The big idea: A voltaic cell lets a spontaneous redox reaction push electrons round a wire. Electrolysis does the opposite: an external power supply forces electrons through a liquid to drive a non-spontaneous redox reaction.
The liquid that conducts is the electrolyte — it must contain mobile ions, so it is either a molten ionic compound or an ionic solution. The two metal/graphite rods dipping into it are the electrodes.
The key wiring fact: the power supply makes one electrode negative (the cathode) and the other positive (the anode).
Which ion goes where: Opposite charges attract, so:
- Cations (positive ions) move to the negative cathode, where they are reduced (gain electrons). - Anions (negative ions) move to the positive anode, where they are oxidised (lose electrons).
A memory hook: PANIC — Positive ANode Is where oxidation happens; reduction is at the Cathode.
For a molten binary salt (just one metal + one non-metal, with no water) the outcome is simple: the metal forms at the cathode and the non-metal forms at the anode.
Electrolysis of molten lead(II) bromide: the external power supply makes the left electrode the cathode (−, reduction) and the right the anode (+, oxidation). Pb²⁺ migrates to the cathode and is deposited as lead; Br⁻ migrates to the anode and is evolved as bromine gas.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
| Electrode | Sign (powered cell) | What happens | Example: molten NaCl |
|---|---|---|---|
| Cathode | Negative (−) | Cations attracted here are REDUCED (gain e⁻) | Na⁺(l) + e⁻ → Na(l) |
| Anode | Positive (+) | Anions attracted here are OXIDISED (lose e⁻) | 2Cl⁻(l) → Cl₂(g) + 2e⁻ |
Worked example — electrolysis of molten lead(II) bromide
Molten lead(II) bromide, PbBr2(l), is electrolysed with inert graphite electrodes. Write the half-equation at each electrode and name the product.
Solution
- Identify the ions present in the melt:
- Cathode (−): the cation Pb²⁺ is reduced — add electrons to balance the 2+ charge:
- Anode (+): the anion Br⁻ is oxidised — two are needed to make one Br2 molecule:
Final answer
Cathode: lead metal, Pb(l). Anode: bromine gas, Br₂(g). The electrons lost at the anode equal those gained at the cathode (2 each).
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When the electrolyte is a solution, water is present as well as the dissolved ions — and water itself can be reduced or oxidised. So at each electrode there is a competition, and only one species is discharged.
The two reactions water can undergo are:
- water — present in every aqueous electrolyte
- hydrogen gas (cathode product when water is reduced)
- oxygen gas (anode product when water is oxidised)
Three factors decide what is discharged: 1. Position in the electrochemical (E°) series. The species that is more easily reduced (more positive E°) is discharged at the cathode; the species more easily oxidised at the anode. So an unreactive metal ion (Cu²⁺, Ag⁺) is deposited as metal, but a reactive metal ion (Na⁺, K⁺, Ca²⁺) is not — water is reduced to H2 instead.
2. Ion concentration. A high concentration of an ion shifts the competition in its favour. A concentrated halide solution releases the halogen at the anode even though E° alone slightly favours O2.
3. Electrode material and overpotential. Inert electrodes (graphite, platinum) just carry the current. A reactive electrode can dissolve instead (e.g. a copper anode goes into solution as Cu²⁺). The extra voltage needed to release a gas — the overpotential — also helps the halide win over O2 in practice.
| Electrode | Competing species | Usually discharged | Why |
|---|---|---|---|
| Cathode (reduction) | metal cation vs H₂O | H₂(g) if the metal is reactive (e.g. Na⁺, K⁺, Ca²⁺); the METAL if it is below H in the E° series (e.g. Cu²⁺, Ag⁺) | the species more easily reduced (more positive E°) wins |
| Anode (oxidation) | non-metal anion vs H₂O | O₂(g) from water for sulfates/nitrates; the HALIDE (Cl₂, Br₂, I₂) when its ions are concentrated | E° order favours O₂, but high halide concentration + overpotential favour the halide |
Worked example — electrolysis of concentrated aqueous potassium iodide
Concentrated aqueous potassium iodide, KI(aq), is electrolysed with inert electrodes. Predict the product at each electrode and write the half-equations.
Solution
- List what is present: the ions K⁺ and I⁻, plus water.
- Cathode: K⁺ is a reactive metal ion (very negative E°), so water is reduced instead, giving H2:
- Anode: the iodide is concentrated, so the halide is discharged in preference to O2:
Final answer
Cathode: hydrogen gas, H₂ (K⁺ is too reactive to be discharged). Anode: iodine, I₂ (favoured because the iodide is concentrated). The solution near the cathode becomes alkaline (OH⁻).
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Electrolysis becomes quantitative when we count the electrons that flow. The total charge passed is the current multiplied by the time it flows.
- charge passed (coulombs, C)
- current (amperes, A)
- time the current flows (seconds, s)
Use SI units — seconds, not minutes: Q = It only works if the current is in amperes (A) and the time is in seconds (s). If a question gives a time in minutes or hours, convert to seconds first (× 60, or × 3600). The charge then comes out in coulombs (C).
One mole of electrons carries a fixed charge — Faraday's constant, F = 96 500 C mol⁻¹ (given in the data booklet). Dividing the charge by F tells you how many moles of electrons have passed.
- amount of electrons transferred (mol)
- charge passed (C)
- Faraday's constant = 96 500 C mol⁻¹
Worked example — moles of electrons from current and time
A current of 1.50 A is passed through a cell for 20.0 minutes. Calculate the amount, in moles, of electrons transferred.
Solution
- Convert the time to seconds:
- Formula first — find the charge:
- Formula first — convert charge to moles of electrons:
- Work it out:
Final answer
n(e⁻) = 0.0187 mol (3 s.f.).
How this is tested: On Paper 2 (HL) this is a multi-step calculation chain. You are given a current and a time and asked for the mass of metal deposited (or the volume of gas released) at an electrode. The route is always the same:
Q = It → n(e⁻) = Q/F → use the half-equation ratio → moles of product → mass (or volume).
The step examiners reward most is using the electrode half-equation to convert moles of electrons into moles of product — e.g. Cu²⁺ + 2e⁻ → Cu means 2 mol e⁻ make 1 mol Cu, so you divide by 2.
Sure marks: Convert time to seconds first. Write Q = It and n(e⁻) = Q/F before substituting. Then read the electron : product ratio straight off the balanced half-equation — that ratio is where most marks are lost. Finish with m = nM and the correct unit and significant figures.
IB-style question — copper deposited in electroplating (a)
In an electroplating cell, a current of 2.00 A is passed through copper(II) sulfate solution for 30.0 minutes. The cathode half-equation is Cu²⁺(aq) + 2e⁻ → Cu(s). (a) Calculate the amount, in moles, of electrons that pass through the cell.
Solution
- Convert the time to seconds:
- Formula first — charge passed:
- Formula first — moles of electrons:
Final answer
n(e⁻) = 0.0373 mol (3 s.f.).
IB-style question — copper deposited in electroplating (b)
(b) Using the half-equation Cu²⁺(aq) + 2e⁻ → Cu(s), calculate the mass of copper deposited at the cathode. (M(Cu) = 63.55 g mol⁻¹.)
Solution
- Read the ratio from the half-equation — 2 mol e⁻ deposit 1 mol Cu, so divide by 2:
- Formula first — mass from amount:
- Substitute and solve:
Final answer
m(Cu) = 1.19 g (3 s.f.).