State Faraday's two laws of electrolysis. A current of 3.86 A is passed through CuSO₄ solution for 25 minutes; find the mass of copper deposited (Cu = 63.5 g mol⁻¹, F = 96500 C mol⁻¹). Predict the products when aqueous NaCl is electrolysed with inert electrodes.

1. 1 · State the first law. The mass of a substance deposited or set free at an electrode is directly proportional to the charge passed: m = Z × I × t.
2. 2 · State the second law. When the same charge passes through different electrolytes, the masses set free are proportional to their equivalent masses (atomic mass ÷ electrons needed).
3. 3 · Find the charge. t = 25 × 60 = 1500 s; Q = 3.86 A × 1500 s = 5790 C = 5790 ÷ 96500 = 0.06 F.
4. 4 · Find the mass of copper. Cu²⁺ + 2e⁻ → Cu: 0.06 mol of electrons gives 0.03 mol of Cu = 0.03 × 63.5 = 1.905 g ≈ 1.91 g.
5. 5 · Predict the products from brine.
   • Cathode: H₂O + e⁻ → ½H₂ + OH⁻ (water is reduced more easily than Na⁺)
   • Anode: Cl⁻ → ½Cl₂ + e⁻ (O₂ needs a high overpotential)
   • Left in solution: NaOH
6. 6 · Link to real life. This is how industry makes NaOH and chlorine, used for soap and for disinfecting drinking water.
7. 7 · Close in one line. So the charge decides how much forms, while E° and overpotential decide what forms.

Scoring tip: Change minutes to seconds first, and write the half-reaction to show how many electrons each ion needs.