Faraday's Laws of Electrolysis Problems
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Direct answer
Pass one faraday of charge — 96,500 coulombs, one mole of electrons — through three electrolytic cells in series containing silver, copper and aluminium salts, and you deposit exactly 108 g Ag, 31.75 g Cu and 9 g Al. That is Faraday's second law: equal charge deposits masses proportional to equivalent weights (molar mass divided by n). The first law packages the arithmetic: m = ZIt, where Z is the electrochemical equivalent, I the current in amperes and t the time in seconds — always seconds, never minutes. The universal intermediate quantity is moles of electrons, Q/F = It/96,500, from which every mass or gas volume follows by stoichiometry of the electrode reaction.
What you must remember
- Charge accounting: Q = I × t (seconds); moles of electrons = Q/96,500; one ampere running one hour delivers 3,600 C.
- Deposition per faraday: Ag 108 g (n = 1), Cu 31.75 g (n = 2), Al 9 g (n = 3); one faraday also liberates 1 g of H2 (half mole) and 8 g of O2 (quarter mole) from water.
- Working formula: m = (M × I × t)/(n × 96,500), M the molar mass, n the electrons per ion.
- Series cells: the same current passes through all, so equivalents deposited are equal — moles are not.
- Current efficiency: actual mass divided by theoretical mass times 100; side reactions eat the difference, and JEE integer questions embed the percentage quietly.
- Medium matters: aluminium cannot be electroplated from aqueous solution because water is reduced first; it comes only from molten alumina in the Hall-Heroult cell.
A single current, three products
Drive 5 A through copper sulphate solution for 965 s. Charge: Q = 5 × 965 = 4,825 C. Moles of electrons: 4,825/96,500 = 0.05. Copper(II) needs two electrons per atom, so moles of Cu = 0.025, mass = 0.025 × 63.5 = 1.59 g. Same 4,825 C through an AgNO3 cell deposits 0.05 × 108 = 5.4 g silver — and 5.4/1.59 = 3.4, which is exactly 108/31.75, the second law verified by arithmetic. Meanwhile the anode of the copper cell releases 0.0125 mol O2, about 280 mL at STP, because four electrons make one oxygen molecule.
The habit to build: never compute masses directly from current. Convert to moles of electrons first, then walk the electrode half-reaction like any stoichiometry problem. Every Faraday numerical, however decorated, is this two-step.
Where students slip
Time units cause half the losses — a question saying "electrolysed for 20 minutes" must become 1,200 seconds before anything else. The second recurring slip is dilute versus concentrated: electrolysis of dilute NaCl gives O2 at the anode, concentrated brine gives Cl2 and is the industrial chlor-alkali process with H2 at the cathode and NaOH in solution. Third, students confuse the n of deposition with the n of a titration redox — the n here is simply electrons per ion reduced or oxidised at that electrode. JEE Advanced occasionally asks for the electrode that gains mass versus loses it: the copper anode dissolves during CuSO4 electrolysis with inert cathode, keeping the solution concentration constant — a detail that upgrades a guess into a reasoned answer.
Frequently asked questions
What does one faraday of charge deposit?
One gram-equivalent of any element — 108 g silver, 31.75 g copper, 9 g aluminium — because a mole of electrons reduces one equivalent.
Why must time be converted to seconds in Faraday problems?
Current is coulombs per second, so Q = It is valid only with t in seconds; leaving minutes inflates answers sixty-fold.
What gas forms at the anode during brine electrolysis and why?
Chlorine, because in concentrated chloride solution the overpotential of oxygen makes Cl- oxidation easier — the basis of the chlor-alkali industry.
Why can't aluminium be obtained by electrolysis of aqueous AlCl3?
Water is reduced to hydrogen at far lower potential than Al^3+, so the metal deposits only from molten salts, as in the Hall-Heroult process.
What is current efficiency in deposition calculations?
The percentage of total charge that actually deposits the metal — actual mass divided by theoretical mass times 100 — with the rest lost to side reactions.