NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements are part of NCERT Solutions for Class 12 Chemistry. Here we have given NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements.
|Chapter Name||General Principles and Processes of Isolation of Elements|
|Number of Questions Solved||31|
NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements
NCERT IN-TEXT QUESTIONS
Name some ores which can be concentrated by magnetic separation method.
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.
What is the significance of leaching in the extraction of aluminium ?
Leaching or chemical separation is quite effective to purify bauxite an ore of aluminium associated with the impurities of iron oxide. The ore is leached with concentrated solution of NaOH to form a soluble complex leaving behind the impurities.
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 ?
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.
Is it true that under certain conditions, Mg can reduce Al2O3 and Al can reduce MgO ? What are those conditions ?
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.
Copper can be extracted by hydrometallurgy but not zinc. Explain.
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.
What is the role of depressant in froth floatation process ?
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.
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.
Why is the extraction of copper from its sulphide ore difficult than that from its oxide through reduction ?
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.
(i) zone refining
(ii) column chromatography. (C.B.S.E. Sample Paper 2015)
(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.
Out of C and CO, which is a better reducing agent at 673 K ?
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).
Name the common elements present in anode mud in the electro-refining of copper. Why are they so present ?
Anode mud contains metals like Ag, Au, Pt etc. which are less reactive than Cu. Actually, they are not in a position
to lose electrons though they constitute the electrode which acts as anode. All these metals are left as residue under anode (known as anode mud) while the entire copper present participates in the oxidation half reaction.
Cu(s) → Cu2+(aq) + 2e–
Write the chemical reactions which take place in different zones in the blast furnace during the extraction of iron.
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.
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.
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.
Further reduction of FeO to Fe occurs at higher temperature (1123 K) by CO gas.
The direct reduction of iron ore left unreacted also occurs with carbon above 1123 K
Write the chemical reactions which take place in the extraction of zinc from zinc blende.
Zinc blende is chemically zinc sulphide (ZnS). It undergoes following reactions :
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.
From anode, an equivalent amount of zinc gets oxidised to Zn2+ ions which migrate to the electrolyte
State the role of silica in the metallurgy of copper.
Silica (SiO2) acts as an acidic flux in the metallurgy of copper and combines with FeO (the main impurity) to form FeSiO3 which is a slag.
What is meant by the term chromatography ?
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.
What criterion is followed for the selection of the stationary phase in chromatography ?
In chromatography, particularly in adsorption chromatography, the stationary phase is the adsorbent. It should fulfil certain criteria for better results.
(i) It should have high but selective adsorption power.
(ii) The particles should be spherical in shape and of uniform size.
(iii) The adsorbent should not react chemically with the solvents used for elution or with the components of the mixture under investigation.
(iv) The adsorbent should contain as small amount of the soluble components as possible.
(v) The adsorbent should be catalytically inactive and must have a neutral surface.
(vi) The adsorbent should be easily available.
(vi) The adsorbent should be perfectly white.
Describe a method for the refining of nickel.
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.
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.
How can you separate alumina from silica in bauxite ore associated with silica ? Give equations if any.
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.
Giving examples, differentiate between calcination and roasting.
How is ‘cast iron’ different from ‘pig iron’ ?
Cast iron differs from pig iron with respect to the carbon contents. Whereas carbon contents in pig iron are nearly four percent (4%), cast iron contains carbon to the extent of nearly three percent (3%).
Differentiate between mineral and ore.
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.
Why is copper matte put in silicon lined converter ?
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.
What is the role of cryolite in the metallurgy of aluminium ?
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.
How is leaching carried out in case of low grade copper ?
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)
Why is zinc not extracted from zinc oxide through reduction using CO ?
The chemical reaction involving the reduction of ZnO by CO is :
ZnO(s) + CO(g) → Zn(s) + CO2(g)
The process is thermodynamically not feasible because there is hardly any change in entropy as a result of the reaction. This is quite evident from the physical states of the reactants and products involved in the reaction
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)
The two thermochemical equations may be written as follows :
Since ∆G° comes out to be negative, the reaction is feasible.
Out of C and CO, which is a better reducing agent for ZnO ?
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.
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.
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.
(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.
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 ?
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.
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 :
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.
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.
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.
What is the role of graphite rod in the electrometallurgy of aluminium ?
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.
Outline the principles of refining of metals by following methods :
(i) Zone refining
(ii) Electrolytic refining
(iii) Vapour phase refining.
(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.
Predict the conditions under which aluminium can be expected to reduce magnesium oxide.
The equations for the formation of the two oxides are :
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.
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.
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