The Faraday constant

The Faraday constant, F, is the electric charge carried by one mole of electrons. Its value is 96 485 C mol⁻¹ (coulombs per mole), often rounded to 9.65 × 10⁴ C mol⁻¹. It links charge to moles, so it lets you work out how much substance an electric current makes in electrolysis, and how much energy a cell can supply.

Value of the Faraday constant

NIST lists the Faraday constant as exact. That is because, since the SI was redefined, both the Avogadro constant and the charge on an electron are fixed numbers, and F is simply their product. Use the value your exam gives you: Cambridge International's data booklet gives F = 9.65 × 10⁴ C mol⁻¹.

F = Le: where the value comes from

L (also written N_A) is the Avogadro constant, the number of particles in a mole, and e is the charge on one electron. One mole of electrons is L electrons, each carrying charge e, so the total charge is L × e:

6.022 140 76 × 10²³ mol⁻¹ × 1.602 176 634 × 10⁻¹⁹ C = 96 485.33 C mol⁻¹.

Cambridge International's A Level syllabus asks students to state and apply F = Le, and the mole guide covers the Avogadro constant itself.

Using the Faraday constant in electrolysis

In electrolysis, every ion that is discharged gains or loses a fixed number of electrons. Count the electrons and you know how much product forms. The method has four steps:

1. Charge: Q = It, with current I in amps and time t in seconds.
2. Moles of electrons: n(e⁻) = Q ÷ F.
3. Moles of product: use the half-equation. Cu²⁺ + 2e⁻ → Cu needs 2 mol of electrons per mole of copper.
4. Mass or volume: mass = moles × Ar (or Mr); gas volume = moles × molar volume.

So the amount of substance made is proportional to the charge passed: double the current or the time and you double the product. Don't mix it up with Faraday's law of electromagnetic induction in physics, which is about magnetic flux and is a different idea.

Worked examples

All the values here come from the Cambridge data booklet: F = 9.65 × 10⁴ C mol⁻¹, Ar of Ag = 107.9 and Cu = 63.5, and 24.0 dm³ mol⁻¹ for one mole of gas at room conditions.

Example 1: mass of silver deposited

A current of 0.50 A flows through aqueous silver nitrate for 20 minutes. What mass of silver forms at the cathode?

1. Q = It = 0.50 × (20 × 60) = 600 C.
2. n(e⁻) = 600 ÷ 96 500 = 6.22 × 10⁻³ mol.
3. Ag⁺ + e⁻ → Ag is 1 : 1, so n(Ag) = 6.22 × 10⁻³ mol.
4. Mass = 6.22 × 10⁻³ × 107.9 = 0.671 g.

Example 2: volume of hydrogen

In the electrolysis of dilute sulfuric acid, hydrogen forms at the cathode: 2H⁺ + 2e⁻ → H₂. A current of 1.20 A flows for 15 minutes. What volume of hydrogen forms at room conditions?

1. Q = 1.20 × (15 × 60) = 1080 C.
2. n(e⁻) = 1080 ÷ 96 500 = 0.0112 mol.
3. 2 mol of electrons give 1 mol of H₂, so n(H₂) = 0.0112 ÷ 2 = 5.60 × 10⁻³ mol.
4. Volume = 5.60 × 10⁻³ × 24.0 = 0.134 dm³ = 134 cm³.

Example 3: how long it takes

How long must a current of 0.80 A flow to deposit 1.00 g of copper from copper(II) sulfate solution?

1. n(Cu) = 1.00 ÷ 63.5 = 0.0157 mol.
2. Cu²⁺ + 2e⁻ → Cu, so n(e⁻) = 2 × 0.0157 = 0.0315 mol.
3. Q = n(e⁻) × F = 0.0315 × 96 500 = 3040 C.
4. t = Q ÷ I = 3040 ÷ 0.80 = 3800 s, which is about 63 minutes.

The Moles Calculator helps with the mass to moles steps.

Finding the Avogadro constant by electrolysis

F = Le can be turned round: measure the charge needed to make a known amount of product, and you can calculate L. Cambridge's syllabus asks students to describe this electrolytic method. A typical version uses copper electrodes in copper(II) sulfate solution: weigh the cathode, pass a steady measured current for a measured time, then dry and reweigh the cathode.

Example: a current of 0.300 A for 40.0 minutes adds 0.237 g to the cathode. Taking e = 1.60 × 10⁻¹⁹ C:

1. Q = 0.300 × 2400 = 720 C, so the number of electrons is 720 ÷ 1.60 × 10⁻¹⁹ = 4.50 × 10²¹.
2. n(Cu) = 0.237 ÷ 63.5 = 3.73 × 10⁻³ mol, so n(e⁻) = 2 × 3.73 × 10⁻³ = 7.46 × 10⁻³ mol.
3. L = electrons ÷ moles of electrons = 4.50 × 10²¹ ÷ 7.46 × 10⁻³ = 6.03 × 10²³ mol⁻¹, close to the accepted 6.02 × 10²³.

The Faraday constant in cells: ΔG = −nFE

The Faraday constant also connects a cell's voltage to its energy. Cambridge's A Level syllabus includes:

Here n is the number of moles of electrons transferred in the balanced equation. For the zinc and copper cell, the data booklet gives E⦵ = +0.34 V for Cu²⁺/Cu and −0.76 V for Zn²⁺/Zn, so E⦵cell = 0.34 − (−0.76) = 1.10 V, with n = 2. Then ΔG⦵ = −2 × 1.10 × 96 500 = −212 300 J mol⁻¹ = −212 kJ mol⁻¹. A negative ΔG means the reaction is feasible, matching the positive cell potential.

Common mistakes

• Leaving time in minutes. Q = It needs seconds: multiply minutes by 60.
• Ignoring the half-equation. Dividing by F gives moles of electrons, not moles of product. Copper needs 2 electrons per atom, aluminium 3.
• Forgetting H₂ and O₂ are diatomic. 2 mol of electrons make 1 mol of H₂; 4 mol make 1 mol of O₂ (4OH⁻ → O₂ + 2H₂O + 4e⁻).
• Units in ΔG = −nFE. The answer comes out in J mol⁻¹. Divide by 1000 for kJ mol⁻¹.
• Mixing up F and e. e is the charge on one electron; F is the charge on a mole of them, L times bigger.

Practise electrolysis and cell calculations on today's A-level Chemistry: Year 2 short timed paper. For the mole calculations underneath, see quantitative chemistry.

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