NCERT solutions for class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements

NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements - Students appearing in Class 12 board exams must refer to this NCERT Chemistry Class 12 chapter 6 to score well in the examination. To develop a grip on every topic you must solve topic wise questions and exercise questions of NCERT book cCass 12 Chemistry chapter 6. The NCERT solutions for Class 12 Chemistry chapter 6 General Principles and Processes of Isolation of Elements are prepared by subject experts to help students in clarifying their doubts.NCERT solutions for class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements

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Also Read| General Principles and Processes of Isolation of Elements Class 12 Notes

General Principles and Processes of Isolation of Elements Class 12 NCERT Exemplar Solutions

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

Solutions to In Text Questions Ex 6.1 to 6.4

Question 6.1 Which of the ores mentioned in Table 6.1 can be concentrated by magnetic separation method?

Table 6.1: Principle Ores of some Important Metals

Metal Ores Composition
Aluminium Bauxite

Kaolinite(a form of clay)
AlO x (OH) 3-2x
[Where 0<x<1]
[Al 2 (OH) 4 Si 2 O 5 ]
Iron Haemtite
Magnetite
Siderite
Iron pyrites
Fe 2 O 3
Fe 3 O 4
FeCO 3
FeS 2
Copper Copper pyrites
Melachite
Cuprite
Copper glance
CuFeS 2
CuCO 3 .Cu(OH) 2
Cu 2 O
Cu 2 S
Zinc Zinc blande or Sphalerite
Calamine
Zincite
ZnS
ZnCO 3
ZnO

Answer :

Ores which contains some magnetic particles or they can attract by the magnetic field only they can be concentrated by magnetic separation method. Among the ores mentioned in table 6.1, the ores of iron such as haematite (Fe_{2}O_{3}) , magnetite (Fe_{3}O_{4}) , siderite (FeCO_{3}) , and iron pyrites (FeS_{2}) can be separated by the process of magnetic separation

Question 6.2 What is the significance of leaching in the extraction of aluminium?

Answer :

In the extraction of aluminium, the significance of leaching is to concentrate the pure alumina ( Al_{2}O_{3} ) from bauxite ore.

\\Al_{2}O_{3}(s)+2NaOH(aq)+3H_{2}O\rightarrow 2Na[Al(OH)_{4}](aq)\\ SiO_{2}(l)+2NaOH(aq)\rightarrow NaSiO_{3}(aq)+H_{2}O

Impurities are then filtered and the solution is neutralized by carbon dioxide gas

2Na[Al(OH)_{4}](aq)+CO_{2}(g)\rightarrow Al_{2}O_{3}.xH_{2}O(s)+2NaHCO_{3}(aq)

(hydrated alumina)

Al_{2}O_{3}.xH_{2}O(s)\overset{\Delta }{\rightarrow} Al_{2}O_{3}+xH_{2}O

Question 6.3 The reaction, Cr_2 O_3 + 2 AI \rightarrow Al_2 O_3 + 2 Cr (\Delta G^{o} = - 421 kJ) is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?

Answer :

Cr_2 O_3 + 2 AI \rightarrow Al_2 O_3 + 2 Cr
This reaction is thermodynamically feasible, but it does not take place in room temperature because at room temperature all the reactants are solids. At high temperature, the chromiums start melting and the reaction becomes faster.

Question 6.4 Is it true that under certain conditions, Mg can reduce Al2O3 and Al can reduce MgO? What are those conditions ?

Answer :

Yes, temperature below 1623K, Mg can reduce Al_{2}O_{3} to Al and temperature above 1623K Aluminium can reduce MgO to Mg

NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Ssolation of Elements- Exercise Questions

Question 6.1 Copper can be extracted by hydrometallurgy but not zinc. Explain.

Answer :

Copper can be extracted by hydrometallurgy but not zinc because the reduction potential of zinc is lower than that of copper. So, that copper can be reduced from its solution by zinc.

Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)

But to reduce zinc from its solution we need higher reactive metal, which reduction potential is lower than zinc like Al,Mg,Ca,K But these metals readily react with water with the evolution of H_{2} gas. As a result, these metals are cannot use to displace zinc ion from its solution.

2K(s)+2H_{2}O(l)\rightarrow KOH(aq)+H_{2}(g)

Question 6.2 What is the role of depressant in froth floatation process?

Answer :

Role of depressant in froth flotation method is to separate two sulphid ores by adjusting one ore to form froth. For example- NaCN is used as a depressant it selectively prevents ZnS to coming to the froth but allows PbS with the froth. NaCN react with ZnS to form Na_{2}[Zn(CN)_{4}]

Reaction-

4NaCN+ ZnS\rightarrow Na_{2}[Zn(CN)_{4}]+ Na_{2}S

Question 6.3 Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?

Answer :

Gibbs free energy of formation (\Delta _{f}G^{0}) of Cu_{2}S is less than that of H_{2}S and CS_{2} so that hydrogen and carbon are unable to reduce

Cu_{2}S \rightarrow Cu . But the Gibbs free energy of formation (\Delta _{f}G^{0}) of Cu_{2}O is higher than that of CO. So, that carbon can able to reduce

Cu_{2}O \rightarrow Cu . Therefore the extraction of copper from its pyrite ore is more difficult than its oxide ore through reduction.

Question 6.4 Explain: (i) Zone refining

Answer :

ZONE REFINING- This method is based on that the impurities are more soluble in the molten state than n the solid state. In this process, a circular mobile heater surrounding the rod of impure metal is fixed at one end. Move the heater from one end to another end so that the impurities present in the rod also moves to the other end of the rod. Repeat the process several times again and again. Impurities get concentrated at separate ends of the rod. This end is cut off.

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Question 6.4 Explain: (ii)Column chromatography.

Answer :

Column chromatography-
The principle of chromatography is based on, the different component of the mixture is getting adsorbed to a different extent on an adsorbent. In this, there are two phases one is mobile and the other is immobile phase. In column chromatography, Al_{2}O_{3} column is generally used as a stationary phase. The mobile phase is may be a gas or liquid. The mobile phase is forced to move over the stationary phase. The adsorbed component is removed with the help of a suitable solvent.
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Question 6.5 Out of C and CO, which is a better reducing agent at 673 K ?

Answer :

At 673K (\Delta G^{0}) for change of (CO\rightarrow CO_{2}) is less than (\Delta G^{0}) of (C\rightarrow CO) . Therefore CO can easily oxidized to CO_{2} than C to CO.

Hence CO is better reducing agent than C at 673K.

Question 6.6 Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?

Answer :

The following elements present in the anode mud in electrolytic refining of copper are selenium, silver, tellurium, gold, platinum and antimony. They are present because they are very less reactive and are not affected by the purification process. So, therefore they settle down below as anode mud.

Question 6.7 Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.

Answer :

At 500-800K

\\3Fe_{2}O_{3}+CO\rightarrow 2Fe_{3}O_{4}+CO_{2}\\ Fe_{3}O_{4}+CO\rightarrow 3Fe+4CO_{2}\\ Fe_{2}O_{3}+CO\rightarrow 2FeO+CO_{2}

limestone also decomposes to calcium oxide which reverses silicate impurity of the ore as slag CaCO_{3}\rightarrow CaO+CO_{2}

At 900-1500K

\\C+CO_{2}\rightarrow 2CO\\ FeO+CO\rightarrow Fe+CO_{2}

CaO+SiO_{2}\rightarrow CaSiO_{3}(slag)

Question 6.8 Write chemical reactions taking place in the extraction of zinc from zinc blende.

Answer :

Chemical reactions in the extraction of zinc from zinc blende -

After concentration of ore (removal of gauge from zinc blend ( ZnS ) by froth floatation method).

Conversion to the oxide b y the roasting method -

2ZnS+3O_{2}\rightarrow ZnO+2SO_{2}

Extraction of zinc from zinc oxide (by reduction)

ZnO+C\overset{673K}{\rightarrow}Zn+CO

Electrolytic refining of impure zinc (electrolyte used is acidified zinc sulphate ZnSO_{4} )

Anode- Zn\rightarrow Zn^{2+}+2e^{-}

Cathode (made of the pure copper strip)- Zn^{2+}+2e^{-}\rightarrow Zn

Question 6.9 State the role of silica in the metallurgy of copper.

Answer :

The role of silica in metallurgy of copper is to remove the FeO (iron oxide) produced during roasting as slag. If the sulphide ore contains iron, it mixed with silica ( SiO_{2} ) as flux before heating. FeO combines with silica and produce iron silicate ( FeSiO_{3} )

FeO+SiO_{2}\rightarrow FeSiO_{3}(slag)

Question 6.10 Which method of refining may be more suitable if element is obtained in minute quantity?

Answer :

Chromatographic method of refining is particularly suitable if the element is obtained in minute quantity.

Question 6.11 Which method of refining will you suggest for an element in which impurities present have chemical properties close to the properties of that elements?

Answer :

Zone refining is the method for an element in which impurities present have chemical properties close to the properties of those elements.

Question 6.12 Describe a method for refining nickel.

Answer :

Nickel is refined by Mond's Process, in which Nickel is heated in steam in presence of Carbon monoxide ( CO ) to form nickel carbonyl complex . (Ni(CO)_{4}) The complex is volatile in nature.

Ni+4CO\overset{330K-350K}{\rightarrow} Ni(CO)_{4})

This complex is decomposed at high temperature to get a pure nickel.

Ni(CO)_{4}\overset{450K-470K}{\rightarrow} Ni+4CO

Question 6.13 How can you separate alumina from silica in a bauxite ore associated with silica? Give equations, if any.

Answer :

The separation of alumina carried out in the following steps-

  1. The powdered ore is treated with the concentrated NaOH solution at 473K-523K and 35-36 bar pressure.
    \\Al_{2}O_{3}+2NaOH(aq)+3H_{2}O\rightarrow 2Na[Al(OH)_{4}](aq)\\ SiO_{2}+NaOH\rightarrow Na_{2}SiO_{3}+H_{2}O
  2. The CO_{2} gas is passed through the resulting solution to neutralise the aluminate, which results in ppt. of hydrated alumina.
    2Na[Al(OH)_{2}](aq)+CO_{2}\rightarrow Al_{2}O_{3}+2NaHCO_{3}(aq)
    The obtained hydrated alumina is filtered, dried and heated to get pure alumina.
    Al_{2}O_{3}.xH_{2}O\rightarrow Al_{2}O_{3}+x.H_{2}O

Question 6.14 Giving examples, differentiate between ‘roasting’ and ‘calcination’.

Answer :

Roasting- The process of converting sulphide ore to oxide. In this process, the ore is heated in the furnace with a regular supply of air, below its melting temperature: example- sulphide ore of zinc, lead, and copper.

\\2ZnS+3O_{2}\rightarrow 2ZnO+2SO_{2}\\ 2PbS+3O_{2}\rightarrow 2PbO+2SO_{2}\\ 2Cu_{2}S+3O_{2}\rightarrow 2Cu_{2}O+2SO_{2}

Calcination- Process of converting hydroxide and carbonate ore to oxide in a limited supply of air. The temperature should be below the metals melting point in this process volatile matter which escapes, leaving behind the metal oxide.

\\Fe_{2}O_{3}.xH_{2}O \overset{\Delta }{\rightarrow}Fe_{2}O_{3}(s)+xH_{2}O(g)\\ ZnCO_{3} \overset{\Delta }{\rightarrow} ZnO(s)+CO_{2}(g)\\ CaCO_{3}.MgCO_{3} \overset{\Delta }{\rightarrow} CaO(s)+MgO(s)+2CO_{2}(g)

Question 6.15 How is ‘cast iron’ different from ‘pig iron”?

Answer :

The iron that obtains from the blast furnace is pig iron . it contains 4% carbon and many impurities like (s,P,Si,Mn) .

Cast iron is obtained from pig iron by melting pig iron with scrap iron and coke using hot air blast. It contains slightly less carbon content around 3% and it is hard an brittle.

Question 6.16 Differentiate between “minerals” and “ores”.

Answer :

Minerals are the naturally occurring substance which contains some amount of metals. They are found in the earth crust and obtained by mining.

Ores are the specific minerals in which metals are economically extracted. It has a definite composition. Each and every ore is a mineral but not every mineral is ore.

Question 6.17 Why copper matte is put in silica lined converter?

Answer :

The copper matte composed of Cu_{2}S and FeS . It is put in a silica-lined converter to remove the remaining ferrous oxide ( FeO ) and ferrous sulphide ( FeS ) present in copper matte as slag ( FeSiO_{3} )

Some silica is also added and hot air blast is blown to convert the remaining (FeS,FeO \&\ Cu_{2}S/Cu_{2}O) to the metallic copper.

\\2FeS+3O_{2}\rightarrow 2FeO+2SO_{2}\\ FeO+SiO_{2}\rightarrow FeSiO_{3}\\ 2Cu_{2}S+3O_{2}\rightarrow 2Cu_{2}O+2SO_{2}\\ 2CU_{2}O+Cu_{2}S\rightarrow 6Cu+SO_{2}

Question 6.18 What is the role of cryolite in the metallurgy of aluminium?

Answer :

The main purpose of adding cryolite (Na_{3}AlF_{6}) in metallurgy of aluminium is-

  • To increase the electrical conductivity of the alumina Al_{2}O_{3}
  • To decrease the melting point of the mixture

Question 6.19 How is leaching carried out in case of low grade copper ores?

Answer :

In case of low-grade ores, copper is extracted by hydrometallurgy. Leaching is done with the help of acid or bacteria in the presence of air. The solution containing Cu^{2+} is treated with scrap iron or dihydrogen (H_{2}) .

Cu^{2+}(aq)+H_{2}(g)\rightarrow Cu(s)+2H^{+}(aq)

Question 6.20 Why is zinc not extracted from zinc oxide through reduction using CO?

Answer :

According to the Ellingham diagram, the standard Gibbs free energy of formation of ZnO from Zn is lower than the formation of CO_{2} from CO. So, therefore, CO cannot reduce ZnO to zinc ( Zn ).

Question 6.21 The value of
\Delta _ f G ^ \theta for formation of Cr _ 2 O _3 is - 540 KJ mol ^{-1} and that of Al_2 O_3 \: \: is \: \: -827 KJ mol^{-1} . Is the reduction of Cr _ 2 O _3 possible with Al?

Answer :

2Al+3/2O_{2}\rightarrow Al_{2}O_{3} ......................... (-827 KJ/mol)
2Cr+3/2O_{2}\rightarrow Cr_{2}O_{3} ......................... (-540 KJ/mol)
_________________________________ (On substracting )
2Al+Cr_{2}O_{3}\rightarrow Al_{2}O_{3}+2Cr ...............(-287 KJ/mol)

In the above reaction, we see that \Delta_f G^0 of formation of Cr is negative (-287 KJ/mol) Hence the reduction of Cr _ 2 O _3 with aluminium is possible.

Question 6 . 22. Out of C and CO, which is a better reducing agent for ZnO ?

Answer :

15948162614391594816259136
The Gibbs free energy formation of CO_{2} from CO is always higher than the Gibbs free energy of formation of ZnO . Thus CO cannot reduce ZnO .
On the other hand, Gibbs free energy of formation of CO_{2} from C is less than gibbs free energy of formation of ZnO . Hence C can easily reduce ZnO to zinc.

Question 6.23 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.

Answer :

gpisc
From the graph we can observe that, a mtal oxide can reduce the oxide of other metals,if the \Delta_f G of the first oxide is more than the second oxide.
example- Mg can reduce ZnO to Zn but Zn cannot reduce MgO because \Delta_f G of MgO from Mg is more negative.

Question 6.24 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 Down's process, we prepare sodium metals in which chlorine is obtained as a by-product. It involves electrolysis of a fused mixture of NaCl and CaCl_{2} . Sodium is deposited at cathode and chlorine at anode as a by-product.

NaCl(l)\rightarrow Na^{+}(melt)+Cl^{-}(melt)

At Cathode - Na^{+}+e^{-}\rightarrow Na

At anode- Cl^{-}\rightarrow \frac{1}{2}Cl_{2}+e^{-}

If an aqueous solution of NaCl is subjected to electrolysis, dihydrogen gas (H_{2}) is evolved at the cathode while chlorine (Cl_{2}) is obtained at the anode.

Question 6.26 Outline the principles of refining of metals by the following methods : (i) Zone refining

Answer :

ZONE REFINING- This method is based on that the impurities are more soluble in the molten state than n the solid state. In this process, a circular mobile heater surrounding the rod of impure metal is fixed at one end. Move the heater from one end to another end so that the impurities present in the rod also moves to the other end of the rod. Repeat the process several times again and again. Impurities get concentrated at separate ends of the rod. This end is cut off.

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Question 6.26 Outline the principles of refining of metals by the following methods : (ii)Electrolytic refining

Answer :

Electrolytic refining- It is a process of refining impure metals by electricity . In this method, impure metal is act as anode and the same pure metal strip is act as cathode. Same metal salt which is soluble is taken as the electrolyte. When electricity is passed the metal ions deposited at the cathode (pure metal) and the impure metal is starting dissolving in the solution in the form of ions. Impurities present in metal (impure metal) gets collected at bottom of the anode (as anode mud )

Question 6.26 Outline the principles of refining of metals by the following methods :(iii)Vapour phase refining

Answer :

Vapour phase refining- In this process metal is converted into its volatile compound which is collected and decomposed to give pure metals.

Two major requirement-

  • The metal should form a volatile compound with a suitable reagent.
  • the volatile compound should be easily decomposable so that to recover easily

Question 6.27 Predict conditions under which Al might be expected to reduce MgO.

Answer :

As per the Ellingham diagram, above 1350, the standard Gibbs free energy of formation (\Delta G^{0}_{f}) of Al_{2}O_{3} is lower than the MgO from Mg . So, that's why above 1350^{0}C , Al can reduce MgO.

Topics of NCERT Syllabus Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements-

6.1 Occurrence of Metals

6.2 Concentration of Ores

6.3 Extraction of Crude Metal from Concentrated Ore

6.4 Thermodynamic Principles of Metallurgy

6.5 Electrochemical Principles of Metallurgy

6.6 Oxidation Reduction1636.7Refining

6.8 Uses of Aluminium, Copper, Zinc, and Iron

NCERT Solutions Class 12 Chemistry

Chapter 1

The Solid State

Chapter 2

Solutions

Chapter 3

Electrochemistry

Chapter 4

Chemical Kinetics

Chapter 5

Surface chemistry

Chapter 6

General Principles and Processes of isolation of elements

Chapter 7

The P-block elements

Chapter 8

The d and f block elements

Chapter 9

Coordination compounds

Chapter 10

Haloalkanes and Haloarenes

Chapter 11

Alcohols, Phenols, and Ethers

Chapter 12

Aldehydes, Ketones and Carboxylic Acids

Chapter 13

Amines

Chapter 14

Biomolecules

Chapter 15

Polymers

Chapter 16

Chemistry in Everyday life

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

  • In NCERT Class 12 Chemistry chapter 6 General Principles and Processes of Isolation of Elements, there are a total of eight sub-topics that cover important concepts of the chapter.

  • After completing this NCERT Class 12th Chemistry solutions, you will be able to explain the terms like ores, minerals, concentration, benefaction, calcination, roasting, refining, etc.

  • You will be able to understand the principles of oxidation and reduction as applied to the process of extraction, to apply the thermodynamic concepts like that of entropy and Gibbs energy of the extraction of Cu, Al, Fe, and Zn etc.

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