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

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

Class 12 Chemistry chapter 6 explains the applications of various metals in our day-to-day life. Different terminologies such as refining, calcination, concentration, benefaction, roasting, etc. are very well explained here. The concepts of thermodynamics for the extraction of copper, aluminium and zinc are also mentioned here.

This chapter contains the details of various chemical processes associated with metallurgy. Being an important chapter, it will help the students score well during the examinations.

NCERT IN-TEXT QUESTIONS

Question 1.
Name some ores which can be concentrated by magnetic separation method.
Answer:
Only those ores can be concentrated by magnetic separation method in which either the ore particles or the impurities associated with it are of magnetic nature. For example, ores of iron such haematite (Fe2O3), magnetite (Fe3O4), siderite (FeCO3) are magnetic and can be concentrated by this method. Similarly, casseterite (SnO2) an ore of tin is non-magnetic while the impurities of tungstates of iron and chromium are of magnetic nature. Magnetic separation is effective in this case also.

Question 2.
What is the significance of leaching in the extraction of aluminium?
Answer:
Aluminium contains silica (SiO2), iron oxide (Fe2O3) and titanium oxide (TiO4) as impurities. These impurities can be removed by the process of leaching. During leaching, the powdered bauxite ore is heated with a concentrated (45%) solution of NaOH at 473-523 K, where alumina dissolves as sodium meta-aluminate and silica as sodium silicate leaving Fe2O3, TiO2 and other impurities behind:
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 1
The impurities are filtered off and solution of sodium meta-aluminate is neutralised by passing CO2 when hydrated alumina separates out while sodium silicate remains in solution. The hydrated aluminathus obtained is filtered, dried and heated to give back pure alumina.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 2
Thus, by leaching, pure alumina can be obtained from bauxite ore.

Question 3.
The reaction : Cr2O2(s) + 2Al(s) → Al2O3(s) + 2Cr(s) ; ∆G° = – 421 kJ is thermodynamically feasible as is apparent from the value of ∆G°. Why does not it take place at room temperature ?
Answer:
Though the reaction is feasible, it does not proceed at room temperature because all the reactants and products are solids. At elevated temperature, when chromium starts melting, the reaction becomes feasible.

Question 4
Is it true that under certain conditions, Mg can reduce Al2O3 and Al can reduce MgO ? What are those conditions ?
Answer:
If we look at the Ellingham diagram, it becomes evident that the plots for Al and Mg cross each other at 1350 °C (1623 K). Below this temperature, Mg can reduce Al2O3 and above this temperature, Al can reduce MgO.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 3

NCERT EXERCISE

Question 1.
Copper can be extracted by hydrometallurgy but not zinc. Explain.
Answer:
E° value of Zn2+/Zn = – 0·76V is less then that of Cu2+/Cu = + 0·34 V. This means that zinc is a stronger reducing agent than copper. In solution Cu2+ ions can be easily displaced by reducing agents like Fe and Zn which donot react with water.
Fe (s) + Cu2+ (aq) → Fe2+ (aq) + Cu (s)
In order to isolate zinc by hydrometallurgy, we require stronger reducing agents like Ca, Mg, Al etc. However, all of them react with water to evolve hydrogen gas. Therefore, these cannot be used for the purpose. Thus, zinc cannot be extracted by hydrometallurgy.

Question 2.
What is the role of depressant in froth floatation process ?
Answer:
A depressant suppresses the formation of froth with a particular compound in the froth floatation process by reacting chemically with it. Thus, it helps in the separation of two metal sulphides present together in a particular ore.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 4
In actual process, the sulphide ore is finely powdered and is mixed with water to form a slurry in a tank as shown in the Fig. 6.3. To this oily component of the sulphide ore particles by water. As a result, ore and oil constitute hydrophobic or water repelling component while gangue and water form a lyophilic or water attracting component.

Question 3.
Why is the extraction of copper from its sulphide ore difficult than that from its oxide through reduction ?
Answer:
Sulphide ore of copper (Cu2S) cannot be directly reduced by either coke or hydrogen because ∆fG° of Cu2S is more than those of CS2 and H2S that will be formed as a result of the reaction.
These reactions are therefore, not feasible. However the ∆fG° of Cu2O is lower than that of CO2. Therefore, the sulphide ore is first roasted to Cu2O which is then reduced.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 5

Question 4.
Explain
(i) zone refining
(ii) column chromatography. (C.B.S.E. Sample Paper 2015)
Answer:
(i) Zone refining (Fractional Crystallisation). This method is used only if a metal in almost pure state is required. Metals like germanium and gallium which are used in semiconductors are purified by this method. The principle of zone refining is based on the fact that impurities are more soluble in molten metal than in solid metal.

(ii) Chromatographic method. This method can also be employed on small scale for the purification of certain metals. Adsorption chromatography is normally used for this purpose. Different components present in a given sample are adsorbed to different extent on the surface of the adsorbent.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 7

Question 5.
Out of C and CO, which is a better reducing agent at 673 K ?
Answer:
If we look carefully at the Ellingham diagram (Fig. 6.7), we find that at 983 K, the curves intersect. AG° involving change of CO to CO2 (CO, CO2) is more as compared to the value involving change of C to CO2 (C, CO2). This means that at this temperature or above it, coke (C) is a better reducing agent. However, below this temperature, the reverse happens. Therefore, carbon monoxide (CO) is a better reducing agent at 673 K than coke (C).

Question 6.
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 the anode mud are antimony, selenium, tellurium, silver, gold and platinum. These elements settle down under anode as anode mud because they are less reactive and are not effected by CuSO4 – H2SO4solution.

Question 7.
Write the chemical reactions which take place in different zones in the blast furnace during the extraction of iron.
Answer:
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 8
The chemical reactions which take place in the blast furnace are briefly discussed.
Zone of combustion. At the bottom of the furnace the blast of hot air causes the combustion of coke into carbon dioxide. The reaction is highly exothermic and a temperature of nearly 2170 K develops. It supplies most of the heat required for the process.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 9

Zone of heat absorption. As CO2 gas rises up, it combines with more of coke to form carbon monoxide. Since the reaction is endothermic, the temperature in the middle of the furnace is nearly 1570 K. It further decreases as the reaction proceeds.

Zone of slag formation. In the middle of the furnace, the temperature is nearly 1123 K. Here lime stone (CaCO3) decomposes to form CaO and CO2. The former (CaO) combines with silica (SiO2) which is an impurity in the haematite ore to form calcium silicate (CaSiO3) that is fusible. This implies that CaO has acted as a basic flux. It has combined with the acidic impurity of Si02 to form slag.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 10

Zone of reduction : This is the upper part of the furnace. The temperature ranges between 500K to 900K. Here haematite (Fe2O3) is reduced to FeO.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 11
Further reduction of FeO to Fe occurs at higher temperature (1123 K) by CO gas.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 12
The direct reduction of iron ore left unreacted also occurs with carbon above 1123 K
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 13

Question 8.
Write the chemical reactions which take place in the extraction of zinc from zinc blende.
Answer:
Zinc blende is chemically zinc sulphide (ZnS). It undergoes following reactions :
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 14
Zinc metal is in crude or impure form. It can be refined with the help of electro-refining. In this method, impure zinc is made anode while a plate of pure metal acts as the cathode. The electrolyte is aqueous zinc sulphate containing a small amount of dilute H2SO4. On passing electric current, Zn2+ ions from the electrolyte migrate towards cathode and are reduced to zinc metal which gets deposited on the cathode.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 15

From anode, an equivalent amount of zinc gets oxidised to Zn2+ ions which migrate to the electrolyte
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 16

Question 9.
State the role of silica in the metallurgy of copper.
Answer:
During roasting, copper pyrites are converted into a mixture of FeO and Cu20. Thus, acidic flux silica is added during smelting to remove FeO (basic). FeO combines with SiO2to form famous silicate (FeSiO3) slag which floats over molten matte.

Question 10.
What is meant by the term chromatography?
Answer:
The term chromatography means colour writing (Greek : Chroma means colour ; graphy means writing). Initially, chromatography was used to identify and separate only the coloured components or constituents. But now, any type of constituents can be separated even if available in small amount.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 17

Question 11.
What criterion is followed for the selection of the stationary phase in chromatography?
Answer:
The stationary phase is selected in such a way that the impurities are more strongly adsorbed or are more soluble in the stationary phase than component to be purified. Thus, when the column is extracted, the impurities will be retained by the stationary phase while the pure component is easily eluted.

Question 12.
Describe a method for the refining of nickel.
Answer:
Nickel is refined by Mond’s process.
Mond’s process. Mond’s process is used for the refining of nickel metal. In the process, impure metal is heated in a current of carbon monoxide (CO) at 330 to 350 K to form nickel carbonyl which is of volatile nature. The vapours of the metal carbonyl escape leaving behind impurities. Upon heating to about 450 K, nickel carbonyl decomposes to give pure nickel.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 18
From the above discussion, we may conclude that the basic requirements for the refining of a metal by Vand Alkel process and Mond’s process are :
(i) The metal should form a volatile compound with the available reagent.
(ii) The volatile compound should be easily decomposable so that metal in the pure form may be regenerated.

Question 13.
How can you separate alumina from silica in bauxite ore associated with silica ? Give equations if any.
Answer:
The purification of bauxite containing silica as the main impurity is done by Serpeck’s process. The powdered ore is mixed with coke and heated to about 2073 K in an atmosphere of nitrogen. Silica (SiO2) is reduced to silicon which being volatile escapes. Alumina (Al2O3) is converted into aluminium nitride (AIN) by reacting with nitrogen. It is hydrolysed upon heating with water to get the precipitate of Al(OH)3. From the precipitate, Al2O3 is recovered upon strong heating.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 19

Question 14.
Giving examples, differentiate between calcination and roasting.
Answer:
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 20

Question 15.
I low is ‘cast iron’ different from ‘pig iron’?
Answer:
The iron obtained from blast furnace is called pig iron. It contains about 4% carbon and many other impurities in smaller amount (eg., S, P, Si and Mn). Cast iron is made by melting pig iron with scrap iron and coke using hot air blast. It has slightly lower carbon content (about 3%) and is extremely hard and brittle.

Question 16.
Differentiate between mineral and ore.
Answer:
The natural substances in which the metal or their compounds occur in the earth are called minerals. The mineral has a definite composition. It may be a single compound or complex mixture. The minerals from which the metals can be conveniently and economically extracted are known as ores. All the ores arc minerals but all minerals cannot be ores, e.g., both bauxite (Al2O3.xH2O) and clay (Al2O3.2SiO2.2H2O) are minerals of aluminium. It is bauxite which is used for extraction of aluminium and not clay. Thus bauxite is an ore of aluminium.

Question 17.
Why is copper matte put in silicon lined converter?
Answer:
Copper matte consists of a mixture of Cu2S and Cu2O. Along with that, it also contains small amount of FeS and FeO. It is put in a silicon lined converter known as Bessemer converter. Some silica (SiO2) is also added and a blast of hot air is blown. As a result, a number of reactions take place.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 21

Question 18.
What is the role of cryolite in the metallurgy of aluminium?
Answer:
In the metallurgy of aluminium, the metal is to be isolated from alumina (Al2O3) by carrying out its electrolytic reduction. The melting point of alumina as such is 2323 K. It is, therefore, mixed with cryolite (Na3AlF6) which lowers its melting point* to 1173 K. Moreover, cryolite also increases the electrical conductivity of alumina which is a poor conductor.

Question 19.
How is leaching carried out in case of low grade copper?
Answer:
Leaching in case of low grade copper is carried out by reacting with acid like H2SO4 in the presence of air when copper is oxidised to Cu2+ ions which pass into the solution. For example.
2 Cu(s) + 2 H2SO4 (aq) + O2 (g) → 2CuSO4 (aq) + 2 H2O(l)
or Cu + 2H+ (aq) + 1/2 O2(g) → Cu2+ (aq) + H2O(l)

Question 20.
Why is zinc not extracted from zinc oxide through reduction using CO?
Answer:
This is because the standard free energy of formation of CO2 from CO is higher than that of standard free energy of formation of ZnO from Zn.

Question 21.
The value of ∆fG° for the formation of Cr2O3 is – 540 kJ mol-1 and that of Al2O3 is – 827 kJ mol-1. Is the reduction of Cr2O3 with Al possible ? (Pb. Board2009)
Answer:
The two thermochemical equations may be written as follows :
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 22
Since ∆G° comes out to be negative, the reaction is feasible.

Question 22.
Out of C and CO, which is a better reducing agent for ZnO ?
Answer:
The two reduction reactions are :
ZnO(s) + C(s) → Zn(s) + CO(g) …(i)
ZnO(s) + CO(g) → Zn(s) + CO2(g) …(ii)
In the first case, there is increase in the magnitude of ∆S° while in the second case, it almost remains the same. In other words, AG° will have more negative value in the first case when C(s) is the reducing than in the second case when CO(g) acts as the reducing agent. Therefore, C(s) is a better reducing agent.

Question 23.
The choice of a reducing agent in a particular case depends on the thermodynamic factors. How do you agree with this statement ? Support your opinion with two examples.
Answer:
Thermodynamic factors have a major role in selecting the reducing agent for a particular reaction.
(i) Only that reagent will be preferred which will lead to decrease in free energy (∆G°) at a certain specific temperature.
(ii) A metal oxide placed lower in the Ellingham diagram cannot be reduced by the metal involved in the formation of the oxide placed higher in the diagram.
Examples:
(0 Al2O3 cannot be reduced by Cr present in Cr2O3 since the curve for Al2O3 is placed below that of Cr2O3 in the Ellingham diagram
(ii) CO cannot reduce ZnO because there is hardly any change in free energy (∆G°) as a result of the reaction.

Question 24.
Name the processes from which chlorine is obtained as the by-product. What will happen if an aqueous solution of NaCl is subjected to electrolysis ?
Answer:
Chlorine is obtained as the by-product in the manufacture of sodium by Down’s process in which molten sodium chloride is subjected to electrolysis.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 23
Sodium obtained by this method is almost pure while chlorine is the by-product.
Chlorine can also be obtained by carrying out the electrolysis of an aqueous solution of sodium chloride. The process is carried in Nelson’s cell. The various reactions which take place are as follows :
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 24
At cathode. Both Na+ and H+ ions migrate towards the cathode but H+ ions are discharged in preference to Na+ ions since their discharge potential is less. Na+ ions remain in the solution.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 25
At anode : Both Cl and OH ions migrate towards the anode but Cl ions are discharged in preference to OH” since their discharge potential is less. OH ions remain in the solution.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 26
Thus, in the electrolysis of an aqueous NaCl solution, H2 gas is evolved at the cathode and chlorine at the anode. The solution contains NaOH and is therefore, basic in nature.
It is always better to prepare chlorine by Down’s process.

Question 25.
What is the role of graphite rod in the electrometallurgy of aluminium ?
Answer:
In the electrometallurgy of aluminium, oxygen gas is evolved at anode. It reacts with graphite or carbon (graphite electrodes) to form carbon monoxide and carbon dioxide. In case some other metal electrodes act as anode, then oxygen will react with aluminium formed during the process to form aluminium oxide (Al2O3) which will pass into the reaction mixture. Since graphite is cheaper than aluminium, its wastage or consumption can be tolerated.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 27

Question 26.
Outline the principles of refining of metals by following methods :
(i) Zone refining
(ii) Electrolytic refining
(iii) Vapour phase refining.
Answer:
(i) Zone refining (Fractional Crystallisation): This method is used only if a metal in almost pure state is required. Metals like germanium and gallium which are used in semiconductors are purified by this method. The principle of zone refining is based on the fact that impurities are more soluble in molten metal than in solid metal.
In other words we can say that when an impure metal in the molten state is allowed to cool, only the metal crystallises while the impurities remain present in the molten mass or melt.

(ii) Electrolytic refining: This method is commonly used for the purification of the metals like Cu, Ag, Zn, Ni etc. The impure metal converted into a block which is made anode in an electrolytic cell The electrolyte is the solution of the soluble salt of the same metal, preferably a double salt. On passing electric current, metal ions from the electrolyte are reduced to the metal which is deposited the cathode. An equivalent amount of the pure metal from the anode gets oxidised and the metal ions (or cations) go into the solution.

(iii) Vapour phase refining: In this method, the impure metal is converted into a volatile compound, by suitable method leaving behind the impurities. The volatile compound formed being unstable decomposes at an elevated temperature to give pure metal. A few cases are discussed.

Question 27.
Predict the conditions under which aluminium can be expected to reduce magnesium oxide.
Answer:
The equations for the formation of the two oxides are :
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 228
If we look at the plots for the formation of the two oxides on the Ellingham diagram, we find that they intersect at certain point. The corresponding value of ∆G° becomes zero for the reduction of MgO by Al metal.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 29
This means that the reduction of MgO by A1 metal can occur below this temperature.
Aluminium (Al) metal can reduce MgO to Mg above this temperature because ∆fG for Al2O3 is less as compared to that of MgO.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 30NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 30
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