Extraction of Metals — Based on Reactivity

Relationship Between Extraction and Reactivity

The reactivity of a metal — determines its method of extraction.

Extraction in Three Categories

1. Less Reactive Metals (bottom of series):

• Hg, Cu, Ag, Au
• Very simple — by heating directly or burning in air.

2. Moderately Reactive Metals (middle of series):

• Zn, Fe, Pb, Sn
• Moderately difficult — reduction by carbon (coke).

3. Highly Reactive Metals (top of series):

• Na, K, Ca, Mg, Al
• Very difficult — by electrolysis.

Reactivity and Method — Summary

   Reactivity Series (top → bottom)

   K, Na, Ca, Mg, Al    →  Electrolysis

   Zn, Fe, Pb, (Cu)     →  Reduction by carbon

   Hg, Ag, (Au)         →  Direct heating

Principle

More reactive metal = more stable compound = more energy needed for extraction.

Highly reactive like Na, K — even carbon cannot reduce them. Only electric current. Hg, Ag — less stable compounds — simple heating is enough.

Why Reactivity Matters?

A more reactive metal → more strongly bonded with non-metal. To 'free' it — more effort.

A less reactive metal → weak bond. To 'free' it — easy.

Two Main Processes of Extraction

1. Roasting: Heating a sulphide ore in the presence of air. 2. Calcination: Heating a carbonate/hydroxide without air.

Both — convert ore into oxide. The oxide is then reduced.

Why this? Because reduction of oxide — easy.

Roasting and Calcination

Two methods to convert ore into oxide.

Roasting

Heating a sulphide ore in the presence of air.

Principle: More oxygen — more vigorous oxidation.

Example: Roasting of ZnS:

$$2ZnS + 3O_2 \xrightarrow{\Delta} 2ZnO + 2SO_2\uparrow$$

ZnS → ZnO (oxide) SO₂ — harmful gas, released into atmosphere.

Other Examples:

Roasting of Cu pyrites: $$2Cu_2S + 3O_2 \xrightarrow{\Delta} 2Cu_2O + 2SO_2$$

Roasting of HgS: $$2HgS + 3O_2 \xrightarrow{\Delta} 2HgO + 2SO_2$$

Roasting of PbS: $$2PbS + 3O_2 \xrightarrow{\Delta} 2PbO + 2SO_2$$

Calcination

Heating carbonate/hydroxide ore in the absence of air (or limited).

Principle: Decomposition — CO₂ or H₂O is released, oxide remains.

Example: Calcination of Calamine (ZnCO₃):

$$ZnCO_3 \xrightarrow{\Delta} ZnO + CO_2\uparrow$$

Other Examples:

Calcination of Limestone (CaCO₃): $$CaCO_3 \xrightarrow{\Delta} CaO + CO_2$$

Calcination of Magnesite (MgCO₃): $$MgCO_3 \xrightarrow{\Delta} MgO + CO_2$$

Calcination of Bauxite (Al(OH)₃): $$2Al(OH)_3 \xrightarrow{\Delta} Al_2O_3 + 3H_2O$$

Comparison — Roasting vs Calcination

Property	Roasting	Calcination
Ore	Sulphide	Carbonate/Hydroxide
Air	Yes (more)	No (or limited)
Product	Oxide + SO₂	Oxide + CO₂/H₂O
Example	ZnS → ZnO	ZnCO₃ → ZnO

Why Both Are Needed?

Basic Rule: Sulphides and carbonates — not used directly for reduction. First convert them to oxide — then carbon or electrical reduction is easier.

Next Step

Now the oxide is ready — now to convert it into metal. This — depends on reactivity.

Extraction of Less Reactive Metals

Hg, Cu, Ag — at the bottom of the series. Their compounds — less stable. Simple methods of extraction.

Extraction of Mercury (Hg)

Ore: Cinnabar (HgS)

Step 1: Roasting $$2HgS + 3O_2 \xrightarrow{\Delta} 2HgO + 2SO_2$$

Step 2: Decomposition of HgO (heat further): $$2HgO \xrightarrow{\Delta} 2Hg + O_2$$

That is — HgO automatically turns into Hg upon heating.

This property of HgO — less stable. Easily decomposes.

Hence — a 'very simple' extraction.

Extraction of Copper (Cu)

Ore: Copper pyrites (CuFeS₂) or Cuprite (Cu₂O)

From Cu₂S (a simple method):

If both Cu₂O and Cu₂S are available — self-reduction:

$$2Cu_2O + Cu_2S \xrightarrow{\Delta} 6Cu + SO_2\uparrow$$

That is — one ore reduced the other! 'Self-reduction' — an interesting process.

Extraction of Silver (Ag)

Ore: Argentite (Ag₂S)

Modern Method — Cyanide Leaching:

1. Ag₂S in NaCN solution: $$Ag_2S + 4NaCN \rightarrow 2Na[Ag(CN)_2] + Na_2S$$

2. Displacement by Zn: $$2Na[Ag(CN)_2] + Zn \rightarrow Na_2[Zn(CN)_4] + 2Ag\downarrow$$

Ag precipitate — in pure form.

An Interesting Fact

Hg, Cu, Ag — methods of extraction for all three are very simple. That is why — these metals were known thousands of years ago. After 'Stone Age' — the 'Copper Age' came directly! Fe, Al — only recently (last 200 years).

Extraction of Moderately Reactive Metals — Reduction by Carbon

Zn, Fe, Pb — middle of the series. Ore → Oxide → Reduction by carbon → Metal.

Basic Principle

Carbon — a strong reducing agent. It pulls oxygen towards itself. Frees the metal from 'metal oxide'.

General Reaction

$$\text{Metal Oxide} + C \rightarrow \text{Metal} + CO_2$$

or

$$\text{Metal Oxide} + CO \rightarrow \text{Metal} + CO_2$$

Extraction of Fe — In Blast Furnace

Ore: Hematite (Fe₂O₃)

Step 1: Concentration (Section 6 — gravity)

Step 2: Calcination/Roasting Heat ore to remove impurities.

Step 3: Reduction (Blast Furnace):

Tall, cylindrical furnace. From top: ore + coal (coke) + limestone. From bottom: hot air.

Reactions in three zones:

1. Upper zone (~500°C): $$CaCO_3 \xrightarrow{\Delta} CaO + CO_2$$

2. Middle zone (~1000°C): $$Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$$

3. Lower zone (~1500°C): $$C + O_2 \rightarrow CO_2$$ $$CO_2 + C \rightarrow 2CO$$

CaO + impurity (SiO₂): $$CaO + SiO_2 \rightarrow CaSiO_3 \text{ (slag)}$$

That is, CaCO₃ — works as flux.

Final Products:

• From bottom: Molten Fe (Pig iron).
• Above: slag (CaSiO₃) — separated.

Extraction of Zn

Ore: Zinc blende (ZnS) or Calamine (ZnCO₃)

Step 1: Roasting of ZnS: $$2ZnS + 3O_2 \xrightarrow{\Delta} 2ZnO + 2SO_2$$

(Or calcination of ZnCO₃: ZnCO₃ → ZnO + CO₂)

Step 2: Reduction of ZnO by carbon: $$ZnO + C \xrightarrow{\Delta} Zn + CO\uparrow$$

Zn metal formed.

Extraction of Pb

Ore: Galena (PbS)

Step 1: Roasting of PbS: $$2PbS + 3O_2 \xrightarrow{\Delta} 2PbO + 2SO_2$$

Step 2: Reduction of PbO: $$PbO + C \xrightarrow{\Delta} Pb + CO\uparrow$$

Flux and Slag — Again

Hidden impurities along with ore:

• Acidic gangue (SiO₂) → basic flux (CaCO₃).
• Basic gangue (CaO) → acidic flux (SiO₂).

Flux + Gangue → Slag — in molten state, separated.

Extraction of Highly Reactive Metals — Electrolysis

Na, K, Ca, Mg, Al — at the top of the series. Reduction by carbon — not enough. Only electrolysis.

Principle

Electric current passed through molten oxide or chloride. Metal deposits at Cathode. Non-metal released at Anode.

Extraction of Al — Hall-Héroult Process

Ore: Bauxite ($Al_2O_3 \cdot 2H_2O$)

Step 1: Concentration (Bayer Process) (Section 6) Pure Al₂O₃ at the end.

Step 2: Electrolysis (Hall-Héroult):

Melting point of Al₂O₃ is very high (~2050°C). Melting alone — very expensive.

Solution: Dissolve Al₂O₃ in cryolite (Na₃AlF₆). Now melting point — ~950°C. Much cheaper.

Setup:

• A steel box — lined with carbon.
• Carbon itself is Cathode (-).
• Carbon rods from above — Anode (+).
• Solution: Al₂O₃ + Na₃AlF₆ + CaF₂ (molten).

Reactions:

At Cathode (-): $$Al^{3+} + 3e^- \rightarrow Al(l)$$

Molten Al — collects at bottom.

At Anode (+): $$2O^{2-} \rightarrow O_2 + 4e^-$$

O₂ released — reacts with carbon to form CO/CO₂.

Hence the carbon anodes slowly burn — replaced from time to time.

Overall reaction: $$2Al_2O_3 \xrightarrow{\text{electricity}} 4Al + 3O_2$$

Extraction of Na — Down's Cell

Ore: Rock salt (NaCl)

Step: Electrolysis of molten NaCl

Add CaCl₂ to molten NaCl — lower melting point (800°C → 600°C).

Reactions:

At Cathode: $$Na^+ + e^- \rightarrow Na$$

At Anode: $$2Cl^- \rightarrow Cl_2 + 2e^-$$

That is — Na metal + Cl₂ gas.

Extraction of Ca

Electrolysis of molten CaCl₂.

Why Not from Aqueous Solution?

Electrolysis of aqueous solutions of highly reactive metals (Na, K, Ca) — Na/K won't be formed!

Why? At Cathode — H⁺ (from water) gets reduced first (H₂ formed).

Therefore — only from molten ore.

Summary Table — Methods of Extraction

Metal	Ore	Method
Hg	HgS	Just heating
Cu	Cu₂S	Self-reduction
Ag	Ag₂S	Cyanide
Pb	PbS	Roasting + carbon
Zn	ZnS/ZnCO₃	Roasting/calcination + carbon
Fe	Fe₂O₃	Blast furnace + carbon
Al	Al₂O₃	Hall-Héroult (electricity)
Mg	MgCl₂	Electrolysis
Na	NaCl	Down's cell (electricity)
K	KCl	Electrolysis

Basic Principle

'Level of reactivity' = difficulty of extraction.

Highly reactive = electrolysis. Moderate = reduction by carbon. Less reactive = direct heating.

[Board Important] This table — every year in board, 5-mark question.

🧠 Memory Capsule

A quick glance just before the board exam.

1. Reactivity and Extraction Method

Series	Metals	Method
High	K, Na, Ca, Mg, Al	Electrolysis
Moderate	Zn, Fe, Pb	Reduction by carbon
Low	Hg, Cu, Ag, Au	Direct heating

2. Roasting vs Calcination

Property	Roasting	Calcination
Ore	Sulphide	Carbonate
Air	Yes	No
Example	ZnS → ZnO + SO₂	ZnCO₃ → ZnO + CO₂

3. Famous Reactions

Roasting of ZnS: $$2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2$$

Calcination of ZnCO₃: $$ZnCO_3 \rightarrow ZnO + CO_2$$

Reduction of ZnO by carbon: $$ZnO + C \rightarrow Zn + CO$$

Fe extraction: $$Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$$

Decomposition of HgO: $$2HgO \xrightarrow{\Delta} 2Hg + O_2$$

4. Al Extraction (Hall-Héroult)

• Electrolysis: Al₂O₃ + cryolite (Na₃AlF₆).
• Melting point 950°C.
• Cathode: $Al^{3+} + 3e^- \rightarrow Al$.
• Anode: $2O^{2-} \rightarrow O_2 + 4e^-$.

5. Na Extraction (Down's cell)

• Molten NaCl + CaCl₂.
• Cathode: Na.
• Anode: Cl₂.

6. Fe in Blast Furnace

• Ore: Fe₂O₃
• Coke + CaCO₃ + hot air.
• Main reaction: $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$.
• Flux (CaO) + Gangue (SiO₂) → Slag (CaSiO₃).

7. Why Carbon Cannot Reduce Na?

Na more reactive than C. For C to release O₂ — Na would have to do work. Answer: electrolysis.

8. Board's 'Golden' Questions

1. Fe extraction — Blast Furnace.
2. Al extraction — Hall-Héroult.
3. Na extraction — why electrolysis?
4. Difference between roasting and calcination.
5. Method for less reactive metals.

Final Formula: 'Reactivity ↑ → extraction difficulty ↑ → energy ↑.'

Solved Examples

Example 1: NCERT — Three Categories of Extraction

How does extraction of metals depend on their reactivity? Divide into three categories.

Solution:

Basic Principle

More reactive metal → more stable compound → more energy needed for extraction.

Three Categories

1. Highly Reactive (top of series)

Metals: K, Na, Ca, Mg, Al

Method: Electrolysis.

Why? Even carbon cannot reduce them. Only electric current — sufficient energy.

Examples:

• Al: Hall-Héroult process (Al₂O₃ + cryolite)
• Na: Down's cell (molten NaCl)
• Mg: molten MgCl₂

2. Moderately Reactive (middle of series)

Metals: Zn, Fe, Pb, Sn, (Cu)

Method: Ore → Oxide → Reduction by carbon.

Steps:

1. Roasting/Calcination (ore → oxide).
2. Reduction by carbon.

Example: Zn:

• $2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2$ (roasting)
• $ZnO + C \rightarrow Zn + CO$ (reduction)

Example: Fe:

• $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$ (in Blast Furnace)

3. Less Reactive (bottom of series)

Metals: Hg, Cu, Ag, Au

Method: Direct heating or self-reduction.

Why? Compounds are less stable — heat alone forms metal.

Example: Hg:

• $2HgS + 3O_2 \rightarrow 2HgO + 2SO_2$
• $2HgO \rightarrow 2Hg + O_2$

Example: Au, Pt:

• Found in free form — just pick and purify.

Summary Table

Series	Metals	Method
High	K-Al	Electrolysis
Moderate	Zn, Fe, Pb	Reduction by carbon
Low	Hg, Cu, Ag, Au	Direct/Natural

An Interesting Point

This is why:

• After 'Stone Age' came directly the Copper Age (Cu — simple extraction).
• Iron Age — after Cu (Fe extraction harder).
• Al — only recently (1825), because electrolysis technology was needed.

[NCERT textbook — fundamental]

Example 2: NCERT — Roasting and Calcination

What are roasting and calcination? Explain the differences with chemical reactions.

Solution:

Roasting

Definition: Heating sulphide ore in presence of air — to convert it into oxide.

General form: $$2MS + 3O_2 \xrightarrow{\Delta} 2MO + 2SO_2$$

Examples:

1. Zinc blende (ZnS): $$2ZnS + 3O_2 \xrightarrow{\Delta} 2ZnO + 2SO_2$$

2. Cinnabar (HgS): $$2HgS + 3O_2 \xrightarrow{\Delta} 2HgO + 2SO_2$$

3. Galena (PbS): $$2PbS + 3O_2 \xrightarrow{\Delta} 2PbO + 2SO_2$$

4. Iron pyrites (FeS₂): $$4FeS_2 + 11O_2 \xrightarrow{\Delta} 2Fe_2O_3 + 8SO_2$$

Calcination

Definition: Heating carbonate or hydroxide ore in absence of air (or limited) — to convert it into oxide.

General forms: $$MCO_3 \xrightarrow{\Delta} MO + CO_2\uparrow$$

or

$$M(OH)_x \xrightarrow{\Delta} \text{oxide} + H_2O\uparrow$$

Examples:

1. Calamine (ZnCO₃): $$ZnCO_3 \xrightarrow{\Delta} ZnO + CO_2$$

2. Limestone (CaCO₃): $$CaCO_3 \xrightarrow{\Delta} CaO + CO_2$$

3. Magnesite (MgCO₃): $$MgCO_3 \xrightarrow{\Delta} MgO + CO_2$$

4. Bauxite hydroxide (Al(OH)₃): $$2Al(OH)_3 \xrightarrow{\Delta} Al_2O_3 + 3H_2O$$

Main Differences

Property	Roasting	Calcination
Ore type	Sulphide	Carbonate / Hydroxide
Air	Yes (more)	No (or limited)
Product	Oxide + SO₂	Oxide + CO₂ or H₂O
Gas	SO₂ (toxic)	CO₂ (relatively harmless)

Why Both Are Needed?

Direct reduction from sulphide or carbonate is difficult. Reduction of oxide — easy.

Hence, first convert these ores to oxide — then perform reduction.

An Interesting Point

SO₂ — environmental problem. SO₂ released during roasting — cause of acid rain. Modern industry — captures SO₂ and converts to $H_2SO_4$. That is, 'waste' becomes 'valuable product'.

[NCERT — asked every year]

Example 3: NCERT — Extraction of Fe

Explain the extraction of Fe in Blast Furnace. Write the main reactions.

Solution:

Blast Furnace (Detailed)

Setup:

• Tall, cylindrical furnace (~25-30 m).
• Top: 'charge' (ore + coal + limestone) added.
• Bottom: hot air blown.

What's in the Charge?

1. Ore: Fe₂O₃ (Hematite) — concentrated. 2. Coke: Pure carbon — fuel and reducing agent. 3. Limestone: CaCO₃ — flux.

Reactions in Three Zones

Zone 1: Bottom (~1500-2000°C) — Combustion zone

1. Coal + air: $$C + O_2 \rightarrow CO_2 + \text{heat}$$

2. CO₂ + more coal (at high temperature): $$CO_2 + C \rightarrow 2CO$$

CO — main reducing agent.

Zone 2: Middle (~1000°C) — Reduction zone

Reduction of Fe (by CO): $$Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$$

This is the main reaction — Fe is released.

Other reactions: $$Fe_3O_4 + 4CO \rightarrow 3Fe + 4CO_2$$

Zone 3: Upper (~500°C) — Pre-preparation

1. Decomposition of CaCO₃: $$CaCO_3 \rightarrow CaO + CO_2$$

2. CaO + impurity (SiO₂): $$CaO + SiO_2 \rightarrow CaSiO_3 \text{ (slag)}$$

That is, CaO — works as flux, combining with gangue (SiO₂).

Final Products

At bottom of furnace:

• Molten Fe (down, heavy).
• Molten slag (CaSiO₃) (up, light).

Both — released through different openings.

Pig Iron — First Product

Fe from Blast Furnace — 'pig iron'. 4% C + other impurities.

Further refined — 'cast iron' or 'wrought iron'.

A Main Diagram (in words):

  Top — Charge (ore + coal + lime)
   ↓
  500°C — CaCO₃ → CaO + CO₂
   ↓
  1000°C — Fe₂O₃ + 3CO → 2Fe + 3CO₂
   ↓ 
  1500°C — C + O₂ → CO₂; CO₂ + C → 2CO
   ↓
  Bottom — Molten Fe (extract)
            Molten slag (separate)

  Hot air in ←

Summary

Blast Furnace = Backbone of modern steel industry. In India: Tata Steel, SAIL, Bhilai Steel Plant — all Blast Furnaces.

[NCERT — every year in board, 5-mark]

Example 4: NCERT — Al Extraction (Hall-Héroult)

Explain the extraction of Aluminium in detail.

Solution:

Ore

Bauxite ($Al_2O_3 \cdot 2H_2O$)

Impurities: Fe₂O₃, SiO₂, TiO₂

Step 1: Concentration — Bayer Process (Section 6)

Brief here:

1. Al₂O₃ dissolves in NaOH: $$Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O$$

2. Filter — remove impurities.

3. Al(OH)₃ from NaAlO₂: $$NaAlO_2 + 2H_2O \rightarrow NaOH + Al(OH)_3\downarrow$$

4. Heat Al(OH)₃ (calcination): $$2Al(OH)_3 \xrightarrow{\Delta} Al_2O_3 + 3H_2O$$

Now pure Al₂O₃.

Step 2: Electrolysis — Hall-Héroult

Problem: Melting point of Al₂O₃ ~2050°C. Very expensive.

Solution: Dissolve Al₂O₃ in cryolite (Na₃AlF₆). Now melting point — 950°C.

Setup:

• Steel box — carbon lining.
• Carbon lining = Cathode (-).
• Carbon rods from above = Anode (+).
• Solution: molten (Al₂O₃ + Na₃AlF₆ + CaF₂).

Reactions:

Al₂O₃ → 2Al³⁺ + 3O²⁻ (in molten state)

At Cathode (-): $$Al^{3+} + 3e^- \rightarrow Al(l)$$

Molten Al — collects at bottom. Removed from bottom of box.

At Anode (+): $$2O^{2-} \rightarrow O_2 + 4e^-$$

O₂ released — but at high temperature reacts with carbon: $$C + O_2 \rightarrow CO_2$$

Hence carbon anodes slowly burn. Replaced from time to time.

Overall reaction: $$2Al_2O_3 \xrightarrow{\text{electricity}} 4Al + 3O_2$$

Important Points

1. Role of Cryolite:

• Lowers melting point (2050 → 950°C).
• Increases electrical conductivity.

2. Role of CaF₂:

• Lowers melting point further.
• Stabilises the solution.

3. Energy:

• Al extraction — very energy-intensive.
• 1 kg Al ≈ 14-15 kWh electricity.
• Hence — Al recycling is important.

An Interesting Fact

Al — most abundant metal in Earth's crust (~8%). Yet, commercial production of Al was not possible until 1886.

Hall (USA) and Héroult (France) — both discovered this method independently in 1886. In honour of both — 'Hall-Héroult' process.

[NCERT — board: 5-mark]

Example 5: NCERT — Extraction of Less Reactive Metals

Explain extraction of Hg and Cu.

Solution:

Extraction of Hg

Ore: Cinnabar (HgS)

Step 1: Roasting

$$2HgS(s) + 3O_2(g) \xrightarrow{\Delta} 2HgO(s) + 2SO_2(g)$$

HgS first converted to HgO.

Step 2: Self-decomposition of HgO

At higher temperatures, HgO — spontaneously decomposes:

$$2HgO(s) \xrightarrow{\Delta} 2Hg(l) + O_2(g)$$

That is, no external reducing agent needed! HgO itself is unstable.

Final product: Molten Hg — collected directly.

An Interesting Demonstration — In One Step

If furnace temperature is kept very high — both steps together: $$HgS + O_2 \xrightarrow{\Delta} Hg + SO_2$$

That is, Hg directly from HgS!

Extraction of Cu

Ore: Copper pyrites (CuFeS₂)

Step 1: Concentration (froth flotation)

Step 2: Roasting

$$2CuFeS_2 + 3O_2 \rightarrow 2Cu_2S + 2FeO + 2SO_2$$

Cu₂S and FeO formed.

Step 3: Smelting with Flux

Cu₂S and FeO + SiO₂ (flux):

$$FeO + SiO_2 \rightarrow FeSiO_3 \text{ (slag)}$$

Fe separated as slag. Cu₂S — settles as 'matte'.

Step 4: Self-reduction

Heat 'matte' (Cu₂S) in air:

1. Some Cu₂S → Cu₂O: $$2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$$

2. Then Cu₂O + Cu₂S (remaining): $$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$$

That is — one part reduced the other! This is 'self-reduction'.

Why Self-reduction?

Reactivity of Cu — moderate/low. Compounds not stable. S provides a 'natural reducing agent'.

Summary

Metal	Ore	Method
Hg	HgS	Roasting + self-decomposition
Cu	CuFeS₂	Roasting + self-reduction
Ag	Ag₂S	Cyanide + Zn displacement

Final Insight

'Less reactive' = 'simple method'. This is why — Cu, Hg, Ag — known for thousands of years.

[NCERT — important]

Example 6: NCERT — Extraction of Na

Why is Na not extracted by electrolysis of aqueous solution? What is the correct method?

Solution:

Why Not from Aqueous NaCl?

If electrolysis of aqueous NaCl is done:

At Cathode — two options:

• $Na^+ + e^- \rightarrow Na$ (Reduction of Na)
• $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ (Reduction of water)

Which will happen?

Look at reactivity series: Na (highly reactive) > H₂O

Rule: At Cathode — less reactive ion/molecule will be reduced.

Hence — H₂O is reduced. Na is not.

That is — H₂ comes out from aqueous NaCl, not Na.

Correct Method — Down's Cell

Electrolysis of molten NaCl.

Setup:

• A steel box.
• Solution: molten NaCl + CaCl₂.
• Why CaCl₂? To lower melting point of NaCl from 800°C (~600°C).
• Cathode: steel ring (to collect Na).
• Anode: carbon rod (for Cl₂ release).

Reactions:

At Cathode (-): $$Na^+ + e^- \rightarrow Na(l)$$

Molten Na — light, floats up. Out through a tube.

At Anode (+): $$2Cl^- \rightarrow Cl_2 + 2e^-$$

Cl₂ gas — out through separate tube.

Overall: $$2NaCl(l) \xrightarrow{\text{electricity}} 2Na(l) + Cl_2(g)$$

Precautions

*1. Na and Cl₂ — must be kept apart. *(Otherwise they would form NaCl back — and explosion.)* 2. Air-tight box — protects Na from air. 3. Molten Na put in kerosene — protection.

Comparison — Molten vs Aqueous

Property	Molten NaCl	Aqueous NaCl
Electrolysis	Na + Cl₂	H₂ + Cl₂ + NaOH
Product	Na metal	NaOH
Melting point	800°C (less with CaCl₂)	room temp
Use	Na extraction	Chlor-alkali process

Broad Principle

For highly reactive metals (K, Na, Ca, Mg, Al): Extraction only by electrolysis of molten ore.

For moderate (Zn, Fe, Pb): Reduction by carbon.

An Interesting Fact

Aqueous electrolysis of Na — Chapter 2's 'chlor-alkali process'. That gives NaOH, not Na. This process — commercially very important.

[NCERT — every year in board]

Example 7: NCERT — Which Method?

Give the suitable method of extraction for the following metals:

(a) Mg, (b) Cu, (c) Au, (d) Zn, (e) Pb

Solution:

First look at reactivity series: K > Na > Ca > Mg > Al > Zn > Fe > Pb > H > Cu > Hg > Ag > Au

(a) Mg

Series: Highly reactive.

Method: Electrolysis.

Ore: MgCl₂ (in molten form).

At Cathode: Mg²⁺ + 2e⁻ → Mg. At Anode: 2Cl⁻ → Cl₂.

Why not carbon reduction? Mg more reactive than C.

(b) Cu

Series: Less reactive.

Method: Roasting + self-reduction.

Ore: CuFeS₂

Roasting: 2CuFeS₂ + O₂ → Cu₂S + … Self-reduction: 2Cu₂O + Cu₂S → 6Cu + SO₂.

(c) Au

Series: Very low reactivity (bottom of series).

Method: In free form (panning or cyanide leaching).

1. Free gold — picked directly. 2. For concentrated form: cyanide leaching + Zn displacement.

Why no 'reduction'? Au compounds — rare.

(d) Zn

Series: Moderately reactive.

Method: Roasting + reduction by carbon.

Ore: ZnS or ZnCO₃

1. Roasting: 2ZnS + 3O₂ → 2ZnO + 2SO₂ (Or calcination: ZnCO₃ → ZnO + CO₂)

2. Reduction: ZnO + C → Zn + CO.

(e) Pb

Series: Moderately reactive.

Method: Roasting + reduction by carbon.

Ore: PbS (Galena)

1. Roasting: 2PbS + 3O₂ → 2PbO + 2SO₂ 2. Reduction: PbO + C → Pb + CO.

Summary Table

Metal	Reactivity	Method	Main Reaction
Mg	High	Electrolysis	$Mg^{2+} + 2e^- \rightarrow Mg$
Cu	Low	Self-reduction	$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$
Au	Very low	Natural	(free form)
Zn	Moderate	Reduction by C	$ZnO + C \rightarrow Zn + CO$
Pb	Moderate	Reduction by C	$PbO + C \rightarrow Pb + CO$

Key Insight

'Reactivity decides — the method.'

[NCERT textbook — application]

Example 8: An Interesting — Self-reduction

How is Cu extracted by the process of self-reduction?

Solution:

Self-reduction — Definition

'Self-reduction' = One part of the ore reduced another part of the ore. No external reducing agent.

Step-by-step for Cu

Ore: Cu₂S ('matte' obtained from Cu pyrites)

Step 1: Partial Roasting

Burn part of Cu₂S in air:

$$2Cu_2S + 3O_2 \xrightarrow{\Delta} 2Cu_2O + 2SO_2\uparrow$$

Now mixture: Cu₂O + remaining Cu₂S.

Step 2: Self-reduction

Cu₂O and Cu₂S — react with each other:

$$2Cu_2O + Cu_2S \xrightarrow{\Delta} 6Cu + SO_2\uparrow$$

That is:

• Cu₂O released O (reduced).
• Cu₂S released S (oxidised).
• Finally: pure Cu is formed.

Analysis

What's what here?

In Cu₂O:

• Cu: +1
• Finally Cu: 0 → Reduction!

In Cu₂S:

• S: -2
• Finally S: +4 (in SO₂) → Oxidation!

That is — one compound of the same metal (Cu) freed Cu.

Why Possible for Cu?

Cu — less reactive. Cu₂O and Cu₂S — both not very stable. Cu freed easily.

Possible for Fe Too?

Fe — more reactive. Fe₂O₃ more stable. Self-reduction won't free Fe.

For Fe — reduction by carbon is necessary.

An Interesting Comparison

Metal	Extraction	Reducing Agent
Cu	Self-reduction	Other part of ore (S²⁻)
Zn	By C	External carbon
Fe	By C (CO)	External carbon
Hg	Self-decomposition	None (just heat)
Al	Electricity	Electric current

Practical Importance

Self-reduction — less energy-intensive. No coal needed. Efficient for industry.

This is why Cu — known since Bronze Age (3300 BC).

[Board: 5-mark]

Example 9: NCERT — Flux and Slag

What are flux and slag? Explain their role in Fe extraction.

Solution:

Definitions

Flux: Chemical added during extraction — to remove gangue.

Slag: Molten compound formed by reaction of flux + gangue.

Principle

Acidic flux + Basic gangue → Slag. Basic flux + Acidic gangue → Slag.

That is — opposite natures.

Types of Flux

1. Acidic Flux:

• $SiO_2$ (silica)
• For basic impurity.

2. Basic Flux:

• $CaO$ (lime — from decomposition of CaCO₃)
• $MgO$
• For acidic impurity.

Flux in Fe Extraction

Ore: Fe₂O₃ (Hematite)

Impurity (gangue): $SiO_2$ (silica — sand) — acidic.

Flux: $CaCO_3$ (limestone) → on heating → CaO — basic.

Reactions:

1. Decomposition of CaCO₃: $$CaCO_3 \xrightarrow{\Delta} CaO + CO_2$$

2. CaO + SiO₂ → CaSiO₃ (slag): $$CaO + SiO_2 \rightarrow CaSiO_3$$

This is slag — molten, light, floats above Fe. Released through separate opening.

Use of Slag in Industry

Don't think slag is 'waste' — many uses:

1. Cement manufacturing: Slag cement.
2. Road construction: Slag aggregate.
3. Fertilisers: Calcium source.
4. Filling material.

Flux in Cu Extraction

Impurity: FeO Flux: SiO₂

Reaction: $$FeO + SiO_2 \rightarrow FeSiO_3 \text{ (slag)}$$

That is — Cu and Fe — separated with help of flux.

Summary Table

Metal	Impurity (gangue)	Flux	Slag
Fe	SiO₂ (acidic)	CaCO₃ (basic)	CaSiO₃
Cu	FeO (basic)	SiO₂ (acidic)	FeSiO₃

A Basic Rule

Impurity + Flux = Slag By 'acidic + basic' combination.

This is the fundamental principle of 'metallurgy'.

[NCERT — important]

Example 10: A Numerical — Al Extraction

How much Al is obtained from 100 kg of Al₂O₃? (Al=27, O=16)

Solution:

Reaction: $$2Al_2O_3 \xrightarrow{\text{electricity}} 4Al + 3O_2$$

Molecular masses:

• $Al_2O_3 = (2 \times 27) + (3 \times 16) = 54 + 48 = 102$ g/mol
• 1 mol Al = 27 g

Ratio:

• 1 mol Al₂O₃ → 2 mol Al
• 102 g Al₂O₃ → 54 g Al

From 100 kg Al₂O₃:

$$\text{Al} = \frac{54}{102} \times 100 = 52.94 \text{ kg}$$

≈ 53 kg Al.

Answer: From 100 kg Al₂O₃, about 53 kg Al is obtained.

Additional Information

Yield:

• 100 kg Al₂O₃ × 53% = 53 kg Al.
• % of Al in Al₂O₃ = 53%.

Bayer Process: ~2 tonnes bauxite → ~1 tonne Al₂O₃.

Hall-Héroult: ~2 tonnes Al₂O₃ → ~1 tonne Al.

Total: ~4 tonnes bauxite → ~1 tonne pure Al.

Energy Calculation

Energy for Al extraction: ~14-15 kWh per kg of Al.

100 kg Al₂O₃ → 53 kg Al → ~750-800 kWh electricity.

This is equivalent to one month's average Indian household consumption!

Why Al Recycling is Important?

1 kg Al recycled → 95% less energy. Only 0.7-0.8 kWh, instead of 14-15 kWh.

That is, recycling = energy savings + environmental benefits.

'Aluminium can recycling' — across the world.

[Board: 3-mark numerical]

Example 11: A Logical — True or False?

State true/false for the following — with reasons:

(a) Al can be reduced by carbon. (b) Hg is extracted by electrolysis. (c) ZnCO₃ is first calcined. (d) Na is obtained from electrolysis of aqueous NaCl.

Solution:

(a) Al reduced by carbon — False ✗

Reason: Al is more reactive than C.

Reactivity: $Al > C$

More reactive — displaces less reactive. Not the other way round.

Hence: C cannot reduce Al from Al₂O₃.

Correct method: Electrolysis (Hall-Héroult).

(b) Hg by electrolysis — False ✗

Reason: Hg — less reactive. HgO very unstable — decomposes just on heating.

Electrolysis — unnecessary cost.

Correct method: $$2HgS + 3O_2 \rightarrow 2HgO + 2SO_2$$ (roasting) $$2HgO \rightarrow 2Hg + O_2$$ (decomposition)

Just heating — sufficient.

(c) ZnCO₃ first calcined — True ✓

Reason: ZnCO₃ is a carbonate ore. Calcination = converting carbonate to oxide.

$$ZnCO_3 \xrightarrow{\Delta} ZnO + CO_2$$

Now reduction of ZnO by carbon is easy.

This is the standard method.

(d) Na from aqueous NaCl — False ✗

Reason: In aqueous NaCl:

• At Cathode: Na vs H₂O — Na more reactive.
• Hence H₂O reduced (H₂ released).
• Na not formed.

Products: H₂ + Cl₂ + NaOH (chlor-alkali process).

Correct method: Electrolysis of molten NaCl (Down's cell).

Summary

Statement	T/F	Correct Answer
Al + C	✗ False	Electrolysis
Hg + electricity	✗ False	Roasting + decomposition
ZnCO₃ + calcination	✓ True	—
Na + aqueous	✗ False	Molten NaCl

Key Insight

Knowledge = reactivity series + ore type.

[Board: 5-mark logical]

Example 12: NCERT — Cu Extraction

Explain extraction of Cu from copper pyrites (CuFeS₂) — with all steps.

Solution:

Ore

Copper pyrites — $CuFeS_2$ (sulphides of both Cu and Fe).

Step 1: Concentration

Froth flotation method (Section 6). Now concentrated CuFeS₂.

Step 2: Roasting

Heat in air:

$$2CuFeS_2 + 3O_2 \xrightarrow{\Delta} 2Cu_2S + 2FeO + 2SO_2$$

Products: Cu₂S, FeO, SO₂.

Step 3: Smelting with Flux

In 'reverberatory furnace':

FeO + SiO₂ (flux) → FeSiO₃ (slag):

$$FeO + SiO_2 \rightarrow FeSiO_3$$

That is — Fe separated as slag. Cu₂S — settles as 'matte'.

Step 4: Self-reduction

Partial roasting of Cu₂S in 'Bessemer converter':

$$2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$$

Now mixture: Cu₂O + Cu₂S.

Self-reduction:

$$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2\uparrow$$

Finally — blister Cu. This is 99% pure, but has impurities.

Step 5: Refining — Section 8

By electrolytic refining → 99.99% pure Cu.

Flow Diagram

CuFeS₂ (Copper pyrites)
    ↓ Concentration (froth flotation)
Concentrated CuFeS₂
    ↓ Roasting (air + heat)
Cu₂S + FeO + SO₂
    ↓ + SiO₂ (flux)
Cu₂S (matte) + FeSiO₃ (slag)
    ↓ Air + heat
Cu₂O
    ↓ + Cu₂S (remaining)
6Cu + SO₂  (Blister Cu)
    ↓ Electrolytic refining
Pure Cu (99.99%)

Overall Reaction (At a glance)

Ore → metal — in many steps.

Purpose of each reaction:

• Roasting: sulphide to oxide.
• Smelting: removing impurities.
• Self-reduction: freeing Cu.
• Refining: purity.

Practical Use of Cu

99.99% pure Cu — for making electrical wires. High conductivity. Durability.

Lakhs of tonnes of Cu produced in India each year.

[NCERT — every year in board]

Example 13: A Comparative — 3 Examples of Carbon Reduction

Reactions for reduction by carbon for Fe, Zn, Pb — all three.

Solution:

General Principle

All three are moderately reactive. For all three: Ore → Oxide → Reduction by carbon.

Fe Extraction

Ore: Fe₂O₃ (Hematite)

Main Reactions (Blast Furnace):

1. Coal + air: $$C + O_2 \rightarrow CO_2$$

2. CO₂ + more coal: $$CO_2 + C \rightarrow 2CO$$

3. Main reduction: $$Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$$

Here reducing agent: CO (not C directly).

Suitable temperature: ~1500°C.

Zn Extraction

Ore: ZnS (Zinc blende)

Reactions:

1. Roasting: $$2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2$$

2. Reduction: $$ZnO + C \rightarrow Zn + CO\uparrow$$

Reducing agent: C directly.

Suitable temperature: ~1100°C.

Pb Extraction

Ore: PbS (Galena)

Reactions:

1. Roasting: $$2PbS + 3O_2 \rightarrow 2PbO + 2SO_2$$

2. Reduction: $$PbO + C \rightarrow Pb + CO\uparrow$$

Reducing agent: C directly.

Suitable temperature: ~1000°C.

Comparison Table

Metal	Ore	First Step	Reducer	Temp
Fe	Fe₂O₃	(None)	CO	1500°C
Zn	ZnS	Roasting → ZnO	C	1100°C
Pb	PbS	Roasting → PbO	C	1000°C

Why CO for Fe, but C directly for others?

In Blast Furnace:

• Tall furnace.
• Hot air inside.
• CO formed — throughout the furnace.
• CO more powerful reducing agent — especially at lower temperatures.

Smaller furnace for Zn, Pb:

• Higher temperature (1000-1100°C).
• C alone is enough.

An Interesting Point

If Zn formed in gas state (above 1000°C): Zn gas cools — liquid → solid. That is, pure Zn by 'distillation'.

Fe — melts to liquid. Collected at bottom.

Key Insight

All three = reduction by carbon. But — details of method differ. Specific conditions for each metal.

[Board: 5-mark comparative]

Example 14: NCERT — A Mixed Question

Answer the following:

(a) What is the formula of bauxite? (b) What is the role of cryolite? (c) Materials of cathode and anode in Hall-Héroult process? (d) What is the overall reaction?

Solution:

(a) Formula of Bauxite

Formula: $Al_2O_3 \cdot 2H_2O$

That is: hydrated aluminium oxide.

'Bauxite' — named after Les Baux village in France. (First discovered there.)

(b) Role of Cryolite

Cryolite ($Na_3AlF_6$) — a 'solvent':

Melting point of Al₂O₃ — ~2050°C. (Very expensive.) Add Al₂O₃ to cryolite — melting point drops to 950°C.

Two functions:

1. Lower melting point.
2. Increase electrical conductivity.

Additional: CaF₂ also added — further lower melting point.

(c) Cathode and Anode

Cathode (-):

• Carbon lining (inside steel box).
• Carbon itself — works as Cathode.

Anode (+):

• Carbon rods (lowered from above).
• Replaced from time to time — because they keep burning.

(d) Overall Reaction

Electrolysis:

$$2Al_2O_3(l) \xrightarrow{\text{electricity}} 4Al(l) + 3O_2(g)$$

At Cathode: $Al^{3+} + 3e^- \rightarrow Al$ At Anode: $2O^{2-} \rightarrow O_2 + 4e^-$

O₂ released at anode reacts with carbon: $$C + O_2 \rightarrow CO_2$$

Hence carbon anodes slowly burn.

Practical Details

Energy use:

• ~14-15 kWh per kg Al.
• In India: state-level power plants.
• Companies in nearby states: Hindalco, NALCO, Vedanta.

Production:

• 1 kg Al → ~2 kg Al₂O₃ needed.
• 1 kg Al₂O₃ → ~2 kg bauxite needed.
• Total: 1 kg Al → ~4 kg bauxite.

An Interesting Fact

Before Hall-Héroult — Al was so rare it was more expensive than gold!

At the 1855 Paris Exposition — Al declared 'World's most expensive metal'.

Hall (USA) and Héroult (France) — both discovered this method independently in 1886.

Since then, the price of Al — dropped 200 times.

Today: Al → everywhere — aircraft, cars, windows, utensils.

[NCERT — important]

Example 15: An Interesting — History of Steel

From Fe extraction to 'steel' — the entire process.

Solution:

Step 1: Pig Iron from Blast Furnace

Fe₂O₃ + 3CO → 2Fe + 3CO₂ (in Blast Furnace)

Product: Pig iron

• 4% C
• Other impurities (S, P, Si, Mn)
• Brittle, weak.

Step 2: Cast Iron from Pig Iron

Melt pig iron in a smaller furnace — remove some impurities.

Product: Cast iron

• 2-4% C
• Hard, but brittle.
• Use: heavy kitchen utensils, pipes.

Step 3: Wrought Iron from Cast Iron

Further purification — remove most carbon.

Product: Wrought iron

• < 0.1% C
• Soft, ductile.
• Use: gates, grills.

Step 4: Manufacturing of Steel

Cast iron in special furnace. Carbon and other impurities controlled.

Product: Steel

• 0.2-2% C
• Very hard and ductile (both).
• Backbone of modern construction.

Bessemer Process (Steel Manufacturing)

Setup:

• 'Bessemer converter' — egg-shaped.
• Air blown through molten pig iron.

Reactions:

1. C + O₂ → CO₂ (carbon removed) 2. Si + O₂ → SiO₂ 3. P → P₂O₅ 4. Mn → MnO

Impurities — as slag. Finally: steel with controlled C.

Types of Steel

Type	C %	Use
Mild Steel	0.2-0.3%	Buildings, vehicles
Medium Carbon Steel	0.3-0.6%	Railway tracks
High Carbon Steel	0.6-1.5%	Tools, knives
Stainless Steel	+ Cr, Ni	Utensils, kitchen

Alloying Elements in Steel

Steel = Fe + C + others.

Other metals:

• Cr (stainless): corrosion-resistant.
• Ni: durable.
• Mn: hardness.
• W: high-temperature.
• V: tool steel.

Fe Industry in India

Major companies:

• TATA Steel (Jamshedpur)
• SAIL (Bhilai, Bokaro)
• JSW Steel
• Vizag Steel

India — second-largest producer of steel in the world.

Summary of the Entire Process

Fe₂O₃ (ore)
   ↓ Blast Furnace + coke + CaCO₃
Pig Iron (4% C)
   ↓ Some refining
Cast Iron (2-4% C)
   ↓ More refining
Wrought Iron (<0.1% C) or Steel (0.2-2% C)

An Interesting Fact

1856 — Bessemer's discovery — a pillar of modern 'Industrial Revolution'. Before that, steel was expensive. After Bessemer — cheap steel — railways, buildings, bridges.

[Board + General Knowledge]

Example 16: A Concluding Question

(a) 3 steps of metal extraction. (b) Al extraction — complete. (c) Main reaction in Fe extraction. (d) Why can't Na be reduced by carbon?

Solution:

(a) 3 Steps of Extraction

Step 1: Concentration

• Removing gangue.
• Methods: hand picking, gravity, magnetic, froth flotation.

Step 2: Reduction

• Ore → metal.
• Methods: by carbon, by electricity, self-reduction.

Step 3: Refining

• Purification.
• Main: electrolytic.

(b) Al Extraction

Ore: Bauxite (Al₂O₃·2H₂O)

1. Concentration (Bayer): $$Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O$$ $$NaAlO_2 + 2H_2O \rightarrow NaOH + Al(OH)_3$$ $$2Al(OH)_3 \xrightarrow{\Delta} Al_2O_3 + 3H_2O$$

2. Electrolysis (Hall-Héroult):

• Al₂O₃ + cryolite (Na₃AlF₆) → melting point 950°C.
• Cathode: $Al^{3+} + 3e^- \rightarrow Al$.
• Anode: $2O^{2-} \rightarrow O_2 + 4e^-$.

Overall: $2Al_2O_3 \xrightarrow{\text{electricity}} 4Al + 3O_2$.

(c) Main Reaction in Fe Extraction

Ore: Fe₂O₃

In Blast Furnace:

Reducing agent: CO (formed from coal).

$$Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$$

This is the main reaction.

Others:

• $C + O_2 \rightarrow CO_2$
• $CO_2 + C \rightarrow 2CO$
• $CaCO_3 \rightarrow CaO + CO_2$
• $CaO + SiO_2 \rightarrow CaSiO_3$ (slag)

(d) Why Can't Na Be Reduced by C?

Principle: More reactive metal — displaces less reactive.

Reactivity: Na > C.

Hence: Na, more strongly bonded with O₂ than C is. C cannot free Na from Na₂O.

If we try: $Na_2O + C \rightarrow ?$ No reaction.

Reverse can happen: $Na + CO_2 \rightarrow Na_2O + C$ — but this is not Na extraction.

Correct method: Electrolysis of molten NaCl (Down's cell).

Here electric current — provides enough energy to 'free' Na.

Summary

Highly reactive metals (Na, K, Ca, Mg, Al) — electrolysis. Moderate (Zn, Fe, Pb) — reduction by carbon. Less reactive (Hg, Cu, Ag) — direct heating.

This is a basic rule — the essence of the entire section.

[Board: 5-mark mixed question]