Electrolysis and Faraday's Laws for JEE Main 2026

Electrolysis converts electrical energy to chemical energy, and Faraday's laws provide the quantitative link between charge passed and mass deposited or dissolved. JEE Main tests Faraday's laws through numericals on electroplating and metal deposition, and through conceptual questions on electrode reactions and product identification. The calculations are formula-driven and highly predictable.

Faraday's First Law

The mass of substance deposited or dissolved at an electrode is directly proportional to the charge passed: m = ZQ = ZIt, where Z is the electrochemical equivalent (mass deposited per coulomb). Electrochemical equivalent Z = M/(nF), where M is molar mass, n is number of electrons transferred per ion, and F = 96485 C/mol ≈ 96500 C/mol (Faraday's constant). A convenient form: m = MIt/(nF). For the broader electrochemistry context see our electrochemistry guide.

Faraday's Second Law

When the same charge passes through several electrolytic cells in series, the masses deposited are proportional to their equivalent weights (M/n). So m₁/m₂ = (M₁/n₁)/(M₂/n₂). This law lets you compare deposition in different cells without calculating the actual charge. Example: the same current deposits silver (M = 108, n = 1) and copper (M = 63.5, n = 2). Masses ratio = 108 : 31.75. Recognise this equivalent-weight ratio pattern; it solves the question in one step.

Worked Numerical: Copper Electroplating

Problem: what mass of copper deposits when 2 A flows for 30 minutes through CuSO₄ solution? Charge Q = It = 2 × 30 × 60 = 3600 C. Moles of electrons = Q/F = 3600/96500 ≈ 0.0373 mol. Cu²⁺ + 2e⁻ → Cu, so n = 2; moles of Cu = 0.0373/2 = 0.01865 mol. Mass = 0.01865 × 63.5 ≈ 1.18 g. This five-step calculation structure handles every Faraday's law numerical — memorise it and slot in new numbers.

Electrode Reactions and Products

At the cathode (reduction): metal ions are preferentially discharged if their electrode potential is more positive. H₂O/H⁺ are reduced when metal ions are absent or very dilute. At the anode (oxidation): anions are discharged in order: S²⁻ > I⁻ > Br⁻ > Cl⁻ > OH⁻ > NO₃⁻ > SO₄²⁻. In dilute H₂SO₄, water is oxidised at the anode (O₂ produced). In concentrated HCl, Cl⁻ is preferentially oxidised (Cl₂ produced). Identifying the electrode products from the electrolyte composition is a frequent JEE conceptual question.

Applications: Electroplating and Refining

In electroplating: the object is the cathode, the plating metal is the anode, and the salt of the plating metal is the electrolyte. In electrolytic refining of copper: impure copper is the anode (dissolves), pure copper is the cathode (deposits), and anode slime (gold, silver) collects below the anode. JEE asks which electrode is which and what happens to the electrolyte composition — in copper refining, the Cu²⁺ concentration stays roughly constant as anode dissolves and cathode deposits at the same rate. After mastering Faraday's laws, take a free mock test on electrochemistry.