Chemistry Inorganic - How to Solve: Metallurgy (Ellingham Diagram, Leaching, Roasting, Calcination, Electrode Potential) – IIT JEE Guide

How to Solve: Metallurgy (Ellingham Diagram, Leaching, Roasting, Calcination, Electrode Potential) – IIT JEE Guide

Introduction

Mastering metallurgy in IIT JEE unlocks 5-7 marks—enough to push you from a 90 to a 99+ percentile. Whether it’s predicting which metal can reduce another using the Ellingham diagram, choosing between roasting vs. calcination, or applying electrode potentials to extract metals, this topic appears every year in both JEE Main and Advanced.

WHAT YOU NEED TO KNOW FIRST

1. Gibbs Free Energy (ΔG) – Understand that ΔG = ΔH – TΔS and that ΔG < 0 means a reaction is spontaneous.
2. Oxidation & Reduction – Know that oxidation = loss of electrons, reduction = gain of electrons, and oxidizing/reducing agents.
3. Basic Thermodynamics – Be comfortable with enthalpy (ΔH), entropy (ΔS), and temperature (T) relationships.

KEY TERMS & FORMULAS

1. Ellingham Diagram

• Definition: A graph of ΔG° vs. T for metal oxide formation reactions.
• Key Points:
• Lower line = more stable oxide (harder to reduce).
• Intersection point = temperature where ΔG = 0 (equilibrium).
• Reduction possible if the reducing agent’s line is below the metal oxide’s line at a given T.
• MEMORISE THIS:
• Carbon (C → CO₂) line is downward sloping (entropy increases).
• Hydrogen (H₂ → H₂O) line is almost flat (small entropy change).
• Aluminium (Al → Al₂O₃) line is very low (strong reducing agent).

2. Leaching

• Definition: Extracting metal from ore using a solvent (e.g., NaCN for Au, H₂SO₄ for Cu).
• Key Reactions:
• Gold (Au): 4Au + 8NaCN + O₂ + 2H₂O → 4Na[Au(CN)₂] + 4NaOH
• Copper (Cu): CuCO₃.Cu(OH)₂ + 2H₂SO₄ → 2CuSO₄ + 3H₂O + CO₂
• MEMORISE THIS:
• Leaching is used for low-grade ores (e.g., bauxite → Al₂O₃ via Bayer’s process).

3. Roasting vs. Calcination

• MEMORISE THIS:
• Roasting = Sulphide ores → Oxides + SO₂
• Calcination = Carbonate/hydrated ores → Oxides + CO₂/H₂O

4. Electrode Potential (E°) Applications

• Definition: Tendency of a metal to lose/gain electrons (measured in volts).
• Key Rules:
• More negative E° = stronger reducing agent (e.g., Na, K, Al).
• More positive E° = stronger oxidizing agent (e.g., Au³⁺, Ag⁺).
• MEMORISE THIS:
• E°(Al³⁺/Al) = -1.66 V (strong reducing agent).
• E°(Cu²⁺/Cu) = +0.34 V (weak reducing agent).
• A metal can reduce another if its E° is more negative (e.g., Al can reduce Fe²⁺ but not Na⁺).

STEP-BY-STEP METHOD

Step 1: Identify the Ore & Required Process

• Check if the ore is:
• Sulphide? → Roasting (e.g., ZnS, PbS).
• Carbonate/Hydrated? → Calcination (e.g., CaCO₃, Fe₂O₃.3H₂O).
• Low-grade oxide? → Leaching (e.g., bauxite, gold ore).

Step 2: Use Ellingham Diagram for Reduction

• If the question asks: "Can metal X reduce metal Y oxide?"
• Find the ΔG° vs. T lines for both X → XO and Y → YO.
• Check if X’s line is below Y’s line at the given T.
  • If yes → Reduction is possible.
  • If no → Reduction is not possible.
• If lines intersect, find the temperature where ΔG = 0 (equilibrium).

Step 3: Apply Electrode Potential for Spontaneity

• If the question asks: "Will metal A reduce metal B ions?"
• Write the half-reactions:
  • A → Aⁿ⁺ + ne⁻ (oxidation)
  • Bⁿ⁺ + ne⁻ → B (reduction)
• Find E° values from the electrochemical series.
• Calculate E°_cell = E°_cathode – E°_anode.
  • If E°_cell > 0 → Spontaneous (reduction occurs).
  • If E°_cell < 0 → Non-spontaneous (no reduction).

Step 4: Choose the Right Extraction Method

WORKED EXAMPLES

Example 1 – Basic (Ellingham Diagram)

Question: Can carbon (C) reduce ZnO at 1000 K? Given: - Zn + ½O₂ → ZnO (ΔG° = -300 kJ/mol at 1000 K) - C + ½O₂ → CO (ΔG° = -200 kJ/mol at 1000 K)

Solution:
1. Write the reduction reaction: ZnO + C → Zn + CO
2. Calculate ΔG° for the reaction: - ΔG° = ΔG°(CO) – ΔG°(ZnO) - ΔG° = (-200) – (-300) = +100 kJ/mol
3. Since ΔG° > 0 → Non-spontaneous. Answer: No, carbon cannot reduce ZnO at 1000 K.

What we did and why: - We combined the two half-reactions to find the overall ΔG°. - Positive ΔG° means the reaction is not spontaneous under standard conditions.

Example 2 – Medium (Electrode Potential)

Question: Will aluminium (Al) reduce copper ions (Cu²⁺) in solution? Given: - E°(Al³⁺/Al) = -1.66 V - E°(Cu²⁺/Cu) = +0.34 V

Solution:
1. Write the half-reactions: - Oxidation: Al → Al³⁺ + 3e⁻ (E° = +1.66 V, reversed sign) - Reduction: Cu²⁺ + 2e⁻ → Cu (E° = +0.34 V)
2. Balance electrons (multiply Al reaction by 2, Cu by 3): - 2Al → 2Al³⁺ + 6e⁻ (E° = +1.66 V) - 3Cu²⁺ + 6e⁻ → 3Cu (E° = +0.34 V)
3. Calculate E°_cell: - E°_cell = E°_cathode – E°_anode = 0.34 – (-1.66) = +2.00 V
4. Since E°_cell > 0 → Spontaneous. Answer: Yes, Al will reduce Cu²⁺ to Cu.

What we did and why: - We balanced electrons to ensure the same number in both half-reactions. - Positive E°_cell means the reaction is spontaneous (Al is a stronger reducing agent than Cu).

Example 3 – Exam-Style (Leaching + Ellingham)

Question: Bauxite (Al₂O₃.2H₂O) is purified via Bayer’s process. Then, Al₂O₃ is reduced to Al using electrolysis. Why is carbon reduction not used for Al₂O₃? Given: - Ellingham diagram shows Al₂O₃ line is very low (ΔG° << 0). - E°(Al³⁺/Al) = -1.66 V.

Solution:
1. Ellingham Diagram Check: - Al₂O₃ line is below C → CO₂ line at all T. - ΔG° for Al₂O₃ formation is very negative → Very stable oxide. - Carbon cannot reduce Al₂O₃ because its line is above Al₂O₃’s line.
2. Electrode Potential Check: - Al has a very negative E° (-1.66 V) → Strong reducing agent. - No common reducing agent (C, H₂) has a more negative E° than Al.
3. Conclusion: - Carbon reduction is not thermodynamically feasible. - Electrolysis is the only viable method (Hall-Héroult process). Answer: Al₂O₃ is too stable; carbon reduction is not spontaneous. Electrolysis is required.

What we did and why: - We used both Ellingham and electrode potential to justify why carbon fails. - Stable oxides (low ΔG°) require electrolysis for reduction.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is how you crush metallurgy in JEE:

1. Ellingham Diagram:
2. Lower line = more stable oxide.
3. If the reducing agent’s line is below the metal oxide’s line → reduction works.

4. Carbon (C → CO₂) line slopes down; Al₂O₃ line is very low → carbon can’t reduce Al₂O₃.

5. Roasting vs. Calcination:

6. Roasting = Sulphide ores → Oxides + SO₂ (e.g., ZnS → ZnO).

7. Calcination = Carbonate/hydrated ores → Oxides + CO₂/H₂O (e.g., CaCO₃ → CaO).

8. Electrode Potential (E°):

9. More negative E° = stronger reducing agent.
10. E°_cell = E°_cathode – E°_anode.

11. If E°_cell > 0 → spontaneous reduction.

12. Leaching:

13. Low-grade ores (Au, Al₂O₃) need leaching (NaCN for Au, NaOH for bauxite).

Exam tricks: - Ellingham questions? Write the reaction first. - E° questions? Balance electrons before calculating. - Al₂O₃? Electrolysis only—carbon won’t work.

Now go ace that exam!