ScienceCBSE Class 10 · Metals and Non-metals

Metallurgy: Extraction of Metals from Ores

Updated 14 October 2026 · Fact-checked

Metallurgy is extracting metals from their ores and refining them. The method depends on reactivity: low-reactivity metals come from their oxides or sulphides by heating alone, middle ones by roasting or calcination followed by reduction with carbon, and highly reactive ones by electrolysis of molten compounds. Impure metals are then refined electrolytically.

The more reactive the metal, the harder it is to extract: electrolysis at the top, carbon reduction in the middle, heat alone at the bottom.

Illustration drawn for QuestionBuddy.

Understand Metallurgy: Extraction of Metals from Ores

Minerals and ores. The earth's crust is the main source of metals; seawater also has some soluble salts. Elements or compounds that occur naturally in the crust are minerals. A mineral from which a metal can be extracted profitably, because it has a high percentage of that metal, is an ore. Mined ore comes mixed with soil, sand and other impurities called gangue, which are removed first (enrichment), using differences in physical or chemical properties between ore and gangue.

Where metals are found. The least reactive metals at the bottom of the series, such as gold, silver, platinum and copper, are often found free (native). Copper and silver also occur as sulphide or oxide ores. Middle metals such as zinc, iron and lead occur mainly as oxides, sulphides or carbonates. The top metals, K, Na, Ca, Mg and Al, are so reactive that they never occur free. Many ores are oxides because oxygen is very reactive and abundant.

1. Low-reactivity metals: heat alone. Their oxides break down on heating. Cinnabar, $H g S$ , is heated in air to form mercuric oxide, which on further heating gives mercury: $2 H g S + 3 O_{2} \to 2 H g O + 2 S O_{2}$ , then $2 H g O \to 2 H g + O_{2}$ . Copper is obtained from $C u_{2} S$ in the same way: $2 C u_{2} S + 3 O_{2} \to 2 C u_{2} O + 2 S O_{2}$ , then $2 C u_{2} O + C u_{2} S \to 6 C u + S O_{2}$ .

2. Middle metals: convert to oxide, then reduce. It is easier to get a metal from its oxide than from its sulphide or carbonate.

Roasting: heating a sulphide ore strongly in excess air: $2 Z n S + 3 O_{2} \to 2 Z n O + 2 S O_{2}$ .
Calcination: heating a carbonate ore strongly in limited air: $Z n C O_{3} \to Z n O + C O_{2}$ .
Reduction: the oxide is heated with a reducing agent such as carbon (coke): $Z n O + C \to Z n + C O$ . Getting a metal from its compound is a reduction.

Highly reactive metals such as sodium, calcium and aluminium can also act as reducing agents by displacement. $3 M n O_{2} + 4 A l \to 3 M n + 2 A l_{2} O_{3}$ releases so much heat that the metal forms molten. The thermit reaction, $F e_{2} O_{3} + 2 A l \to 2 F e + A l_{2} O_{3}$ , is used to join railway tracks and cracked machine parts.

3. Highly reactive metals: electrolytic reduction. Carbon cannot reduce the oxides of sodium, magnesium, calcium or aluminium, because these metals have a greater affinity for oxygen than carbon has. They are obtained by electrolysis of their molten chlorides (aluminium from its oxide). The metal is deposited at the negative electrode (cathode), $N a^{+} + e^{-} \to N a$ , and chlorine is given off at the positive electrode (anode), $2 C l^{-} \to C l_{2} + 2 e^{-}$ .

4. Refining. Metals from these processes are impure. In electrolytic refining (used for copper, zinc, tin, nickel, silver, gold and others), the impure metal is the anode, a thin strip of pure metal is the cathode, and a solution of a salt of the metal is the electrolyte (acidified copper sulphate for copper). On passing current, metal from the anode dissolves into the electrolyte and an equal amount of pure metal is deposited on the cathode. Soluble impurities stay in solution; insoluble ones settle below the anode as anode mud.

Metallurgy: Extraction of Metals from Ores: formulas and key facts

Mercury from cinnabar: $2 H g S + 3 O_{2} heat 2 H g O + 2 S O_{2}$; Then $2 H g O heat 2 H g + O_{2}$ .
Copper from copper glance: $2 C u_{2} O + C u_{2} S heat 6 C u + S O_{2}$; After $2 C u_{2} S + 3 O_{2} \to 2 C u_{2} O + 2 S O_{2}$ .
Roasting (excess air): $2 Z n S + 3 O_{2} heat 2 Z n O + 2 S O_{2}$; For sulphide ores.
Calcination (limited air): $Z n C O_{3} heat Z n O + C O_{2}$; For carbonate ores.
Reduction by carbon: $Z n O + C \to Z n + C O$; Carbon (coke) is the reducing agent.
Thermit reaction: $F e_{2} O_{3} + 2 A l \to 2 F e + A l_{2} O_{3}$; Highly exothermic; iron forms molten; joins railway tracks.
Electrolysis of molten NaCl: $N a^{+} + e^{-} \to N a; 2 C l^{-} \to C l_{2} + 2 e^{-}$; Metal at cathode (negative); chlorine at anode (positive).

How to solve Metallurgy: Extraction of Metals from Ores questions

For 'how is metal X extracted' questions, decide the route from the reactivity series, then write each step.

1Place the metal: top (K to Al), middle (Zn, Fe, Pb, and Cu in many questions), or bottom (Hg, Ag, Au, Cu from sulphide).
2Top: write 'electrolytic reduction of the molten chloride (or oxide)' and the electrode reactions.
3Middle: identify the ore type. Sulphide: roasting in excess air. Carbonate: calcination in limited air. Then reduction of the oxide with carbon or a more reactive metal.
4Bottom: heating in air converts the sulphide to oxide, and further heating reduces it.
5Finish with electrolytic refining if purity is asked: anode impure metal, cathode pure strip, electrolyte a salt solution of the metal, anode mud below the anode.
6Balance every equation and write 'heat' over the arrow where it applies.

Metallurgy: Extraction of Metals from Ores in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Thermit reaction

Suppose you see railway workers fill a gap between two rail sections with a powder mixture in a small mould. They ignite it, and within seconds glowing liquid metal flows into the gap and joins the rails.

Name the reaction, write its equation, identify the substance oxidised and the substance reduced, and explain why the iron is produced as a liquid.

Try it first, then show the solutionSolution

The powder is iron(III) oxide mixed with aluminium: the thermit reaction.
$F e_{2} O_{3} (s) + 2 A l (s) \to 2 F e (l) + A l_{2} O_{3} (s)$ + heat (check: Fe 2 = 2, Al 2 = 2, O 3 = 3).
Aluminium gains oxygen (oxidised); iron(III) oxide loses oxygen (reduced). Aluminium, being more reactive than iron, displaces it.
The displacement is highly exothermic; so much heat is released that iron forms in the molten state and fills the gap.

Answer: Thermit reaction, $F e_{2} O_{3} + 2 A l \to 2 F e + A l_{2} O_{3}$ ; Al oxidised, $F e_{2} O_{3}$ reduced; the large heat released melts the iron.

Memory hook: Aluminium steals iron's oxygen so violently that the iron comes out melted.

Tricks to remember Metallurgy: Extraction of Metals from Ores

Short memory aids. Say each one aloud once, then use it on a question the same day.

Roast the Sulphide, Calcine the Carbonate
RS-CC: Roasting for Sulphides (excess air); Calcination for Carbonates (limited air).
Use it when: Never mixing up the two processes.
Three bands of the series
Top: electrolysis. Middle: carbon. Bottom: heat alone or found native.
Use it when: Choosing an extraction route.
Refining PAN
Pure at Negative (cathode). Impure metal is the anode.
Use it when: Labelling electrolytic refining.

Metallurgy: Extraction of Metals from Ores: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Definitions · 2 marks · Very short answerBoard level

Define (a) mineral, (b) ore, (c) gangue. Name two metals found in nature in the free state.

Try it first, then show the solutionSolution

(a) An element or compound occurring naturally in the earth's crust (½ mark).
(b) A mineral from which a metal can be extracted profitably because it has a high percentage of that metal (½ mark).
(c) Impurities such as soil and sand mixed with the mined ore (½ mark).
Free state: gold and silver (also platinum, copper) (½ mark).

Answer: Mineral: natural element/compound in the crust; ore: mineral giving metal profitably; gangue: earthy impurities. Gold, silver.

Example 2: Zinc from two ores · 5 marks · Long answerHarder board level

Zinc occurs as zinc blende (ZnS) and as calamine ( $Z n C O_{3}$ ). (a) Name the process that converts each ore into zinc oxide and state the air condition. (b) Write the equations. (c) Write the equation for obtaining zinc from zinc oxide and name the reducing agent. (d) Why are sulphide and carbonate ores converted to oxides first?

Try it first, then show the solutionSolution

(a) ZnS: roasting, heating strongly in excess air; $Z n C O_{3}$ : calcination, heating strongly in limited air (1 mark).
(b) $2 Z n S + 3 O_{2} heat 2 Z n O + 2 S O_{2}$ ; $Z n C O_{3} heat Z n O + C O_{2}$ (2 marks).
(c) $Z n O + C \to Z n + C O$ ; carbon (coke) (1 mark).
(d) It is easier to obtain a metal from its oxide than from its sulphide or carbonate (1 mark).

Answer: (a) Roasting (excess air), calcination (limited air). (b) Equations above. (c) $Z n O + C \to Z n + C O$ , carbon. (d) Oxides are easier to reduce.

Example 3: Why not carbon for sodium? · 3 marks · Short answerHarder board level

(a) Why can sodium not be obtained by heating sodium oxide with carbon? (b) How is sodium obtained? Write the reactions at each electrode. (c) Why must the chloride be molten and not solid?

Try it first, then show the solutionSolution

(a) Sodium has a greater affinity for oxygen than carbon has, so carbon cannot remove oxygen from sodium oxide (1 mark).
(b) Electrolysis of molten sodium chloride: cathode $N a^{+} + e^{-} \to N a$ ; anode $2 C l^{-} \to C l_{2} + 2 e^{-}$ (1 mark).
(c) Ions in solid NaCl cannot move; in the molten state they move to the electrodes, so current flows and electrolysis occurs (1 mark).

Answer: (a) Na holds oxygen more strongly than carbon. (b) Electrolysis of molten NaCl; Na at cathode, $C l_{2}$ at anode. (c) Ions must be free to move.

Example 4: Electrolytic refining of copper · 4 marks · Short answerHOTS / competency

In the electrolytic refining of copper: (a) what is taken as the anode, cathode and electrolyte? (b) What happens at each electrode when current is passed? (c) What is anode mud? (d) Suppose the anode loses 50 g in mass while the cathode gains 48 g. Account for the 2 g difference.

Try it first, then show the solutionSolution

(a) Anode: impure copper; cathode: thin strip of pure copper; electrolyte: acidified copper sulphate solution (1 mark).
(b) Copper from the anode dissolves into the electrolyte; an equal amount of pure copper from the electrolyte is deposited on the cathode (1 mark).
(c) Insoluble impurities that settle at the bottom below the anode (1 mark).
(d) The anode's loss includes impurities as well as copper: soluble impurities pass into the solution and insoluble ones fall as anode mud, while only pure copper is deposited on the cathode; so the cathode gains less than the anode loses (1 mark).

Answer: (a) Impure Cu anode, pure Cu cathode, acidified $C u S O_{4}$ . (b) Anode dissolves, cathode gains pure Cu. (c) Insoluble impurities below the anode. (d) The 2 g is impurities that went into solution or the anode mud.

Where students lose marks in Metallurgy: Extraction of Metals from Ores

The slips examiners see most often, and what to write instead.

Mistake: Swapping roasting and calcination
Both involve strong heating.
Wrong working
$Z n C O_{3}$ is roasted in excess air.
Right working
$Z n C O_{3}$ is calcined in limited air; ZnS is roasted in excess air.
Do this: Link the ore type and the air condition together (RS-CC).
Usually costs: 1 mark.
Mistake: Wrong electrodes in refining
Students think pure metal must be the source.
Wrong working
Pure copper is the anode; impure copper is the cathode.
Right working
Impure copper is the anode; a thin strip of pure copper is the cathode.
Do this: Pure metal collects at the negative electrode (cathode).
Usually costs: 1 mark in diagram and labelling questions.
Mistake: Unbalanced roasting equation
Students stop at the skeletal equation and never balance the oxygen.
Wrong working
Skeletal form only, not balanced: $Z n S + O_{2} \to Z n O + S O_{2}$
Right working
$2 Z n S + 3 O_{2} \to 2 Z n O + 2 S O_{2}$
Do this: Oxygen on the right is 2 + 4 = 6 for two ZnS; so three $O_{2}$ .
Usually costs: ½ to 1 mark.
Mistake: Saying carbon reduces all metal oxides
Reduction by carbon is learnt as a general rule.
Wrong working
Aluminium is obtained by heating aluminium oxide with carbon.
Right working
Aluminium is obtained by electrolytic reduction of aluminium oxide.
Do this: Check the series: K to Al need electrolysis.
Usually costs: 1-2 marks.

Exam tips

Start any extraction answer by placing the metal in the reactivity series; it justifies your route.
Write the air condition with the process name: roasting in excess air, calcination in limited air.
In refining answers, name all three parts: anode, cathode, electrolyte, then what happens and anode mud.
Use the thermit reaction as your example of displacement used in metallurgy, with its use.
For electrolysis, say 'molten' every time; solid compounds do not allow ion movement.

Practice questions on Metallurgy: Extraction of Metals from Ores

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Heating a carbonate ore strongly in limited air is called:
(a) roasting (b) calcination (c) refining (d) reduction
Hint
RS-CC.
Answer
(b) calcination.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Aluminium is used to extract manganese from manganese dioxide.
Reason (R): Aluminium is more reactive than manganese and the reaction is highly exothermic.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Which metal displaces which?
Answer
(a). Aluminium, being more reactive, displaces manganese: $3 M n O_{2} + 4 A l \to 3 M n + 2 A l_{2} O_{3}$ , releasing much heat.
Q3 · 2 marks · Very short answerBoard level
Write balanced equations for the extraction of mercury from cinnabar and state why mercury can be obtained by heating alone.
Hint
Mercury is low in the reactivity series.
Answer
$2 H g S + 3 O_{2} \to 2 H g O + 2 S O_{2}$ ; $2 H g O \to 2 H g + O_{2}$ . Mercury is very unreactive, so its oxide breaks down to the metal on heating alone.
Q4 · 5 marks · Long answerHarder board level
A metal M is found as its sulphide in nature. M is moderately reactive. Outline, with general equations using M (valency 2), the steps you would use to get pure M, naming each process.
Hint
Roasting, reduction, refining.
Answer
Enrichment: remove gangue. Roasting in excess air: $2 M S + 3 O_{2} \to 2 M O + 2 S O_{2}$ . Reduction with carbon: $M O + C \to M + C O$ . Electrolytic refining: impure M as anode, pure M strip as cathode, a solution of a salt of M as electrolyte; pure M is deposited on the cathode and impurities collect as anode mud or stay in solution.
Q5 · 4 marks · Case-basedHOTS / competency
Read the passage and answer.
Suppose a mining company finds three deposits: deposit 1 has a carbonate of a metal from the middle of the reactivity series; deposit 2 has nuggets of a shiny yellow metal that does not tarnish; deposit 3 has a chloride of a metal that reacts violently with cold water.
(a) Which deposit needs no chemical reduction at all? Why? (1 mark)
(b) Name the first process for deposit 1 and write a general equation using metal M (valency 2). (1 mark)
(c) Why can deposit 3's metal not be extracted using carbon, and what method must be used? (2 marks)
Hint
Relate each deposit to a band of the reactivity series.
Answer
(a) Deposit 2: the metal (like gold) is so unreactive that it is found in the free state. (b) Calcination: $M C O_{3} \to M O + C O_{2}$ . (c) A metal that reacts violently with cold water is near the top of the series and has a greater affinity for oxygen than carbon, so carbon cannot reduce its compounds; it must be obtained by electrolytic reduction of its molten chloride.
Q6 · 3 marks · Short answerHOTS / competency
Copper can be extracted from $C u_{2} S$ by heating in air alone, while zinc from ZnS needs carbon after roasting. Explain this difference using the reactivity series, and write the two equations for copper.
Hint
Copper is less reactive than zinc.
Answer
Copper is low in the series, so its oxide can be reduced by its own sulphide on heating: $2 C u_{2} S + 3 O_{2} \to 2 C u_{2} O + 2 S O_{2}$ , then $2 C u_{2} O + C u_{2} S \to 6 C u + S O_{2}$ . Zinc is more reactive; its oxide is not reduced by heating alone, so after roasting a reducing agent such as carbon is needed.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is gangue?Show answer
Impurities like soil and sand in the mined ore.
2.Roasting is used for which ores?Show answer
Sulphide ores, heated strongly in excess air.
3.Ore of mercury named in NCERT?Show answer
Cinnabar, HgS.
4.Use of the thermit reaction?Show answer
Joining railway tracks or cracked machine parts.
5.Where is sodium deposited during electrolysis of molten NaCl?Show answer
At the cathode (negative electrode).
6.What is anode mud?Show answer
Insoluble impurities settling below the anode in electrolytic refining.

Metallurgy: Extraction of Metals from Ores: frequently asked questions

What is metallurgy?

The extraction of metals from their ores and their refining for use.

What is the difference between roasting and calcination?

Roasting heats sulphide ores strongly in excess air; calcination heats carbonate ores strongly in limited air. Both give the metal oxide.

Why are highly reactive metals extracted by electrolysis?

They hold oxygen more strongly than carbon does, so carbon cannot reduce their oxides. Electrolysis of the molten compound is used instead.

Which metals are found in the free state?

The least reactive ones, such as gold, silver, platinum and copper.

Why are many ores oxides?

Oxygen is very reactive and very abundant on the earth, so many metals occur combined with it.