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

NCERT Solutions for Class 12 Chemistry Chapter 6 – General Principles and Processes of Isolation of Elements – FREE PDF Download

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Revision Notes Class 12 chemistry
NCERT Solutions Class 12 chemistry

Subtopics for Class 12 Chemistry Chapter 6 – General Principles and Processes of Isolation of Elements

1. Occurrence of Metals
2. Concentration of Ores
3. Extraction of Crude Metal from Concentrated Ore
4. Thermodynamic Principles of Metallurgy
5. Electrochemical Principles of Metallurgy
6. Oxidation-Reduction
7. Refining
8. Uses of Aluminium, Copper, Zinc, and Iron.

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

Chemistry NCERT Solutions for Class 12  Chapter 6 INTEXT Questions

Question 2.
What is the significance of leaching in the extraction of aluminium?

Solution:
Leaching is significant as it helps in removing the impurities like SiO₂, Fe₂O₃, etc. from the bauxite ore.

Question 1.
Copper can be extracted by hydrometallurgy but not zinc. Explain.
Solution:
The E° of zinc (Zn²⁺/Zn = – 0.76 V) is lower luau that of copper (Cu²⁺/Cu = +0.34 V). Hence, zinc can displace Cu from solutions of Cu²⁺ ions.

But all these metals react with water forming their corresponding ions with the evolution of H₂ gas. Hence, Al, Mg, etc. cannot be used to displace zinc from the solution of Zn²⁺ ions. Thus, copper can be extracted by hydrometal-lurgy but not zinc.

Question 2.
What is the role of depressant in froth-floatation process?
Solution:
In froth -floatation process, the role of the depressant is to prevent one type of sulphide ore particles from forming the froth with air bubbles. For example, NaCN is used as a depressant to separate lead sulphide (PbS) ore from zinc sulphide (ZnS) ore. The reason is that NaCN forms a zinc complex, Na₂[Zn(CN)₄] on the surface of ZnS preventing it from the formation of the froth. Under these conditions, only PbS forms froth and therefore, it can be separated from ZnS ore.

Question 3.
Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?
Solution:
In the Ellingham diagram, the Cu₂O line is almost at the top. So it is quite easy to reduce oxide ores of copper directly to the metal by heating with coke (as can be seen in graph that the lines of C, CO and C, CO₂ are at much lower positions). But most of the ores are sulphide and some may also contain iron. So, the sulphide ores are roasted/smelted to give oxides :

Question 4.
Explain : (i) Zone refining (ii) Column chroma-tography.
Solution:
(i) This method is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal. A circular mobile heater is fixed at one end of a rod of the impure metal. The molten zone moves along with the heater which is moved forward. As the heater moves forward, the pure metal crystallises out of the melt and the impurities pass on into the adjacent molten zone. The process is repeated several times and the heater is moved in the same direction. At one end, impurities get concentrated. This end is cut off. This method is very useful for producing semiconductor and other metals of very high purity, c.g., germanium, silicon, boron, gallium and indium.

(ii) Chromatographic method is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. The adsorbed components are removed (eluted) by using suitable (eluant). There are several chromatogrpahic techniques such as paper chromatography, column chromatography, gas chromatography etc.

Column chromatography : Column chroma-tography involves separation of a mixture over a column of adsorbent (stationary phase) packed in a glass tube. The column is fitted with a stopcock at its lower end (fig). The mixture adsorbed on adsorbent is placed on the top of the adsorbent column packed in a glass tube.

An appropriate eluant which is a liquid or a mixture of liquids is allowed to flow the column slowly. Depending upon the degree to which the compounds are adsored, complete separation takes place. The most readily adsorbed substances are retained near the top and others come down to various distance in the column.

Question 5.
Out of C and CO, which is a better reducing agent at 673 K?
Solution:
When carbon acts as a reducing agent, it is either converted into CO or CO₂ or both.
2C + O₂ → 2CO
C + O₂ → CO₂
CO is oxidised to CO₂ when it is used as a reducing agent.
2CO + O₂ → 2CO₂
From the Ellingham diagram (refer answer number 3), it is clear that at the temperature 673 K, the AG° of the formation of CO₂ from CO is more negative than the formation of CO or CO₂ from carbon. Hence, at temperature 673 K, CO is a better reducing agent than C.

Question 6.
Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?
Solution:
Many of the metals such as copper, silver, gold, aluminium, lead, etc., are purified by this method. This is perhaps the most important method. The impure metal is made anode while a thin sheet of pure metal acts as a cathode. The electrolytic solution consists of generally an aqueous solution of a salt or a complex of the metal. On passing the current, the pure metal is deposited on the cathode and equivalent amount of the metal gets dissolved from the anode. Thus, the metal is transferred from anode to cathode through solution. The soluble impurities pass into the solution while the insoluble one, especially less electropositive impurities collect below the anode as anodic mud or anode sludge. Impurities from the blister copper deposit as anode mud which contains antimony, selenium, tellurium, silver, gold and platinum.

Question 7.
Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.
Solution:
Near the bottom of the furnace (zone of combustion, 2170 K), coke first combines with air to form CO₂ which then combines with more coke (zone of heat absorption, 1423 K) to form CO. The CO thus produced acts as the reducing agent and reduces iron oxide to spongy iron near the top of the furnace (zone of reduction, 823 K).

At the lower part of the furnace (zone of fusion, 1423-1673 K) the spongy iron melts and dissolves some carbon, S, P, SiO₂, Mn, etc.

The molten slag being less dense floats over the surface of the molten iron. The molten iron is then tapped off from the furnace and is then solidified to give blocks of iron called cast iron or pig iron.

Question 8.
Write chemical reactions taking place in the extraction of zinc from zinc blende.
Solution:
Concentration : When zinc blende is used, the powdered ore is concentrated by froth-floatation process.

Roasting : The concentrated ore is heated in excess of oxygen at about 900°C. Zinc sulphide is oxidised to zinc oxide. If some of the ore is oxidised to zinc sulphate, it also decomposes at 900°C into ZnO.

For roasting, a reverberatory furnace may be used.

Reduction : The principal reaction that takes place during reduction is the conversion of the oxide into the metal with the help of carbon.

Electrolytic refining : Purification of zinc is done by electrolytic refining using pure Zn as cathode and impure Zn as anode. The electrolyte is ZnSO₄.

Reaction at cathode : Zn2⁺_(aq)+ 2e^– → Zn_(s)
Reaction at anode : 2H₂O_(l) → O_2(g)+ 4H⁺_(aq)+ 4e^–
ZnSO₄ electrolyte is added from time to time.

Question 9.
State the role of silica in the metallurgy of copper.
Solution:
Iron present in pyrites has greater affinity for oxygen than copper. The copper oxide formed reacts with unchanged iron sulphide to form iron oxide so, most of the iron sulphide is oxidised to ferrous oxide.
2FeS + 3O₂ → 2FeO + 2sO₂
Ferrous oxide combines with silica which acts as flux and forms ferrous silicate. By this reaction most of the iron is removed as slag.

Question 10.
What is meant by the term “chromatography”?
Solution:
Chromatographic method is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. The adsorbed components are removed (eluted) by using suitable (eluant). There are several chromatogrpahic techniques such as paper chromatography, column chromatography, gas chromatography, etc.

Question 11.
What criterion is followed for the selection of the stationary phase in chromatography?
Solution:
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 element to be purified. Under these conditions, when the column is extracted, the impurities will be retained by the stationary phase whereas the pure component is easily removed.

Question 12.
Describe a method for refining nickel.
Solution:
Nickel is purified by Mond’s process. Impure nickel is treated with carbon monoxide at 60-80°C when volatile compound nickel carbonyl is formed. Nickel carbonyl decomposes at 180°C to form pure nickel and carbon monoxide.

Question 13.
How can you separate alumina from silica in a bauxite ore associated with silica? Give equations, if any.
Solution:
Serpeck’s process is used when silica is present in considerable amounts in bauxite ore. The ore is mixed with coke and heated at 1800°C in presence of nitrogen, where AIN is formed.
Al₂O₃ + 3C + N₂ → 2AlN + 3CO
Silica is reduced to silicon which volatilises off at this temperature.
SiO₂ + 2C → Si + 2CO

Question 14.
Giving examples, differentiate between ‘roasting’ and ‘calcination’.
Solution:
Calcination : It is the process of converting an ore into its oxide by heating it strongly below its melting point either in absence or limited supply of air.

Roasting : In roasting, the ore is heated in a regular supply of air in a furnace at a temperature below the melting point of the metal.
2 ZnS + 3O₂ → 2ZnO + 2SO₂
2 PbS + 3O₂ → 2PbO + 2SO₂

Question 15.
How is’cast iron’different from ‘pig iron’?
Solution:
The iron obtained from Blast furnace contains about 4% carbon and many impurities in smaller amount (e.g., S, P, Si, Mn). This is known as pig iron and cast into variety of shapes.

Cast iron is different from pig iron and 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 “minerals”and “ores’.
Solution:
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 copper matte is put in silica lined converter ?
Solution:
Copper matte consists of Cu2S and FeS. Wheia a blast of hot air is passed through molten matte taken in a silica lined converter, FeS present in matte is oxidised to FeO which combines with silica (SiO2) to form FeSiO3 (slag).

When complete iron has been removed as slag, some of the Cu2S undergoes oxidation to form Cu2O which then reacts with more Cu2S to form copper metal.
2CU₂S + 3O₂ → 2CU₂O + 2SO₂
2CU₂O + Cu₂S → 6Cu + SO₂
Thus, copper matte is heated in silica lined converter to remove FeS present in matte as FeSiO₃ (slag).

Question 18.
What is the role of cryolite in the metallurgy of aluminium?
Solution:
The role of cryolite is as follows :

1. It makes alumina a good conductor of electricity.
2. It lowers the fusion temperature of the bath from 2323 K to about 1140 K.

Question 19.
How is leaching carried out in case of low grade copper ores?
Solution:
The leaching of the low grade copper ores is carried out with acids in the presence of air when copper goes into solution as Cu²⁺ ions. Therefore,

Question 20.
Why is zinc not extracted from zinc oxide through reduction using CO?
Solution:
The standard free energy of formation (∆_fG°) of CO₂ from CO is higher than that of the formation of ZnO from Zn. Hence, CO cannot be used to reduce ZnO to Zn.

Question 21.
The value of ∆_fG° for formation of Cr₂O₃ is – 540 kJ mol^-1 and that of Al₂O₃ is – 827 kJ mol^-1. Is the reduction of Cr₂O₃ possible with Al?
Solution:

Question 22.
Out of C and CO, which is a better reducing agent for ZnO?
Solution:
The free energy of formation (∆_f G°) of CO from C becomes lower at temperatures above 1120 K whereas that of CO₂ from C becomes lower above 1323 K than ∆_fG° of ZnO. However, ∆_fG° 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.
Reduction of ZnO is usually carried out around

Question 23.
The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.
Solution:
Some basic concepts of thermodynamics lielp us in understanding the theory of metallurgical transformations. Gibb’s energy is the most significant term.
The graphical representation of Gibb’s energy was first used by H.J.T. Ellingham. This
provides a sound basis for considering the k choice of reducing agent in the reduction of oxides. This is known as Ellingham diagram. (Tor diagram refer answer number 3) Such diagrams help us in predicting the feasibility of thermal reduction of an ore. The criterion of feasibility is that at a given temperature Gibb’s energy of the reaction must be negative.
Examples :
(i) Thermodynamics helps us to understand how coke reduces the oxide and why blast furnace is chosen. One of the main reduction steps in this process is :
FeO_(s) + C_(s) → Fe_(s/l) + CO_(g)… (1)
It can be seen as a couple of two simpler reactions. In one, the reduction of FeO is taking place and in the other, C is being oxidised to CO :

When both the reactions take place to yield the equation (1), the net Gibb’s energy change becomes :

Naturally, the resultant reaction will take place when the right hand side in equation (3) is negative. In ∆G° vs T plot representing reaction 2, the plot goes upward and that representing the change C → CO (C, CO) goes downward. At temperatures above 1073 K (approx), the C, CO line comes below the Fe, FeO line [∆G_(c,co) < ∆G_(Fe,FeO)] So in this range, coke will be reducing the FeO and will itself be oxidised to CO. In a similar way the reduction of Fe₃O₄ and Fe₂O₃ at relatively lower temperatures by CO can be explained on the basis of lower lying points of intersection of their curves with the CO, CO₂ curve in Ellinghan diagram.

(ii) In the graph of ∆r.G° vs T for formation of oxides, the Cu₂O line is almost at the top. So it is quite easy to reduce oxide ores of copper directly to the metal by heating with coke (both the lines of C, CO and C, CO₂ are at much lower position in the graph particularly after 500 – 600 K). However most of the ores are sulphide and some may also contain iron. The sulphide ores are roasted/smelted to give oxides:
2CU₂S + 3O₂ → 2Cu₂O + 2SO₂
The oxide can then be easily reduced to metallic copper using coke.
Cu₂O + C → 2Cu + CO

Question 24.
Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous solution of NaCI is subjected to electrolysis?
Solution:
Sodium metal is prepared by Down’s process. This process involves the electrolysis of a fused mixture of NaCI and CaCl₂ at 873 K. During electrolysis, sodium is discharged at the cathode and Cl₂ is obtained at the anode as a by-product.

Question 25.
What is the role of graphite rod in the electrometallurgy of aluminium?
Solution:
In this process, a fused mixture of alumina, cryolite and fluorspar (CaF₂) is electrolysed using graphite as anode and steel as cathode. During electrolysis, Al is liberated at the cathode whereas CO and CO₂ are liberated at the anode.

If some other metal is used as the anode other than graphite, then O₂ liberated will not only oxidise the metal of the electrode but would also convert some of the Al liberated at the cathode back to Al₂O₃. Since graphite is much cheaper than any metal, graphite is used as the anode. So, the role of graphite in electrometallurgy of Al is to prevent the liberation of O₂ at the anode which may otherwise oxidise some of the liberated Al back to Al₂O₃.

Question 26.
Outline the principles of refining of metals by the following methods :

1. Zone refining
2. Electrolytic refining
3. Vapour phase refining

Solution:
(i) Refer answer number 4(i).

(ii) In this method, the impure metal is made to act as anode. A strip of the same metal in pure form is used as cathode. They are put in a suitable electrolytic bath containing soluble salt of the same metal. The more basic metal remains in the solution and the less basic ones go to the anode mud.
Anode : M → Mⁿ⁺ ne^–
Cathode : Mⁿ⁺ + ne^– → M
Copper is refined using an electrolytic method. Anodes are of impure copper and pure copper strips are taken as cathode. The electrolyte is acidified solution of copper sulphate and the net result of electrolysis is the transfer of copper in pure form from the anode to the cathode :
Anode : Cu → Cu²⁺ + 2e^–
Cathode : Cu²⁺ + 2e^– → Cu
Impurities from the blister copper deposit as anode mud which contains antimony, selenium, tellurium, silver, gold and platinum; recovery of these elements may meet the cost of refining.
Zinc may also be refined this way.

(iii) In this method, the metal is converted into its volatile compound and collected elsewhere. It is then decomposed to give pure metal. So, the two requirements are :

(a) The metal should form a volatile compound with an available reagent.
(b) The volatile compound should be easily decomposable, so that the recovery is easy.

Following example will illustrate this technique.
Mond process for Refining Nickel : In this process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetracarbonyl :

The carbonyl is subjected to higher temperature so that it is decomposed giving the pure metal :

Question 27.
Predict conditions under which Al might be expected to reduce MgO.
Solution:
The two equations are :