Extraction of Copper

Q: How is copper extracted from copper pyrites?

Copper pyrites is concentrated by froth flotation, partly roasted in air, and smelted with silica so that iron leaves as ferrous silicate slag and copper collects as matte. Air blown through the matte in a converter removes the remaining iron and frees copper by self-reduction. The blister copper is refined by poling and electrolysis.

Q: What is copper matte?

Copper matte is the molten mixture of copper(I) sulphide with a smaller amount of iron(II) sulphide that forms when roasted copper pyrites is smelted with silica. It sinks below the iron silicate slag, is tapped off, and is then blown with air in a converter to give blister copper.

Q: Why is it called blister copper?

When molten copper from the converter cools and solidifies, sulphur dioxide dissolved in it escapes and leaves blisters on the surface. This copper, about 98 percent pure, is therefore called blister copper. It still contains iron, sulphur and precious metals and must be refined.

Q: What is self-reduction in copper extraction?

Self-reduction, or auto-reduction, frees copper without an added reducing agent. In the converter part of the copper(I) sulphide is oxidised to copper(I) oxide, which then reacts with the remaining sulphide to give copper metal and sulphur dioxide. It becomes feasible above about 860 K.

Q: Why can carbon not reduce copper sulphide?

Carbon disulphide has a positive Gibbs energy of formation and is much less stable than copper sulphide, so the reaction of copper sulphide with carbon is strongly uphill, about plus 238 kJ. Hydrogen fails for the same reason. Sulphide ores are therefore roasted, and copper is freed by self-reduction instead.

Q: What is anode mud in the refining of copper?

Anode mud is the sludge that settles below the impure copper anodes during electrolytic refining. It contains metals less reactive than copper, such as silver, gold and platinum, together with selenium, tellurium and antimony. These do not dissolve, and their recovery pays for much of the refining.

Q: Is the extraction of copper in the NEET syllabus?

No. NEET dropped metallurgy in 2024, so matte smelting and the converter are not examined. NEET students still meet the chemistry behind them: copper's +1 and +2 oxidation states and its positive electrode potential in the d-block and electrochemistry chapters, which explain why iron displaces copper from solution.

ChemistryGeneral Principles And Processes Of Isolation Of ElementsFor JEE aspirants

The extraction of copper from copper pyrites, $CuFe S_{2}$ , follows five steps: froth flotation, partial roasting, smelting with silica to copper matte, Bessemerisation, in which copper sulphide reduces itself to blister copper, and refining by poling and electrolysis. The extraction of copper shows why sulphide ores are roasted instead of being reduced by carbon. JEE Advanced lists copper among the thermodynamic principles of metallurgy.

On this page1Ores2Route3Roasting4Matte5Converter6Thermodynamics7Refining8Wet process9Uses and alloys10Mind map11Examples

Key Formulas - Quick Reference

Ores: copper pyrites $CuFe S_{2}$ (chief), copper glance $C u_{2} S$ , cuprite $C u_{2} O$ , malachite $CuC O_{3} \cdot Cu (OH)_{2}$ , azurite $2CuC O_{3} \cdot Cu (OH)_{2}$
★ Must learnPartial roasting: $2CuFe S_{2} + O_{2} \to C u_{2} S + 2FeS + S O_{2}$
★ Must learnSmelting: $C u_{2} O + FeS \to C u_{2} S + FeO$ ; $FeO + Si O_{2} \to FeSi O_{3}$ (slag); matte = $C u_{2} S$ + FeS
Slag blow: $2FeS + 3 O_{2} \to 2FeO + 2S O_{2}$ , then FeO removed as $FeSi O_{3}$
★ Must learnCopper blow (self-reduction): $2C u_{2} S + 3 O_{2} \to 2C u_{2} O + 2S O_{2}$ ; $C u_{2} S + 2C u_{2} O \to 6Cu + S O_{2}$
★ Must learnBlister copper: about 98 % Cu; blisters come from escaping $S O_{2}$
Poling: $4C u_{2} O + C H_{4} \to 8Cu + C O_{2} + 2 H_{2} O$
★ Must learnElectrorefining: anode $Cu \to C u^{2 +} + 2 e^{-}$ ; cathode $C u^{2 +} + 2 e^{-} \to Cu$ ; anode mud: Ag, Au, Pt, Se, Te, Sb
Low-grade ores: $CuS O_{4} + Fe \to FeS O_{4} + Cu$ ; $C u^{2 +} + H_{2} \to Cu + 2 H^{+}$

1. Occurrence and Ores of Copper

Copper makes up only about 0.006 % of the Earth's crust, but its deposits are concentrated, which makes mining worthwhile. It occurs native in large amounts near Lake Superior (USA), in the Ural mountains and in Sweden. In India, copper ores are mined at Khetri (Rajasthan), Singhbhum (Jharkhand) and Malanjkhand (Madhya Pradesh). Most copper, however, comes from sulphide ores.

Ore	Formula	Type	Cu (% by mass)
Copper pyrites (chalcopyrite)	$CuFe S_{2}$	sulphide; chief ore	34.6
Copper glance (chalcocite)	$C u_{2} S$	sulphide	79.9
Bornite (peacock ore)	$C u_{5} Fe S_{4}$	sulphide	63.3
Cuprite (ruby copper)	$C u_{2} O$	oxide; red	88.8
Malachite	$CuC O_{3} \cdot Cu (OH)_{2}$	basic carbonate; green	57.5
Azurite	$2CuC O_{3} \cdot Cu (OH)_{2}$	basic carbonate; blue	55.3

The percentages are for the pure minerals; actual ores contain only 0.5-2 % copper, which is why concentration is essential.

2. Choosing the Route

Ore	Process	Idea
Sulphide ore, rich enough to smelt	Dry process (smelting)	roast, smelt to matte, convert to blister copper, refine
Poor sulphide ore	Wet process (hydrometallurgy)	turn Cu into soluble $CuS O_{4}$ , then precipitate Cu with Fe or by electrolysis
Oxide and carbonate ores	Reduction or leaching	calcine, reduce with carbon, or leach with dilute $H_{2} S O_{4}$

Figure 1 turns the table into two questions.

Figure 1: Choosing the copper route. The ore type and its copper content decide between the dry process, the wet process and simple reduction.

The dry process from copper pyrites is the main route, and it is outlined in Figure 2.

Figure 2: Route map for copper. Iron is removed in two stages as iron silicate slag, and copper is finally set free by self-reduction, with no carbon needed.

3. Concentration and Partial Roasting

3.1 Froth flotation

The ore is crushed, sieved and concentrated by froth flotation: the powdered ore is stirred with water, pine oil and a collector such as a xanthate while air is blown through. The sulphide particles rise with the froth and are skimmed off; the siliceous gangue sinks.

3.2 Partial roasting

The concentrate is heated in a reverberatory furnace in a current of air, below its melting point. Free sulphur, arsenic and antimony burn off as volatile oxides, and the pyrites breaks down into the sulphides of copper and iron:

S + O_{2} S O_{2} ↑

4As + 3 O_{2} 2A s_{2} O_{3} ↑

4Sb + 3 O_{2} 2S b_{2} O_{3} ↑

2CuFe S_{2} + O_{2} C u_{2} S + 2FeS + S O_{2} ↑

Part of each sulphide is then oxidised:

2FeS + 3 O_{2} 2FeO + 2S O_{2} ↑

2C u_{2} S + 3 O_{2} 2C u_{2} O + 2S O_{2} ↑

The air supply is kept limited so that roasting stays partial. Iron sulphide is oxidised more readily than copper sulphide, and enough $C u_{2} S$ must survive for the next two steps.

Exam Trick

Roast copper only halfway. Full roasting to CuO would force you to use carbon later. Copper sulphide is kept, because in the converter it reduces its own oxide.

4. Smelting to Copper Matte

The roasted ore is mixed with silica (and a little coke as fuel) and heated strongly. NCERT describes this step in a reverberatory furnace; lump ore is also smelted in a small water-jacketed blast furnace, about 5-6 m high, fed with an air blast through tuyeres (Figure 3).

Two changes take place. First, any $C u_{2} O$ formed during roasting is turned back into sulphide by iron sulphide, because copper holds sulphur more strongly and iron holds oxygen more strongly:

C u_{2} O + FeS C u_{2} S + FeO

Second, the iron oxide combines with silica to form a fusible slag:

2FeS + 3 O_{2} 2FeO + 2S O_{2}

FeO + Si O_{2} FeSi O_{3}

Two liquid layers form. The lighter slag of ferrous silicate floats and is run off. The heavier layer is copper matte, a molten mixture of $C u_{2} S$ with a smaller amount of FeS.

Figure 3: Smelting in a reverberatory furnace. The charge is heated by gases reflected from the roof, never touching the fuel; iron leaves as

FeSi O_{3}

slag above the heavier matte.

JEE Advanced

Why does copper end up in the matte and iron in the slag? Using $Δ_{f} G^{\circ}$ values (kJ mol^-1): $C u_{2} O$ $- 146$ , FeS $- 100$ , $C u_{2} S$ $- 86$ , FeO $- 251$ ,

Δ_{r} G^{\circ} = [(- 86) + (- 251)] - [(- 146) + (- 100)] = - 91 kJ

The exchange $C u_{2} O + FeS \to C u_{2} S + FeO$ is strongly downhill. Iron grabs the oxygen (and then the silica), while copper keeps the sulphur, so every bit of copper is gathered into the sulphide matte and the iron is slagged off.

5. Bessemerisation: Self-Reduction to Blister Copper

The molten matte is transferred to a Bessemer converter, a little silica is added, and a blast of air is blown through the melt from tuyeres (Figure 4). The heat of the reactions keeps the charge molten. Blowing takes place in two stages.

Stage 1: slag blow. The remaining iron sulphide is oxidised and removed as slag, which is poured off by tilting the converter. After about three hours all the iron is gone:

2FeS + 3 O_{2} 2FeO + 2S O_{2} ↑

FeO + Si O_{2} FeSi O_{3}

Stage 2: copper blow. The blast is continued (about two hours more). Part of the copper sulphide is oxidised, and the cuprous oxide then reacts with the remaining sulphide. This is self-reduction (auto-reduction): no separate reducing agent is needed.

2C u_{2} S + 3 O_{2} 2C u_{2} O + 2S O_{2} ↑

C u_{2} S + 2C u_{2} O 6Cu + S O_{2} ↑

The molten copper is poured out and allowed to solidify. $S O_{2}$ dissolved in the liquid escapes as it sets, leaving blisters on the surface, so the product is called blister copper. It is about 98 % pure.

NCERT describes a silica-lined converter, in which the lining itself serves as flux. Modern converters have a basic (magnesite) lining, with silica added as flux; the excess silica is taken up by the slag.

Figure 4: Bessemerisation of matte. First the iron is blown out as slag; then copper sulphide reduces itself, and the escaping

S O_{2}

leaves blisters on the solidified metal.

Exam Trick

Iron leaves twice, both times as $FeSi O_{3}$ . Some goes as slag in smelting, the rest in the slag blow. Copper never meets carbon: it is freed by $C u_{2} S + 2C u_{2} O \to 6Cu + S O_{2}$ .

Quick Recall: tap to check

What is copper matte?

Molten

C u_{2} S

with a little FeS.

Which reaction frees copper in the converter?

C u_{2} S + 2C u_{2} O \to 6Cu + S O_{2}

(self-reduction).

Why is the product called blister copper?

Escaping

S O_{2}

leaves blisters as it solidifies.

Key idea

Iron is slagged off twice as

FeSi O_{3}

; copper is then freed by its own sulphide, with no carbon at all.

6. The Thermodynamics of Copper Extraction

After iron, copper is the classic example of thermodynamic principles in metallurgy. Three facts explain the route (Figure 5).

Cuprous oxide is easy to reduce. The Cu/ $C u_{2} O$ line lies near the top of the Ellingham diagram, so hot coke reduces it readily: $C u_{2} O + C \to 2Cu + CO$ .

Copper sulphide cannot be reduced by carbon or hydrogen. The $Δ_{f} G^{\circ}$ of $C S_{2}$ is positive and that of $H_{2} S$ is less negative than that of $C u_{2} S$ , so these reductions are uphill (298 K values):

2C u_{2} S + C 4Cu + C S_{2} Δ_{r} G^{\circ} \approx + 238 kJ

C u_{2} S + H_{2} 2Cu + H_{2} S Δ_{r} G^{\circ} \approx + 53 kJ

That is why sulphide ores are roasted rather than reduced directly.

Self-reduction becomes downhill when hot. For $C u_{2} S + 2C u_{2} O \to 6Cu + S O_{2}$ , $Δ H^{\circ} \approx + 120$ kJ and $Δ S^{\circ} \approx + 140$ J K^-1 (a gas is formed). $Δ G^{\circ}$ is $+ 78$ kJ at 298 K but becomes negative above $T = Δ H^{\circ} /Δ S^{\circ} \approx 860$ K. The converter works near 1500 K, where $Δ G^{\circ} \approx - 90$ kJ.

Figure 5: Why copper is made by self-reduction. Carbon and hydrogen cannot reduce

C u_{2} S

at furnace temperatures, but

C u_{2} S + 2C u_{2} O

turns downhill above about 860 K because a gas (

S O_{2}

) is formed. Lines are approximate, from 298 K data.

Key idea

Carbon cannot take sulphur from

C u_{2} S

, but the self-reduction turns downhill above about 860 K because it releases gaseous

S O_{2}

7. Refining of Blister Copper

7.1 Poling (fire refining)

Blister copper is melted in a reverberatory furnace lined with silica, and air is blown over it. Part of the copper is oxidised to $C u_{2} O$ , which dissolves in the melt and passes its oxygen to the more reactive impurities (Fe, Zn, Pb, Ni); their oxides form a slag, while S and As leave as volatile oxides. The dissolved $C u_{2} O$ is then removed by stirring the melt with poles of green wood. The hydrocarbons released, such as methane, reduce it:

4C u_{2} O + C H_{4} 8Cu + C O_{2} + 2 H_{2} O

Figure 6 shows the two stages.

Figure 6: Poling works in two stages: oxidise the impurities with air, then remove the extra

C u_{2} O

with the gases from green wood.

The product is about 99.5 % pure. Copper for electrical work must be purer still, so it is refined electrolytically.

7.2 Electrolytic refining

Thick plates of impure copper are the anodes, thin sheets of pure copper are the cathodes, and acidified copper sulphate solution is the electrolyte (Figure 7). On passing current, copper dissolves from the anode and deposits on the cathode:

CuS O_{4} ⇌ C u^{2 +} + S O_{4}^{2 -}

Cu anode C u^{2 +} + 2 e^{-}

C u^{2 +} + 2 e^{-} cathode Cu

Impurities more reactive than copper (Fe, Zn, Ni) dissolve at the anode and stay in solution; the voltage is too low to deposit them at the cathode.
Impurities less reactive than copper (Ag, Au, Pt), together with Se, Te and Sb, do not dissolve. They fall below the anode as anode mud, whose recovery pays for much of the refining.
The cathode copper is 99.9 % pure or better.

Figure 7: Electrolytic refining. Only copper moves from anode to cathode; more reactive metals stay dissolved and less reactive ones drop out as valuable anode mud.

Poling

Fire refining of molten blister copper.
Air oxidises impurities; green poles reduce $C u_{2} O$ .
Gives about 99.5 % Cu.
No precious metals recovered.

Electrolytic refining

Impure anode, pure cathode, acidified $CuS O_{4}$ .
Fe, Zn stay dissolved; Ag, Au, Pt form anode mud.
Gives 99.9 % Cu or better.
Precious metals recovered.

Quick Recall: tap to check

Which is the anode in copper refining?

The impure copper plate.

What removes

C u_{2} O

in poling?

Hydrocarbon gases from green wood, such as methane.

Key idea

Poling cleans the melt; electrolysis gives electrical-grade copper and pays for itself through the anode mud.

8. Wet Process and Non-Sulphide Ores

8.1 Wet process (hydrometallurgy) for poor ores

Low-grade sulphide ores and mine waste are not worth smelting. The copper is turned into a soluble compound, either by roasting in air or by leaching with dilute acid or with bacteria:

2C u_{2} S + 5 O_{2} Δ 2CuS O_{4} + 2CuO

Copper is then precipitated from the solution with scrap iron, or with hydrogen:

CuS O_{4} + Fe FeS O_{4} + Cu ↓

C u^{2 +} (aq) + H_{2} (g) Cu (s) + 2 H^{+} (aq)

Figure 8 shows the wet route.

Figure 8: The wet process. Acid or bacteria dissolve the copper; iron, being more reactive, puts it back as metal.

8.2 Oxide and carbonate ores

These ores are crushed, concentrated by gravity and calcined in a reverberatory furnace, which decomposes the carbonate to the oxide:

CuC O_{3} \cdot Cu (OH)_{2} Δ 2CuO + C O_{2} + H_{2} O

The oxide is then either reduced with carbon:

CuO + C Δ Cu + CO

C u_{2} O + C Δ 2Cu + CO

or leached with dilute sulphuric acid, and copper is recovered from the solution by electrolysis (copper cathode, inert lead anode) or with scrap iron:

CuO + H_{2} S O_{4} CuS O_{4} + H_{2} O

C u^{2 +} + 2 e^{-} copper cathode Cu

2 H_{2} O lead anode O_{2} + 4 H^{+} + 4 e^{-}

9. Properties, Uses and Alloys of Copper

Copper is a reddish-brown, malleable and ductile metal (melting point 1358 K) and, after silver, the best conductor of electricity. With $E^{\circ} (C u^{2 +} /Cu) = + 0.34$ V it does not release hydrogen from dilute non-oxidising acids. In moist air it slowly gains a green coating of basic copper carbonate.

Electrical wires and cables; water and steam pipes.
Alloys such as brass (with zinc), bronze (with tin) and cupronickel coins (with nickel).

Alloy	Composition (%)	Uses
Brass	Cu 60-80, Zn 20-40	utensils, fittings, cartridges
Bronze	Cu 75-90, Sn 10-25	statues, medals, coins
Bell metal	Cu 80, Sn 20	bells
German silver	Cu 50-60, Zn 20-25, Ni 20-25	cutlery, ornaments
Gun metal	Cu 87, Sn 10, Zn 3	gears, castings
Aluminium bronze	Cu 90, Al 10	imitation gold jewellery, coins
Cupronickel	Cu 75, Ni 25	coins

10. Summary Mind Map

The whole extraction at a glance (Figure 9).

Figure 9: Mind map of copper: roast partly, smelt to matte, blow to blister copper, refine.

11. Solved Examples

Solved Example 1

In the extraction of copper from copper pyrites, iron is finally removed as
(A) FeO
(B)

FeSi O_{3}

(C)

F e_{2} O_{3}

(D) FeS

Solution:

Answer: (B). Iron sulphide is oxidised to FeO, which combines with the silica flux: $FeO + Si O_{2} \to FeSi O_{3}$ . The ferrous silicate slag floats and is run off.

Solved Example 2

Copper matte contains
(A)

C u_{2} O

and FeS
(B)

C u_{2} S

and FeS
(C)

C u_{2} S

and

FeSi O_{3}

(D) Cu and

C u_{2} O

Solution:

Answer: (B). Matte is molten $C u_{2} S$ with a little FeS. Any $C u_{2} O$ is converted back to sulphide by FeS: $C u_{2} O + FeS \to C u_{2} S + FeO$ .

Solved Example 3

Why is copper pyrites only partially roasted, and what is the role of silica in smelting?

Solution:

Enough $C u_{2} S$ must remain, because in the converter it reduces cuprous oxide (self-reduction), so no carbon is needed. Partial roasting removes S, As and Sb and oxidises part of the FeS to FeO. Silica is the acidic flux that turns this basic FeO into fusible $FeSi O_{3}$ slag.

Solved Example 4

Write the reactions in the Bessemer converter during the extraction of copper and explain the name blister copper.

Solution:

Slag blow:

2FeS + 3 O_{2} 2FeO + 2S O_{2}

FeO + Si O_{2} FeSi O_{3}

Copper blow (self-reduction):

2C u_{2} S + 3 O_{2} 2C u_{2} O + 2S O_{2}

C u_{2} S + 2C u_{2} O 6Cu + S O_{2}

$S O_{2}$ dissolved in the molten copper escapes as the metal solidifies and leaves blisters on its surface.

Solved Example 5

A copper ore contains 20.0 % copper pyrites by mass. What mass of copper can be obtained from 1.00 tonne of ore? (Cu = 63.5, Fe = 55.8, S = 32.1)

Solution:

Molar mass of $CuFe S_{2}$ $= 63.5 + 55.8 + 2 \times 32.1 = 183.5$ g mol^-1, so copper is $\frac{63.5}{183.5} \times 100 = 34.6 %$ of it.

Mass of pyrites $= 0.200 \times 1000 = 200$ kg. Copper $= 200 \times 0.346 \approx 69.2$ kg.

Solved Example 6

Using

Δ_{f} G^{\circ}

values (kJ mol^-1)

C S_{2}

+ 65.3

and

C u_{2} S

- 86.2

, show that carbon cannot reduce copper glance at 298 K.

Solution:

2C u_{2} S + C 4Cu + C S_{2}

Δ_{r} G^{\circ} = (+ 65.3) - 2 (- 86.2) = + 237.7 kJ

$Δ_{r} G^{\circ}$ is large and positive, so the reaction is not feasible. Carbon disulphide is far less stable than copper sulphide; the sulphide must be roasted instead.

Solved Example 7

For the self-reduction

C u_{2} S + 2C u_{2} O \to 6Cu + S O_{2}

Δ H^{\circ} = + 120

kJ and

Δ S^{\circ} = + 140

J K^-1. Above what temperature is it feasible? Find

Δ G^{\circ}

at 1500 K.

Solution:

Feasible when $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} < 0$ :

T > \frac{Δ H ^{\circ}}{Δ S ^{\circ}} = \frac{120 000}{140} \approx 857 K

At 1500 K: $Δ G^{\circ} = 120 - 1500 \times 0.140 = - 90$ kJ, so the reaction runs well in the converter. The positive $Δ S^{\circ}$ (gaseous $S O_{2}$ is formed) is what makes heating help.

Solved Example 8

A current of 10.0 A is passed for 1.00 h through a copper refining cell. What mass of copper deposits on the cathode? (Cu = 63.5, F = 96 500 C mol^-1)

Solution:

C u^{2 +} + 2 e^{-} Cu

m = \frac{I tM}{n F} = \frac{10.0 \times 3600 \times 63.5}{2 \times 96 500} = 11.8 g

Practice Questions

Name the flux used in the extraction of copper and the slag it forms.Answer: Silica; it removes FeO as ferrous silicate, $FeSi O_{3}$ .
Why does copper not need a reducing agent in the converter?Answer: It is freed by self-reduction: $C u_{2} S + 2C u_{2} O \to 6Cu + S O_{2}$ .
Name two metals recovered from the anode mud of copper refining.Answer: Any two of silver, gold and platinum (also selenium and tellurium).
How is copper obtained from malachite?Answer: Calcination to CuO, then reduction with carbon ( $CuO + C \to Cu + CO$ ) or leaching with dilute $H_{2} S O_{4}$ and electrolysis.
What is the purity of blister copper, and what does poling remove?Answer: About 98 %; poling reduces dissolved $C u_{2} O$ after the other impurities have been oxidised into slag.
In the electrolytic refining of copper, which of Fe, Zn, Ag and Ni collects in the anode mud?Answer: Ag (with Au and Pt); Fe, Zn and Ni are more reactive, dissolve and stay in solution.
What mass of scrap iron precipitates 1.00 kg of copper from $CuS O_{4}$ solution? (Cu = 63.5, Fe = 55.8)Answer: 1 mol Fe per mol Cu: $1000 \times \frac{55.8}{63.5} \approx 879$ g.

Common Mistakes to Avoid

Watch out

Writing matte as $C u_{2} O$ + FeS. Matte is $C u_{2} S$ + FeS; iron sulphide turns any $C u_{2} O$ back into $C u_{2} S$ .
Roasting copper pyrites completely. Roasting is only partial, so that $C u_{2} S$ is left for self-reduction.
Using coke to reduce $C u_{2} S$ . Neither carbon nor hydrogen reduces copper sulphide; copper comes from self-reduction.
Calling blister copper pure. It is about 98 % Cu and must still be refined.
Saying the blisters are caused by air or oxygen. They are caused by escaping $S O_{2}$ .
Making the impure copper the cathode. Impure copper is the anode; the cathode is a thin sheet of pure copper.
Putting Fe and Zn in the anode mud. More reactive metals dissolve and stay in solution; only less reactive ones (Ag, Au, Pt) form anode mud.
Saying iron is removed as FeO. It leaves as $FeSi O_{3}$ slag.

Frequently Asked Questions

How is copper extracted from copper pyrites?

Copper pyrites is concentrated by froth flotation, partly roasted in air, and smelted with silica so that iron leaves as ferrous silicate slag and copper collects as matte. Air blown through the matte in a converter removes the remaining iron and frees copper by self-reduction. The blister copper is refined by poling and electrolysis.

What is copper matte?

Copper matte is the molten mixture of copper(I) sulphide with a smaller amount of iron(II) sulphide that forms when roasted copper pyrites is smelted with silica. It sinks below the iron silicate slag, is tapped off, and is then blown with air in a converter to give blister copper.

Why is it called blister copper?

When molten copper from the converter cools and solidifies, sulphur dioxide dissolved in it escapes and leaves blisters on the surface. This copper, about 98 percent pure, is therefore called blister copper. It still contains iron, sulphur and precious metals and must be refined.

What is self-reduction in copper extraction?

Self-reduction, or auto-reduction, frees copper without an added reducing agent. In the converter part of the copper(I) sulphide is oxidised to copper(I) oxide, which then reacts with the remaining sulphide to give copper metal and sulphur dioxide. It becomes feasible above about 860 K.

Why can carbon not reduce copper sulphide?

Carbon disulphide has a positive Gibbs energy of formation and is much less stable than copper sulphide, so the reaction of copper sulphide with carbon is strongly uphill, about plus 238 kJ. Hydrogen fails for the same reason. Sulphide ores are therefore roasted, and copper is freed by self-reduction instead.

What is anode mud in the refining of copper?

Anode mud is the sludge that settles below the impure copper anodes during electrolytic refining. It contains metals less reactive than copper, such as silver, gold and platinum, together with selenium, tellurium and antimony. These do not dissolve, and their recovery pays for much of the refining.

Which copper extraction questions are asked in JEE Advanced?

JEE Advanced lists copper among the thermodynamic principles of metallurgy. Questions ask for the composition of matte, the role of silica, the converter reactions and self-reduction, why carbon cannot reduce copper sulphide, and which impurities go to the anode mud during electrolytic refining.

Is the extraction of copper in the NEET syllabus?

No. NEET dropped metallurgy in 2024, so matte smelting and the converter are not examined. NEET students still meet the chemistry behind them: copper's +1 and +2 oxidation states and its positive electrode potential in the d-block and electrochemistry chapters, which explain why iron displaces copper from solution.

Previous year questions on Extraction of Copper

1 question from past papers, each with a step-by-step solution.

JEE Advanced 2023 Paper 1, Chemistry Section 1 Q1

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