In electrolysis, a negative electrode attracts positive ions (cations) from the solution and supplies electrons to them — causing reduction. The anode is the positive electrode, which attracts negative ions (anions) and removes electrons from them — causing oxidation. A useful check: Cathode = reduction = gains electrons = negative electrode.
Why does molten NaCl produce sodium at the cathode but aqueous NaCl does not?
In molten NaCl, the only cation present is Na⁺, which is reduced to sodium metal at the cathode. In aqueous NaCl, water molecules are also present and provide H⁺ ions (from the equilibrium: 2H₂O ⇌ H₃O⁺ + OH⁻). Sodium is so reactive (high in the reactivity series) that H⁺ ions are preferentially reduced instead: 2H⁺ + 2e⁻ → H₂. Sodium would react violently with water anyway, so hydrogen gas is produced at the cathode from aqueous NaCl, not sodium metal.
Do half equations always need balancing for both atoms and charge?
Yes — a correct half equation must be balanced for both atoms and overall charge. A common mistake is to balance atoms but forget the charge, or to add the wrong number of electrons. After writing the half equation, always check: (1) count each type of atom on both sides; (2) calculate the total charge on each side, including the electrons. Both must match. If one is wrong, re-examine the equation rather than adding water or H⁺ randomly without justification.
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