Menu Top
Non-Rationalised Science NCERT Notes and Solutions (Class 6th to 10th)
6th 7th 8th 9th 10th
Non-Rationalised Science NCERT Notes and Solutions (Class 11th)
Physics Chemistry Biology
Non-Rationalised Science NCERT Notes and Solutions (Class 12th)
Physics Chemistry Biology

Class 12th (Chemistry) Chapters
1. The Solid State 2. Solutions 3. Electrochemistry
4. Chemical Kinetics 5. Surface Chemistry 6. General Principles And Processes Of Isolation Of Elements
7. The P-Block Elements 8. The D-And F-Block Elements 9. Coordination Compounds
10. Haloalkanes And Haloarenes 11. Alcohols, Phenols And Ethers 12. Aldehydes, Ketones And Carboxylic Acids
13. Amines 14. Biomolecules 15. Polymers
16. Chemistry In Everyday Life



Chapter 6 General Principles And Processes Of Isolation Of Elements



Occurrence Of Metals

Metals have played a significant role in human history and civilization, with different periods even named after metals. The ability to extract metals from their natural sources transformed society, providing materials for tools, weapons, ornaments, and utensils.

While some elements like carbon, sulfur, gold, and noble gases exist in their free state, most metals are found in the Earth's crust in combined forms as various chemical compounds.

Metals vary in their abundance. Aluminum is the most abundant metal in the Earth's crust, ranking as the third most abundant element overall (around 8.3% by weight). It is found in minerals like mica and clays and is the main component of bauxite ore. Iron is the second most abundant metal, forming a variety of compounds and being essential in biological systems.

A mineral is a naturally occurring chemical substance in the Earth's crust obtained by mining. An ore is a mineral from which a metal can be extracted economically and efficiently.

The selection of an ore for metal extraction depends on factors like its abundance, ease of extraction, and environmental considerations (e.g., avoiding ores that produce polluting gases).

Principal ores of some important metals:

Metal Ores Composition
AluminiumBauxite
Kaolinite (a form of clay)
AlO$_x$(OH)$_{3-2x}$ [where 0 < x < 1]
Al$_2$(OH)$_4$Si$_2$O$_5$
IronHaematite
Magnetite
Siderite
Iron pyrites
Fe$_2$O$_3$
Fe$_3$O$_4$
FeCO$_3$
FeS$_2$
CopperCopper pyrites
Malachite
Cuprite
Copper glance
CuFeS$_2$
CuCO$_3$.Cu(OH)$_2$
Cu$_2$O
Cu$_2$S
ZincZinc blende or Sphalerite
Calamine
Zincite
ZnS
ZnCO$_3$
ZnO

The scientific and technological process of isolating a metal from its ore is called metallurgy. Metallurgical processes are based on chemical principles and aim to be chemically feasible and commercially viable.


Concentration Of Ores

Ores are typically contaminated with unwanted earthly or rocky materials called gangue. Removing gangue from the ore is known as concentration, dressing, or benefaction. Ores are usually crushed before concentration.

The choice of concentration method depends on the physical properties of the ore and gangue particles. Some important methods:


Hydraulic Washing

Based on the difference in specific gravities of the ore and gangue particles (gravity separation). Powdered ore is washed with an upward stream of running water. Lighter gangue particles are washed away, leaving heavier ore particles behind.


Magnetic Separation

Used when either the ore or the gangue has magnetic properties. Powdered ore is dropped onto a conveyor belt passing over a magnetic roller. Magnetic particles are attracted to the roller and fall closer to it, separating them from non-magnetic particles.

Schematic diagram of magnetic separation process using a magnetic roller

Froth Floatation Method

This method is mainly used for concentrating sulphide ores. A suspension of powdered ore is made in water with collectors (e.g., pine oils, fatty acids) and froth stabilisers (e.g., cresols, aniline).

Diagram illustrating the froth floatation process

Collectors make the mineral particles water-repellent (non-wettable by water) and oil-wettable. Froth stabilisers make the froth stable. A rotating paddle agitates the mixture, drawing in air, which forms froth. The oil-wet mineral particles are carried by the froth to the surface and skimmed off. Gangue particles, being water-wet, settle down. Depressants (e.g., NaCN for ZnS/PbS ore) can be used to selectively prevent one sulphide ore from forming froth.


Leaching

Leaching is used when the ore is soluble in a suitable solvent. It involves treating the powdered ore with a reagent that dissolves the metal compound, leaving impurities undissolved. The solution is then processed to recover the metal.

Intext Questions

6.1 Which of the ores mentioned in Table 6.1 can be concentrated by magnetic separation method?

6.2 What is the significance of leaching in the extraction of aluminium?

Answer:

6.1 Magnetic separation is used for ores where either the ore particle or the gangue particle is magnetic. From Table 6.1, the iron ores Haematite ($\textsf{Fe}_2\text{O}_3$) and Magnetite ($\textsf{Fe}_3\text{O}_4$) contain iron compounds which are magnetic. Iron pyrites ($\textsf{FeS}_2$) is weakly magnetic. Siderite ($\textsf{FeCO}_3$) also contains iron and is magnetic. Therefore, iron ores can be concentrated by magnetic separation. Among other ores listed, Zinc blende ($\textsf{ZnS}$) and Cuprite ($\textsf{Cu}_2\text{O}$) generally do not have significant magnetic properties. Copper pyrites ($\textsf{CuFeS}_2$) contains iron, so it might be magnetic to some extent, but it is usually concentrated by froth floatation. Bauxite ($\textsf{Al}_2\text{O}_3$.xH$_2\text{O}$) contains iron oxides as impurities which are magnetic, so magnetic separation can be used to remove these impurities from bauxite. Calamine ($\textsf{ZnCO}_3$) and Zincite ($\textsf{ZnO}$) are not typically concentrated by magnetic separation.

The ores that can be concentrated by magnetic separation method are primarily the iron ores (Haematite, Magnetite, Siderite, Iron pyrites), and magnetic impurities can be removed from Bauxite by this method.

6.2 Leaching is the primary method for concentrating the bauxite ore in the extraction of aluminium (Bayer's process). Its significance is to selectively dissolve alumina ($\textsf{Al}_2\text{O}_3$) from the bauxite ore using a strong base ($\textsf{NaOH}$), while leaving behind most of the impurities like iron oxides and titanium oxide, and separating silica (which also dissolves) in a controlled manner. This process effectively purifies the alumina, which is then precipitated and calcined to obtain pure $\textsf{Al}_2\text{O}_3$, the compound that is subsequently reduced electrolytically to obtain aluminium metal. Leaching is crucial for obtaining high-purity alumina required for efficient electrolytic reduction.



Extraction Of Crude Metal From Concentrated Ore

After concentration, the ore must be converted into a form suitable for reduction to the metal. Usually, sulfide ores are converted to oxides because oxides are easier to reduce than sulfides.

This stage involves two main steps:


Conversion To Oxide

Section of a modern reverberatory furnace used for roasting

Reduction Of Oxide To Metal

Metal oxides are reduced to crude metal, typically by heating with a reducing agent like carbon (coke), carbon monoxide ($\textsf{CO}$), or another metal. The reducing agent removes oxygen from the metal oxide.

$\textsf{M}_\text{x}\textsf{O}_\text{y} + \textsf{yC} \to \textsf{xM} + \textsf{yCO}$

The ease of reduction varies for different metal oxides.


Thermodynamic Principles Of Metallurgy

Thermodynamics provides a framework to understand the feasibility of metallurgical reduction processes and choose suitable reducing agents. The criterion for a reaction to be spontaneous (feasible) at a given temperature (T) is that the change in Gibbs energy ($\Delta \textsf{G}$) must be negative:

$\Delta \textsf{G} = \Delta \textsf{H} - \textsf{T}\Delta \textsf{S}$

Where $\Delta \textsf{H}$ is the enthalpy change and $\Delta \textsf{S}$ is the entropy change. For thermal reduction of a metal oxide (decomposition), the reaction is typically $\textsf{M}_\text{x}\textsf{O} \to \textsf{xM} + \frac{1}{2}\textsf{O}_2\text{(g)}$. This reaction involves an increase in entropy (formation of gas), so $\Delta \textsf{S}$ is positive. At high temperatures, the $-\textsf{T}\Delta \textsf{S}$ term becomes large and negative, making $\Delta \textsf{G}$ more negative and favoring decomposition.

Reduction of a metal oxide $\textsf{M}_\text{x}\textsf{O}$ by a reducing agent ($\textsf{A}_{\text{red}}$) involves two coupled reactions:

  1. Reduction of the metal oxide: $\textsf{M}_\text{x}\textsf{O} \to \textsf{xM} + \frac{1}{2}\textsf{O}_2\text{(g)}$ ($\Delta \textsf{G}_1$)
  2. Oxidation of the reducing agent: $\textsf{A}_{\text{red}} + \frac{1}{2}\textsf{O}_2\text{(g)} \to \textsf{A}_{\text{red}}\textsf{O}$ ($\Delta \textsf{G}_2$)

Overall reaction: $\textsf{M}_\text{x}\textsf{O} + \textsf{A}_{\text{red}} \to \textsf{xM} + \textsf{A}_{\text{red}}\textsf{O}$ ($\Delta \textsf{G}_{\text{overall}} = \Delta \textsf{G}_1 + \Delta \textsf{G}_2$)

The reduction is feasible if $\Delta \textsf{G}_{\text{overall}}$ is negative. This occurs when the $\Delta \textsf{G}$ for the oxidation of the reducing agent ($\Delta \textsf{G}_2$) is sufficiently negative to make the sum ($\Delta \textsf{G}_1 + \Delta \textsf{G}_2$) negative. A reducing agent is chosen such that its oxidation reaction has a more negative $\Delta \textsf{G}$ than the reduction reaction of the metal oxide at the chosen temperature.


Applications

Ellingham Diagram: These are plots of $\Delta_\text{f}\textsf{G}^0$ (standard Gibbs energy of formation of metal oxides per mole of oxygen) vs temperature (T). They are used to predict the feasibility of reducing a metal oxide with a particular reducing agent at a given temperature.

Schematic Ellingham diagram showing DrG0 vs T plots for the formation of various metal oxides and carbon oxides

Limitations of Ellingham Diagram:

(a) Extraction of iron from its oxides (e.g., Haematite, $\textsf{Fe}_2\text{O}_3$) in a Blast furnace:

The iron obtained (pig iron) contains $\sim 4\%$ C and impurities. Cast iron has $\sim 3\%$ C. Wrought iron is the purest form of commercial iron ($\sim 0.15\%$ C), produced by oxidising impurities from cast iron.

Example 6.1 Suggest a condition under which magnesium could reduce alumina.

Answer:

Reduction of alumina ($\textsf{Al}_2\text{O}_3$) by magnesium ($\textsf{Mg}$) is represented by the reaction: $\textsf{Al}_2\text{O}_3 + 3\textsf{Mg} \to 2\textsf{Al} + 3\textsf{MgO}$. This reaction is feasible if its Gibbs energy change ($\Delta \textsf{G}$) is negative.

In the Ellingham diagram (Fig. 6.4), we compare the curves for the formation of $\textsf{Al}_2\text{O}_3$ and $\textsf{MgO}$ (per mole of $\textsf{O}_2$). The reaction $\textsf{Al}_2\text{O}_3 + 3\textsf{Mg} \to 2\textsf{Al} + 3\textsf{MgO}$ corresponds to the difference between the formation of $\textsf{MgO}$ and the formation of $\textsf{Al}_2\text{O}_3$. Specifically, it is related to $( \Delta \textsf{G}^0 \text{ for } 3\textsf{MgO} \text{ formation} ) - ( \Delta \textsf{G}^0 \text{ for } \textsf{Al}_2\text{O}_3 \text{ formation} )$. Since the Ellingham diagram plots are for formation per mole of $\textsf{O}_2$, we are comparing the curves $2\textsf{Mg} + \textsf{O}_2 \to 2\textsf{MgO}$ and $\frac{4}{3}\textsf{Al} + \textsf{O}_2 \to \frac{2}{3}\textsf{Al}_2\text{O}_3$.

Reduction of $\textsf{Al}_2\text{O}_3$ by $\textsf{Mg}$ is feasible when the $\textsf{Mg}$, $\textsf{MgO}$ curve is below the $\textsf{Al}$, $\textsf{Al}_2\text{O}_3$ curve in the Ellingham diagram. The point of intersection of these two curves (marked as "A" in Fig. 6.4) is approximately at $\sim 1350^\circ\textsf{C}$ or $\sim 1623$ K. Below this temperature, the $\Delta \textsf{G}$ for $\textsf{MgO}$ formation is more negative than that for $\textsf{Al}_2\text{O}_3$ formation, making $\textsf{Mg}$ a better reducing agent for $\textsf{Al}_2\text{O}_3$.

Therefore, magnesium could reduce alumina under conditions where the temperature is below approximately 1350$^\circ$C.

Example 6.2 Why is the reduction of a metal oxide easier if the metal formed is in liquid state at the temperature of reduction?

Answer:

The feasibility of a reaction is determined by the change in Gibbs energy, $\Delta \textsf{G} = \Delta \textsf{H} - \textsf{T}\Delta \textsf{S}$. For reduction of a metal oxide $\textsf{M}_\text{x}\textsf{O}$ to metal M and oxygen (e.g., $\textsf{M}_\text{x}\textsf{O(s)} \to \textsf{xM} + \frac{1}{2}\textsf{O}_2\text{(g)}$), entropy generally increases ($\Delta \textsf{S} > 0$) due to the formation of gas. This positive $\Delta \textsf{S}$ term makes $\Delta \textsf{G}$ more negative as temperature increases, favoring reduction.

If the metal formed is in the liquid state at the reduction temperature, the entropy change ($\Delta \textsf{S}$) for the process $\textsf{M}_\text{x}\textsf{O(s)} \to \textsf{xM(liquid)} + \frac{1}{2}\textsf{O}_2\text{(g)}$ will be higher (more positive) than if the metal formed is in the solid state ($\textsf{M}_\text{x}\textsf{O(s)} \to \textsf{xM(solid)} + \frac{1}{2}\textsf{O}_2\text{(g)}$). This is because the entropy of the liquid state is higher than that of the solid state.

A larger positive $\Delta \textsf{S}$ value leads to a larger negative contribution from the $-\textsf{T}\Delta \textsf{S}$ term in the $\Delta \textsf{G}$ equation, especially at high temperatures. This makes $\Delta \textsf{G}$ more negative, indicating that the reduction reaction is more thermodynamically favorable or "easier" when the metal is formed in the liquid state compared to the solid state at the same temperature.

(b) Extraction of copper from cuprous oxide ($\textsf{Cu}_2\text{O}$):

(c) Extraction of zinc from zinc oxide ($\textsf{ZnO}$):

Intext Questions

6.3 The reaction, Cr$_2$O$_3$+2Al $\to$ Al$_2$O$_3$+2Cr (DrG$^0$= – 421kJ) is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?

6.4 Is it true that under certain conditions, Mg can reduce Al$_2$O$_3$ and Al can reduce MgO? What are those conditions?

Answer:

6.3 The reaction $\textsf{Cr}_2\text{O}_3 + 2\textsf{Al} \to \textsf{Al}_2\text{O}_3 + 2\textsf{Cr}$ has a negative standard Gibbs energy change ($\Delta \textsf{G}^0 = -421$ kJ), indicating it is thermodynamically feasible under standard conditions. However, thermodynamics only predicts the possibility of a reaction, not its rate. Even though the products are more stable than the reactants, the reaction does not take place at room temperature because there is a significant activation energy barrier that needs to be overcome for the reaction to start. At room temperature, very few reactant molecules possess enough kinetic energy to reach the transition state and react. Heating provides the necessary activation energy for a sufficient number of molecules to react, allowing the reaction to proceed at a noticeable rate.

6.4 Yes, it is true that under certain conditions, Mg can reduce Al$_2$O$_3$ and Al can reduce MgO. This can be understood by looking at the Ellingham diagram (Fig. 6.4) and the curves for the formation of $\textsf{MgO}$ and $\textsf{Al}_2\text{O}_3$ (per mole of $\textsf{O}_2$).

  • Mg can reduce Al$_2$O$_3$: This reaction ($\textsf{Al}_2\text{O}_3 + 3\textsf{Mg} \to 2\textsf{Al} + 3\textsf{MgO}$) is feasible when the $\textsf{Mg}$, $\textsf{MgO}$ formation curve is below the $\textsf{Al}$, $\textsf{Al}_2\text{O}_3$ formation curve. In Fig. 6.4, these curves intersect at point "A" (approximately 1350$^\circ$C). Below this intersection temperature (at temperatures below approximately 1350$^\circ$C), the $\Delta \textsf{G}$ for $\textsf{MgO}$ formation is more negative than that for $\textsf{Al}_2\text{O}_3$ formation, making $\textsf{Mg}$ a stronger reducing agent than Al. So, Mg can reduce Al$_2$O$_3$.
  • Al can reduce MgO: This reaction ($\textsf{MgO} + 2/3\textsf{Al} \to \textsf{Mg} + 1/3\textsf{Al}_2\text{O}_3$) is feasible when the $\textsf{Al}$, $\textsf{Al}_2\text{O}_3$ formation curve is below the $\textsf{Mg}$, $\textsf{MgO}$ formation curve. This occurs at temperatures above the intersection point A (above approximately 1350$^\circ$C). Above this temperature, Al becomes a stronger reducing agent than Mg, and Al can reduce MgO.


Electrochemical Principles Of Metallurgy

Electrochemical methods (electrolysis) are used for reducing metal ions in solution or molten state to obtain metals, especially for highly reactive metals where thermal reduction with common reducing agents is difficult or impossible. These methods are based on electrochemical principles, related to electrode potentials and Gibbs energy.

The feasibility of reducing a metal ion ($\textsf{M}^{\text{n}+}$) by electrolysis is related to the standard electrode potential ($\textsf{E}^0$) of the $\textsf{M}^{\text{n}+}$/M redox couple and the Gibbs energy change ($\Delta \textsf{G}^0$) of the reduction process:

$\Delta \textsf{G}^0 = - \textsf{nF} \textsf{E}^0$

Where n is the number of electrons transferred, and F is Faraday's constant. A more negative $\textsf{E}^0$ (meaning the element is more reactive or a stronger reducing agent) corresponds to a more positive $\Delta \textsf{G}^0$ for reduction, indicating that the reduction is thermodynamically less favorable. To drive the reduction of a highly electropositive metal ion, a large negative external potential must be applied at the cathode.

In electrolysis, $\textsf{M}^{\text{n}+}$ ions are discharged (reduced) at the cathode (negative electrode) and deposited as metal. Precautions are needed based on the reactivity of the metal produced. Fluxes may be added to make the molten mass more conducting.


Oxidation Reduction

Some extraction processes involve oxidation, particularly for non-metals or when recovering metals from complexes formed after leaching. For example, chlorine is extracted from brine (aqueous $\textsf{NaCl}$) by electrolysis, which involves the oxidation of $\textsf{Cl}^{-}$ ions:

$2\textsf{Cl}^{-}\text{(aq)} + 2\textsf{H}_2\text{O(l)} \to 2\textsf{OH}^{-}\text{(aq)} + \textsf{H}_2\text{(g)} + \textsf{Cl}_2\text{(g)}$ ($\Delta \textsf{G}^0 = + 422$ kJ, $\textsf{E}^0 = -2.2$ V)

This non-spontaneous reaction requires an external voltage greater than 2.2 V for electrolysis to occur, producing $\textsf{H}_2$ and $\textsf{NaOH}$ as by-products.

Extraction of gold and silver often involves leaching with $\textsf{CN}^{-}$ in the presence of $\textsf{O}_2$, which is an oxidation of the metal:

$4\textsf{Au(s)} + 8\textsf{CN}^{-}\text{(aq)} + 2\textsf{H}_2\text{O(aq)} + \textsf{O}_2\text{(g)} \to 4[\textsf{Au(CN)}_2]^{-}\text{(aq)} + 4\textsf{OH}^{-}\text{(aq)}$

The metal is then recovered by reducing the complex using a more electropositive metal, like zinc (displacement reaction):

$2[\textsf{Au(CN)}_2]^{-}\text{(aq)} + \textsf{Zn(s)} \to 2\textsf{Au(s)} + [\textsf{Zn(CN)}_4]^{2-}\text{(aq)}$ (Zinc acts as a reducing agent).

Example 6.4 At a site, low grade copper ores are available and zinc and iron scraps are also available. Which of the two scraps would be more suitable for reducing the leached copper ore and why?

Answer:

Low grade copper ores can be leached to obtain $\textsf{Cu}^{2+}$ ions in solution. Reducing these $\textsf{Cu}^{2+}$ ions requires a metal that is more electropositive (a stronger reducing agent) than copper. Both zinc and iron are more electropositive than copper, as seen from their standard reduction potentials ($\textsf{E}^0_{\text{Zn}^{2+}|\text{Zn}} = -0.76$ V, $\textsf{E}^0_{\text{Fe}^{2+}|\text{Fe}} = -0.44$ V, $\textsf{E}^0_{\text{Cu}^{2+}|\text{Cu}} = +0.34$ V). Both can displace copper from solution:

$\textsf{Cu}^{2+}\text{(aq)} + \textsf{Zn(s)} \to \textsf{Cu(s)} + \textsf{Zn}^{2+}\text{(aq)}$

$\textsf{Cu}^{2+}\text{(aq)} + \textsf{Fe(s)} \to \textsf{Cu(s)} + \textsf{Fe}^{2+}\text{(aq)}$

Zinc is more electropositive than iron (its reduction potential is more negative). This means zinc is a stronger reducing agent than iron. Therefore, the reduction of $\textsf{Cu}^{2+}$ ions would be thermodynamically more favorable and potentially faster using zinc scraps compared to iron scraps.

However, the question asks which is more suitable considering availability of scraps. While zinc might be thermodynamically better for reduction kinetics, iron scraps are generally much more readily available and significantly cheaper than zinc scraps. Therefore, from a commercial and practical point of view, using iron scraps would be more suitable and economically advantageous for reducing leached copper ore, despite zinc being a stronger reducing agent.



Refining

The crude metal obtained after extraction usually contains impurities. Refining is the process used to obtain metals of high purity. Various techniques are used, chosen based on the differences in properties between the metal and its impurities.


Distillation

Useful for low boiling metals like zinc, mercury, and cadmium. The impure metal is heated to evaporate the pure metal, which is then condensed to obtain the distillate.


Liquation

Used for metals that melt at a lower temperature than their impurities, such as tin, lead, and bismuth. The impure metal is heated on a sloping hearth. The low-melting metal flows down, separating from the higher-melting impurities.


Electrolytic Refining

This is a widely used method for obtaining high-purity metals like copper, zinc, nickel, silver, gold, and aluminum. The impure metal is made the anode, and a thin strip of the pure metal is the cathode. Both electrodes are immersed in an electrolytic bath containing a soluble salt of the same metal.

During electrolysis:

For copper refining: Impure copper anode, pure copper cathode, acidified $\textsf{CuSO}_4$ electrolyte. $\textsf{Cu}$ dissolves from anode ($\textsf{Cu} \to \textsf{Cu}^{2+} + 2\textsf{e}^{-}$), $\textsf{Cu}^{2+}$ ions deposit on cathode ($\textsf{Cu}^{2+} + 2\textsf{e}^{-} \to \textsf{Cu}$). Impurities like Sb, Se, Te, Ag, Au, Pt collect as anode mud.


Zone Refining

Based on the principle that impurities are generally more soluble in the molten state (melt) of the metal than in the solid state. Used for producing very high purity semiconductors (Ge, Si, B, Ga, In).

Diagram illustrating the zone refining process with a mobile heater

A mobile circular heater is slowly moved along a rod of the impure metal. A narrow molten zone is created and moves with the heater. As the heater moves, the pure metal solidifies behind it, and the impurities, being more soluble in the melt, preferentially stay in the molten zone and move forward with the heater. Repeating this process concentrates impurities at one end of the rod, which is then cut off.


Vapour Phase Refining

The impure metal is converted into a volatile compound, separated, and then decomposed to obtain the pure metal. Requirements: Metal forms a volatile compound with a reagent, and the volatile compound can be easily decomposed to pure metal.


Chromatographic Methods

Chromatography is useful for purifying elements available in minute quantities or when impurities have chemical properties very similar to the element to be purified (where other methods are less effective). Different chromatographic techniques (column chromatography, gas chromatography, etc.) can be used based on differences in adsorption, partition, or exchange properties of the metal and impurities.


Uses Of Aluminium, Copper, Zinc And Iron

Metals are widely used in various industries and applications:

Summary table for selected metals:

Metal Occurrence Common method of extraction Remarks
AluminiumBauxite, CryoliteElectrolysis of Al$_2$O$_3$ dissolved in molten Na$_3$AlF$_6$ (Hall-Heroult process)Requires large amount of electricity. Conductive cathode lining (carbon), sacrificial anode (graphite).
IronHaematite, Magnetite, Siderite, Iron pyritesReduction of oxide with CO and coke in Blast furnaceHigh temperature involved. CO is main reducing agent at lower T, C at higher T. Produces pig iron.
CopperCopper pyrites, Copper glance, Malachite, CupriteRoasting of sulphide partially and reduction (Pyrometallurgy)
Sulphuric acid leaching (Hydrometallurgy for low grade ores)
Self-reduction occurs in a converter. Easily reduced oxide.
ZincZinc blende (Sphalerite), Calamine, ZinciteRoasting followed by reduction of oxide with cokeReduction at higher T than copper. Zinc vapor is collected. Purified by fractional distillation.

Intext Questions

6.1 Which of the ores mentioned in Table 6.1 can be concentrated by magnetic separation method?

6.2 What is the significance of leaching in the extraction of aluminium?

6.3 The reaction, Cr$_2$O$_3$+2Al $\to$ Al$_2$O$_3$+2Cr (DrG$^0$= – 421kJ) is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?

6.4 Is it true that under certain conditions, Mg can reduce Al$_2$O$_3$ and Al can reduce MgO? What are those conditions?

Answer:

6.1 Magnetic separation is used for ores where either the ore particle or the gangue particle is magnetic. From Table 6.1, the iron ores Haematite ($\textsf{Fe}_2\text{O}_3$) and Magnetite ($\textsf{Fe}_3\text{O}_4$) contain iron compounds which are magnetic. Iron pyrites ($\textsf{FeS}_2$) is weakly magnetic. Siderite ($\textsf{FeCO}_3$) also contains iron and is magnetic. Therefore, iron ores can be concentrated by magnetic separation. Among other ores listed, Zinc blende ($\textsf{ZnS}$) and Cuprite ($\textsf{Cu}_2\text{O}$) generally do not have significant magnetic properties. Copper pyrites ($\textsf{CuFeS}_2$) contains iron, so it might be magnetic to some extent, but it is usually concentrated by froth floatation. Bauxite ($\textsf{Al}_2\text{O}_3$.xH$_2\text{O}$) contains iron oxides as impurities which are magnetic, so magnetic separation can be used to remove these impurities from bauxite.

The ores that can be concentrated by magnetic separation method are primarily the iron ores (Haematite, Magnetite, Siderite, Iron pyrites), and magnetic impurities can be removed from Bauxite by this method.

6.2 Leaching is the primary method for concentrating the bauxite ore in the extraction of aluminium (Bayer's process). Its significance is to selectively dissolve alumina ($\textsf{Al}_2\text{O}_3$) from the bauxite ore using a strong base ($\textsf{NaOH}$), while leaving behind most of the impurities like iron oxides and titanium oxide, and separating silica (which also dissolves) in a controlled manner. This process effectively purifies the alumina, which is then precipitated and calcined to obtain pure $\textsf{Al}_2\text{O}_3$, the compound that is subsequently reduced electrolytically to obtain aluminium metal. Leaching is crucial for obtaining high-purity alumina required for efficient electrolytic reduction.

6.3 The reaction $\textsf{Cr}_2\text{O}_3 + 2\textsf{Al} \to \textsf{Al}_2\text{O}_3 + 2\textsf{Cr}$ has a negative standard Gibbs energy change ($\Delta \textsf{G}^0 = -421$ kJ), indicating it is thermodynamically feasible under standard conditions. However, thermodynamics only predicts the possibility of a reaction, not its rate. Even though the products are more stable than the reactants, the reaction does not take place at room temperature because there is a significant activation energy barrier that needs to be overcome for the reaction to start. At room temperature, very few reactant molecules possess enough kinetic energy to reach the transition state and react. Heating provides the necessary activation energy for a sufficient number of molecules to react, allowing the reaction to proceed at a noticeable rate.

6.4 Yes, it is true that under certain conditions, Mg can reduce Al$_2$O$_3$ and Al can reduce MgO. This can be understood by looking at the Ellingham diagram (Fig. 6.4) and the curves for the formation of $\textsf{MgO}$ and $\textsf{Al}_2\text{O}_3$ (per mole of $\textsf{O}_2$).

  • Mg can reduce Al$_2$O$_3$: This reaction ($\textsf{Al}_2\text{O}_3 + 3\textsf{Mg} \to 2\textsf{Al} + 3\textsf{MgO}$) is feasible when the $\textsf{Mg}$, $\textsf{MgO}$ formation curve is below the $\textsf{Al}$, $\textsf{Al}_2\text{O}_3$ formation curve. In Fig. 6.4, these curves intersect at point "A" (approximately 1350$^\circ$C). Below this intersection temperature (at temperatures below approximately 1350$^\circ$C), the $\Delta \textsf{G}$ for $\textsf{MgO}$ formation is more negative than that for $\textsf{Al}_2\text{O}_3$ formation, making $\textsf{Mg}$ a stronger reducing agent than Al. So, Mg can reduce Al$_2$O$_3$.
  • Al can reduce MgO: This reaction ($\textsf{MgO} + 2/3\textsf{Al} \to \textsf{Mg} + 1/3\textsf{Al}_2\text{O}_3$) is feasible when the $\textsf{Al}$, $\textsf{Al}_2\text{O}_3$ formation curve is below the $\textsf{Mg}$, $\textsf{MgO}$ formation curve. This occurs at temperatures above the intersection point A (above approximately 1350$^\circ$C). Above this temperature, Al becomes a stronger reducing agent than Mg, and Al can reduce MgO.


Intext Questions



Question 6.1. Which of the ores mentioned in Table 6.1 can be concentrated by magnetic separation method?

Answer:

Question 6.2. What is the significance of leaching in the extraction of aluminium?

Answer:

Question 6.3. The reaction,

$Cr_2O_3 + 2Al \rightarrow Al_2O_3 + 2Cr$ ($\Delta_rG^\circ = – 421$ kJ)

is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?

Answer:

Question 6.4. Is it true that under certain conditions, Mg can reduce $Al_2O_3$ and Al can reduce MgO? What are those conditions?

Answer:



Exercises



Question 6.1. Copper can be extracted by hydrometallurgy but not zinc. Explain.

Answer:

Question 6.2. What is the role of depressant in froth floatation process?

Answer:

Question 6.3. Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?

Answer:

Question 6.4. Explain: (i) Zone refining (ii) Column chromatography.

Answer:

Question 6.5. Out of C and CO, which is a better reducing agent at 673 K ?

Answer:

Question 6.6. Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present ?

Answer:

Question 6.7. Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.

Answer:

Question 6.8. Write chemical reactions taking place in the extraction of zinc from zinc blende.

Answer:

Question 6.9. State the role of silica in the metallurgy of copper.

Answer:

Question 6.10. Which method of refining may be more suitable if element is obtained in minute quantity?

Answer:

Question 6.11. Which method of refining will you suggest for an element in which impurities present have chemical properties close to the properties of that elements?

Answer:

Question 6.12. Describe a method for refining nickel.

Answer:

Question 6.13. How can you separate alumina from silica in a bauxite ore associated with silica? Give equations, if any.

Answer:

Question 6.14. Giving examples, differentiate between ‘roasting’ and ‘calcination’.

Answer:

Question 6.15. How is ‘cast iron’ different from ‘pig iron”?

Answer:

Question 6.16. Differentiate between “minerals” and “ores”.

Answer:

Question 6.17. Why copper matte is put in silica lined converter?

Answer:

Question 6.18. What is the role of cryolite in the metallurgy of aluminium?

Answer:

Question 6.19. How is leaching carried out in case of low grade copper ores?

Answer:

Question 6.20. Why is zinc not extracted from zinc oxide through reduction using CO?

Answer:

Question 6.21. The value of $\Delta_fG^\circ$ for formation of $Cr_2O_3$ is – 540 kJmol$^{-1}$ and that of $Al_2O_3$ is – 827 kJmol$^{-1}$. Is the reduction of $Cr_2O_3$ possible with Al ?

Answer:

Question 6.22. Out of C and CO, which is a better reducing agent for ZnO ?

Answer:

Question 6.23. The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.

Answer:

Question 6.24. Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous solution of NaCl is subjected to electrolysis?

Answer:

Question 6.25. What is the role of graphite rod in the electrometallurgy of aluminium?

Answer:

Question 6.26. Outline the principles of refining of metals by the following methods:

(i) Zone refining

(ii) Electrolytic refining

(iii) Vapour phase refining

Answer:

Question 6.27. Predict conditions under which Al might be expected to reduce MgO. (Hint: See Intext question 6.4)

Answer: