Question

Why carbonate or sulphide ores are converted to oxides before extraction of metal from it?

Answer 1

Step 1: Aim of metal extraction.
The main objective in metallurgy is to obtain the metal in its free state from its ore.
For this, the ore must be converted into a form that can be easily reduced to metal.

Step 2: Stability of carbonate and sulphide ores.
Carbonate and sulphide ores are chemically more stable in their original form.
They are not easily reduced directly by common reducing agents like carbon or carbon monoxide.
Direct reduction may either require very high temperature or produce unwanted side products.

Step 3: Conversion of carbonate ores into oxides (Calcination).
Carbonate ores decompose on strong heating in limited or no air to form metal oxides and carbon dioxide gas.
This process is called calcination.
Example:
CaCO₃ → CaO + CO₂
The oxide formed is more reactive towards reduction than the carbonate.

Step 4: Conversion of sulphide ores into oxides (Roasting).
Sulphide ores are heated in excess air to convert them into metal oxides.
This process is called roasting.
Example:
2ZnS + 3O₂ → 2ZnO + 2SO₂
Roasting removes sulphur as sulphur dioxide gas and leaves behind the oxide.

Step 5: Why oxides are preferred for reduction.
Metal oxides can be easily reduced by carbon, carbon monoxide, or other suitable reducing agents.
The reduction of oxides is thermodynamically more feasible and economical.
It also gives purer metal compared to direct reduction of sulphides or carbonates.

Final Answer:
Carbonate and sulphide ores are first converted into oxides because oxides are easier and more economical to reduce to metals. Carbonates are converted by calcination and sulphides by roasting before reduction.

Answer 2

Step 1: Principle of the reaction.
Aluminium is a very reactive metal and has a stronger affinity for oxygen than many other metals.
Because of this high reactivity, aluminium can remove oxygen from the oxides of less reactive metals.
This type of reduction reaction is known as the thermite process or Goldschmidt process.

Step 2: Explanation of displacement.
When aluminium powder is mixed with the oxide of a less reactive metal and ignited, aluminium combines with oxygen to form aluminium oxide.
The less reactive metal is displaced from its oxide and obtained in molten form.

Step 3: Example – Reduction of iron(III) oxide.
The most common example is the reduction of iron(III) oxide using aluminium:
Fe₂O₃ + 2Al → 2Fe + Al₂O₃
In this reaction, aluminium reduces iron(III) oxide to iron while itself getting oxidized to aluminium oxide.

Step 4: Nature and applications of the reaction.
The reaction is highly exothermic and produces a large amount of heat.
The heat generated is sufficient to melt the iron formed.
This molten iron is used in thermite welding, especially for joining railway tracks.

Step 5: Another example.
Aluminium can also reduce chromium(III) oxide:
Cr₂O₃ + 2Al → 2Cr + Al₂O₃
Here also, aluminium forms aluminium oxide and chromium metal is obtained.

Final Answer:
A common example of reduction of a metal oxide by aluminium is:
Fe₂O₃ + 2Al → 2Fe + Al₂O₃

Answer 3

Step 1: Nature of the ore.
Copper(I) sulphide (Cu₂S) is a sulphide ore of copper, commonly known as chalcocite.
Since it is a sulphide ore, it cannot be reduced directly by carbon and must first be converted into oxide form.

Step 2: Concentration of ore.
The powdered ore is concentrated by froth flotation process.
This method removes gangue particles and increases the percentage of Cu₂S in the ore.

Step 3: Partial roasting.
The concentrated ore is heated in a limited supply of air.
During roasting, part of Cu₂S gets oxidized to Cu₂O and sulphur dioxide gas is released.
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂

Step 4: Formation and removal of slag.
Impurities like iron sulphide (FeS) are also present.
FeS reacts with silica (SiO₂) to form iron silicate (FeSiO₃), which is removed as slag.
FeS + SiO₂ → FeSiO₃ (slag)

Step 5: Auto-reduction (Self-reduction).
The mixture of Cu₂O and Cu₂S formed during roasting reacts together on further heating.
No external reducing agent is required.
2Cu₂O + Cu₂S → 6Cu + SO₂
This step produces molten copper known as blister copper.

Step 6: Refining of copper.
The blister copper (about 98% pure) is purified by electrolytic refining.
Pure copper is deposited at the cathode and impurities settle as anode mud.

Final Answer:
Copper is extracted from Cu₂S by partial roasting followed by auto-reduction:
2Cu₂O + Cu₂S → 6Cu + SO₂

Answer 4

Step 1: Position of highly reactive metals.
Highly reactive metals such as sodium, potassium, calcium, magnesium and aluminium are placed at the top of the reactivity series.
They have a very strong tendency to lose electrons and form stable positive ions.

Step 2: Condition for carbon reduction.
For carbon to reduce a metal oxide, carbon must be more reactive than that metal.
Only then can carbon remove oxygen from the metal oxide and form carbon monoxide or carbon dioxide.
This is possible only for metals placed below carbon in the reactivity series.

Step 3: Stability of oxides of highly reactive metals.
Oxides of highly reactive metals such as MgO, Al₂O₃ and CaO are extremely stable.
They have very large negative values of Gibbs free energy of formation.
Because of this high stability, carbon cannot break these strong metal–oxygen bonds.

Step 4: Thermodynamic explanation.
The reduction reaction can be written as:
MO + C → M + CO
For this reaction to occur, the Gibbs free energy change must be negative.
In the case of highly reactive metals, the formation of their oxides is more favorable than the formation of CO or CO₂.
Therefore, the reaction has positive ΔG and is not feasible.

Step 5: Ellingham diagram reasoning.
In the Ellingham diagram, the lines representing oxides of highly reactive metals lie below the carbon oxidation line.
This shows that carbon cannot reduce these oxides at practical temperatures.

Step 6: Conclusion.
Hence, highly reactive metals cannot be obtained from their oxides by reduction with carbon.
They are usually extracted by electrolysis of their molten compounds instead.

Final Answer:
Highly reactive metals cannot be obtained from their oxides using carbon because their oxides are extremely stable and carbon is not reactive enough to reduce them under normal metallurgical conditions.

Answer 5

Step 1: Composition of solder.
Solder is an alloy mainly made of tin (Sn) and lead (Pb).
A common composition is 60% tin and 40% lead, though other proportions are also used.
It is specially designed to join metal surfaces, particularly electrical wires.

Step 2: Low melting point.
Solder has a much lower melting point than most metals used in electrical wiring.
Pure lead melts at 327°C and pure tin melts at 232°C.
However, solder melts at about 183–190°C.
Because of this low melting point, it can join wires without overheating or damaging them.

Step 3: Good electrical conductivity.
Solder conducts electricity effectively.
Although its conductivity is less than copper, it is sufficient to maintain a proper electrical connection between wires.

Step 4: Good wetting property and strong bonding.
When heated, molten solder spreads easily over copper surfaces.
It forms a strong metallic bond after cooling.
This ensures both mechanical strength and reliable electrical contact.

Step 5: Resistance to corrosion.
Solder resists corrosion and oxidation.
This helps in maintaining long-lasting electrical joints without deterioration.

Step 6: Ease of use.
Solder is easy to apply using a soldering iron.
It melts and solidifies quickly, making it convenient for assembling and repairing circuits.

Final Answer:
Solder is used for welding electrical wires because it has a low melting point, good electrical conductivity, strong adhesion to metals like copper, corrosion resistance, and is easy to work with.