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Electrolysis questions come in two kinds: how much is deposited, and what is formed at each electrode. The first is arithmetic built on Faraday's laws. The second needs reasoning about which species is easiest to reduce or oxidise, and that is where many students go wrong. This post works three calculations, explains the products of four common electrolyses and ends with a practice routine.
What does the syllabus ask for?
The JEE (Advanced) 2026 brochure names "Faraday's laws of electrolysis". The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list "electrolytic and galvanic cells", and the CBSE Chemistry curriculum for 2026-27 lists "Electrolytic Cells and Electrolysis" in Class 12. Check the current documents for your exam.
What do Faraday's two laws say?
The NCERT chapter Electrochemistry states them as Faraday published them in 1833-34:
First law: the amount of chemical reaction at an electrode is proportional to the quantity of electricity passed.
Second law: the amounts of different substances liberated by the same quantity of electricity are proportional to their chemical equivalent weights (atomic mass divided by the number of electrons needed to reduce the cation).
The quantity of electricity is Q = It, in coulombs when I is in amperes and t in seconds. The charge on one mole of electrons is the Faraday constant. NIST gives the Faraday constant as 96 485.332 12 C mol⁻¹; NCERT uses 96 487 C mol⁻¹ and approximates it to 96 500 for quick work.
Every calculation follows one chain: current and time → charge → moles of electrons → moles of product → mass.
Worked example 1: silver
A current of 2.00 A passes through silver nitrate solution for 30.0 minutes. What mass of silver is deposited? (Ag = 107.87 g/mol, from the IUPAC standard atomic weights.)
Q = 2.00 × 1800 = 3600 C.
Moles of electrons = 3600/96 485 = 0.03731 mol.
Ag⁺ + e⁻ → Ag, so moles of Ag = 0.03731 mol.
Mass = 0.03731 × 107.87 = 4.03 g.
Worked example 2: the second law with cells in series
The same charge flows through a copper sulphate cell connected in series. What mass of copper is deposited? (Cu = 63.55 g/mol.)
Cu²⁺ + 2e⁻ → Cu, so moles of Cu = 0.03731/2 = 0.01866 mol.
Mass = 0.01866 × 63.55 = 1.19 g.
Check with the second law: the ratio of masses, 4.03/1.19 ≈ 3.4, should equal the ratio of equivalent weights, 107.87/(63.55/2) ≈ 3.4. It does.
Worked example 3: how long will it take?
How long must 5.00 A flow through molten Al₂O₃ (with cryolite) to produce 0.500 g of aluminium? (Al = 26.98 g/mol.)
Moles of Al = 0.500/26.98 = 0.01853 mol.
Al³⁺ + 3e⁻ → Al, so moles of electrons = 3 × 0.01853 = 0.05559 mol.
Q = 0.05559 × 96 485 ≈ 5364 C.
t = Q/I = 5364/5.00 ≈ 1073 s, about 17.9 minutes.
How do you predict the products?
NCERT explains that products depend on the material being electrolysed, the type of electrode and the electrode potentials of the species present, and that some reactions need an extra voltage called overpotential.
Molten NaCl: only Na⁺ and Cl⁻ are present. Sodium forms at the cathode and chlorine at the anode.
Aqueous NaCl: at the cathode, water is reduced in preference to Na⁺ (E° for Na⁺/Na is −2.71 V), giving H₂ and OH⁻. At the anode, water oxidation (1.23 V) should win over chloride (1.36 V), but the overpotential of oxygen means Cl₂ forms. The products are H₂, Cl₂ and NaOH.
Aqueous CuSO₄ with copper electrodes: copper dissolves at the anode and deposits at the cathode. NCERT notes this is the basis of copper refining.
Sulphuric acid: NCERT says dilute acid gives oxygen at the anode, while concentrated acid gives peroxodisulphate (S₂O₈²⁻).
Why do students get electrolysis wrong?
Two classic interview studies explain the usual errors. Garnett and Treagust (1992) interviewed 32 final-year school students and found that many believed current in a solution is carried by drifting electrons, and that many did not link the anode and cathode in an electrolytic cell to the polarity of the applied voltage. Sanger and Greenbowe (1997) repeated the questions with college students and found, among other ideas, that students treated water as unreactive in the electrolysis of aqueous solutions.
The fixes are clear. In the solution, ions move and carry the current. Reduction always happens at the cathode, whatever its sign. And in any aqueous electrolysis, write water as a possible reactant at both electrodes.
Which mistakes cost marks?
Using minutes instead of seconds in Q = It.
Forgetting to divide by the number of electrons per ion.
Taking Na from aqueous NaCl.
Mixing up anode and cathode signs between galvanic and electrolytic cells. Our post on cell notation and the Nernst equation covers the galvanic side.
A practice routine
Write the chain Q → moles of e⁻ → moles of product → mass on a card, and use it for every problem this week.
Solve NCERT Example 2.10 (copper, 1.5 A for 10 minutes) and in-text question 2.12 on dichromate.
For five electrolytes, write the competing reactions at each electrode before deciding the products.
Link conductivity to this chapter with our post on conductance and Kohlrausch's law.
If you keep losing the thread between charge and moles, the Fermi Problem Library has a step-by-step solver that explains the reasoning, so you can see each conversion in order.
Before your next test, redo worked example 3 for magnesium from molten MgCl₂ and compare the time.
