Lesson 6 of 7

Electrolysis products and calculations

Choose a half-reaction, then convert charge into yield.

A-Level 9476 (2026-2027)

I can recognise limits of standard-potential predictions.Syllabus
Syllabus 9476, 12(g). Recognise limits of standard-potential predictions.
I can relate Faraday, Avogadro and electron-charge constants.Syllabus
Syllabus 9476, 12(l). Relate Faraday, Avogadro and electron-charge constants.
I can predict electrolytic products.Syllabus
Syllabus 9476, 12(m). Predict electrolytic products.
I can calculate charge and electrolysis yield.Syllabus
Syllabus 9476, 12(n). Calculate charge and electrolysis yield.

Choose the products from the species and conditions

An external supply drives reduction at the cathode and oxidation at the anode.

Make a guess. You are not marked.

Aqueous copper(II) sulfate is electrolysed with inert graphite electrodes. What forms at the anode?

Sulfur dioxide, from sulfate ions
Oxygen gas, from oxidising water
Copper metal, from copper(II) ions

Show the answer

Oxygen gas, from oxidising water

Sulfate ions are very hard to oxidise, so water is oxidised instead: 2H₂O -> O₂ + 4H⁺ + 4e^-.

Electrolysis uses electrical energy to drive a reaction. The supply delivers electrons to the negative cathode and removes them from the positive anode. Cathode still means reduction and anode still means oxidation: those definitions do not change when the electrode signs differ from a galvanic cell.

Predict an electrolytic product

List the actual species
A molten salt contains its ions; an aqueous electrolyte also contains water and its acid-base species.
Check the electrode
An inert electrode provides a surface. An active metal anode may itself oxidise.
Compare plausible half-reactions
At the cathode compare reductions; at the anode compare reversed reduction equations. Use potentials under the relevant conditions.
Account for concentration and kinetics
Competing discharge depends on ion concentration, pH and electrode surface. Standard potentials alone can be insufficient.

Representative products with their conditions
Electrolyte and electrodes	Cathode reduction	Anode oxidation
Molten NaCl; inert electrodes	Na⁺(l) + e^- → Na(l).	2Cl^-(l) → Cl₂(g) + 2e^-.
Aqueous CuSO₄; inert electrodes	Cu²⁺(aq) + 2e^- → Cu(s), while sufficient Cu²⁺ remains.	2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e^-.
Dilute aqueous NaCl; suitable inert electrodes	2H₂O(l) + 2e^- → H₂(g) + 2OH^-(aq).	Oxygen is favoured in the usual dilute-solution model; chloride oxidation can compete as conditions change.
Concentrated aqueous NaCl; suitable inert electrodes	Water gives H₂; sodium metal is not deposited.	2Cl^-(aq) → Cl₂(g) + 2e^- is favoured under the usual concentrated-brine conditions.
Aqueous CuSO₄; copper electrodes	Cu²⁺(aq) + 2e^- → Cu(s).	Cu(s) → Cu²⁺(aq) + 2e^-; the anode dissolves.

In aqueous sodium chloride, water reduction is much easier than Na⁺ reduction, so hydrogen forms at the cathode. At an inert anode, water/oxygen and chloride/chlorine compete. Increasing chloride concentration favours chloride oxidation; oxygen formation also has a kinetic barrier at many electrode surfaces. This is why a simple comparison of the standard +1.23 V oxygen and +1.36 V chlorine reduction values cannot, by itself, correctly predict every brine experiment.

Electrolysis can change its own conditions. As Cu²⁺ is depleted near an inert cathode, hydrogen evolution can become important; changes in concentration and pH may alter the products. An electrolyte conducts because its ions move, not because electrons pass through the bulk liquid.

Check your understanding

Why does changing the anode from platinum to copper change the products in aqueous CuSO₄?

Copper is less reactive than platinum, so sulfate ions are oxidised at its surface instead.
Copper is a better conductor than platinum, so it makes the oxygen form at the cathode instead.
Copper is oxidised more readily than water, so the copper anode dissolves instead of forming O₂.

Show the answer

Copper is oxidised more readily than water, so the copper anode dissolves instead of forming O₂.

Platinum is inert, so water is oxidised and oxygen forms. A copper anode offers Cu -> Cu²⁺ + 2e^-, which happens more easily. The electrode material can decide the product.

Turn charge into product through the half-equation

One mole of electrons carries one Faraday of charge.

The Faraday constant F is the charge per mole of electrons: F = Le, where L is the Avogadro constant and e is the magnitude of the charge on one electron. The Data Booklet uses L = 6.02 × 10²³ mol^-1, e = 1.60 × 10^-19 C and F = 9.65 × 10⁴ C mol^-1. The small difference in multiplying the printed rounded constants comes from rounding.

For a constant current, Q = It, with current in amperes, time in seconds and charge in coulombs. Then n(e^-) = Q/F. Divide by the number of electrons required per product particle in the half-equation. Finally use mass = amount × molar mass, or gas volume = amount × the stated molar volume. If current varies, Q is the area under the current-time graph.

The charge-to-product calculation

Do not convert charge directly to moles of metal without the electron ratio: Ag⁺, Cu²⁺ and Al³⁺ require different electron amounts.

Worked exampleOne current, two electrode products

Pass 2.50 A for 1930 s through aqueous CuSO₄ using inert electrodes. Find copper mass and oxygen volume at a molar gas volume of 24.0 dm³ mol^-1. Use F = 9.65 × 10⁴ C mol^-1, M(Cu) = 63.5 g mol^-1, sufficient Cu²⁺, and 100% current efficiency.

Try it first. Then show the worked stepsHide the worked steps

Q = 2.50 × 1930 = 4825 C. Hence n(e^-) = 4825/96500 = 0.0500 mol.
Cu²⁺ + 2e^- → Cu, so n(Cu) = 0.0500/2 = 0.0250 mol.
m(Cu) = 0.0250 × 63.5 = 1.5875 g.
2H₂O → O₂ + 4H⁺ + 4e^-, so n(O₂) = 0.0500/4 = 0.0125 mol.
V(O₂) = 0.0125 × 24.0 = 0.300 dm³.

Answer

Copper mass = 1.59 g; oxygen volume = 0.300 dm³ = 300 cm³. The same charge passes both electrodes, but the product amounts differ because their electron ratios differ.

Try it step by step

Work it out in steps: Charge, Q = It, then Moles of electrons, Q/F, then Moles of product from the half-equation, then Mass of product.

Problem 1. A current of 1.50 A passes for 3860 s through aqueous silver nitrate. Ag⁺ + e^- -> Ag. Find the mass of silver deposited, in g to 3 significant figures. Use F = 96 500 C mol^-1 and Ag = 107.9.
Problem 2. A current of 2.00 A passes for 3860 s through aqueous copper(II) sulfate. Cu²⁺ + 2e^- -> Cu. Find the mass of copper deposited, in g to 3 significant figures. Use F = 96 500 C mol^-1 and Cu = 63.5.
Problem 3. A current of 0.500 A passes for 5790 s through molten aluminium oxide. Al³⁺ + 3e^- -> Al. Use F = 96 500 C mol^-1 and Al = 27.0. Give each answer to 3 significant figures.

Show the answer

Problem 1.

Charge, Q = It: Q = 1.50 x 3860 = 5790 C.

Moles of electrons, Q/F: n(e^-) = 5790 / 96 500 = 0.0600 mol.

Moles of product from the half-equation: One electron per Ag, so n(Ag) = 0.0600 mol.

Mass of product: m = 0.0600 x 107.9 = 6.47 g.

Problem 2.

Charge, Q = It: Q = 2.00 x 3860 = 7720 C.

Moles of electrons, Q/F: n(e^-) = 7720 / 96 500 = 0.0800 mol.

Moles of product from the half-equation: Two electrons per Cu, so n(Cu) = 0.0800 / 2 = 0.0400 mol.

Mass of product: m = 0.0400 x 63.5 = 2.54 g.

Problem 3.

Charge, Q = It: Q = 0.500 x 5790 = 2895 C, which is 2.90 x 10³ C to 3 significant figures.

Moles of electrons, Q/F: n(e^-) = 2895 / 96 500 = 0.0300 mol.

Moles of product from the half-equation: Three electrons per Al, so n(Al) = 0.0300 / 3 = 0.0100 mol.

Mass of product: m = 0.0100 x 27.0 = 0.270 g.

Check your understanding

The same charge deposits 0.0100 mol of silver from Ag⁺ and some copper from Cu²⁺. How many moles of copper form?

0.0100 mol, because the charge passed is the same
0.00500 mol, because each Cu needs two electrons
0.0200 mol, because copper ions carry twice the charge

Show the answer

0.00500 mol, because each Cu needs two electrons

0.0100 mol Ag needs 0.0100 mol of electrons. The same electrons deposit 0.0100 / 2 = 0.00500 mol Cu. Equal charge means equal electrons, not equal moles of metal.