Class 12th Chemistry Part I CBSE Solution
Intext Questions Pg-150- Which of the ores mentioned in Table 6.1 can be concentrated by magnetic separation…
- What is the significance of leaching in the extraction of aluminium?…
Intext Questions Pg-157- The reaction, Cr2O3 + 2 Al → Al2 O3 + 2 Cr (ΔG^0 = - 421 kJ) is thermodynamically feasible…
- Is it true that under certain conditions, Mg can reduce Al2O3 and Al can reduce MgO? What…
Exercises- Copper can be extracted by hydrometallurgy but not zinc. Explain.…
- What is the role of depressant in froth floatation process?
- Why is the extraction of copper from pyrites more difficult than that from its oxide ore…
- Zone refining Explain:
- Column chromatography. Explain:
- Out of C and CO, which is a better reducing agent at 673 K?
- Name the common elements present in the anode mud in electrolytic refining of copper. Why…
- Write down the reactions taking place in different zones in the blast furnace during the…
- Write chemical reactions taking place in the extraction of zinc from zinc blende.…
- State the role of silica in the metallurgy of copper.
- What is meant by the term “chromatography”?
- What criterion is followed for the selection of the stationary phase chromatography?…
- Describe a method for refining nickel.
- How can you separate alumina from silica in a bauxite ore associated with silica? Give…
- Giving examples, differentiate between ‘roasting’ and ‘calcination’.…
- How is ‘cast iron’ different from ‘pig iron”?
- Differentiate between “minerals” and “ores”.
- Why is copper matte put in the silica lined converter?
- What is the role of cryolite in the metallurgy of aluminium?
- How is leaching carried out in case of low-grade copper ores?
- Why is zinc not extracted from zinc oxide through reduction using CO?…
- The value of for the formation of Cr2 O3 is - 540 kJmol−1and that of Al2 O3 is - 827…
- Out of C and CO, which is a better reducing agent for ZnO?
- The choice of a reducing agent in a particular case depends on the thermodynamic factor.…
- Name the processes from which chlorine is obtained as a by-product. What will happen if an…
- What is the role of graphite rod in the electrometallurgy of aluminium?…
- Outline the principles of refining of metals by the following methods: (i) Zone refining…
- Predict conditions under which Al might be expected to reduce MgO. (Hint: See Intext…
- Which of the ores mentioned in Table 6.1 can be concentrated by magnetic separation…
- What is the significance of leaching in the extraction of aluminium?…
- The reaction, Cr2O3 + 2 Al → Al2 O3 + 2 Cr (ΔG^0 = - 421 kJ) is thermodynamically feasible…
- Is it true that under certain conditions, Mg can reduce Al2O3 and Al can reduce MgO? What…
- Copper can be extracted by hydrometallurgy but not zinc. Explain.…
- What is the role of depressant in froth floatation process?
- Why is the extraction of copper from pyrites more difficult than that from its oxide ore…
- Zone refining Explain:
- Column chromatography. Explain:
- Out of C and CO, which is a better reducing agent at 673 K?
- Name the common elements present in the anode mud in electrolytic refining of copper. Why…
- Write down the reactions taking place in different zones in the blast furnace during the…
- Write chemical reactions taking place in the extraction of zinc from zinc blende.…
- State the role of silica in the metallurgy of copper.
- What is meant by the term “chromatography”?
- What criterion is followed for the selection of the stationary phase chromatography?…
- Describe a method for refining nickel.
- How can you separate alumina from silica in a bauxite ore associated with silica? Give…
- Giving examples, differentiate between ‘roasting’ and ‘calcination’.…
- How is ‘cast iron’ different from ‘pig iron”?
- Differentiate between “minerals” and “ores”.
- Why is copper matte put in the silica lined converter?
- What is the role of cryolite in the metallurgy of aluminium?
- How is leaching carried out in case of low-grade copper ores?
- Why is zinc not extracted from zinc oxide through reduction using CO?…
- The value of for the formation of Cr2 O3 is - 540 kJmol−1and that of Al2 O3 is - 827…
- Out of C and CO, which is a better reducing agent for ZnO?
- The choice of a reducing agent in a particular case depends on the thermodynamic factor.…
- Name the processes from which chlorine is obtained as a by-product. What will happen if an…
- What is the role of graphite rod in the electrometallurgy of aluminium?…
- Outline the principles of refining of metals by the following methods: (i) Zone refining…
- Predict conditions under which Al might be expected to reduce MgO. (Hint: See Intext…
Intext Questions Pg-150
Question 1.Which of the ores mentioned in Table 6.1 can be concentrated by magnetic separation method?
Answer:The ore can be concentrated by the process of magnetic separation, only if either the ore or the gangue can be attracted in presence of magnetic field. In table 6.1, the ores of iron such as haematite (Fe2O3), magnetite (Fe3O4), siderite (FeCO3), and iron pyrite (FeS2) can be separated by the process of magnetic separation.
Question 2.What is the significance of leaching in the extraction of aluminium?
Answer:The method of leaching consists of treating the powdered ore with a suitable reagent which can selectively dissolve the ore but not the impurities. The impurities are filtered out and are recovered from the solution. For example, bauxite ore containing SiO2, iron oxide, and titanium oxides impurities are concentrated by this method.
Leaching is significant as it helps in removing the impurities like SiO2, FeO2, TiO2, etc from the bauxite ore.
Which of the ores mentioned in Table 6.1 can be concentrated by magnetic separation method?
Answer:
The ore can be concentrated by the process of magnetic separation, only if either the ore or the gangue can be attracted in presence of magnetic field. In table 6.1, the ores of iron such as haematite (Fe2O3), magnetite (Fe3O4), siderite (FeCO3), and iron pyrite (FeS2) can be separated by the process of magnetic separation.
Question 2.
What is the significance of leaching in the extraction of aluminium?
Answer:
The method of leaching consists of treating the powdered ore with a suitable reagent which can selectively dissolve the ore but not the impurities. The impurities are filtered out and are recovered from the solution. For example, bauxite ore containing SiO2, iron oxide, and titanium oxides impurities are concentrated by this method.
Leaching is significant as it helps in removing the impurities like SiO2, FeO2, TiO2, etc from the bauxite ore.
Intext Questions Pg-157
Question 1.The reaction,
Cr2O3 + 2 Al → Al2 O3 + 2 Cr (ΔG0 = – 421 kJ)
is thermodynamically feasible as is apparent from the Gibbs energy value.
Why does it not take place at room temperature?
Answer:The reaction,
Cr2O3 + 2 Al → Al2 O3 + 2 Cr (ΔG0 = – 421 kJ)
Is thermodynamically feasible as is apparent from Gibbs energy value. The change in Gibbs free energy is related to the equilibrium constant, k as
ΔG = - RT ln K
A certain amount of energy activation is required even for such reactions which are thermodynamically feasible, therefore heating is required.
ΔG = ΔH + S
Increasing the temperature increases the value of TΔS, making the value of ΔG more and more negative. Therefore, the reaction becomes more and more feasible as the temperature is increased.
Question 2.Is it true that under certain conditions, Mg can reduce Al2O3 and Al can reduce MgO? What are those conditions?
Answer: It is true because when the given has less affinity for oxygen than magnesium and alumnium reduce oxides of each other.
The reaction,
Cr2O3 + 2 Al → Al2 O3 + 2 Cr (ΔG0 = – 421 kJ)
is thermodynamically feasible as is apparent from the Gibbs energy value.
Why does it not take place at room temperature?
Answer:
The reaction,
Cr2O3 + 2 Al → Al2 O3 + 2 Cr (ΔG0 = – 421 kJ)
Is thermodynamically feasible as is apparent from Gibbs energy value. The change in Gibbs free energy is related to the equilibrium constant, k as
ΔG = - RT ln K
A certain amount of energy activation is required even for such reactions which are thermodynamically feasible, therefore heating is required.
ΔG = ΔH + S
Increasing the temperature increases the value of TΔS, making the value of ΔG more and more negative. Therefore, the reaction becomes more and more feasible as the temperature is increased.
Question 2.
Is it true that under certain conditions, Mg can reduce Al2O3 and Al can reduce MgO? What are those conditions?
Answer: It is true because when the given has less affinity for oxygen than magnesium and alumnium reduce oxides of each other.
Exercises
Question 1.Copper can be extracted by hydrometallurgy but not zinc. Explain.
Answer:Copper can be extracted by hydrometallurgy but not zinc as per the below-given reason:
The Eo of zinc ( Zn2+/Zn = - 0.76V) is lower than that of copper (Cu2+/Cu = 0.34V). this means that zinc is a stronger reducing agent and can displace copper from a solution of Cu2+ ions.
Zn (s) + Cu2+ (aq) → Zn2+ + Cu (s)
In order to displace zinc by hydrometallurgy, we need stronger reducing agent like K ( EoK+/K = - 2.93V)
2K (s) + 2H2O (l) → 2KOH (aq) + H2 (g)
As a result, these metals cannot be used in hydrometallurgy to extract zinc. Hence, copper can be extracted by hydrometallurgy but not zinc.
Question 2.What is the role of depressant in froth floatation process?
Answer:In the froth floatation method, the role of depressants is to prevent certain types of particles from forming the froth with the air bubbles. For example, NaCN is used as a depressant to separate lead sulfide ore (PbS) from zinc Sulphide ore (ZnS).NaCN forms a zinc complex Na2[Zn(CN)4] on the surface of ZnS thereby preventing it from the formation of froth.
Question 3.Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?
Answer:The Gibbs free energy formation ( ΔfG) of Cu2S is less than that of H2S and CS2. Therefore, H2 and C cannot reduce Cu2S to Cu.
On the other hand, the Gibbs free energy formation of Cu2O is greater than that of CO. hence, C can reduce Cu2O to Cu.
C (s) + Cu2O (s) → 2Cu (s) + CO (g)
Hence, the extraction of copper from its pyrite ore is difficult than from its oxide ore through reduction.
Question 4.Explain:
Zone refining
Answer:Zone refining method is based on the principle that impurities are more soluble in the molten state of metal (the melt) than in the solid state. The impure metal is heated with the help of a circular mobile heater at one end. This results in the formation of the molten zone or melt. As the heater is removed along with the length of the rod, the pure metal crystallizes out of the melt and impurities pass into the adjacent molten zone. This process is repeated several times till the impurities are completely driven to the end of the rod which is then cut off and discarded. The method is very useful for semiconductor and other metals of very high purity, e.g., silicon, germanium, boron, and gallium.
Question 5.Explain:
Column chromatography.
Answer:It is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. The mixture to be separated is put in a liquid or gaseous medium which is moved through the adsorbent. Different components are adsorbed at different levels on the column. Later, the adsorbed components are removed (eluted) by using suitable solvents (eluents). There are several chromatographic techniques such as paper chromatography, column chromatography, gas chromatography, etc. chromatography is used for the purification of elements which are available in minute quantities and the impurities are not very different in chemical properties from the element to be purified.
Question 6.Out of C and CO, which is a better reducing agent at 673 K?
Answer:
At 673K, the value of G(CO,Co2) is less than that of G(C,Co). Therefore, CO can be reduced more easily to Co2 than C to CO. hence, CO is a better reducing agent than C at 673K.
Question 7.Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?
Answer:The common elements present in anode mud in electrolytic refining are antimony, selenium, tellurium, silver, gold, and platinum. These elements being less reactive, are not affected by CuSO4 + H2SO4 solution and hence settle down under anode as anode mud.
Question 8.Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.
Answer:During the extraction of iron, the reduction of iron oxides takes place in the blast furnace. In the process, hot air is blown from the bottom of the furnace and coke is burnt to raise the temperature up to 2200K in the lower portion itself. The temperature is lower in the upper part. Thus, it is the lower part where the reduction of iron oxides (Fe2O3 and Fe3O4) takes place.
The reactions taking place in the lower temperature range (500K – 800K ) in the blast furnace are:-
3 Fe2O3 + CO → 2 Fe3O4 + CO2
Fe3O4 + 4 CO → 3 Fe + 4 CO2
Fe2O3 + CO → 2 Fe3 + CO2
The reactions taking place in the higher temperature range ( 900 – 1500K ) in the blast furnace are:-
C + CO2→ 2 CO
FeO + CO → Fe + CO2
The silicate impurity of the ore is removed as a slag by calcium oxide ( CaO), which is formed by the decomposition of limestone (CaCO3).
CaCO3→ CaO + CO2
CaO + SiO2→ CaSiO3 (Calcium silicate/ slag)
Question 9.Write chemical reactions taking place in the extraction of zinc from zinc blende.
Answer:The steps involved in the extraction of zinc from zinc blende (ZnS) are as followed:-
1) Concentration: the ore is crushed and then concentrated by froth floatation process.
2) Roasting: the concentrated ore is heated in the presence of an excess of air at about 1200K to form zinc oxide.
ZnS + 3 O2→ 2 ZnO + 2 SO2
3) Reduction: ZnO obtained above is mixed with powdered coke and heated to 1673K in a fireclay resort.
4) Electrolytic Refining: Zinc is refined by the process of electrolytic refining. In this process, impure zinc is made the anode and a pure copper strip is made the cathode. The electrode used in an acidified solution of zinc sulphate (ZnSO4). Electrolysis results in the transfer of zinc in pure form from the anode to the cathode.
Question 10.State the role of silica in the metallurgy of copper.
Answer:During the roasting of pyrite ore, a mixture of FeO and Cu2O is obtained.
2 CuFeS2 + O2→ Cu2S + 2 FeS + SO2
2 Cu2S + 3 O2 → 2 Cu2O + 2 SO2
2 FeS + 3 O2 → 2 FeO + 2 SO2
The role of silica in the metallurgy of copper is to remove the iron oxide obtained during the process of roasting as ‘slag’. If the sulphide ore of copper iron, then silica (SiO2) is added as a flux before roasting. Then, FeO combines with silica to form silicate, FeSiO3 (slag).
FeO + SiO2→ FeSiO3 (slag)
Question 11.What is meant by the term “chromatography”?
Answer:Chromatography is a term used for numerous laboratory techniques for the separation of mixtures. The term is derived from Greek word ‘chroma’ meaning ‘color’ and ‘graphein’ meaning ‘to write’. Chromatographic techniques are based on the principles that different components are adsorbed differently on an adsorbent. There are various chromatographic techniques used such as paper chromatography, column chromatography, gas chromatography, etc.
Question 12.What criterion is followed for the selection of the stationary phase chromatography?
Answer:The components selected for the stationary phase should have different solubility’s in the phase, only then it is selected as the stationary phase. Hence, different components have different rates of movement through the stationary phase, as a result, can be separated from each other.
Question 13.Describe a method for refining nickel.
Answer:Nickel is refined by Mond’s process. In this process, nickel is heated in the presence of carbon monoxide to form nickel tetracarbonyl, which is a volatile complex.
Question 14.How can you separate alumina from silica in a bauxite ore associated with silica? Give equations, if any.
Answer:To separate alumina from silica in bauxite ore associated with silica, firstly, the powdered ore is digested with a concentrated NaOH solution at a temperature of 473 – 523K and 35 – 36 bar pressure. This results in the leaching out of alumina (Al2O3) as sodium aluminate and silica (SiO2) as sodium silicate leaving the impurities behind.
Al2O3 (s) + 2NaOH (aq) + 3H2O (l) → 2 Na[Al(OH)4] (aq)
Alumina
SiO2 + 2NaOH (aq)→ Na2SiO3 (aq) + H2O (l)
Silica
After this, CO2 gas is passed through the resultant solution to neutralize the alumina present in the solution, this results in the precipitation of hydrated samples of alumina. To include precipitation, the solution is seeded with freshly prepared samples of hydrated alumina.
During this process, sodium silicate remains in the solution. The hydrated alumina obtained is then filtered, dried, and heated to get back pure alumina.
Question 15.Giving examples, differentiate between ‘roasting’ and ‘calcination’.
Answer:The difference between roasting and calcination is: 
Question 16.How is ‘cast iron’ different from ‘pig iron”?
Answer:The iron obtained from blast furnaces is known as pig iron. It contains around 4% carbon and many impurities such as S, P, Si, Mn in smaller amounts.
Cast iron is obtained by melting pig iron and coke using a hot air blast. It contains a lower amount of carbon (3%) than pig iron, cast iron is extremely hard and brittle.
Question 17.Differentiate between “minerals” and “ores”.
Answer:The naturally occurring substances in the form of which metals occur in the earth’s crust along with impurities are called as minerals. The mineral from which the metal can be extracted conveniently and profitably is called as an ore. Thus, all ores are minerals but all minerals are not ores. For example, aluminium occurs in earth’s crust in the form of bauxite and clay. Out of these two minerals, aluminium can be conveniently and economically extracted from the bauxite. Therefore, bauxite is the ore of aluminium. Similarly, Zinc can be obtained from blende (ZnS), calamine (ZnCO3), Zincite (ZnO) etc.
Thus these minerals are called ores of zinc.
Question 18.Why is copper matte put in the silica lined converter?
Answer:Copper matte contains Cu2S and FeS. When a blast of hot air is passed through molten matte took in a silica lined convertor, FeS present in matte is oxidized to FeO which combines with Silica (SiO2) to form FeSiO3, slag.
When the whole of iron has been removed as slag, some of the Cu2S undergoes oxidation to form Cu2O which then reacts with more Cu2O to form copper metal.
Question 19.What is the role of cryolite in the metallurgy of aluminium?
Answer:Cryolite (Na3AlF6) has two roles in the metallurgy of aluminium:
1) It lowers the melting point of the mixture to about 1140K.
2) It increases the electrical conductivity of the mixture.
Question 20.How is leaching carried out in case of low-grade copper ores?
Answer:For the group of low-grade copper ores, leaching is carried out using acid or bacteria in the presence of air or oxygen. In this process, copper goes into the solution as Cu2+ ions.
The resulting solution is then treated with scrap iron or H2 to obtain metallic copper.
Question 21.Why is zinc not extracted from zinc oxide through reduction using CO?
Answer:The standard Gibbs free energy formation of ZnO from Zn is lower than that of CO2 from CO. That is why CO cannot reduce ZnO to Zn. thus, Zn is not extracted from ZnO through reduction using CO.
Question 22.The value of 
for the formation of Cr2 O3 is – 540 kJmol−1and that of Al2 O3 is – 827 kJmol−1. Is the reduction of Cr2 O3 possible with Al?
Answer:The value of Δ fG0 for the formation of Cr2O3 is – 540 kJ mol-1 which are higher than that of Al2O3 is – 827 Kjmol-1. Thus, Al can reduce Cr2O3 to Cr. Hence, the reduction of Cr2O3 with Al is possible.
Question 23.Out of C and CO, which is a better reducing agent for ZnO?
Answer:
The free energy formation ( ΔfGo ) of CO from C becomes lower at temperatures above 1120K whereas that of CO2 from C becomes lower above 1323K than ΔfG0 of ZnO. However, Δ fG0 of CO2 from CO is always higher than that of ZnO. Therefore, C can reduce ZnO to Zn but not CO. therefore, out of C and CO, C is a better reducing agent than CO for ZnO.
Question 24.The choice of a reducing agent in a particular case depends on the thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.
Answer:
Thermodynamic factors help us in choosing a suitable reducing agent for the reduction of a particular metal state as described below.
From Ellingham diagram, it is evident that metals for which the standard free energy of formation of their oxides is more negative can reduce those metal oxides for which the standard free energy of formation of their respective oxides is less negative. In other words, any metal will reduce the oxides of other metals which lie above in the Ellingham Diagram because the standard free energy change of the combined redox reaction will be negative by any amount of equal to the difference in Δ fG0 of the two metal oxides. Hence, both Al, Zn can reduce FeO to Fe, but Fe cannot reduce Al2O3 to Al and ZnO to Zn.
Question 25.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:In the electrolysis of NaCl by Down’s process, chlorine is obtained as a by-product. This process involves the electrolysis of a fused mixture of NaCl and CaCl2 at 873K. during electrolysis, sodium is liberated at the cathode and Cl2 is liberated at the anode.
If an aqueous solution of NaCl is electrolysed, H2 is evolved at the cathode and Cl2 is obtained at the anode. the reason being that E0 of Na+/Na redox couple is much lower (E0 = - 2.71 V) than that of H2O (E0H2O/H2) = - 0.83V ) and hence water is reduced to H2 in presence of Na+ ions. However, NaOH is obtained in the solution.
Question 26.What is the role of graphite rod in the electrometallurgy of aluminium?
Answer:Graphite rod acts as anode and graphite lined iron acts as a cathode in the electrometallurgy of aluminium. Carbon reacts with oxygen liberated at anode producing CO and CO2 otherwise oxygen liberated at the anode may oxidize some of the liberated aluminium back to Al2O3.
Question 27.Outline the principles of refining of metals by the following methods:
(i) Zone refining
(ii) Electrolytic refining
(iii) Vapour phase refining
Answer:(i) Zone refining Is the method based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.
(ii) Electrolytic refining works on the principle of refining impure metals by the use of electricity. In Electrolytic refining, the impure metal is made the anode and a strip of pure metal is made as 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 a 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 called as the anode mud.
(iii) Vapour 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.
iii) Nickel, Zircinium. And titanium are refined using this method.
Question 28.Predict conditions under which Al might be expected to reduce MgO. (Hint: See Intext question 6.4)
Answer:Above 1350oC, the standard Gibbs free energy formation of Al2O3 from Al is less than that of MgO from Mg. Therefore, above 1350oC, Al can reduce MgO.
Copper can be extracted by hydrometallurgy but not zinc. Explain.
Answer:
Copper can be extracted by hydrometallurgy but not zinc as per the below-given reason:
The Eo of zinc ( Zn2+/Zn = - 0.76V) is lower than that of copper (Cu2+/Cu = 0.34V). this means that zinc is a stronger reducing agent and can displace copper from a solution of Cu2+ ions.
Zn (s) + Cu2+ (aq) → Zn2+ + Cu (s)
In order to displace zinc by hydrometallurgy, we need stronger reducing agent like K ( EoK+/K = - 2.93V)
2K (s) + 2H2O (l) → 2KOH (aq) + H2 (g)
As a result, these metals cannot be used in hydrometallurgy to extract zinc. Hence, copper can be extracted by hydrometallurgy but not zinc.
Question 2.
What is the role of depressant in froth floatation process?
Answer:
In the froth floatation method, the role of depressants is to prevent certain types of particles from forming the froth with the air bubbles. For example, NaCN is used as a depressant to separate lead sulfide ore (PbS) from zinc Sulphide ore (ZnS).NaCN forms a zinc complex Na2[Zn(CN)4] on the surface of ZnS thereby preventing it from the formation of froth.
Question 3.
Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?
Answer:
The Gibbs free energy formation ( ΔfG) of Cu2S is less than that of H2S and CS2. Therefore, H2 and C cannot reduce Cu2S to Cu.
On the other hand, the Gibbs free energy formation of Cu2O is greater than that of CO. hence, C can reduce Cu2O to Cu.
C (s) + Cu2O (s) → 2Cu (s) + CO (g)
Hence, the extraction of copper from its pyrite ore is difficult than from its oxide ore through reduction.
Question 4.
Explain:
Zone refining
Answer:
Zone refining method is based on the principle that impurities are more soluble in the molten state of metal (the melt) than in the solid state. The impure metal is heated with the help of a circular mobile heater at one end. This results in the formation of the molten zone or melt. As the heater is removed along with the length of the rod, the pure metal crystallizes out of the melt and impurities pass into the adjacent molten zone. This process is repeated several times till the impurities are completely driven to the end of the rod which is then cut off and discarded. The method is very useful for semiconductor and other metals of very high purity, e.g., silicon, germanium, boron, and gallium.
Question 5.
Explain:
Column chromatography.
Answer:
It is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. The mixture to be separated is put in a liquid or gaseous medium which is moved through the adsorbent. Different components are adsorbed at different levels on the column. Later, the adsorbed components are removed (eluted) by using suitable solvents (eluents). There are several chromatographic techniques such as paper chromatography, column chromatography, gas chromatography, etc. chromatography is used for the purification of elements which are available in minute quantities and the impurities are not very different in chemical properties from the element to be purified.
Question 6.
Out of C and CO, which is a better reducing agent at 673 K?
Answer:
At 673K, the value of G(CO,Co2) is less than that of G(C,Co). Therefore, CO can be reduced more easily to Co2 than C to CO. hence, CO is a better reducing agent than C at 673K.
Question 7.
Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?
Answer:
The common elements present in anode mud in electrolytic refining are antimony, selenium, tellurium, silver, gold, and platinum. These elements being less reactive, are not affected by CuSO4 + H2SO4 solution and hence settle down under anode as anode mud.
Question 8.
Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.
Answer:
During the extraction of iron, the reduction of iron oxides takes place in the blast furnace. In the process, hot air is blown from the bottom of the furnace and coke is burnt to raise the temperature up to 2200K in the lower portion itself. The temperature is lower in the upper part. Thus, it is the lower part where the reduction of iron oxides (Fe2O3 and Fe3O4) takes place.
The reactions taking place in the lower temperature range (500K – 800K ) in the blast furnace are:-
3 Fe2O3 + CO → 2 Fe3O4 + CO2
Fe3O4 + 4 CO → 3 Fe + 4 CO2
Fe2O3 + CO → 2 Fe3 + CO2
The reactions taking place in the higher temperature range ( 900 – 1500K ) in the blast furnace are:-
C + CO2→ 2 CO
FeO + CO → Fe + CO2
The silicate impurity of the ore is removed as a slag by calcium oxide ( CaO), which is formed by the decomposition of limestone (CaCO3).
CaCO3→ CaO + CO2
CaO + SiO2→ CaSiO3 (Calcium silicate/ slag)
Question 9.
Write chemical reactions taking place in the extraction of zinc from zinc blende.
Answer:
The steps involved in the extraction of zinc from zinc blende (ZnS) are as followed:-
1) Concentration: the ore is crushed and then concentrated by froth floatation process.
2) Roasting: the concentrated ore is heated in the presence of an excess of air at about 1200K to form zinc oxide.
ZnS + 3 O2→ 2 ZnO + 2 SO2
3) Reduction: ZnO obtained above is mixed with powdered coke and heated to 1673K in a fireclay resort.
4) Electrolytic Refining: Zinc is refined by the process of electrolytic refining. In this process, impure zinc is made the anode and a pure copper strip is made the cathode. The electrode used in an acidified solution of zinc sulphate (ZnSO4). Electrolysis results in the transfer of zinc in pure form from the anode to the cathode.
Question 10.
State the role of silica in the metallurgy of copper.
Answer:
During the roasting of pyrite ore, a mixture of FeO and Cu2O is obtained.
2 CuFeS2 + O2→ Cu2S + 2 FeS + SO2
2 Cu2S + 3 O2 → 2 Cu2O + 2 SO2
2 FeS + 3 O2 → 2 FeO + 2 SO2
The role of silica in the metallurgy of copper is to remove the iron oxide obtained during the process of roasting as ‘slag’. If the sulphide ore of copper iron, then silica (SiO2) is added as a flux before roasting. Then, FeO combines with silica to form silicate, FeSiO3 (slag).
FeO + SiO2→ FeSiO3 (slag)
Question 11.
What is meant by the term “chromatography”?
Answer:
Chromatography is a term used for numerous laboratory techniques for the separation of mixtures. The term is derived from Greek word ‘chroma’ meaning ‘color’ and ‘graphein’ meaning ‘to write’. Chromatographic techniques are based on the principles that different components are adsorbed differently on an adsorbent. There are various chromatographic techniques used such as paper chromatography, column chromatography, gas chromatography, etc.
Question 12.
What criterion is followed for the selection of the stationary phase chromatography?
Answer:
The components selected for the stationary phase should have different solubility’s in the phase, only then it is selected as the stationary phase. Hence, different components have different rates of movement through the stationary phase, as a result, can be separated from each other.
Question 13.
Describe a method for refining nickel.
Answer:
Nickel is refined by Mond’s process. In this process, nickel is heated in the presence of carbon monoxide to form nickel tetracarbonyl, which is a volatile complex.
Question 14.
How can you separate alumina from silica in a bauxite ore associated with silica? Give equations, if any.
Answer:
To separate alumina from silica in bauxite ore associated with silica, firstly, the powdered ore is digested with a concentrated NaOH solution at a temperature of 473 – 523K and 35 – 36 bar pressure. This results in the leaching out of alumina (Al2O3) as sodium aluminate and silica (SiO2) as sodium silicate leaving the impurities behind.
Al2O3 (s) + 2NaOH (aq) + 3H2O (l) → 2 Na[Al(OH)4] (aq)
Alumina
SiO2 + 2NaOH (aq)→ Na2SiO3 (aq) + H2O (l)
Silica
After this, CO2 gas is passed through the resultant solution to neutralize the alumina present in the solution, this results in the precipitation of hydrated samples of alumina. To include precipitation, the solution is seeded with freshly prepared samples of hydrated alumina.
During this process, sodium silicate remains in the solution. The hydrated alumina obtained is then filtered, dried, and heated to get back pure alumina.
Question 15.
Giving examples, differentiate between ‘roasting’ and ‘calcination’.
Answer:
The difference between roasting and calcination is:
Question 16.
How is ‘cast iron’ different from ‘pig iron”?
Answer:
The iron obtained from blast furnaces is known as pig iron. It contains around 4% carbon and many impurities such as S, P, Si, Mn in smaller amounts.
Cast iron is obtained by melting pig iron and coke using a hot air blast. It contains a lower amount of carbon (3%) than pig iron, cast iron is extremely hard and brittle.
Question 17.
Differentiate between “minerals” and “ores”.
Answer:
The naturally occurring substances in the form of which metals occur in the earth’s crust along with impurities are called as minerals. The mineral from which the metal can be extracted conveniently and profitably is called as an ore. Thus, all ores are minerals but all minerals are not ores. For example, aluminium occurs in earth’s crust in the form of bauxite and clay. Out of these two minerals, aluminium can be conveniently and economically extracted from the bauxite. Therefore, bauxite is the ore of aluminium. Similarly, Zinc can be obtained from blende (ZnS), calamine (ZnCO3), Zincite (ZnO) etc.
Thus these minerals are called ores of zinc.
Question 18.
Why is copper matte put in the silica lined converter?
Answer:
Copper matte contains Cu2S and FeS. When a blast of hot air is passed through molten matte took in a silica lined convertor, FeS present in matte is oxidized to FeO which combines with Silica (SiO2) to form FeSiO3, slag.
When the whole of iron has been removed as slag, some of the Cu2S undergoes oxidation to form Cu2O which then reacts with more Cu2O to form copper metal.
Question 19.
What is the role of cryolite in the metallurgy of aluminium?
Answer:
Cryolite (Na3AlF6) has two roles in the metallurgy of aluminium:
1) It lowers the melting point of the mixture to about 1140K.
2) It increases the electrical conductivity of the mixture.
Question 20.
How is leaching carried out in case of low-grade copper ores?
Answer:
For the group of low-grade copper ores, leaching is carried out using acid or bacteria in the presence of air or oxygen. In this process, copper goes into the solution as Cu2+ ions.
The resulting solution is then treated with scrap iron or H2 to obtain metallic copper.
Question 21.
Why is zinc not extracted from zinc oxide through reduction using CO?
Answer:
The standard Gibbs free energy formation of ZnO from Zn is lower than that of CO2 from CO. That is why CO cannot reduce ZnO to Zn. thus, Zn is not extracted from ZnO through reduction using CO.
Question 22.
The value of for the formation of Cr2 O3 is – 540 kJmol−1and that of Al2 O3 is – 827 kJmol−1. Is the reduction of Cr2 O3 possible with Al?
Answer:
The value of Δ fG0 for the formation of Cr2O3 is – 540 kJ mol-1 which are higher than that of Al2O3 is – 827 Kjmol-1. Thus, Al can reduce Cr2O3 to Cr. Hence, the reduction of Cr2O3 with Al is possible.
Question 23.
Out of C and CO, which is a better reducing agent for ZnO?
Answer:
The free energy formation ( ΔfGo ) of CO from C becomes lower at temperatures above 1120K whereas that of CO2 from C becomes lower above 1323K than ΔfG0 of ZnO. However, Δ fG0 of CO2 from CO is always higher than that of ZnO. Therefore, C can reduce ZnO to Zn but not CO. therefore, out of C and CO, C is a better reducing agent than CO for ZnO.
Question 24.
The choice of a reducing agent in a particular case depends on the thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.
Answer:
Thermodynamic factors help us in choosing a suitable reducing agent for the reduction of a particular metal state as described below.
From Ellingham diagram, it is evident that metals for which the standard free energy of formation of their oxides is more negative can reduce those metal oxides for which the standard free energy of formation of their respective oxides is less negative. In other words, any metal will reduce the oxides of other metals which lie above in the Ellingham Diagram because the standard free energy change of the combined redox reaction will be negative by any amount of equal to the difference in Δ fG0 of the two metal oxides. Hence, both Al, Zn can reduce FeO to Fe, but Fe cannot reduce Al2O3 to Al and ZnO to Zn.
Question 25.
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:
In the electrolysis of NaCl by Down’s process, chlorine is obtained as a by-product. This process involves the electrolysis of a fused mixture of NaCl and CaCl2 at 873K. during electrolysis, sodium is liberated at the cathode and Cl2 is liberated at the anode.
If an aqueous solution of NaCl is electrolysed, H2 is evolved at the cathode and Cl2 is obtained at the anode. the reason being that E0 of Na+/Na redox couple is much lower (E0 = - 2.71 V) than that of H2O (E0H2O/H2) = - 0.83V ) and hence water is reduced to H2 in presence of Na+ ions. However, NaOH is obtained in the solution.
Question 26.
What is the role of graphite rod in the electrometallurgy of aluminium?
Answer:
Graphite rod acts as anode and graphite lined iron acts as a cathode in the electrometallurgy of aluminium. Carbon reacts with oxygen liberated at anode producing CO and CO2 otherwise oxygen liberated at the anode may oxidize some of the liberated aluminium back to Al2O3.
Question 27.
Outline the principles of refining of metals by the following methods:
(i) Zone refining
(ii) Electrolytic refining
(iii) Vapour phase refining
Answer:
(i) Zone refining Is the method based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.
(ii) Electrolytic refining works on the principle of refining impure metals by the use of electricity. In Electrolytic refining, the impure metal is made the anode and a strip of pure metal is made as 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 a 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 called as the anode mud.
(iii) Vapour 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.
iii) Nickel, Zircinium. And titanium are refined using this method.
Question 28.
Predict conditions under which Al might be expected to reduce MgO. (Hint: See Intext question 6.4)
Answer:
Above 1350oC, the standard Gibbs free energy formation of Al2O3 from Al is less than that of MgO from Mg. Therefore, above 1350oC, Al can reduce MgO.