Extractive Metallurgy (Pyrometallurgy, Hydrometallurgy, Electrometallurgy)
Extractive metallurgy questions test whether you can choose and justify a route from ore to metal, and whether you can execute the mass, energy and Faraday calculations that go with it. Expect coverage of roasting, smelting, converting and refining (pyro), leaching, purification and recovery (hydro), and electrowinning/electrorefining (electro).
Route selection rules
- Sulphide concentrates: roast or smelt to matte, then convert to oxidise FeS and blow off sulphur as SO2.
- Oxide ores: carbothermic or CO reduction (blast furnace, rotary kiln) or direct acid/alkaline leaching.
- Low-grade or complex ores: heap, dump, pressure or bio-leaching followed by solvent extraction and electrowinning, because smelting energy per tonne of contained metal is prohibitive.
- Slag basicity (%CaO/%SiO2) drives refining: high basicity plus oxidising conditions removes phosphorus; strongly reducing basic slag in the ladle removes sulphur.
Electrochemistry
Faraday's law: m = I·t·M/(n·F) with F = 96,485 C/mol. Current efficiency is actual mass divided by theoretical mass, expressed as a percentage. Cell voltage equals reversible decomposition voltage plus anode and cathode overpotentials plus the IR drop through electrolyte, busbars and contacts.
Worked example
Copper electrowinning at 10,000 A for 24 h at 92% current efficiency. Charge = 10,000 × 86,400 = 8.64 × 10^8 C. With n = 2 and M = 63.55 g/mol: m = 8.64 × 10^8 × 63.55/(2 × 96,485) = 284.5 kg theoretical. Actual deposit = 0.92 × 284.5 = 262 kg Cu.
Traps that cost marks
Electrorefining uses a soluble impure anode that dissolves and replates; electrowinning uses an inert anode and recovers metal from purified solution. Swapping the two loses the whole question. Apply current efficiency to the theoretical mass, not to the charge. Remember converting is oxidation of the matte, not simple melting, and that a fluid slag is not automatically a selective slag.