NCERT Solutions for Class 12 Chemistry Chapter 6
NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements
NCERT Class 12 Chemistry Chapter 6 details the Principles and Processes of Isolation of Elements. Students will learn various terminologies such as concentration, calcination, roasting, refining, etc., in this chapter. In addition, they will understand the methodologies of oxidation and reduction as used within the extraction procedures. With chapter6 class 12 Chemistry, students will easily be able to apply the thermodynamic concepts to the extraction of Aluminium, Copper, Zinc, etc. Moreover, they will understand why CO is a reducing agent at some specific temperatures. All in all, chapter 6 class 12 Chemistry, will be helpful to understand the large number of chemical processes associated with metallurgy.
Keeping in mind the growing curiosity of students, Extramarks has curated NCERT solutions class 12 Chemistry chapter 6 to aid students to understand the various concepts detailed in the chapter. Students may refer to the NCERT class 12 Chemistry chapter 6 study material for free on Extramarks.
Key Topics Covered In NCERT Solutions Class 12 Chemistry Chapter 6
The key topics covered in NCERT solutions class 12 Chemistry chapter 6 – Principles and Processes of Isolation of Elements are in the table below.
|6.2||Extraction of metal|
|6.3||Gibbs free energy|
|6.4||Gibbs energy in Chemistry|
|6.7||Application of Chromatography|
Students can click on the above topics under NCERT solutions class 12 Chemistry chapter 6 to refer to the study material, revision notes, and essential questions that can help them during their examination.
The key topics covered in NCERT solutions class 12 Chemistry chapter 6 in brief:
The first exercise under class 12 Chemistry chapter 6 NCERT solutions is an introduction to metals. Metals are essential for various purposes. Thus extracting them from the mineral resource in a commercially feasible way is necessary. The minerals in which the metal is present and extracted are called ores. Usually, the ores contain impurities that can be removed to a specific limit in the concentration steps. After this step, the ore is chemically treated to obtain metals. With the help of reducing agents such as CO, carbon, etc., the metal compound is decreased to metal. Then,
- The metal oxide reacts with a reducing agent;
- A. The oxide reduces to metal
- Oxidation of reducing agent
The Isolation of elements as defined in the NCERT solutions class 12 Chemistry chapter 6 aims to teach the students different extraction processes of metals from ores. Some metals, such as noble metals, Gold, Silver, Platinum etc., are present in their original metallic state. Metallurgy is the branch of science that works with extracting metals from ores that are naturally available in the environment.
All of the elements, especially metals, are presently combined with other elements, and these are called minerals. An element may combine with various other elements to make various minerals, but out of them, only a few are feasible sources of that metal. Such sources are known as Ores.
6.2 Extraction of metals
Under this exercise of NCERT solutions class 12 Chemistry chapter 6, students learn the technique of extracting metal ores buried deep underground, known as Mining. The metal ores are available in the earth’s crust in varied abundance. The extraction of metals from ores allows us to use the minerals in the ground! The ores are very dissimilar from the finished metals in buildings and bridges. Simply Ores consist of the desired metal compound, Gangue’s impurities, and earthly substances.
Extraction of metals and their Isolation occurs over some significant steps:
- Concentration of Ore
- Isolation of metal from concentrated ore
- Purification of the metal
- Bauxite AlOx(OH)3-2x [where 0 < x < 1]
- Kaolinite (a form of clay) [Al2 (OH)4 Si2O5]
- Haematite Fe2O3
- Magnetite Fe3O4
- Siderite FeCO3
- Iron pyrites FeS2
- Copper pyrites CuFeS2
- Malachite CuCO3.Cu(OH)2
- Cuprite Cu2O
- Copper glance Cu2S
- Zinc blende/Sphalerite ZnS
- Calamine ZnCO3
- Zincite ZnO
6.3 Gibbs free energy:
As per the definition compiled in NCERT solutions class 12 Chemistry chapter 6, Gibbs free energy, also called the Gibbs function, Gibbs energy, or free enthalpy, is a quantity used to estimate the maximum amount of work done in a thermodynamic system when the temperature and pressure are kept constant. Gibbs’s free energy is represented by the symbol ‘G’, and the value is expressed in Joules or Kilojoules. It can be defined as the maximum amount of work extracted from a closed system.
Gibbs’s free energy is equal to the system’s enthalpy minus the product of the temperature and entropy. The equation is shown as follows;
G = H – TS
G = Gibbs free energy
H = enthalpy
T = temperature
S = entropy
G = U + PV – TS
- U = internal energy(unit is joule)
- P = pressure ( pascal)
- V = volume ( m3)
- T = temperature (kelvin)
- S = entropy (kelvin)
Variations of the Equation: The Gibbs free energy is a state function; therefore, it is not dependent on the path. As defined in NCERT solutions class 12 Chemistry chapter 6, a change in Gibbs free energy gives a change in enthalpy minus the product of temperature and entropy change of the system.
ΔG = ΔH – Δ(TS)
If the reaction happens under constant temperature ΔT=O
ΔG = ΔH – TΔS
This equation shows as the Gibbs Helmholtz equation.
ΔG > 0; the reaction is nonspontaneous as well as endergonic
ΔG < 0; the reaction is spontaneous as well as exergonic
ΔG = 0; reaction is at equilibrium
The Spontaneity of a process Gibbs equation suggests we conclude the Spontaneity of reaction due to enthalpy and entropy values directly. The system enthalpy is negative if the reaction is exothermic, making Gibbs’s free energy negative. So, we can say that all exothermic reactions are spontaneous.
Spontaneity can only be suggested if a reaction can occur, not necessarily if a reaction will occur. For E.g. the conversion of diamond to graphite is a spontaneous process at STP, but it is a slow step. It will take so many years for the transformation to occur.
6.4 Gibbs Energy in Chemistry
Under NCERT solutions class 12 Chemistry Chapter 6 Exercise 6.4, students learn about Gibbs energy. In the energetics of processes for systems at constant temperature and pressure, the appropriate quantity is the Gibbs free energy. The Gibbs free energy has a beneficial property, decreasing for a spontaneous process at constant temperature and pressure. Under these conditions, Gibbs’s free energy decrease equals the maximum amount of energy available for work. Whether it increases for some transition, the change in Gibbs free energy constitutes the minimum amount of work required.
The transformation of a system from one stage to another, at constant temperature and pressure, is spontaneous if Gibbs’s free energy decreases. If the transformation’s Gibbs free energy is unchanged, the two states are in equilibrium. Gibbs energy is sometimes called the thermodynamic potential at constant pressure to specify its analogy with the potential energy of the mechanical system, which also has a minimum value under equilibrium conditions.
Refining plays a crucial role in metallurgy, which has been explained in NCERT solutions class 12 Chemistry chapter 6. Any metal extracted from its ore is usually impure. This impure metal that is extracted is known as crude metal. Refining is a method of removing impurities to obtain metals of high purity. The impurities are removed from hard metal by various techniques based on the properties of the metal and the properties of impurities.
Some processes involved in the purification of crude metal are shown below:
- Zone refining
- Vapour phase refining
- Chromatographic methods
Detailed information given by refining the technique is explained in the chapter.
Distillation: This technique is widely used to purify metals with low boiling points, like mercury and zinc. In this process, the impure metal is heated above its boiling point so that it can produce vapour. The impurities are not vapourised, and hence they are isolated. The mists of the pure metal are then condensed, leaving the impurities behind.
Liquation: In this process, the melting point of the metals is taken into consideration. Metals having low melting points are purified using this method. The melting point of the impurities is maximized in comparison to the metal. The metals are converted into liquid states by supplying heat slightly above their melting point. So pure metal melts and flows down from the furnace, leaving the impurities behind.
Vapour phase refining: Under this topic of NCERT solutions class 12 Chemistry chapter 6, students learn that in this type of purification, the metal should form a volatile compound in the presence of a reagent, and it should decompose quickly to recover the metal. The metal is transformed into its volatile compound. This volatile compound then undergoes decomposition to give pure metal. For example, nickel is purified in this way.
In this exercise of NCERT solutions class 12 Chemistry chapter 6, students learn about Chromatography as a technique for separating, purifying, and testing metal and organic compounds. The term “chromatography” is derived from Greek, chroma meaning “colour,” and graphein meaning “to write.”
In this method, we apply the mixture to be isolated on a stationary phase (solid or liquid). A pure solvent like water or any gas can pass slowly over the stationary phase, moving the components separately as per their solubility in the pure solvent.
Chromatography principle: This is a separation method where the analyte is mixed within a liquid or gaseous mobile phase, pumped by a stationary phase. Normally, one phase is hydrophilic, and another phase is lipophilic. Components of the analyte communicate differently with these two phases. Depending on their polarity, they spend more or less time interacting with the stationary phase and are therefore retarded to a greater or lesser extent. The separation of the various components present in the sample. Each sample component elutes from the stationary phase at a specific retention time. As components pass through the detector, their signal is recorded, and the graph is plotted in a chromatogram.
In addition to the Chromatography principle, students also learn the types of Chromatography under NCERT solutions class 12 Chemistry chapter 6. There are four types of chromatography.
- Adsorption Chromatography
- Thin layer Chromatography
- Column Chromatography
- Partition Chromatography
Students may learn more about it in NCERT solutions class 12 Chemistry chapter 6, provided by Extramarks. Extramarks provides detailed study material and revision notes. Students may click here to access notes on Exercise 6.6.
6.7 Applications of Chromatography
In bioanalytical Chemistry, chromatography is widely used to separate, isolate, and purify proteins from complex sample matrices. In cells, for example, proteins occur besides numerous other compounds such as lipids and nucleic acid. These proteins must be isolated from all the different cell components to be analyzed.
Students may refer to various study materials based on NCERT solutions class 12 Chemistry chapter 6 to ace their exams.
NCERT solutions class 12 Chemistry chapter 6 Exercise & Answer Solutions
The NCERT solutions class 12 Chemistry chapter 6 exercise and answer solutions are based on NCERT books. Every exercise is compiled to add more value to the chapter. Students may refer to various study materials such as revision notes, past year questions papers, important questions, and more pertaining to NCERT solutions on Extramarks.
NCERT solutions class 12 Chemistry chapter 6 is explained in detail by the experts of Extramarks. In addition to chapter 6, students can access NCERT Solution for all other Chemistry chapters of class 12. Furthermore, students can click on the links provided below to access the study material of other classes and subjects.
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Q.1 Copper can be extracted by hydrometallurgy but not zinc. Explain.
The reduction potentials of zinc and iron are lower than that of copper. In hydrometallurgy, zinc and iron can be used to displace copper from their solution.
Fe(s) + Cu2+ (aq) → Fe2+(aq) + Cu(s)
But to displace zinc, more reactive metals i.e., metals having lower reduction potentials than zinc such as Mg, Ca, K, etc. are required. But all these metals react with water with the evolution of H2 gas.
2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)
As a result, these metals cannot be used in hydrometallurgy to extract zinc.
Hence, copper can be extracted by hydrometallurgy but not zinc.
Q.2 What is the role of depressant in froth floatation process?
In the froth floatation process, the role of the depressants is to separate two sulphide ores by selectively preventing one ore from forming froth. For example, to separate two sulphide ores (ZnS and Pbs), NaCN is used as depressant which selectively allows PbS to come with froth, but prevents ZnS from coming to froth. This happens because NaCN reacts with ZnS to form Na2[Zn(CN)4].
4NaCN + Zns → Na2 [Zn(CN)4] + Na2S
Q.3 Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?
The Gibbs free energy of formation (ΔfG) of CS2 and H2S is higher than that of Cu2S. Therefore, C and H2 cannot reduce Cu2S to Cu. On the other hand, the Gibbs free energy of formation (ΔfG) of Cu2O is higher than that of CO and thus C can reduce Cu2O to Cu.
Q.4 Explain: (i) Zone refining (ii) Column chromatography.
(i) Zone refining:
This method is based on the principle that impurities are more soluble in the molten state of metal (the melt) than in the solid state. In the process of zone refining, a circular mobile heater is fixed at one end of a rod of impure metal. As the heater moves, the molten zone of the rod also moves with it. As a result, pure metal crystallizes out of the melt and the impurities pass onto the adjacent molten zone. This process is repeated several times, which leads to the segregation of impurities at one end of the rod. Then, the end with the impurities is cut off. Silicon, boron, gallium, indium etc. can be purified by this process.
(ii) Column chromatography:
Column chromatography is a technique used to separate different components of a mixture. It is a very useful technique used for the purification of elements available in minute quantities. It is also used to remove the impurities that are not very different in chemical properties from the element to be purified. Chromatography is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. In chromatography, there are two phases: mobile phase and stationary phase. The stationary phase is immobile and immiscible. Al2O3 column is usually used as the stationary phase in column chromatography. The mobile phase may be a gas, liquid, or supercritical fluid in which the sample extract is dissolved. Then, the mobile phase is forced to move through the stationary phase. The component that is more strongly adsorbed on the column takes a longer time to travel through it than the component that is weakly adsorbed. The adsorbed components are then removed (eluted) using a suitable solvent (eluant).
Q.5 Out of C and CO, which is a better reducing agent at 673 K?
At 673 K, the value of
is less than that of ∆G(C,CO) . Therefore, CO can be reduced more easily to CO2 than C to CO. Hence, CO is a better reducing agent than C at 673 K.
Q.6 Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?
In electrolytic refining of copper, the common elements present in anode mud are selenium, tellurium, silver, gold, platinum, and antimony.
These elements are very less reactive and are not affected during the purification process. Hence, they settle down below the anode as anode mud.
Q.7 Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.
During the extraction of iron, the reduction of iron oxides takes place in the blast furnace. In this process, hot air is blown from the bottom of the furnace and coke is burnt to raise the temperature up to 2200 K in the lower portion itself. The temperature is lower in the upper part. Thus, it is the lower part where the reduction of iron oxides (Fe2O3 and Fe3O4) takes place. The reactions taking place in the lower temperature range (500 – 800 K) in the blast furnace are:
3Fe2O3 + CO → 2Fe3O4 + CO2
Fe3O4 + 4CO → 3FeO + 4CO2
Fe2O3 + CO → 2Fe3 + CO2
The reactions taking place in the higher temperature range (900 – 1500 K) in the blast furnace are:
C + CO2 → 2CO
FeO + CO → Fe + CO2
The silicate impurity of the one is removed as slag by calcium oxide (CaO), which is formed by the decomposition of limestone (CaCO3).
CaCO3 → CaO + CO2
CaO + SiO2 → CaSiO3
Q.8 Which method of refining may be more suitable if element is obtained in minute quantity?
Q.9 Write chemical reactions taking place in the extraction of zinc from zinc blende.
The different steps involved in the extraction of zinc from zinc blende (ZnS) are given below:
(i) Concentration of ore
First, the gangue from zinc blende is removed by the froth floatation method.
(ii) Conversion to oxide (Roasting)
Sulphide one is converted into oxide by the process of roasting. In this process, ZnS is heated in a regular supply of air in a furnace at a temperature, which is below the melting point of Zn.
2ZnS + 3O2 → 2ZnO + 2SO2
(iii) Extraction of zinc from zinc oxide (Reduction)
Zinc is extracted from zinc oxide by the process of reduction. The reduction of zinc oxide is carried out by mixing it with powdered coke and then, heating it at 673 K.
(iv) Electrolytic Refining
Zinc can be refined by the process of electrolytic refining. In this process, impure zinc is made the anode while a pure copper strip is made the cathode. The electrolyte used is an acidified solution of zinc sulphate (ZnSO4). Electrolysis results in the transfer of zinc In pure from the anode to the cathode.
Anode: Zn → Zn2+ + 2e–
Cathode: Zn2+ + 2e– → Zn
Q.10 State the role of silica in the metallurgy of copper.
During the roasting of pyrite one, a mixture of FeO and Cu2O is obtained.
Q.11 Which method of refining will you suggest for an element in which impurities present have chemical properties close to the properties of that element?
Q.12 Describe a method for refining nickel.
Nickel is refined by Mond’s process. In this process, nickel is heated in the presence of carbon monoxide to form nickel tetracarbonyl, which is a volatile complex.
Q.13 How can you separate alumina from silica in bauxite are associated with silica? Give equations, if any?
To separate alumina from silica in bauxite are associated with silica, first the powdered one is digested with a concentrated NaOH solution at 473 – 523 K and 35 – 36 bar pressure. This results in the leaching out of alumina (Al2O3) as sodium aluminate and silica (SiO2) as sodium silicate leaving the impurities behind.
Q.14 Giving examples, differentiate between ‘roasting’ and ‘calcination’.
Roasting is the process of converting sulphide ores to oxides by heating the ores in a regular supply of air at a temperature below the melting point of the metal. For example, sulphide ores of Zn, Pb, and Cu are converted to their respective oxides by this process.
Q.15 How is ‘cast iron’ different from ‘pig iron”?
The iron obtained from blast furnaces is known as pig iron. It contains around 4% carbon and many impurities such as S, P, Si, Mn in smaller amounts.
Cast iron is obtained by melting pig iron and coke using a hot air blast. It contains a lower amount of carbon (3%) than pig iron. Unlike pig iron, cast iron is extremely hard and brittle.
Q.16 Differentiate between “minerals” and “ores”.
Minerals are naturally occurring chemical substances containing metals. They are found in the Earth’s crust and are obtained by mining.
Ores are rocks and minerals viable to be used as a source of metal.
For example, there are many minerals containing zinc, but zinc cannot be extracted profitably (conveniently and economically) from all these minerals. Zinc can be obtained from zinc blende (ZnS), calamine (ZnCO3), Zincite (ZnO) etc. Thus, these minerals are called ores of zinc.
Q.17 Why copper matte is put in silica lined converter?
Copper matte contains Cu2S and FeS. Copper matte is put in a silica-lined converter to remove the remaining FeO and FeS present in the matte as slag (FeSiO3). Also, some silica is added to the silica-lined converter. Then, a hot air blast is blown. As a result, the remaining FeS and FeO are converted to iron silicate (FeSiO3) and Cu2S is converted into metallic copper.
Q.18 What is the role of cryolite in the metallurgy of aluminium?
Cryolite (Na3AlF6) has two roles in the metallurgy of aluminium:
1. To decrease the melting point of the mixture from 2323 K to 1140 K.
2. To increase the electrical conductivity of Al2O3.
Q.19 How is leaching carried out in case of low grade copper ores?
In case of low grade copper ores, leaching is carried out using acid or bacteria in the presence of air. In this process, copper goes into the solution as Cu2+ ions.
Q.20 Why is zinc not extracted from zinc oxide through reduction using CO?
The standard Gibbs free energy of formation of zinc oxide from zinc is lower than that of CO2 from CO. Therefore, CO cannot reduce zinc oxide to zinc. Hence, zinc is not extracted from zinc oxide through reduction using CO.
Q.21 The value of ∆fGΦ for formation of Cr2O3 is − 540 kJmol-1 and that of Al2 O3 is – 827 kJmol-1. Is the reduction of Cr2O3 possible with Al?
The value of ∆fGΦ for the formation of Cr2O3 from Cr (−540 kJmol-1) is higher than that of Al2O3 from Al (−827 kJmol-1). Therefore, Al can reduce Cr2O3 to Cr. Hence, the reduction of Cr2O3 with Al is possible.
As ∆fGΦ for the reduction reaction of Cr2O3 by Al is negative, this reaction is possible.
Q.22 Out of C and CO, which is a better reducing agent for ZnO?
Reduction of ZnO to Zn is usually carried out at 1673 K. From the above figure, it can be observed that above 1073 K, the Gibbs free energy of formation of CO from C and above 1273 K, the Gibbs free energy of formation of CO2 from C is lesser than the Gibbs free energy of formation of ZnO. Therefore, C can easily reduce ZnO to Zn. On the other hand, the Gibbs free energy of formation of CO2 from CO is always higher than the Gibbs free energy of formation of ZnO. Therefore, CO cannot reduce ZnO. Hence, C is a better reducing agent than CO for reducing ZnO.
Q.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.
The above figure is a plot of Gibbs energy (∆GΦ) vs. T for formation of some oxides. It can be observed from the above graph that a metal can reduce the oxide of other metals, if the standard free energy of formation (∆fGΦ) of the oxide of the former is more negative than the latter. For example, since ∆fGΦ(Al, Al2,O3) is more negative than, ∆fGΦ(Cu, Cu2,O), Al can reduce Cu2O to Cu, but Cu cannot reduce Al2O3. Similarly, Mg can reduce ZnO to Zn, but Zn cannot reduce MgO because ∆fGΦ(Mg, MgO) is more negative than ∆fGΦ(Zn, ZnO)
Q.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?
In the electrolysis of molten NaCl, Cl2 is obtained at the anode as a by product.
Q.25 What is the role of graphite rod in the electrometallurgy of aluminium?
In the electrometallurgy of aluminium, a fused mixture of purified alumina (Al2O3), cryolite (Na3AlF6) and fluorspar (CaF2) is electrolysed. In this electrolysis, graphite is used as the anode and graphite-lined iron is used as the cathode. During the electrolysis, Al is liberated at the cathode, while CO and CO2 are liberated at the anode, according to the following equation.
Cathode: Al3+(melt) + 3e–→ Al(l)
Anode: C(s) + O2- (melt) → CO(g) + 2e–
C(s) + 2O2- (melt) → CO2(g) + 4e–
Q.26 Outline the principles of refining of metals by the following methods:
(i) Zone refining
(ii) Electrolytic refining
(iii) Vapour phase refining
(i) Zone refining:
This method is based on the principle that impurities are more soluble in the molten state of metal (the melt) than in the solid state. In the process of zone refining, a circular mobile heater is fixed at one end of a rod of impure metal. As the heater moves, the molten zone of the rod also moves along with it. As a result, pure metal crystallizes out of the melt and the impurities pass to the adjacent molten zone. This process is repeated several times, which leads to the segregation of impurities at one end of the rod. Then, the end with the impurities is cut off. Silicon, boron, gallium, indium etc. can be purified by this process.
(ii) Electrolytic refining:
Electrolytic refining is the process of refining impure metals by using electricity. In this process, impure metal is made the anode and a strip of pure metal is made the cathode. A solution of a soluble salt of the same metal is taken as the electrolyte. When an electric current is passed, metal ions from the electrolyte are deposited at the cathode as pure metal and the impure metal from the anode dissolves into the electrolyte in the form of ions. The impurities present in the impure metal gets collected below the anode. This is known as anode mud.
(iii) Vapour phase refining :
Vapour phase refining is the process of refining metal by converting it into its volatile compound and then, decomposing it to obtain a pure metal. To carry out this process,
(i) the metal should form a volatile compound with an available reagent, and
(ii) the volatile compound should be easily decomposable so that the metal can be easily recovered.
Nickel, zirconium, and titanium are refined using this method.
Q.27 Predict conditions under which Al might be expected to reduce MgO.
Above 1350°C, the standard Gibbs free energy of formation of Al2O3 from Al is less than that of MgO from Mg. Therefore, above 1350°C, Al can reduce MgO.
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The Chemistry class 12 chapter 6 NCERT solutions help students understand the basic concepts and give a detailed explanation of various laws and formulae under the chapter. It allows students to study for the exams without depending on someone to explain. Students may refer to NCERT solutions class 12 Chemistry chapter 6 for free on Extramarks.
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