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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 Cass 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.
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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 , magnetite , siderite , and iron pyrites 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 ( ) from bauxite ore.
Impurities are then filtered and the solution is neutralized by carbon dioxide gas
(hydrated alumina)
Answer :
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 to Al and temperature above 1623K Aluminium can reduce to
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.
But to reduce zinc from its solution we need higher reactive metal, which reduction potential is lower than zinc like But these metals readily react with water with the evolution of gas. As a result, these metals are cannot use to displace zinc ion from its solution.
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- is used as a depressant it selectively prevents to coming to the froth but allows PbS with the froth. NaCN react with to form
Reaction-
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 of is less than that of and so that hydrogen and carbon are unable to reduce
. But the Gibbs free energy of formation of is higher than that of CO. So, that carbon can able to reduce
. 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.
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, 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.
Question 6.5 Out of C and CO, which is a better reducing agent at 673 K ?
Answer :
At 673K for change of is less than of . Therefore CO can easily oxidized to 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
limestone also decomposes to calcium oxide which reverses silicate impurity of the ore as slag
At 900-1500K
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 ( ) by froth floatation method).
Conversion to the oxide b y the roasting method -
Extraction of zinc from zinc oxide (by reduction)
Electrolytic refining of impure zinc (electrolyte used is acidified zinc sulphate )
Anode-
Cathode (made of the pure copper strip)-
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 (iron oxide) produced during roasting as slag. If the sulphide ore contains iron, it mixed with silica ( ) as flux before heating. combines with silica and produce iron silicate ( )
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.
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 ( ) to form nickel carbonyl complex . The complex is volatile in nature.
This complex is decomposed at high temperature to get a pure nickel.
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-
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.
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.
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 .
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 and . It is put in a silica-lined converter to remove the remaining ferrous oxide ( ) and ferrous sulphide ( ) present in copper matte as slag ( )
Some silica is also added and hot air blast is blown to convert the remaining to the metallic copper.
Question 6.18 What is the role of cryolite in the metallurgy of aluminium?
Answer :
The main purpose of adding cryolite in metallurgy of aluminium is-
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 is treated with scrap iron or dihydrogen .
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 from is lower than the formation of from CO. So, therefore, CO cannot reduce to zinc ( ).
Question 6.21 The value of
for formation of is and that of . Is the reduction of possible with Al?
Answer :
......................... (-827 KJ/mol)
......................... (-540 KJ/mol)
_________________________________ (On substracting )
...............(-287 KJ/mol)
In the above reaction, we see that of formation of Cr is negative (-287 KJ/mol) Hence the reduction of with aluminium is possible.
Question 6 . 22. Out of C and CO, which is a better reducing agent for ZnO ?
Answer :
The Gibbs free energy formation of from is always higher than the Gibbs free energy of formation of . Thus CO cannot reduce .
On the other hand, Gibbs free energy of formation of from C is less than gibbs free energy of formation of . Hence C can easily reduce to zinc.
Answer :
From the graph we can observe that, a mtal oxide can reduce the oxide of other metals,if the of the first oxide is more than the second oxide.
example- can reduce to Zn but Zn cannot reduce because of from Mg is more negative.
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 and . Sodium is deposited at cathode and chlorine at anode as a by-product.
At Cathode -
At anode-
If an aqueous solution of NaCl is subjected to electrolysis, dihydrogen gas is evolved at the cathode while chlorine 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.
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-
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 of is lower than the from . So, that's why above , 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 |
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.
The step-by-step solutions given in the NCERT class 12 Chemistry solutions chapter 6 will aid you in understanding the chapter easily.
The revision will be a lot much easier because with the help of solutions you will always remember the concepts and get very good marks in your class.
Homework will not be a problem now, all you need to do is check the detailed CBSE Class 12 Chemistry Chapter 6 NCERT solutions and you are good to go.
NCERT Solutions for Class 12 Chemistry Chapter 6 PDF Download option can be used to download the complete page and read offline.
NCERT Solutions for Class 12 Subject wise
One question can be expected for JEE Main from the chapter General Principles and Processes of Isolation of Elements
Around 3 marks of questions will be asked in the CBSE exam from General Principles and Processes of Isolation of Elements. Follow NCERT book for a good score. More problems can be solved using NCERT exemplar for Class 12 Chemistry.
The important topics of this chapter are
Leaching reactions
Blast furnace reactions
Extraction processes
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