Topic 6 of 7
Electrolysis products and calculations
Choose a half-reaction, then convert charge into yield.
A-Level 9476 (2026-2027)
Choose the products from the species and conditions
An external supply drives reduction at the cathode and oxidation at the anode.
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.
- 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.
| 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 understandingWhy does changing the anode from platinum to copper alter the products in aqueous CuSO4?Think it through, then reveal the answer
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
Current multiplied by time gives charge. Dividing by the Faraday constant gives moles of electrons. The balanced half-equation converts electron amount into product amount, which can then be converted to mass or gas volume.
Worked example
One 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.
- 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.
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.