24.1 Electrolysis
- Predict the identities of substances liberated during electrolysis from the state of the electrolyte (molten or aqueous), its position in the electrochemical (redox) series, and its concentration.
- State and apply the relationship F = Le between the Faraday constant, F, the Avogadro constant, L, and the charge on the electron, e.
- Calculate (a) the quantity of charge passed during electrolysis using Q = It, and (b) the mass and/or volume of substance liberated during electrolysis.
- Describe the determination of the value of the Avogadro constant by an electrolytic method.
1What Is Electrolysis?▾
Electrolysis is the decomposition of a compound into its elements by passing an electric current through it. For example, molten zinc chloride is split into zinc and chlorine:
Electrolysis: the decomposition of an ionic compound, when molten or in aqueous solution, by an electric current.
Electrolyte: a molten ionic compound, or an aqueous solution of ions, that is decomposed during electrolysis.
Electrode: a rod of metal or carbon (graphite) that conducts electricity into or out of the electrolyte.
Cathode: the electrode at which reduction occurs. Anode: the electrode at which oxidation occurs.
An electrolytic cell is the electrochemical cell in which electrical energy is used to bring about a chemical reaction. Two electrodes are dipped into the electrolyte and connected to an external d.c. source (a battery). Direct current is essential — it fixes which electrode stays positive and which stays negative throughout, so ions migrate consistently in one direction.
- The electrode joined to the positive terminal of the battery is the anode.
- The electrode joined to the negative terminal of the battery is the cathode.
- When current flows, cations (positive ions) migrate to the cathode and anions (negative ions) migrate to the anode.
At the cathode, ions gain electrons from the electrode → reduction. At the anode, ions lose electrons to the electrode → oxidation. Electrolysis is therefore always a redox process. Memory aid: An Ox, Red Cat — Anode = Oxidation, Reduction = Cathode.
2Electrolysis of a Molten Compound▾
Molten electrolytes are the simplest case: only one type of cation and one type of anion is free to move, so there is no competition between ions at either electrode. Consider molten sodium chloride, NaCl(l).
Molten NaCl consists only of free Na⁺ and Cl⁻ ions: NaCl (l) → Na⁺ (l) + Cl⁻ (l)
At the anode (oxidation): 2Cl⁻ (l) → Cl₂ (g) + 2e⁻
At the cathode (reduction): Na⁺ (l) + e⁻ → Na (l)
Overall, chlorine gas bubbles off at the anode and molten sodium collects at the cathode — exactly reversing the formation of NaCl from its elements.
3Electrolysis of Aqueous Solutions — The Role of Water▾
Electrolysing a solution rather than a melt is more complicated because water is present. Water is itself a very weak electrolyte — it ionises to a tiny extent:
This means more than one type of ion can arrive at each electrode, and there is a choice over which one is actually discharged. For example, electrolysing aqueous sodium chloride:
- At the cathode, both Na⁺ and H⁺ (from water) are attracted and compete to be reduced.
- At the anode, both Cl⁻ and OH⁻ (from water) are attracted and compete to be oxidised.
So which ion actually wins? That’s decided by the electrochemical series and, in borderline cases, by concentration — covered in the next two sections.
4The Electrochemical Series & the Two Golden Rules▾
The electrochemical series ranks equilibria by their standard electrode potential, E°. The more negative the E° value, the further left the equilibrium position lies — i.e. the greater the tendency of the metal to lose electrons and form its ion. The more positive the E° value, the greater the tendency of the ion to gain electrons and form the neutral element.
| Electrode equilibrium | E° / V |
|---|---|
| Li⁺ (aq) + e⁻ ⇌ Li (s) | −3.03 |
| K⁺ (aq) + e⁻ ⇌ K (s) | −2.92 |
| Ca²⁺ (aq) + 2e⁻ ⇌ Ca (s) | −2.87 |
| Na⁺ (aq) + e⁻ ⇌ Na (s) | −2.71 |
| Mg²⁺ (aq) + 2e⁻ ⇌ Mg (s) | −2.37 |
| Al³⁺ (aq) + 3e⁻ ⇌ Al (s) | −1.66 |
| Zn²⁺ (aq) + 2e⁻ ⇌ Zn (s) | −0.76 |
| Fe²⁺ (aq) + 2e⁻ ⇌ Fe (s) | −0.44 |
| Pb²⁺ (aq) + 2e⁻ ⇌ Pb (s) | −0.13 |
| 2H⁺ (aq) + 2e⁻ ⇌ H₂ (g) | 0.00 |
| Cu²⁺ (aq) + 2e⁻ ⇌ Cu (s) | +0.34 |
| Ag⁺ (aq) + e⁻ ⇌ Ag (s) | +0.80 |
| Au³⁺ (aq) + 3e⁻ ⇌ Au (s) | +1.50 |
Lithium (E° = −3.03 V) has almost no tendency to pick up electrons once ionised — it stays as Li⁺. Gold (E° = +1.50 V), by contrast, is reluctant to lose electrons to form ions, but its ions pick up electrons very readily to re-form the metal — which is exactly why gold is so unreactive and Au³⁺ is so easily discharged.
The Two Golden Rules
Electrolysis is simply a competition between ions for discharge at each electrode:
- At the cathode (−), reduction occurs — the ion that is least reactive (most positive E°, i.e. lowest down the reactivity series) is discharged preferentially.
- At the anode (+), oxidation occurs — for anions this is governed by a separate, simpler rule set (Section 5).
REACTIVE LEAST
REACTIVE
5Predicting Products at Each Electrode▾
🔻 At the Cathode (cations compete)
Rule 1 — metals below hydrogen (e.g. Cu, Ag) are discharged as the metal:
Rule 2 — metals above hydrogen (e.g. Na, K, Mg, Ca) are not discharged from aqueous solution. Water is reduced instead, releasing hydrogen gas:
Metals such as Pb²⁺ and Zn²⁺ sit close to hydrogen in the series, so their discharge depends on concentration: concentrated solution → metal deposited; dilute solution → hydrogen produced; intermediate concentrations → a mixture of both.
🔺 At the Anode (anions compete)
Rule 1 — halide ions. If Cl⁻, Br⁻ or I⁻ are present, the corresponding halogen is usually produced:
Rule 2 — other anions. Common oxoanions such as SO₄²⁻ and NO₃⁻ are not discharged. Water is oxidised instead, releasing oxygen gas:
A concentrated chloride solution behaves differently from a dilute one — as the solution becomes more dilute, oxygen is produced more readily than chlorine at the anode.
| Solution | Cathode | Anode |
|---|---|---|
| Concentrated NaCl (aq) | H₂ | Cl₂ |
| Dilute NaCl (aq) | H₂ | Mainly O₂ |
🎯 Quick memory trick
- Cathode: metal below H → metal deposited | metal above H → hydrogen gas
- Anode: halide ions → halogen | sulfate / nitrate / hydroxide → oxygen gas
6Worked Examples▾
Cathode: Cu²⁺ ions are discharged because copper lies below hydrogen in the series.
Copper metal coats the cathode.
Anode: sulfate ions are not discharged (Rule 2 for anions), so water is oxidised instead.
Oxygen gas is produced.
Cathode: Na⁺ is far too reactive to be discharged, so water is reduced instead.
Hydrogen gas is produced.
Anode: in a concentrated solution, chloride ions are discharged preferentially:
In a dilute solution, oxygen is formed instead, since the concentration of Cl⁻ is no longer high enough to compete with OH⁻.
7Faraday’s Constant & Electrolysis Calculations▾
During electrolysis, the amount of substance liberated at an electrode depends directly on the quantity of charge passed through the cell.
Coulomb (C): the amount of charge passed when a current of one ampere flows for one second.
Faraday constant, F: the quantity of electric charge (in coulombs) carried by one mole of electrons, or one mole of singly-charged ions. F = 96 500 C mol⁻¹.
One mole of electrons carries a charge of 96 500 C. This is the charge on a single electron, e = 1.60 × 10⁻¹⁹ C, multiplied by the Avogadro constant, L = 6.02 × 10²³ mol⁻¹:
The charge, Q, passed in a time t by a steady current I is:
One mole of an ion with charge z+ carries a charge of zF. So the number of moles of substance liberated is:
In a cell where zinc is deposited and copper dissolves:
Cathode: Zn²⁺ (aq) + 2e⁻ → Zn (s) | Anode: Cu (s) → Cu²⁺ (aq) + 2e⁻
To deposit one mole of zinc (or dissolve one mole of copper), two moles of electrons must pass through the cell — because each ion carries a charge of 2+.
A current of 2.5 A is passed through aqueous CuSO₄ using copper electrodes for 40 minutes. Calculate the mass of copper deposited. (Ar Cu = 63.5, F = 96 500 C mol⁻¹)
- 1Charge passed: Q = It = 2.5 × (40 × 60) = 6000 C
- 2Moles of electrons: Q ⁄ F = 6000 ⁄ 96 500 = 0.0622 mol
- 3Moles of Cu (Cu²⁺ + 2e⁻ → Cu, so z = 2): 0.0622 ⁄ 2 = 0.0311 mol
- 4Mass of Cu: 0.0311 × 63.5 = 1.98 g
Using the same charge passed above (6000 C) with inert electrodes in dilute H₂SO₄, calculate the volume of hydrogen gas produced at the cathode at r.t.p. (molar gas volume = 24 000 cm³ mol⁻¹ at r.t.p.)
- 1Half-equation: 2H⁺ + 2e⁻ → H₂, so z = 2 per mole of H₂
- 2Moles of electrons: 6000 ⁄ 96 500 = 0.0622 mol
- 3Moles of H₂: 0.0622 ⁄ 2 = 0.0311 mol
- 4Volume: 0.0311 × 24 000 = 746 cm³
Determining the Avogadro constant by electrolysis
Because F = Le, measuring the Faraday constant experimentally gives a route to the Avogadro constant, provided the electronic charge e is already known (from other experiments).
- 1Pass a measured, steady current I for a measured time t through a cell depositing a metal of known charge z+ (e.g. Cu²⁺) at an electrode.
- 2Weigh the electrode before and after to find the mass of metal deposited, then convert to moles using its molar mass.
- 3Since moles = It ⁄ zF, rearrange to find the experimental value of F.
- 4Use L = F ⁄ e to calculate the Avogadro constant.
Exam tips
- Always identify the electrode first — cathode (reduction) or anode (oxidation).
- Check whether the electrolyte is molten or aqueous — molten means no water competition at all.
- Compare ions against hydrogen in the electrochemical series to decide what is discharged at the cathode.
- Remember that concentration mainly affects chlorine production from chloride solutions, and the discharge of borderline metal ions (e.g. Pb²⁺, Zn²⁺).
- Always convert time into seconds before using Q = It.
- z in moles = It ⁄ zF is the charge on the discharged ion — not necessarily the number of electrons transferred per formula unit, so check the half-equation carefully.
Key equations
Quick summary
- Electrolysis decomposes an ionic compound using electrical energy; reduction always occurs at the cathode, oxidation at the anode.
- Molten electrolytes: one cation, one anion — no competition.
- Aqueous electrolytes: water supplies extra H⁺ and OH⁻, so ions must compete for discharge based on their position in the electrochemical series and their concentration.
- Cathode: metals below hydrogen deposit as metal; metals above hydrogen leave water to be reduced to H₂.
- Anode: halides discharge as halogens; other common anions leave water to be oxidised to O₂.
- The Faraday constant (96 500 C mol⁻¹) links charge passed to moles of product via F = Le, Q = It, and moles = It ⁄ zF.