24.1 Electrolysis
🎯 Learning Outcomes
Candidates should be able to:
- ✓ Predict the identities of substances liberated during electrolysis, from the state of the electrolyte (molten or aqueous), position in the redox series (electrode potential), and 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:
- the quantity of charge passed during electrolysis, using Q = It
- the mass and/or volume of substance liberated during electrolysis
- ✓ Describe the determination of the value of the Avogadro constant by an electrolytic method
1 ⚡ Introduction to Electrolysis ▶
Every time aluminium is extracted from its ore, or a spoon is silver-plated, or a battery is recharged, the same trick is being used: forcing electricity through a compound to pull it apart into its elements. That’s electrolysis — and it’s one of the most direct, practical demonstrations that chemical bonds are really about the movement of electrons.
Electrolysis is the decomposition of an ionic compound, when molten or dissolved in water, by an electric current. For example, molten zinc chloride can be split into its elements:
⚙️ The Electrolytic Cell
- An electrochemical cell in which electrical energy is used to bring about a chemical reaction is called an electrolytic cell.
- Two electrodes are dipped into the electrolyte and connected to an external d.c. source, such as a battery.
- The electrode connected to the positive terminal is the anode; the electrode connected to the negative terminal is the cathode.
- When current flows, cations migrate towards the cathode and anions migrate towards the anode.
Reactions involved:
Reduction always occurs at the cathode (ions gain electrons from it); oxidation always occurs at the anode (ions lose electrons to it). Electrolysis is a redox reaction.
2 💧 Water & the Electrochemical Series ▶
💧 The Role of Water in Aqueous Electrolysis
Electrolysing a solution is more complicated than electrolysing a melt, because of the presence of water. Water is itself a very weak electrolyte — it splits, to a tiny extent, into hydrogen ions and hydroxide ions. This means more than one ion can arrive at each electrode, and there’s a choice over which one is actually discharged.
For example, if you electrolyse sodium chloride solution: sodium ions and hydrogen ions (from the water) are both attracted to the cathode, while chloride ions and hydroxide ions (from the water) are both attracted to the anode.
📊 The Electrochemical Series
The table below lists some metals (and hydrogen) by their tendency to lose electrons. The more negative the E⦵ value, the further the position of equilibrium lies to the left — i.e. the greater the tendency of that element to lose electrons and form its ion.
| 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 |
- The more negative the E⦵ value, the greater the tendency to lose electrons and form ions.
- Something like lithium has little tendency to pick up electrons once it has ionised.
- Something with a positive E⦵ value is reluctant to lose electrons, but its ion picks up electrons easily to re-form the neutral element.
That’s why gold isn’t very reactive: its very positive E⦵ value means it’s hard to remove electrons to make gold ions, but easy to convert gold ions back to gold metal. The electrochemical series can be thought of as an extended, and slightly modified, reactivity series.
📋 The Full Standard Electrode Potential Series
| Electrode reaction | E⦵ / V |
|---|---|
| F₂ + 2e⁻ ⇌ 2F⁻ | +2.87 |
| S₂O₈²⁻ + 2e⁻ ⇌ 2SO₄²⁻ | +2.01 |
| H₂O₂ + 2H⁺ + 2e⁻ ⇌ 2H₂O | +1.77 |
| MnO₄⁻ + 8H⁺ + 5e⁻ ⇌ Mn²⁺ + 4H₂O | +1.52 |
| PbO₂ + 4H⁺ + 2e⁻ ⇌ Pb²⁺ + 2H₂O | +1.47 |
| Cl₂ + 2e⁻ ⇌ 2Cl⁻ | +1.36 |
| Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O | +1.33 |
| O₂ + 4H⁺ + 4e⁻ ⇌ 2H₂O | +1.23 |
| Br₂ + 2e⁻ ⇌ 2Br⁻ | +1.07 |
| ClO⁻ + H₂O + 2e⁻ ⇌ Cl⁻ + 2OH⁻ | +0.89 |
| NO₃⁻ + 10H⁺ + 8e⁻ ⇌ NH₄⁺ + 3H₂O | +0.87 |
| NO₃⁻ + 2H⁺ + e⁻ ⇌ NO₂ + H₂O | +0.81 |
| Ag⁺ + e⁻ ⇌ Ag | +0.80 |
| Fe³⁺ + e⁻ ⇌ Fe²⁺ | +0.77 |
| I₂ + 2e⁻ ⇌ 2I⁻ | +0.54 |
| O₂ + 2H₂O + 4e⁻ ⇌ 4OH⁻ | +0.40 |
| Cu²⁺ + 2e⁻ ⇌ Cu | +0.34 |
| SO₄²⁻ + 4H⁺ + 2e⁻ ⇌ SO₂ + 2H₂O | +0.17 |
| Sn⁴⁺ + 2e⁻ ⇌ Sn²⁺ | +0.15 |
| S₄O₆²⁻ + 2e⁻ ⇌ 2S₂O₃²⁻ | +0.09 |
| 2H⁺ + 2e⁻ ⇌ H₂ | 0.00 |
| Pb²⁺ + 2e⁻ ⇌ Pb | −0.13 |
| Sn²⁺ + 2e⁻ ⇌ Sn | −0.14 |
| Fe²⁺ + 2e⁻ ⇌ Fe | −0.44 |
| Zn²⁺ + 2e⁻ ⇌ Zn | −0.76 |
| 2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻ | −0.83 |
| V²⁺ + 2e⁻ ⇌ V | −1.20 |
| Mg²⁺ + 2e⁻ ⇌ Mg | −2.38 |
| Ca²⁺ + 2e⁻ ⇌ Ca | −2.87 |
| K⁺ + e⁻ ⇌ K | −2.92 |
“Please Stop Calling Me A Careless Zebra Instead Try Learning How Copper Saves Gold”
| Mnemonic | Element |
|---|---|
| Please | Potassium |
| Stop | Sodium |
| Calling | Calcium |
| Me | Magnesium |
| A | Aluminium |
| Careless | (Carbon) |
| Zebra | Zinc |
| Instead | Iron |
| Try | Tin |
| Learning | Lead |
| How | (Hydrogen) |
| Copper | Copper |
| Saves | Silver |
| Gold | Gold |
↑ Most reactive Least reactive ↓
3 🚦 Predicting the Products: Two Golden Rules ▶
Electrolysis is simply a competition between ions — the products formed depend on which ions are discharged more easily.
🔽 What Happens at the Cathode?
Positive ions (cations) move towards the negative electrode.
Rule 1 — Metals below hydrogen (e.g. Cu, Ag) are discharged. Metal is deposited.
Rule 2 — Metals above hydrogen (e.g. Na, K, Mg, Ca) are not discharged from aqueous solution. Water is reduced instead, producing hydrogen gas.
Metals such as Pb²⁺ and Zn²⁺ sit close to hydrogen in the electrochemical series — their discharge depends on concentration. Concentrated solution → metal deposited. Dilute solution → hydrogen produced. Intermediate concentrations → both may form.
🔼 What Happens at the Anode?
Negative ions (anions) move towards the positive electrode.
Rule 1 — Halide ions. If Cl⁻, Br⁻ or I⁻ are present, the corresponding halogen is usually produced.
Rule 2 — Other anions such as SO₄²⁻ and NO₃⁻ are not discharged. Water is oxidised instead, producing oxygen gas.
A concentrated sodium chloride solution behaves differently from a dilute one:
| Solution | Cathode | Anode |
|---|---|---|
| Concentrated NaCl | H₂ | Cl₂ |
| Dilute NaCl | H₂ | Mainly O₂ |
As the solution becomes more dilute, oxygen is produced more readily than chlorine.
4 🧪 Worked Examples ▶
Cathode: Cu²⁺ ions are discharged (copper is below hydrogen).
✔ Copper coats the cathode.
Anode: sulfate ions are not discharged; water is oxidised instead.
✔ Oxygen gas is produced.
Cathode: Na⁺ is too reactive to be discharged; water is reduced instead.
✔ Hydrogen gas is produced.
Anode: in concentrated solution:
✔ Chlorine gas is produced (in dilute solution, oxygen forms instead).
Cathode (reduction): metal below hydrogen → metal deposited. Metal above hydrogen → hydrogen gas.
Anode (oxidation): halide ions → halogen. Sulfate, nitrate, hydroxide/water → oxygen.
✔ Always identify the electrode first (cathode or anode). ✔ Check whether the electrolyte is molten or aqueous. ✔ Compare the ions with hydrogen in the electrochemical series. ✔ Remember that concentration mainly affects chlorine production from chloride solutions, and some borderline metal ions.
5 🧮 Calculations: F = Le and Q = It ▶
During electrolysis of a Zn/Cu cell, current is forced through in the non-spontaneous direction:
To deposit one mole of zinc (or dissolve one mole of copper), two moles of electrons must pass through the cell. One mole of electrons carries a charge of 96 500 C — the same charge as one electron (1.602 × 10⁻¹⁹ C) multiplied by the Avogadro constant (6.022 × 10²³ mol⁻¹). This quantity is called the Faraday constant, F.
where F = the Faraday constant, L = the Avogadro constant, e = the charge on the electron.
The coulomb (C) is the unit of electrical charge — the charge passed when one ampere flows for one second. The charge passed, Q, in time t is:
The charge on one mole of z⁺ ions is zF, so the number of moles deposited or dissolved is:
where z is the charge on the ion, I is the current in amperes, and t is the time in seconds.
A current of 2.00 A is passed through aqueous CuSO₄ for 30.0 minutes using inert electrodes. Calculate the mass of copper deposited at the cathode. (Cu²⁺ + 2e⁻ → Cu, so z = 2; Ar(Cu) = 63.5)
Q: One Faraday of electricity is equivalent to which quantity of charge? A: 96 500 coulombs.
Q: In an electrolytic cell, where does oxidation take place? A: At the anode.
6 🔍 Determining the Avogadro Constant ▶
Because F = Le, an electrolysis experiment that measures the Faraday constant — combined with an independently known value for the charge on the electron, e — gives a direct experimental route to the Avogadro constant, L.
- Set up an electrolytic cell with copper electrodes in aqueous copper(II) sulfate (or silver electrodes in silver nitrate), connected in series with an ammeter.
- Clean and weigh the cathode accurately before starting.
- Pass a steady, measured current, I, for an accurately measured time, t.
- Remove, wash, dry, and re-weigh the cathode to find the mass of metal deposited.
- Convert this mass to moles of metal deposited, n, using its molar mass.
- Calculate the charge passed: Q = It.
- Calculate the Faraday constant: F = Q / (z × n), where z is the charge on the metal ion.
- Using the independently known charge on the electron, e, calculate L = F / e.
This method links a macroscopic, measurable quantity (mass of metal deposited, current, time) to a microscopic constant (the number of particles in a mole) — a beautiful example of how careful quantitative experiments can reveal the scale of the atomic world. With modern apparatus, this method gives a value close to the accepted L = 6.022 × 10²³ mol⁻¹.