Chapter 6: Isolation of Elements Class 12 Chemistry NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 6 – General Principles and Processes of Isolation of Elements (In-text Questions)

Question 6.1: Which of the ores mentioned can be concentrated by magnetic separation method?

Answer: Magnetic separation is used to separate magnetic ores from non-magnetic impurities. Ores like haematite (Fe₂O₃), magnetite (Fe₃O₄), siderite (FeCO₃), and iron pyrites (FeS₂) are magnetic in nature. These can be effectively separated from impurities like silica using magnetic separation.

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

Answer: Leaching is used to purify bauxite, the primary ore of aluminium, which is often mixed with iron oxides. The ore is treated with a concentrated NaOH solution that dissolves alumina as sodium aluminate, leaving behind iron oxides as impurities. This process helps in extracting pure aluminium oxide.

Question 6.3: Why do some redox reactions used in metallurgy occur only at high temperatures?

Answer: This is due to thermodynamic principles, specifically the Gibbs free energy change (ΔG). At room temperature, if ΔG is positive or not sufficiently negative, the reaction doesn’t occur. However, at high temperatures, entropy (ΔS) increases, making ΔG more negative, enabling the reaction to proceed.

Question 6.4: Is it true that under certain conditions, Mg can reduce Al₂O₃ and Al can reduce MgO? What are those conditions?

Answer: Yes, it is true. This depends on the Ellingham diagram which plots ΔG vs. T. At higher temperatures, the element lower in the diagram can reduce the oxide of the one above it.
Mg, being below Al in the diagram, can reduce Al₂O₃. However, under normal conditions, Al cannot reduce MgO because its ΔG is less negative than that for the formation of MgO.

NCERT Solutions for Class 12 Chemistry Chapter 6 – General Principles and Processes of Isolation of Elements (Exercises)

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

Answer: Copper can be extracted from low-grade ores using hydrometallurgy because Cu²⁺ ions can be easily displaced by more reactive metals like Fe or Zn. However, zinc is a highly reactive metal and cannot be displaced from its salt solution by any metal placed above it in the reactivity series. Hence, zinc cannot be extracted by this method.

Question 6.2: What is the role of depressant in the froth flotation process?

Answer: Depressants help to suppress the flotation of certain ores. For example, NaCN is used to prevent ZnS from forming froth in a mixture of PbS and ZnS by forming a complex Na₂[Zn(CN)₄]. This allows PbS to float and be separated from ZnS.

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

Answer: The ΔG° of formation of Cu₂S is less negative than that of other sulphides. Thus, reducing copper from its sulphide ore (pyrite) is difficult with common reducing agents. Copper is more easily extracted from its oxide ores.

Question 6.4: Explain the following:

Answer: (i) Zone refining: This technique purifies metals like Si, Ge, and Ga. A moving heater melts a small region of impure metal. Impurities concentrate in the molten zone and are pushed to one end, which is then removed.
(ii) Column chromatography: Components of a mixture are separated based on different adsorption to a stationary phase like Al₂O₃ or SiO₂, using a moving solvent (mobile phase).

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

Answer: At 673 K, CO is a better reducing agent. According to the Ellingham diagram, CO’s line lies below that of carbon at this temperature, indicating CO can reduce metal oxides more effectively.

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

Answer: Anode mud contains Ag, Au, and Pt. These less reactive metals do not dissolve during electrolysis and settle at the bottom.

Question 6.7: Write the reactions occurring in different zones of the blast furnace during iron extraction.

Answer: At bottom (1200–1700 K): Fe₂O₃ + 3C → 2Fe + 3CO
Middle zone (900–1500 K): Fe₂O₃ + 3CO → 2Fe + 3CO₂
Upper zone (500–900 K): 3Fe₂O₃ + CO → 2Fe₃O₄ + CO₂

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

Answer: Roasting: 2ZnS + 3O₂ → 2ZnO + 2SO₂
Reduction: ZnO + C → Zn + CO
Refining (Electrolytic): ZnSO₄; Zn (anode) → Zn²⁺ + 2e⁻; Zn²⁺ + 2e⁻ → Zn (cathode)

Question 6.9: What is the role of silica in the metallurgy of copper?

Answer: Silica acts as a flux to remove FeO impurities by forming iron silicate (FeSiO₃), which floats as slag and is removed.

Question 6.10: What is meant by the term “chromatography”?

Answer: Chromatography is a separation technique based on different adsorption or solubility of components of a mixture on a stationary phase.

Question 6.11: What criterion is followed for the selection of the stationary phase in chromatography?

Answer: The stationary phase is chosen based on differential adsorption of components. Stronger adsorbing components move slower, aiding separation.

Question 6.12: Describe a method for refining nickel.

Answer: Mond’s Process:
Ni + 4CO → Ni(CO)₄ (volatile compound at 330–350 K)
Ni(CO)₄ → Ni + 4CO (pure Ni formed at 450–470 K)

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

Answer: Baeyer’s Process:
Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O (silica remains undissolved)
NaAlO₂ + 2H₂O → Al(OH)₃ ↓ + NaOH

Question 6.14: Giving examples, differentiate between roasting and calcination.

Answer: Roasting: Heating in presence of air. Example: 2ZnS + 3O₂ → 2ZnO + 2SO₂
Calcination: Heating in absence of air. Example: CaCO₃ → CaO + CO₂

Question 6.15: How is the extraction of aluminium different from iron?

Answer: Aluminium is extracted by electrolyzing alumina in molten cryolite, while iron is extracted by reduction of iron oxides using coke in a blast furnace.

Question 6.16: Explain the significance of leaching in the extraction of aluminium.

Answer: Leaching helps remove Fe₂O₃ and SiO₂ impurities by dissolving alumina in NaOH solution, thus purifying the ore before electrolysis.

Question 6.17: What is the role of cryolite in the extraction of aluminium?

Answer: Cryolite lowers the melting point of alumina and increases electrical conductivity, making electrolysis more efficient.

Question 6.18: How is cast iron different from pig iron?

Answer: Pig iron contains ~4% carbon, whereas cast iron, made from pig iron, contains ~3% carbon and is tougher and more malleable.

Question 6.19: Differentiate between ‘minerals’ and ‘ores’.

Answer: Minerals are naturally occurring compounds, while ores are minerals from which metals can be economically extracted.

Question 6.20: Why copper matte is put in a silica-lined converter during extraction?

Answer: Silica reacts with FeO forming slag (FeSiO₃), and Cu₂S reacts with oxygen to form Cu₂O, which further reacts with Cu₂S to give copper metal and SO₂ gas.

Question 6.21: What is the role of zinc in the extraction of silver?

Answer: Zinc displaces silver from [Ag(CN)₂]⁻ complex:
2[Ag(CN)₂]⁻ + Zn → 2Ag + [Zn(CN)₄]²⁻

Question 6.22: What type of metals are generally extracted by electrolytic methods? Why?

Answer: Highly electropositive metals like Na, K, Al, Ca, and Mg are extracted by electrolysis as they form very stable oxides that cannot be reduced chemically.

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?

Answer: The choice depends on Gibbs free energy (ΔG°). Only if ΔG° is negative, the reduction is feasible. Examples: Al reduces FeO; Zn reduces CuO.

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: Down’s process produces chlorine gas as a by-product.
On electrolysis of aqueous NaCl, H₂ gas is liberated at cathode and Cl₂ gas at anode.

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

Answer: Graphite rods act as anodes. Oxygen gas liberated would otherwise oxidize aluminium; graphite reacts with oxygen to form CO or CO₂ instead.

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

Answer: (i) Zone Refining: Impurities concentrate at one end as the heater moves.
(ii) Electrolytic Refining: Metal dissolves at anode and deposits at cathode.
(iii) Vapour-Phase Refining: Metal forms a volatile compound which decomposes to give pure metal.

Question 6.27: Predict conditions under which Al might be expected to reduce MgO.

Answer: At temperatures where Al’s line lies below Mg’s line in the Ellingham diagram (high temperatures), Al can reduce MgO.