Metallurgy

Metallurgy

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Metallurgy is the science and technology of extracting pure metals from their naturally occurring compounds (ores) and purifying them for practical use. The four sequential steps are: (i) concentration of ore (removing gangue by methods like froth flotation, leaching, magnetic separation), (ii) conversion to oxide (calcination for carbonates, roasting for sulphides), (iii) reduction to crude metal (using C, CO, Al, or electrolysis), and (iv) refining (electrolytic, zone, vapour-phase, poling, liquation). The Ellingham diagram plots ΔG° of oxide formation vs temperature; a metal reduces the oxide of any metal lying above it in the graph because that oxide is less stable. Flux + gangue → slag (fusible product); acidic flux (SiO₂) is used for basic gangue, basic flux (CaO) for acidic gangue.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Step 1 — Concentration of Ore

Ores contain the desired mineral plus earthy impurities called gangue. The chosen concentration method depends on the difference between ore and gangue properties:

• Froth flotation — for sulphide ores (e.g., ZnS, PbS, CuFeS₂). The ore is powdered, wetted with pine oil/collectors, and agitated with air; sulphide particles attach to froth and float, while gangue settles.
• Leaching — for bauxite (Al₂O₃·2H₂O) using Baeyer’s process (concentrated NaOH at 150–200 °C dissolves Al₂O₃ as NaAlO₂; Fe₂O₃ and SiO₂ remain insoluble). For silver ore, leaching uses NaCN (MacArthur–Forrest process).
• Hydraulic washing, magnetic separation, leaching with acids are used for oxide ores.

Step 2 — Conversion to Metal Oxide

• Calcination = heating ore in absence of air (carbonate/hydrated ores), e.g. ZnCO₃ → ZnO + CO₂ ; Al₂O₃·2H₂O → Al₂O₃ + 2H₂O
• Roasting = heating ore in presence/limited air (sulphide ores), e.g. 2ZnS + 3O₂ → 2ZnO + 2SO₂

Step 3 — Reduction

The reducing agent depends on the metal’s position in the activity series. C/CO reduce Fe, Zn, Sn, Pb; Al (thermit process) reduces Fe, Cr, Mn oxides: 2Al + Fe₂O₃ → Al₂O₃ + 2Fe. Highly reactive metals (Al, Na, K) are obtained by electrolytic reduction of fused salts (e.g., fused NaCl + CaCl₂ for Na; cryolite-fluxed Al₂O₃ for Al).

Step 4 — Refining

Crude metal still contains impurities. Methods: electrolytic refining (Cu, Zn — impure metal anode, pure metal cathode, acidified salt solution as electrolyte), zone refining (used for semiconductors Si, Ge, B, Ga — based on difference in solubility of impurity in molten vs solid state), vapour phase refining (Mond process for Ni: Ni + 4CO → Ni(CO)₄ → Ni + 4CO at 450 K and 1800 K), poling (for Cu — green logs of wood stir molten metal; hydrocarbon gases reduce Cu₂O impurities), liquation (for low-melting metals Sn, Pb — metal melts and flows out leaving higher-melting impurities).

Ellingham Diagram

ΔG° = ΔH° − TΔS° for oxide formation. A negative ΔG° means a thermodynamically favourable oxide. The line for C → CO slopes downward (ΔS° > 0 since 1 mol gas becomes 2 mol), so above ~1073 K carbon can reduce almost all metal oxides. A metal lying lower on the diagram (more negative ΔG°) can reduce the oxide of any metal lying above it.

Exam Pointers

• 3% of CUET Chemistry paper = ~3–4 questions, mostly on Baeyer’s process, blast furnace reactions, thermite process, Ellingham interpretation, and refining methods.
• Match-the-following and assertion-reason questions frequently test flux/slag/roasting/calcination distinctions.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Worked Example — Iron Extraction in the Blast Furnace

Charge: haematite Fe₂O₃ + coke (C) + limestone (CaCO₃). Reactions in order:

1. CaCO₃ → CaO + CO₂ (at top, calcination)
2. C + O₂ → CO₂ ; CO₂ + C → 2CO (zone of combustion)
3. Fe₂O₃ + 3CO → 2Fe + 3CO₂ (at ~900 K, reduction)
4. CaO + SiO₂ → CaSiO₃ (slag formation; SiO₂ is the acidic gangue in haematite)
5. C + CO₂ → 2CO (recycled as reducing agent)

Worked Example — Electrolytic Refining

For copper refining, m = (I × t × M) / (n × F). If 5 A passes through CuSO₄ solution for 9650 s, mass of Cu deposited at cathode = (5 × 9650 × 63.5) / (2 × 96500) = 15.875 g. (n = 2 for Cu²⁺; F = 96500 C mol⁻¹.) At anode, copper dissolves: Cu → Cu²⁺ + 2e⁻; at cathode Cu²⁺ + 2e⁻ → Cu.

Common Mistakes

• Confusing roasting vs calcination: roasting is in air and targets sulphides; calcination is without air and targets carbonates/hydrates.
• Reversing the role of flux (added substance that reacts with gangue) and slag (fusible product formed by their reaction). Slag must be fusible at the furnace temperature.
• Misreading the Ellingham diagram: a metal whose line lies above (less negative ΔG°) cannot reduce the oxide of a metal below it — it would be thermodynamically uphill.
• Thinking all refining gives 100% pure metal; only zone refining routinely achieves ultra-high purity (impurities at ppb level) needed for semiconductors.
• Forgetting that poling uses wood, not metal rods, and is specific to removing oxide impurities of Cu (Cu₂O).

Connections to Other Topics

• Electrochemistry: Faraday’s laws of electrolysis underpin electrolytic refining.
• Thermodynamics: Ellingham diagram is a direct application of ΔG = ΔH − TΔS.
• p-Block: extraction of Al links metallurgy with the unique amphoteric behaviour of Al₂O₃ (dissolves in both acid and base — this is why NaOH leaching works).
• Environmental chemistry: roasting of sulphide ores releases SO₂, requiring desulphurisation or contact process for H₂SO₄ manufacture.

Practice Prompts

1. Explain why aluminium metal can reduce Cr₂O₃ to chromium but magnesium cannot reduce Al₂O₃ to aluminium. Use the Ellingham diagram.
2. A bauxite ore contains 60% Al₂O₃. After Baeyer’s process the mass drops to 25% Al₂O₃ due to removal of impurities. If 1000 kg of ore yields 200 kg of pure Al (assuming all Al₂O₃ is reduced electrolytically with 100% efficiency in the Hall–Héroult cell), calculate the percentage yield of Al from the Al₂O₃ present. [(% Al in Al₂O₃ = 54/102 × 100 ≈ 52.94%; theoretical Al = 1000 × 0.60 × 0.5294 ≈ 317.6 kg; yield = 200/317.6 × 100 ≈ 62.97%.]

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