# 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- Are you having doubts in chapter 6 of chemistry? Here you will find all the solutions of NCERT class 12 chemistry chapter 6 General Principles and Processes of isolation of elements. In this chapter, you will explore the occurrence of metals, the concentration of ores and their cleaning processes and the extraction of crude metal from ores and many refining processes. To develop a grip on every topic you must solve topic wise questions and exercise questions of this chapter. In this chapter, 4 topic wise questions are given between the chapter and 27 exercise questions are given at the end of the chapter. 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. Also, the NCERT solutions for other classes and subjects can assist you in your preparation.

In NCERT solutions for class 12 chemistry chapter 6 General Principles and Processes of Isolation of Elements, there are a total of eight sub-topics that covers important concepts of the chapter. After completing this chapter, you will be able to explain the terms like ores, minerals, concentration, benefaction, calcination, roasting, refining, etc. Also 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. Please scroll down to get all the NCERT solutions for class 12 chemistry chapter 6 General Principles and Processes of isolation of elements.

The history of civilisation is linked to the use of metals in different ways, different periods of human civilisations have been named after metals like bronze age, iron age, etc and the extraction of metals brought about several changes in society. So, the class 12 chemistry chapter 6 General Principles and Processes of Isolation of Elements is an important chapter from the historical point of view. Aluminium (Al) is the most abundant metal found in the earth's crust,  approximately 8.3% by weight. Cleaning of ore of metals by the removal of particles like clay, sand, etc. from the ore is called concentration of ore. This is done in the following steps: Hydraulic washing, Magnetic separation, froth floatation method, Leaching. You will read all these methods in the chapter.

## Topics and Sub-topics of NCERT 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

## 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

Table 6.1: Principle Ores of some Important Metals

 Metal Ores Composition Aluminium Bauxite   Kaolinite(a form of clay) AlOx(OH)3-2x [Where 0

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

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)

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

$Cr_2 O_3 + 2 AI \rightarrow Al_2 O_3 + 2 Cr$
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.

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 isolation of elements- Exercise Questions

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.

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

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-

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

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

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.

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.

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.

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.

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

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

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

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 by 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)-   $Zn^{2+}+2e^{-}\rightarrow Zn$

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 ($FeSiO_{3}$

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

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

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.

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.

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

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.
$Al_{2}O_{3}.xH_{2}O\rightarrow Al_{2}O_{3}+x.H_{2}O$

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”?

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”.

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.

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.

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

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

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 $(H_{2})$

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

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 ($Zn$).

$2Al+3/2O_{2}\rightarrow Al_{2}O_{3}$......................... (-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.

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.

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.

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-  $\dpi{100} Cl^{-}\rightarrow \frac{1}{2}Cl_{2}+e^{-}$

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.

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.

Electrolytic refining- It is a process of refining impure metals by electricityIn 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)

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

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.

## NCERT Solutions Class 12 Chemistry

 Chapter 1 CBSE NCERT solutions for class 12 chapter 1 The Solid State Chapter 2 NCERT solutions for class 12 chemistry chapter 2 Solutions Chapter 3 Solutions of NCERT class 12 chemistry chapter 3 Electrochemistry Chapter 4 CBSE NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics Chapter 5 Solutions of NCERT class 12 chemistry chapter 5 Surface chemistry Chapter 6 NCERT solutions for class 12 chemistry chapter 6 General Principles and Processes of isolation of elements Chapter 7 CBSE NCERT solutions for class 12 chemistry chapter 7 The P-block elements Chapter 8 Solutions of NCERT class 12 chemistry chapter 8 The d and f block elements Chapter 9 NCERT solutions for class 12 chemistry chapter 9 Coordination compounds Chapter 10 Solutions of NCERT class 12 chemistry chapter 10 Haloalkanes and Haloarenes Chapter 11 CBSE NCERT solutions for class 12 chemistry Alcohols, Phenols, and Ethers Chapter 12 Solutions of NCERT class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic Acids Chapter 13 NCERT solutions for class 12 chemistry chapter 13 Amines Chapter 14 CBSE NCERT solutions for class 12 chemistry chapter 14 Biomolecules Chapter 15 Solutions of NCERT class 12 chemistry chapter 15 Polymers Chapter 16 NCERT solutions for class 12 chemistry chapter 16 Chemistry in Everyday life

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