Question

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

Answer 1

Step 1: Understand the nature of carbonate and sulphide ores.
Carbonate ores (e.g., limestone CaCO₃, magnesite MgCO₃) and sulphide ores (e.g., galena PbS, zinc blende ZnS) are naturally occurring compounds of metals. To extract the metal, we need to reduce the ore to obtain the free metal.
Step 2: Difficulty in direct reduction of carbonate and sulphide ores.
Carbonate and sulphide ores cannot be directly reduced by common reducing agents like carbon because:

• Carbonates decompose at high temperatures rather than getting reduced directly.
• Sulphides form undesirable byproducts if reduced directly (like metal sulphides instead of metal).

Step 3: Reason for converting to oxides.
Oxides are more suitable for reduction because:

• Oxides can be easily reduced using carbon (smelting) or other reducing agents.
• The reduction of oxides is thermodynamically more feasible and gives pure metal.
• The Gibbs free energy change for oxide reduction is more favorable.

Step 4: Specific reasons for each type.
For carbonate ores:

• Carbonates decompose on heating to give oxides (calcination): \[ \text{CaCO}_3 \xrightarrow{\Delta} \text{CaO} + \text{CO}_2 \]
• The oxide formed can then be easily reduced.

For sulphide ores:

• Sulphides are heated in presence of excess air (roasting) to convert them to oxides: \[ 2\text{ZnS} + 3\text{O}_2 \xrightarrow{\Delta} 2\text{ZnO} + 2\text{SO}_2 \]
• Roasting also removes impurities like sulphur and arsenic as volatile oxides.
• Direct reduction of sulphides would produce toxic SO₂ and less pure metal.

Step 5: Advantages of oxide reduction.

• Oxides can be reduced with common reducing agents like carbon (coke), CO, or more reactive metals.
• The process is economical and yields pure metal.
• Gangue can be easily separated during oxide reduction.

Step 6: Final answer.
Carbonate and sulphide ores are converted to oxides because oxides are easier to reduce with common reducing agents like carbon. Carbonates decompose to oxides on heating (calcination), and sulphides are oxidized to oxides by heating in air (roasting). Direct reduction of carbonates or sulphides is not feasible.

Answer 2

Step 1: Understand the process.
Aluminium is a highly reactive metal and acts as a powerful reducing agent. It has a strong affinity for oxygen. When aluminium is mixed with the oxide of a less reactive metal and heated, it displaces the metal from its oxide. This process is called the Goldschmidt process or thermite process.
Step 2: Example reaction.
A common example is the reduction of iron(III) oxide (Fe₂O₃) with aluminium to obtain iron metal: \[ \boxed{\text{Fe}_2\text{O}_3\text{ (s) + 2Al (s) \(\Rightarrow\) 2Fe (s) + Al}_2\text{O}_3\text{ (s)}} \]
Step 3: Observations in this reaction.

• The reaction is highly exothermic (releases a large amount of heat).
• The heat generated melts the iron produced, which can be used for welding railway tracks (thermite welding).
• Aluminium oxide (Al₂O₃) is formed as a byproduct.

Step 4: Another example.
Aluminium can also reduce chromium(III) oxide to obtain chromium metal: \[ \text{Cr}_2\text{O}_3 + 2\text{Al} \(\Rightarrow\) 2\text{Cr} + \text{Al}_2\text{O}_3 \]
Step 5: Final answer.
\[ \boxed{\text{Fe}_2\text{O}_3 + 2\text{Al} \(\Rightarrow\) 2\text{Fe} + \text{Al}_2\text{O}_3} \]

Answer 3

Step 1: Identify the ore.
Copper(I) sulphide (Cu₂S) is an important ore of copper, commonly found as chalcocite (copper glance).
Step 2: Overview of extraction process.
Copper from Cu₂S is extracted through a series of steps:

• Concentration of ore (by froth flotation)
• Roasting (partial)
• Smelting in a reverberatory furnace
• Bessemerization (conversion)
• Refining (electrolytic)

Step 3: Step 1 - Roasting.
The concentrated ore is roasted in a limited supply of air. Part of Cu₂S is oxidized to Cu₂O: \[ 2\text{Cu}_2\text{S} + 3\text{O}_2 \xrightarrow{\Delta} 2\text{Cu}_2\text{O} + 2\text{SO}_2 \]
Step 4: Step 2 - Smelting.
The roasted ore is mixed with silica (sand, SiO₂) and smelted in a reverberatory furnace. The FeS present in the ore reacts with silica to form slag (FeSiO₃), which is removed: \[ \text{FeS} + \text{SiO}_2 \rightarrow \text{FeSiO}_3 \text{ (slag)} \]
Step 5: Step 3 - Bessemerization (Auto-reduction).
The mixture of Cu₂S and Cu₂O is heated in a Bessemer converter. They react to form copper metal and SO₂: \[ \boxed{2\text{Cu}_2\text{O} + \text{Cu}_2\text{S} \xrightarrow{\Delta} 6\text{Cu} + \text{SO}_2} \] This is called auto-reduction because no external reducing agent is needed.
Step 6: Step 4 - Refining.
The blister copper obtained (about 98% pure) is refined electrolytically to get 99.98% pure copper.
Step 7: Summary of key reactions.
The main reaction for obtaining copper from Cu₂S is: \[ \boxed{2\text{Cu}_2\text{O} + \text{Cu}_2\text{S} \rightarrow 6\text{Cu} + \text{SO}_2} \]
Step 8: Final answer.
\[ \boxed{\text{Copper is obtained from Cu₂S by roasting followed by auto-reduction: } 2\text{Cu}_2\text{O} + \text{Cu}_2\text{S} \rightarrow 6\text{Cu} + \text{SO}_2} \]

Answer 4

Step 1: Recall the reactivity series of metals.
Metals are arranged in the reactivity series based on their tendency to lose electrons and form positive ions. Highly reactive metals like sodium, potassium, calcium, magnesium, and aluminum are at the top of the reactivity series.
Step 2: Principle of reduction using carbon.
Carbon can reduce metal oxides only if carbon is more reactive than the metal. In other words, carbon can displace a metal from its oxide only if the metal is below carbon in the reactivity series.
Step 3: Position of carbon in reactivity series.
Carbon is placed between aluminum and zinc in the reactivity series. It can reduce oxides of metals that are below it (like zinc, iron, lead, copper) but cannot reduce oxides of metals above it (like sodium, potassium, calcium, magnesium, aluminum).
Step 4: Thermodynamic reason.
The reduction of a metal oxide by carbon involves the reaction: \[ \text{MO} + \text{C} \rightarrow \text{M} + \text{CO} \] For this reaction to be spontaneous, the Gibbs free energy change (\(\Delta G\)) must be negative. The \(\Delta G\) for formation of oxides of highly reactive metals is very high (highly negative), meaning these oxides are very stable. Carbon cannot provide enough energy to break these stable oxides.
Step 5: Ellingham diagram explanation.
According to Ellingham diagram, the \(\Delta G\) vs temperature curves for oxides of highly reactive metals (like Al₂O₃, MgO, CaO) lie below the curve for CO formation. This means that carbon cannot reduce these oxides at any practical temperature because the reaction would have positive \(\Delta G\).
Step 6: Example.
For magnesium oxide: \[ 2\text{MgO} + \text{C} \rightarrow 2\text{Mg} + \text{CO}_2 \] This reaction is not feasible because Mg is more reactive than carbon and MgO is very stable.
Step 7: Final answer.
Highly reactive metals cannot be obtained from their oxides using carbon because these metals are more reactive than carbon. Their oxides are very stable and carbon cannot reduce them due to unfavorable thermodynamics (positive }\Delta G\text{).

Answer 5

Step 1: Understand what solder is.
Solder is an alloy of lead (Pb) and tin (Sn). The most common composition is 60% tin and 40% lead (60-40 solder), though other ratios exist. It is a fusible alloy used for joining metal surfaces.
Step 2: Property 1 - Low melting point.
The melting point of solder is much lower than that of its constituent metals:

• Pure lead melts at 327°C
• Pure tin melts at 232°C
• Solder (60% Sn, 40% Pb) melts at about 183-190°C (eutectic temperature)

This low melting point allows soldering without damaging the electrical wires or components.
Step 3: Property 2 - Good electrical conductivity.
Solder provides good electrical conductivity, ensuring proper electrical connection between wires. Though not as conductive as pure copper, it is sufficient for electrical joints.
Step 4: Property 3 - Good wetting and adhesion.
When molten, solder flows easily and "wets" the surfaces of copper wires, forming a strong metallic bond upon solidification. This ensures a mechanically strong and electrically reliable joint.
Step 5: Property 4 - Corrosion resistance.
Solder is resistant to corrosion, which helps maintain the integrity of the electrical connection over time.
Step 6: Property 5 - Workability.
Solder is easy to work with using a soldering iron. It solidifies quickly, allowing efficient assembly of electrical circuits.
Step 7: Final answer.
\text{Solder is used for welding electrical wires because it has a low melting point (does not damage wires), provides good electrical conductivity, adheres well to copper, and is corrosion-resistant.}}