NCERT Solutions Class 12 Chemistry Chapter 6 General Principles of Isolation of Elements

Question1. Explain why hydro-metallurgy can extract copper , but cannot extract zinc.
Solution : Copper has a higher reduction potential than zinc or iron. Thus, in hydro-metallurgy iron and zinc can be used to displace copper from its solution.
Fe(s) + Cu²⁺_(aq)
⇒ Fe²⁺ (aq) + Cu(s)
However, in-order to displace zinc a metal more reactive than it is required e.g. K, Ca , etc. The problem arising now is that all these metals react with water to liberate hydrogen gas. Thus, they cannot be used for the extraction of zinc using hydro-metallurgy.

Question2. What role does a depressant play in the process of froth floatation ?
Solution : The role of a depressant in forth floatation is to set apart two sulphide ores through selective prevention of one ore from frothing. E.g. NaCN is used to separate two sulphide ores , PbS and ZnS. NaCN forms a complex, Na₂ [Zn( CN )₄], with zinc on the surface of zinc sulphide, thus selectively preventing ZnS from frothing.
4NaCN + ZnS ⇒ Na₂ [Zn( CN )₄] + Na₂S

Question3. Why is it more difficult to extract copper from pyrites than it is to extract copper from its oxide ore ?
Solution : Hydrogen sulfite and carbon disulfite (CS₂) has a larger ∆_fG (Gibbs free energy of formation) than copper (I) sulfite. Thus, C and H₂S will not be able to reduce Cu₂S to Cu.
Whereas, ∆_fG of Cu₂O is larger than carbon monoxide’s , thus C can reduce copper (I) oxide to copper.
Therefore, it is more difficult to extract copper from pyrites than it is to extract from its oxide.

Question4. Write notes on :
(a) Zone refining
(b) Column chromatography.
Solution : (a) Zone refining:
This method makes use of a principle according to which impurities are more soluble in molten metal than in a solid metal. In zone refining, a moving circular heater slowly moves over an impure metal rod/ bar. As the circular heater moves, the impurities move into the adjacent molten zone and the pure metal crystallizes out. This is repeated over several times, causing the segregation of impurities at one end of the bar. This impure end is then cut off. Boron, silicon, gallium etc. are purified this way.

(ii) Column chromatography:
Column chromatography is a technique used to separate different components of a mixture. It is a very useful technique used for the purification of elements available in minute quantities. It is also used to remove the impurities that are not very different in chemical properties from the element to be purified. Chromatography is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. In chromatography, there are two phases: mobile phase and stationary phase. The stationary phase is immobile and immiscible. Al2O3 column is usually used as the stationary phase in column chromatography. The mobile phase may be a gas, liquid, or supercritical fluid in which the sample extract is dissolved. Then, the mobile phase is forced to move through the stationary phase. The component that is more strongly adsorbed on the column takes a longer time to travel through it than the component that is weakly adsorbed. The adsorbed components are then removed (eluted) using a suitable solvent (eluant).

Question5. At 673 K which is a more effective reducing agent, CO or C ?
Solution : At 673 K ,CO is a more effective reducing agent because, at 673 K the value of ∆G_{( C,CO )} is greater than the value of ∆G_{( CO,CO2 )}, which means that CO is more easily reduced to CO₂ than C to CO.

Question6. In the process of electrolytic refining of copper what are the common elements in the anode mud ? What is the reason for their presence ?
Solution : Selenium, silver, tellurium, platinum, antimony and gold are the common elements present in the anode mud during the electrolytic refining of copper.
As these metals are almost unreactive they are not affected during the purification method. Thus, they are present in the anode mud.

Question7. State the various reactions that take place inside the different zones of a blast furnace during iron extraction.
Solution : Inside a blast furnace iron oxide is reduced at different temperature ranges, as given below:
At 500 – 800 K
3Fe₂O₃ +CO → 2Fe₃O₄ + CO₂
Fe₂O₃ +CO → 2FeO + CO₂
At 850 K
Fe₃O₄ +CO → 3FeO + CO₂
At 900 – 1500 K
FeO +CO → Fe + CO₂
CO₂ + C → 2CO
Above 1570 K
C+ O₂ → CO₂
FeO +C → Fe + CO
CaCO_3(s) → CaO_(s) + CO_2(g)
SiO₂ + CaO → CaSiO₃ ( slag )

Question8.Write the chemical reactions involved in zinc extraction from zinc blend.
Solution : The various steps involved in zinc extraction are :
(1) Concentration of ore:
Firstly, to remove the impurities of the ore, zinc blende is crushed and concentration is done by froth floatation.
(2) Conversion to oxide (Roasting):
In this step ZnS is roasted in excess of air in a furnace at a temperature below the melting point of zinc. This results in the formation of zinc oxide.
2ZnS + 3O₂ ⇒ 2ZnO + 2SO₂
(3) Extraction of Zn from ZnO (Reduction):
To extract zinc from zinc oxide it is mixed with powdered coke and then heated.
ZnO + C ⇒ Zn + CO
(4) Electrolytic Refining:
In this method, impure zinc becomes the anode while the cathode is a pure copper strip. The electrolyte is an acidic solution of zinc sulphate (ZnSO₄) with dilute sulfuric acid. On passing a current, pure zinc deposits on the copper strip.

Question9. What role does silica play in copper metallurgy ?
Solution : Copper pyrites are converted to a mixture of Cu₂O and FeO during roasting. Acidic silica is added during smelting to separate the FeO from the mixture. SiO₂ and FeO forms FeSiO₃ ( slag )which floats on the molten matter.
FeO + SiO₂ ⇒ FeSiO₃ (slag)

Question10. “Chromatography”, What do you understand by this term?
Solution : Chromatography is a collective term used for a range of laboratory techniques for the purification, separation, characterization and identification of the species of a mixtures wether colored or colorless. The word has been obtained from two Greek words; ‘graphy’ meaning ‘to write’ and ‘chroma’ meaning ‘colour’. There are numerous chromatographic techniques like column chromatography, paper chromatography, gas chromatography, etc.

Question11. What is the criterion followed while selecting the stationary phase of chromatography ?
Solution : The choice of stationary phase is done in a way that the elements of the mixture have different solubility in the phase. Thereby, different elements have different movement speed through the phase , because of which they can be separated from each other.

Question12. Write about a technique used to refine nickel ?
Solution : Mond’s process is technique used to refine nickel. In this method, heat is supplied to nickel in the presence of carbon monoxide to produce nickel tetracarbonyl, which is a volatile complex.
Ni + 4CO ⇒ Ni( CO)₄ [330k – 350k]
The nickel tetracarbonyl so obtained is then decomposed by heating it at a higher temperature (450 − 470 K) to form pure nickel.
Ni( CO)₄ ⇒ Ni + CO [450k – 470k]

Question13. How is alumina separated from silica in a bauxite ore associated with silica? Provide equations, if present.
Solution : Separation of alumina from silica in a bauxite ore associated with silica is as follows :
Firstly, concentrated NaOH solution is used to digest the powdered ore at 473 − 523 K and at a pressure of 35 − 36 bar. This causes alumina ( Al₂O₃ ) and silica ( SiO₂ ) to leach as sodium aluminate and sodium silicate respectively, abandoning the impurities behind.
Al₂O₃ (s) + 2NaOH(aq) + 3H₂O (f) ⇒ 2Na [ Al ( OH)₄ ](aq)
Alumina sodium aluminate
SiO₂ + 2NaOH(aq) ⇒ Na₂SiO(aq) + H₂O(l)
Silica Sodium silicate
Now, to neutralize the aluminate present in the solution, carbon dioxide gas is passed through the solution we obtained. This causes the sodium meta – aluminate to precipitate as hydrated alumina.
2Na [ Al ( OH)₄ ](aq) + CO₂
⇒ Al₂O₃.xH₂O(s) + 2NaHCO₃(aq)
hydrated alumina
Sodium silicate cannot be precipitated so it is filtered off. The hydrated alumina is dried, heated and filtered to obtain pure alumina.
Al₂O₃.xH₂O(s) ⇒ Al₂O₃(s) + xH₂O( g) [1470 k]

Question14. Differentiate between ‘calcination’ and ‘roasting’ with the help of examples.
Solution : Calcination is the conversion of carbonate and hydroxide ores to oxides by heating them at a temperature below their melting points and in the absence or in a very limited supply of air.
For example, carbonates of Ca, Mg, Zn are turned into their respective oxides through this process.
ZnCO₃ ZnO + CO₂
Fe₂O₃.3H₂O ⇒ Fe₂O₃ + 3H₂ [Heat]
Roasting is the conversion of sulfide ores into their metallic oxides by heating at a temperature below their melting points in excess presence of air.
For e.g, sulfide ores of Pb, and Zn are turned into their respective oxides through this process.
2Zns + 3O₂ ⇒ 2ZnO + 2SO₂
2Pbs + 3O₂ ⇒ 2PbO + 2SO₂

Question15. What is the difference between ‘cast iron’ and ‘pig iron’ ? 
Solution : Pig iron is the iron obtained from blast furnace. It is 4% carbon and contains many other impurities like Si, P, S, Mn in lesser amounts.
Cast iron is made by melting pig iron and coke with the help of a hot air blast. It has lower levels of carbon in it (3%). Unlike pig iron, cast iron is brittle and extremely hard.

Question16. What is the difference between “ores” and “minerals” ?
Solution : Minerals are naturally available substances having metals or their compounds.
Ores are rocks and minerals from which metals are economically and conveniently obtained.
For e.g., there are a lot of minerals that contain, but zinc cannot be conveniently and economically extracted from all of them. Zinc is extracted from calamine (ZnCO3), zinc blende (ZnS), etc. Hence, these minerals qualify as ores of zinc.

Question17. Why is copper matte put in converters lined with silica ?
Solution : Copper matte has Cu₂S and some FeS. So, when a hot blast of air is blown through a molten matte placed in a silica lined converter , FeS of the matte oxidizes to FeO. This FeO combines with SiO₂ ( silica ) to produce FeSiO₃ (slag).
2FeS + 3O₂ ⇒ 2FeO + 2SO₂
FeO + SiO₂ ⇒ FeSiO₃
When all of the iron has been removed as slag Cu₂S is oxidized to Cu₂O which then undergoes a reaction with Cu₂S to give copper metal.
2Cu₂S + 3O₂ ⇒ 2CuO₂ + 2SO₂
2Cu₂O + Cu₂S ⇒ 6Cu + SO₂

Question18. In the metallurgy of aluminium, what role does Cryolite play ?
Solution : Na₃AlF₆ ( Cryolite ) has two basic roles:
1. The melting point of the solution is decreased from 2323 to 1140 K due to the presence of cryolite.
2. Cryolite makes alumina a good conductor of electricity.

Question19. How is leaching done in when dealing with low grade copper ores?
Solution : When dealing with low grade copper ores, bacteria or acids are used in the presence of air to leach the copper. In this method, copper enters the solution as Cu2+ ions :
Cu + 2H⁺ + ½ O₂ ⇒ Cu2+ + 2H₂O
The solution so obtained is then treated with H₂ or scrap iron to get copper metal.
Cu + 2H+ ⇒ 2H²⁺ + Cu

Question20. Why isn’t zinc oxide reduction (using carbon monoxide) used to extract zinc ?
Solution : As the standard Gibbs free energy of formation of CO to CO₂ is higher than that of Zn to ZnO, CO will not be able to reduce ZnO to Zn. Thereby, zinc oxide reduction using CO isn’t used to extract zinc.

Question21. ∆_fGθ has a value of − 540 kJ mol⁻¹ for Cr₂O₃ formation and − 827 kJ mol ⁻¹for Al₂ O₃ formation .Can Cr₂O₃ be reduced by Al ?
Solution : As Cr₂O₃ has a higher value of formation ( −540 kJ mol⁻¹ ) than Al₂O₃ ( −827 kJ /mol ), Al can thus reduce Cr₂O₃ to Cr.
Alternatively,
2Al + (3/2)O₂ ⇒ Al₂O₃
2Cr + (3/2)O₂ ⇒ Cr₂O₃
Subtracting equation (ii) from (i) , we have
2Al + Cr₂O₃⇒ Al₂O₃ +2Cr; ∆_fGθ = -827 – ( -540 ) = -287kJ / mol
As ∆_fGθ is negative for the reduction reaction of Cr₂O₃ by Al , the reaction can take place.

Question22. Which one is better at reducing ZnO, C or CO₂ ?
Solution :

nO is reduced to Zn at around 1673 K. From the figure below, we can observe that beyond 1073 K the Gibbs free energy of formation of CO from C and beyond 1273 K, the Gibbs free energy of formation of CO₂ from C is smaller than the Gibbs free energy of formation of ZnO. Thus, C can reduce ZnO to Zn. However, the Gibbs free energy of formation of CO₂ from CO is greater than the Gibbs free energy of formation of ZnO. Thus, CO is not able to reduce ZnO making C a better reducing agent for ZnO.

Question23. In some particular case thermodynamic factor dictates the choice of the reducing agent. To what extent do you believe this statement to be true ? Provide two examples to back your opinion.
Solution :

The figure given below is a graph of Gibbs energy ∆Gθ vs Temperature for the formation of solid oxides. We can see from this graph that if ∆_fGθ of a metal’s oxide is more negative than another metal’s oxide, then the former metal can reduce the oxide of the latter metal ( i.e., the oxide with less negative ∆_fGθ ). For example, ∆_fGθ ( Al, Al₂O₃ ) is more negative than ∆_fGθ ( Cu, Cu₂O ), thus Al can readily reduce Cu₂O to Cu but it is not possible for Cu to reduce Al₂O₃. Likewise, it isn’t possible for Zn to reduce MgO but Mg can reduce ZnO to Zn. This is because ∆_fGθ ( Mg, MgO ) is more negative than ∆_fGθ ( Zn,ZnO ).

Question24. What are the processes that give chlorine as a by-product ? What happens when an aqueous of NaCl is electolyzed ?
Solution : In Down process, chlorine is obtained as a byproduct. In this process fused mixture of CaCl₂ and NaCl is electrolyzed at 873 K. Sodium is obtained the cathode and Cl₂ is discharged at the anode.
When an aqueous solution of NaCl is subjected to electrolysis, Cl₂ is discharged at the
anode. At the cathode, H₂ is obtained ( in place of Na ). This is because the standard reduction potential of H₂O (E° = − 0.83 V) is less negative than that of Na (E°= − 2.71 V). Thus, H₂O will get a higher preference to get reduced at the cathode thereby discharging H₂.
NaCl ⇒ Na⁺ + Cl ^–
At cathode : 2H₂O + 2e ^– ⇒ H₂ + 2OH ^–
At anode : Cl ^– ⇒ Cl + e^–
2Cl ⇒ Cl₂

Question25. What role does a graphite rod play in the electrometallurgy of aluminium?
Solution : In the electrometallurgy of aluminum, the anode is graphite and the cathode iron is graphite lined .The O₂ being liberated due to electrolysis reacts with the graphite anode to liberate CO₂ and CO. Had it been any other metal besides graphite, the oxygen would have reacted with the aluminum during the electrolysis to form Al₂O₃. This would have resulted in the wastage of aluminium and as graphite is cheaper than aluminum it is expendable.
This is the role of aluminum.

Question26. What are the principles of refining a metal by the following processes :
(a) Zone refining
(b) Electrolytic refining
(c) Vapour phase refining
Solution : (a) Zone refining :
This method makes use of a principle according to which impurities are more soluble in molten metal than in a solid metal. In zone refining, a moving circular heater slowly moves over an impure metal rod/ bar. As the circular heater moves, the impurities move into the adjacent molten zone and the pure metal crystallizes out. This is repeated over several times, causing the segregation of impurities at one end of the bar. This impure end is then cut off. Boron, silicon, gallium etc. are purified this way.

(ii) Electrolytic refining; Electrolytic refining is the process of refining impure metals by using electricity. In this process, impure metal is made the anode and a strip of pure metal is made the cathode. A solution of a soluble salt of the same metal is taken as the electrolyte. When an electric current is passed, metal ions from the electrolyte are deposited at the cathode as pure metal and the impure metal from the anode dissolves into the electrolyte in the form of ions. The impurities present in the impure metal gets collected below the anode.
This is known as anode mud.

(iii) Vapour phase refining
Vapour phase refining is the process of refining metal by converting it into its volatile compound and then, decomposing it to obtain a pure metal. To carry out this process,
(i) the metal should form a volatile compound with an available reagent, and
(ii) the volatile compound should be easily decomposable so that the metal can be easily recovered.
Nickel, zirconium, and titanium are refined using this method.'

Question 6.28: Predict conditions under which Al might be expected to reduce MgO.
Answer Above 1350°C, the standard Gibbs free energy of formation of Al2O3 from Al is less than that of MgO from Mg. Therefore, above 1350°C, Al can reduce MgO.