Faraday's Laws of Electrolysis — Calculations
Faraday's laws connect an ammeter reading and a stopwatch to a mass on a balance. Almost every numerical in this part of electrochemistry reduces to one working formula, and the marks are lost not in the chemistry but in three small places: converting minutes to seconds, choosing the right number of electrons, and remembering that charge — not current — is what actually deposits the metal.
The two laws
First law: the mass of a substance liberated at an electrode is proportional to the quantity of electricity passed through the solution.
Second law: when the same quantity of electricity passes through different electrolytes, the masses liberated are proportional to their equivalent masses.
moles of electrons = Q ÷ F where F = 96 485 C mol⁻¹
m = (M × I × t) ÷ (n × F)
| Symbol | Meaning | Unit |
|---|---|---|
| m | mass deposited or liberated | g |
| M | molar mass of the substance | g mol⁻¹ |
| I | current | A |
| t | time — in seconds | s |
| n | electrons per ion in the half-reaction | — |
| F | Faraday constant, charge on one mole of electrons | C mol⁻¹ |
School problems often use F = 96 500 C mol⁻¹ while data books give 96 485 C mol⁻¹. Both are in circulation; the difference is about 0.016 per cent, which never changes a three-significant-figure answer. Use whichever value your question supplies, and do not switch mid-problem.
The quantity M ÷ n is the equivalent mass E, and M ÷ (nF) is the electrochemical equivalent Z — the mass deposited by one coulomb. The formula above is just m = ZQ written out in full.
Worked example 1 — mass deposited
A current of 2.0 A is passed through CuSO₄ solution for 30 minutes. What mass of copper is deposited? (Cu = 63.546 g mol⁻¹)
Step 1 — the half-reaction gives n.
Cu²⁺ + 2e⁻ → Cu, so n = 2.
Step 2 — time in seconds. t = 30 × 60 = 1800 s
Step 3 — charge. Q = I × t = 2.0 × 1800 = 3600 C
Step 4 — moles of electrons.
n(e⁻) = 3600 ÷ 96 485 = 0.037 311 mol
Step 5 — moles of copper. Two electrons per atom, so
n(Cu) = 0.037 311 ÷ 2 = 0.018 656 mol
Step 6 — mass.
m = 0.018 656 × 63.546 = 1.185 g
Check with the one-line formula:
m = (63.546 × 2.0 × 1800) ÷ (2 × 96 485) = 228 765.6 ÷ 192 970 = 1.1855 g ✔
With F = 96 500 the answer is 228 765.6 ÷ 193 000 = 1.1853 g — identical to three significant figures, as promised.
Worked example 2 — how long will it take?
How long must a current of 0.500 A flow to deposit 1.00 g of silver from AgNO₃ solution? (Ag = 107.868 g mol⁻¹)
Ag⁺ + e⁻ → Ag, so n = 1.
Moles of silver: 1.00 ÷ 107.868 = 9.2705 × 10⁻³ mol
Since n = 1, the same number of moles of electrons is needed.
Charge: Q = 9.2705 × 10⁻³ × 96 485 = 894.5 C
Time: t = Q ÷ I = 894.5 ÷ 0.500 = 1789 s = 29.8 minutes (29 minutes 49 seconds)
Rearranging the master formula gives the same thing in one line: t = (m × n × F) ÷ (M × I) = (1.00 × 1 × 96 485) ÷ (107.868 × 0.500) = 96 485 ÷ 53.934 = 1789 s ✔
Worked example 3 — the second law, in a series circuit
When cells are connected in series the same current flows through all of them for the same time, so every cell receives an identical charge. This is the classic test of the second law.
Exactly one faraday (96 485 C) is passed in series through solutions of AgNO₃, CuSO₄ and AlCl₃. What mass of each metal is deposited?
One faraday is one mole of electrons, so:
| Metal | Half-reaction | n | Moles deposited | Mass |
|---|---|---|---|---|
| Ag | Ag⁺ + e⁻ → Ag | 1 | 1 ÷ 1 = 1.000 | 1.000 × 107.868 = 107.9 g |
| Cu | Cu²⁺ + 2e⁻ → Cu | 2 | 1 ÷ 2 = 0.500 | 0.500 × 63.546 = 31.77 g |
| Al | Al³⁺ + 3e⁻ → Al | 3 | 1 ÷ 3 = 0.3333 | 0.3333 × 26.982 = 8.994 g |
Compare those masses with the equivalent masses M ÷ n: 107.868 ÷ 1 = 107.9, 63.546 ÷ 2 = 31.77, 26.982 ÷ 3 = 8.994. They are identical — which is precisely what the second law states.
Worked example 4 — gases, and a genuine textbook divergence
A charge of 9650 C is passed through acidified water. Find the volumes of oxygen and hydrogen liberated at STP.
Moles of electrons: 9650 ÷ 96 485 = 0.1000 mol
At the anode: 2H₂O → O₂ + 4H⁺ + 4e⁻, so n = 4
n(O₂) = 0.1000 ÷ 4 = 0.02500 mol
At the cathode: 2H⁺ + 2e⁻ → H₂, so n = 2
n(H₂) = 0.1000 ÷ 2 = 0.05000 mol
The 2 : 1 ratio of hydrogen to oxygen falls straight out of the electron counting, which is a useful check.
Now the volumes — and here the answer depends on which STP your book uses. The older definition is 273.15 K and 1 atm, where the molar volume is 22.414 L mol⁻¹. The current IUPAC definition is 273.15 K and 1 bar, where it is 22.711 L mol⁻¹.
| Gas | Moles | Volume at 1 atm (22.414 L/mol) | Volume at 1 bar (22.711 L/mol) |
|---|---|---|---|
| O₂ | 0.02500 | 0.5604 L = 560 mL | 0.5678 L = 568 mL |
| H₂ | 0.05000 | 1.121 L | 1.136 L |
Neither is wrong. State which molar volume you used, and you cannot lose the mark.
Current efficiency — the industrial correction
In a real electroplating or refining cell, part of the current does something other than the reaction you want: it may liberate hydrogen, oxidise an impurity, or leak. Current efficiency is the fraction that did the intended job.
If the copper cell in example 1 ran at 90 per cent current efficiency, the copper actually deposited would be 1.185 × 0.90 = 1.067 g. Faraday's law is never wrong here — it gives the maximum, and the efficiency tells you how much of that maximum the plant achieved.
Common mistakes that cost marks
- Leaving time in minutes or hours. The ampere is a coulomb per second. Convert first, every single time.
- Using the wrong n. Fe²⁺ + 2e⁻ → Fe needs n = 2, but Fe³⁺ + 3e⁻ → Fe needs n = 3. Write the half-reaction before you write any number.
- Dividing by M instead of by E. Forgetting the n in M ÷ n gives an answer that is n times too large — the commonest single error in this topic.
- Mixing 96 485 and 96 500 inside one question. Pick one and stay with it.
- Assuming the metal is the only cathode product. In aqueous solution hydrogen may be discharged instead, which is why real cells are below 100 per cent efficient.
- Quoting a gas volume without saying which STP. 22.4 and 22.7 L mol⁻¹ are both defensible; an unlabelled number is not.
- Forgetting that charge, not current, is conserved. Halving the current and doubling the time deposits exactly the same mass.
Where this appears in exams
| Level | Typical question |
|---|---|
| CBSE/ICSE Class 12 | Mass deposited for a given current and time; state both laws |
| JEE/NEET | Series cells, gas volumes, combined stoichiometry and electrolysis |
| IIT-JAM / CUET-PG | Equivalent mass reasoning, mixed-product electrolysis |
| GATE / CSIR-NET | Coulometry, current efficiency and industrial extraction calculations |
Every Faraday's-law sum needs the equivalent mass E = M ÷ n, and that is exactly what the Equivalent Weight calculator returns — enter the species and the number of electrons and it gives E, ready to drop into m = (E × I × t) ÷ F.
Open the Equivalent Weight Calculator →Need the molar mass first? The Molar Mass calculator takes any formula and gives M in one step.
Electrochemistry is a full-marks chapter once the electron bookkeeping is automatic. ABC Chemistry teaches Class 11–12 chemistry at its Gurugram centre and in online batches across India, with home tuition available in Delhi-NCR — abcchemistry.in.