Electrolysis and Faraday's Law: Study Guide (Chemistry 9.11)

Counting electrons with an ammeter and a clock

In Topic 9.8 you saw that an electrolytic cell uses a power supply to force an unfavourable reaction, such as depositing a metal on an electrode. You also linked the two electrodes through the electrons they share. This topic answers the next question: how much product do you get?

The key idea is simple. Every particle that is reduced at the cathode needs a fixed number of electrons, set by its half-reaction. If you can count the electrons that flowed, you can count the particles. You cannot count electrons one by one, but you can measure two things in the lab:

• the current, I, with an ammeter, in amperes (A);
• the time, t, with a clock, in seconds (s).

Current is the rate at which charge flows. One ampere means one coulomb of charge passes each second (1 A = 1 C s⁻¹). So:

I = q / t, which rearranges to q = I × t

Here q is the charge in coulombs (C). Always convert minutes or hours into seconds first. A current of 1.50 A for 2.0 minutes passes 1.50 × 120 = 180 C, not 1.50 × 2.0 = 3.0 C.

The Faraday constant: from coulombs to moles of electrons

One electron carries a charge of 1.602 × 10⁻¹⁹ C. One mole of electrons is 6.022 × 10²³ electrons, so a mole of electrons carries:

6.022 × 10²³ mol⁻¹ × 1.602 × 10⁻¹⁹ C ≈ 96,500 C mol⁻¹

The precise value is F = 96,485 C mol⁻¹, the Faraday constant. It is the charge on one mole of electrons. It works exactly like a molar mass, but for charge instead of mass:

To convert	Use	Like
grams → moles of substance	n = m / M	dividing by grams per mole
coulombs → moles of electrons	n(e⁻) = q / F	dividing by coulombs per mole

So the full chain from the lab measurements to the amount of product is:

The half-reaction sets the electron count

The step from moles of electrons to moles of substance comes from the balanced half-reaction. The charge on the ion decides how many electrons each atom needs.

Half-reaction	Electrons per particle formed	Moles formed by 1 mol e⁻
Ag⁺ + e⁻ → Ag	1 per Ag	1 mol Ag
Ni²⁺ + 2e⁻ → Ni	2 per Ni	0.5 mol Ni
Al³⁺ + 3e⁻ → Al	3 per Al	0.333 mol Al
2Cl⁻ → Cl₂ + 2e⁻	2 per Cl₂	0.5 mol Cl₂
2H₂O → O₂ + 4H⁺ + 4e⁻	4 per O₂	0.25 mol O₂

So the same charge gives more moles of a metal whose ion has a smaller charge. For example, making 1.00 mol of Cl₂ at an anode needs 2.00 mol of electrons, which is 2 × 96,485 = 192,970 C.

The table also works for oxidation at the anode. If the anode is a metal that dissolves, as in electroplating with a silver or nickel anode, the same electron count tells you how much metal is removed from it.

Worked example 1: mass deposited in electroplating

Question. A steel key is nickel-plated. It is made the cathode in a solution of nickel(II) sulfate, and a current of 1.80 A flows for 40.0 minutes. What mass of nickel is deposited? (Ni = 58.69 g mol⁻¹)

1. Time in seconds: 40.0 min × 60 s min⁻¹ = 2400 s.
2. Charge: q = I × t = 1.80 A × 2400 s = 4320 C.
3. Moles of electrons: n(e⁻) = 4320 C ÷ 96,485 C mol⁻¹ = 0.044774 mol.
4. Half-reaction: Ni²⁺ + 2e⁻ → Ni, so n(Ni) = 0.044774 ÷ 2 = 0.022387 mol.
5. Mass: m = 0.022387 mol × 58.69 g mol⁻¹ = 1.3139 g.

Answer. 1.31 g of nickel (3 significant figures, matching the data).

Check. If you forget the “÷ 2”, you get 2.63 g, twice the right answer. If you leave the time in minutes, you get about 0.022 g, which is 60 times too small. Both are common slips, so check the units and the half-reaction before you round.

Worked example 2: how long will it take?

Question. A jeweller wants to deposit 0.500 g of gold on a ring from a solution containing Au³⁺ ions, using a steady current of 0.250 A. How long will this take? (Au = 196.97 g mol⁻¹)

This time you run the chain backwards (the dashed arrows in Figure 1).

1. Moles of gold: n(Au) = 0.500 g ÷ 196.97 g mol⁻¹ = 0.0025385 mol.
2. Half-reaction: Au³⁺ + 3e⁻ → Au, so n(e⁻) = 3 × 0.0025385 = 0.0076154 mol.
3. Charge: q = n(e⁻) × F = 0.0076154 mol × 96,485 C mol⁻¹ = 734.8 C.
4. Time: t = q / I = 734.8 C ÷ 0.250 A = 2939 s.

Answer. 2.94 × 10³ s, which is about 49.0 minutes.

Interpretation. Gold needs three electrons per atom, so it takes three times as long as plating the same number of moles of silver at the same current. Using one electron per atom would give about 16.3 minutes, which is far too short.

Worked example 3: cells in series and the charge on an ion

When two electrolytic cells are joined in series (one after the other in a single loop), every electron that passes through one cell also passes through the other. The charge, and so the moles of electrons, is the same in both.

Question. The two cells in Figure 2 run for 20.0 minutes. The cathode in cell A gains 1.079 g of silver and the cathode in cell B gains 0.594 g of tin. (a) Find the charge on the tin ions in cell B. (b) Find the average current. (Ag = 107.87, Sn = 118.71 g mol⁻¹)

(a)

1. Silver: Ag⁺ + e⁻ → Ag, so n(e⁻) = n(Ag) = 1.079 g ÷ 107.87 g mol⁻¹ = 0.010003 mol.
2. The same 0.010003 mol of electrons passed through cell B.
3. Moles of tin: n(Sn) = 0.594 g ÷ 118.71 g mol⁻¹ = 0.0050038 mol.
4. Electrons per tin atom: 0.010003 ÷ 0.0050038 = 1.999, which is 2.

So each tin ion gained two electrons: the ions are Sn²⁺, and the cathode half-reaction is Sn²⁺ + 2e⁻ → Sn.

(b)

1. Charge: q = n(e⁻) × F = 0.010003 mol × 96,485 C mol⁻¹ = 965.1 C.
2. Current: I = q / t = 965.1 C ÷ 1200 s = 0.804 A.

Why this works. The silver cell acts as a “charge meter”: silver ions always take one electron each, so the mass of silver tells you the moles of electrons directly. Comparing that with the moles of another metal reveals the charge on its ion.

The same counting works in a galvanic cell

Faraday’s law is not only for electrolysis. In a galvanic cell (a battery), the current comes out of the cell instead of being pushed in, but every electron still comes from the anode half-reaction.

For example, a zinc-based battery delivers 0.200 A for 5.00 hours. The charge is 0.200 A × (5.00 × 3600 s) = 3600 C, which is 3600 ÷ 96,485 = 0.037311 mol of electrons. Zinc is oxidised at the anode: Zn → Zn²⁺ + 2e⁻. So 0.018656 mol of zinc is used, a mass of 0.018656 × 65.38 = 1.22 g. The same reasoning tells you how long a battery with a given mass of anode metal can run at a given current.

What changes the amount of product, and what does not

The mass of product depends only on the charge passed and the half-reaction:

• Doubling the current (same time) doubles the charge, so it doubles the product.
• Doubling the time (same current) also doubles the product.
• Doubling the current and halving the time leaves the charge, and the product, unchanged.
• A metal whose ion has a higher charge gives fewer moles for the same charge.
• Concentration and electrode size do not change the amount deposited for a given charge, as long as the ion does not run out. (They can affect how easy it is to keep the current steady, but the calculation uses the current that actually flowed.)

Real cells sometimes give less product than the calculation predicts. In water-based solutions, some electrons may reduce water to hydrogen gas instead of depositing the metal, or some deposit may flake off before weighing. The Faraday calculation gives the maximum product that the charge could make.

Common misconceptions

• “q = I × t works with the time in minutes.” Amperes are coulombs per second. Convert minutes and hours to seconds first.
• “The charge in coulombs is the number of moles of electrons.” You must divide by F. 1930 C is about 0.0200 mol of electrons, not 1930 mol.
• “One mole of electrons deposits one mole of any metal.” Only for 1+ ions. Cu²⁺ needs two moles of electrons per mole of copper; Al³⁺ needs three.
• “Both electrodes change by the same mass.” They share the same moles of electrons. Different metals, different ion charges and different molar masses usually give different masses.
• “A stronger solution plates faster.” For a fixed current and time, the amount deposited is fixed by the charge. Concentration does not appear in the calculation.
• “In a series circuit, the current splits between the cells.” In series there is one path, so the same charge passes through every cell. (Current splits only between branches of a parallel circuit.)
• “The calculated mass is what you always get.” It is the maximum the charge can produce. Side reactions or losses make the measured mass smaller.

Where this leads

This is the last topic of the course framework. Faraday’s law ties together moles (Topic 1.1), redox half-reactions (Topic 4.9) and the cells from Topic 9.8, so it is a good place to check that unit 9 fits together. If cell potentials still feel shaky, go back to Topic 9.10. Try the practice questions now, then use the revision notes and the checklist to consolidate, and finish with the course roadmap for whole-course revision.