NCERT Solutions for Class 12 Chemistry General Principles and Processes of Isolation of Elements [Old Book]

NCERT
SOLUTIONS
CLASS - 12th

aglase .co

Class : 12th
Subject : Chemistry
Chapter : 6
Chapter Name : General Principles and Processes of Isolation of Elements

Q6.1 Which of the ores mentioned in the table can be concentrated by magnetic separation
method?

Answer. If the ore or the gangue can be attracted by the magnetic eld, then the ore can be
concentrated by the process of magnetic separation. Among the ores mentioned in table , the ores
of iron such as haematite (Fe O ), magnetite (Fe O ), siderite (FeCO ), and iron pyrites (FeS )
2 3 3 4 3 2

can be separated by the process of magnetic separation.

Page : 155 , Block Name : Intext Questions

Q6.2 What is the signi cance of leaching in the extraction of aluminium?

Answer. In the extraction of aluminium, the signi cance of leaching is to concentrate pure
alumina (Al O ) from bauxite ore. Bauxite usually contains silica, iron oxide, and titanium oxide
2 3

as impurities. In the process of leaching, alumina is concentrated by digesting the powdered ore
with a concentrated solution of NaOH at 473-523 K and 35-36 bar. Under these conditions,
alumina (Al O ) dissolves as sodiummeta-aluminate and silica (SiO ) dissolves as sodium silicate
2 3 2

leaving the impurities behind.

The impurities are then ltered and the solution is neutralized by passing C02 gas. In this process,
hydrated Al O gets precipitated and sodium silicate remains in the solution. Precipitation is
2 3

induced by seeding the solution with freshly prepared samples of hydrated Al O .
2 3

Page : 155 , Block Name : Intext Questions

Q6.3 The reaction Cr O
2 3 + 2Al ⟶ Al2 O3 + 2Cr (ΔGo = −421kJ) is thermodynamically
feasible as is apparent from the Gibbs energy value. Why does it not take place at room
temperature?

Answer. The change in Gibbs energy is related to the equilibrium constant, K as
ΔG = −RT ln K

At room temperature, all reactants and products of the given reaction are in the solid state. As a
result, equilibrium does not exist between the reactants and the products. Hence, the reaction
does not take place at room temperature. However, at a higher temperature, chromium melts and
the reaction takes place.
We also know that according to the equation
ΔG = ΔH − T ΔS

Increasing the temperature increases the value of T ΔS making the value of ΔG more and more
negative. Therefore, the reaction becomes more and more feasible as the temperature is increased.

Page : 162 , Block Name : Intext Questions

Q6.4 Is it true that under certain conditions, Mg can reduce Si02 and Si can reduce MgO? What are
those conditions?

Answer.

The temperature at which Δ G curves of these two substances intersect is 1966 K. Thus, at
r

temperatures less than 1966 K, Mg can reduce SiO and above 1966 K, Si can reduce Mgo.
2

Page : 162 , Block Name : Intext Questions

Q6.1 Copper can be extracted by hydrometallurgy but not zinc. Explain.

Answer. 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.
2+ 2+
Fe(s) + Cu ⟶ Fe + Cu(s)
(α) (aq)

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 H
2

gas.
2K(s) + 2H2 O(i) ⟶ 2KOH(aq)) + H2(s)

As a result, these metals cannot be used in hydrometallurgy to extract zinc.
Hence, copper can be extracted by hydrometallurgy but not zinc.

Page : 168 , Block Name : Exercise

Q6.2 What is the role of depressant in froth oatation process?

Answer. In the froth oatation 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 a 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
Na [Zn(CN) ].
2 4

4NaCN + ZnS ⟶ Na2 [Zn(CN)4 ] + Na2 S

Page : 168 , Block Name : Exercise

Q6.3 Why is the extraction of copper from pyrites more dif cult than that from its oxide ore
through reduction?

Answer. The Gibbs free energy of formation (Δ G) of Cu S is less than that of H S and CS
r 2 2 2

Therefore, and C cannot reduce Cu S to Cu.
2

On the other hand, the Gibbs free energy of formation of Cu O is greater than that of
2

CO . Hence, C can reduce Cu O to Cu.
2

C(n) + Cu2 O(s) ⟶ 2Cu(s) + CO(g)

Hence, the extraction of copper from its pyrite ore is dif cult than from its oxide ore through
reduction.

Page : 168 , Block Name : Exercise

Q6.4 Explain: (i) Zone re ning (ii) Column chromatography.

Answer. (i) Zone re ning:
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 re ning, a circular mobile heater is
xed 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 puri ed 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 puri cation 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 puri ed. 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. Al O 2 3

column iS usually used as the stationary Phase in column Chromatography. The mobile Phase may
be a gas, liquid, or supercritical uid 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).

Page : 168 , Block Name : Exercise

Q6.5 Out Of C and CO. which is a better reducing agent at 673 K?

Answer. At 673 K, the value of ΔG (co,co2 )
is less than that of ΔG (C,CO)
. Therefore, CO can be
reduced more easily to CO than C to CO. Hence, CO is a better reducing agent than C at 673 K.
2

Page : 168 , Block Name : Exercise

Q6.6 Name the common elements present in the anode mud in electrolytic re ning of copper. Why
are they so present ?

Answer. In electrolytic re ning 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 puri cation process. Hence, they settle down below the anode as
anode mud.

Page : 168 , Block Name : Exercise

Q6.7 Write down the reactions taking place in different zones in the blast furnace during the
extraction of iron.

Answer. 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 (Fe O and Fe O ) takes place. The
2 3 3 4

reactions taking place in the lower temperature range (500 — 800 K) in the blast furnace are:

3Fe2 O3 + CO ⟶ 2Fe3 O4 + CO2

Fe3 O4 + 4CO ⟶ 3Fe + 4CO2

Fe2 O3 + 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 ore is removed as slag by calcium oxide (CaO), which is formed by the
decomposition of limestone (CaCO ). 3

CaCO3 ⟶ CaO + CO2

CaO + SiO2 ⟶ CaSiO3

Page : 168 , Block Name : Exercise

Q6.8 Write chemical reactions taking place in the extraction of zinc from zinc blende.

Answer. 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 oatation method.
(ii) Conversion to oxide (Roasting)
Sulphide ore 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.
cote,67k

ZnO + C ⟶ Zn + CO

(iv) Electrolytic Re ning
Zinc can be re ned by the process of electrolytic re ning. In this process, impure zinc is made the
anode while a pure copper strip is made the cathode. The electrolyte used is an acidi ed solution
of zinc sulphate (ZnSO ). Electrolysis results in the transfer of zinc in pure from the anode to the
4

cathode.
2+ −
Anode: Zn ⟶ Zn + 2e

2+ −
Cathode: Zn + 2e ⟶ Zn

Page : 168 , Block Name : Exercise

Q6.9 State the role of silica in the metallurgy of copper.

Answer. During the roasting of pyrite ore, a mixture of Feo and Cu₂0 is obtained.
Δ

2CuFeS2 + O2 ⟶ Cu2 S + 2FeS + SO2

Δ

2Cu2 S + 3O2 ⟶ 2Cu2 O + 2SO2

2FeS + 3O2 ⟶ 2FeO + 2SO2

The role of silica in the metallurgy of copper is to remove the iron oxide obtained during the
process of roasting as 'slag'. If the sulphide ore of copper contains iron, then silica
(SiO ) is added as ux before roasting. Then, Feo combines with silica to form iron silicate,
2

FeSiO (slag).
3

Δ

FeO + SiO2 ⟶ FeSiO3

(Slag)

Page : 168 , Block Name : Exercise

Q6.10 What is meant by the term "chromatography"?

Answer. Chromatography is a collective term used for a family of laboratory techniques for the
separation of mixtures. The term is derived from Greek words 'chroma' meaning 'colour' and
'graphein' meaning 'to write'. Chromatographic techniques are based on the principle that
different components are absorbed differently on an absorbent. There are several chromatographic
techniques such as paper chromatography, column chromatography, gas chromatography, etc.

Page : 168 , Block Name : Exercise

Q6.11 What criterion is followed for the selection of the stationary phase in chromatography?

Answer. The stationary phase is selected in such a way that the components of the sample have
different solubilities in the phase. Hence, different components have different rates of movement
through the stationary phase and as a result, can be separated from each other.

Page : 169 , Block Name : Exercise

Q6.12 Describe a method for re ning nickel.

Answer. Nickel is re ned 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.
330−350k

Ni + 4CO ⟶ Ni(CO)4

Then, the obtained nickel tetracarbonyl is decomposed by subjecting it to a higher temperature
(450 — 470 K) to obtain pure nickel metal.
450−470K

Ni(co)4 ⟶ Ni + 4CO

Page : 169 , Block Name : Exercise

Q6.13 How can you separate alumina from silica in bauxite ore associated with silica? Give
equations, if any.

Answer. To separate alumina from silica in bauxite ore associated with silica, rst the powdered
ore is digested with a concentrated NaOH solution at 473 - 523 K and 35 - 36 bar pressure. This
results in the leaching out of alumina (Al O ) as sodium aluminate and silica (SiO ) as sodium
2 3 2

silicate leaving the impurities behind.
Al2 Oy(s) + 2NaOH(aq) + 3H2 O(j) ⟶ 2Na[Al(OH)4 ]
(aq)

Alumina Sodium aluminate

SiO2 + 2NaOH(aq) ⟶ Na2 SiO3(aq) + H2 O(l)

Then, CO gas is passed through the resulting solution to neutralize the aluminate in the
2

solution, which results in the precipitation of hydrated alumina. To induce precipitation, the
solution is seeded with freshly prepared samples of hydrated alumina.
2Na[Al(OH)4 ] + CO2(g) ⟶ Al2 O3 ⋅ xH2 O(s) + 2NaHCO3(aq)
(aq)

During this process, sodium silicate remains in the solution. The obtained hydrated alumina is
ltered, dried, and heated to get back pure alumina.
140k

Al2 O3 ⋅ xH2 O(s) ⟶ Al2 O3(s) + xH2 O(g)

Page : 169 , Block Name : Exercise

Q6.14 Giving examples, differentiate between 'roasting' and 'calcination'.

Answer. 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.

Δ

2Zns + 3O2 ⟶ 2ZnO + 2SO2

Zinc blende

Δ

2PbS + 3O2 ⟶ 2PbO + 2SO2

Galena

Δ

2Cu2 S + 3O2 ⟶ 2Cu2 O + 2SO2

Copper

glance

On the other hand, calcination is the process of converting hydroxide and carbonate ores to oxides
by heating the ores either in the absence or in a limited supply of air at a temperature below the
melting point of the metal. This process causes the escaping of volatile matter leaving behind the
metal oxide. For example, hydroxide of Fe, carbonates of Zn, Ca, Mg are converted to their
respective oxides by this process.
Fe2 O3 ⋅ 3H2 O ⟶ Fe2 O3 + 3H2 O

Limonite

ZnCO3(s) ⟶ ZnO(s) + CO2(g)

Calamine
Δ

CaMg (CO3 ) ⟶ CaO(s) + MgO + 2CO
2 (s)

Dolomite

Page : 169 , Block Name : Exercise

Q6.15 How is 'cast iron' different from 'pig iron"?

Answer. 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.

Page : 169 , Block Name : Exercise

Q6.16 Differentiate between "minerals" and "ores".

Answer. 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 pro tably (conveniently and economically) from all these minerals. Zinc can be obtained
from zinc blende (ZnS), calamine (ZnCO ), Zincite (ZnO) etc. Thus, these minerals are called ores
3

of zinc.

Page : 169 , Block Name : Exercise

Q6.17 Why copper matte is put in silica lined converter?

Answer. 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 (FeSiO ). Also, some silica is 3

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 (FeSiO ) and Cu S is converted into metallic copper.
3 2

2FeS + 3O2 ⟶ 2FeO + 2SO2

FeO + SiO2 ⟶ FeSiO3

2Cu2 S + 3O2 ⟶ 2Cu2 O + 2SO2

2Cu2 O + Cu2 S ⟶ 6Cu + SO2

Page : 169 , Block Name : Exercise

Q6.18 What is the role of cryolite in the metallurgy of aluminium?

Answer. Cryolite 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 Al O . 2 3

Page : 169 , Block Name : Exercise

Q6.19 How is leaching carried out in case of low grade copper ores?

Answer. 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.
+ 1 2+
Cu(s) + 2H + O2(g) ⟶ Cu + 2H2 O(j)
(αq) 2 (a))

The resulting solution is treated with scrap iron or H to get metallic copper. 2
2+ +
Cu + H2(g) ⟶ Cu(s) + 2H
(aq) (aq)

Page : 169 , Block Name : Exercise

Q6.20 Why is zinc not extracted from zinc oxide through reduction using CO?

Answer. The standard Gibbs free energy of formation of ZnO from Zn is lower than that of CO 2

from CO. Therefore, CO cannot reduce ZnO to Zn. Hence, Zn is not extracted from through
reduction using CO.

Page : 169 , Block Name : Exercise

Q6.21 The value of Δ G for formation of Cr O is −540kJmol
f
⊖
2 3
−1
and that of Al O is – 827
2 3

kJmol−1 . Is the reduction of Cr O possible with Al ?
2 3

Answer. The value of Δ G for formation of Cr O is −540kJmol
f
⊖
2 3
−1
is higher than that of

Al2 O3 from Al (−827kJmol −1
) . Therefore Al cxan reduce Cr O
2 3 to Cr . Hence, the reduction of
Cr2 O3 with Al is possible.
Alternatively,
3 θ −1
2A1 + O2 ⟶ Al2 O3 Δ1 G = −827kJmol
2
3 9 −1
2Cr + O2 ⟶ Cr2 O3 Δ1 G = −540kJmol
2

Substracting equation (ii) from(i), we have
2Al + Cr2 O3 ⟶ Al2 O3 + 2Cr

As Δ G for the reduction of Cr O by Al is negative, this reaction is possible
f
⊖
2 3

Page : 169 , Block Name : Exercise

Q6.22 Out of C and CO, which is a better reducing agent for ZnO ?

Answer.

Reduction of ZnO to Zn is usually carried out at 1673 K. From the above gure, 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 CO from C is lesser than the Gibbs free energy of formation of ZnO.
2

Therefore, C can easily reduce ZnO to Zn. On the other hand, the Gibbs free energy of formation of
CO from CO is always higher than the Gibbs free energy of formation of ZnO. Therefore, CO
2

cannot reduce ZnO. Hence, C is a better reducing agent than CO for reducing ZnO.

Page : 169 , Block Name : Exercise

Q6.23 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.

The above gure is a plot of Gibbs energy (ΔG ) ⊖

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 (Δ G ) of the oxide of the former is more negative than the latter. For
f
⊖

example, since Δ G f
⊖
(Al, A2 , O3 ) is more negative than Δ G f
⊖
(cu, Cu2 , O) can reduce Cu:O to
Cu, but Cu cannot reduce Al O . Similarly, Mg can reduce ZnO to Zn, but Zn cannot reduce MgO
2 3

because Δ G more negative than Δ G .
∘ ∘
f (Mg,MgO) f (Zn,ZnO)

Page : 169 , Block Name : Exercise

Q6.24 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 the electrolysis of molten NaCl, Cl is obtained at the anode as a by product.
2
+ −
NaCl(math) ⟶ Na (melt) + Cl (melt)

At cathode : Na +

(melt)
+ e
−
⟶ Na(s)

At anode: Cl −

(melt)
⟶ Cl(g) + e
−

2Cl(g) ⟶ Cl2(g)

The overall reaction is as follows:
Electrolysis 1
NaCl( melt) Na(s) + Cl2(g)
⟶ 2

If an aqueous solution of NaCl is electrolyzed, Cl will be obtained at the anode but at the
2

cathode, H will be obtained (instead of Na). This is because the standard reduction potential Of
2

Na (E = −2.71V ) is more negative than that Of H O ((E = −0.83V ). Hence, H O will get
∘
2
∘
2

preference to get reduced at the cathode and as a result, H is evoIved. 2
+ −
NaCl(aq) ⟶ Na + C
(aq) (aq)

At cathode : 2H O 2 (l)
+ 2e
−
⟶ H2(g) + 2OH
−

(aq)

At anode: Cl
− −
⟶ Cl(g) + e
(melt)

2Cl(g) ⟶ Cl2(g)

Page : 169 , Block Name : Exercise

Q6.25 What is the role of graphite rod in the electrometallurgy of aluminium?

Answer. In the electrometallurgy Of aluminium, a fused mixture Of puri ed alumina (Al O ), 2 3

cryolite (Na AlF ) and uorspar (CaF ) is electrolysed. In this electrolysis, graphite is used as
3 6 2

the anode and graphite-lined iron is used as the cathode. During the electrolysis, Al is liberated at
the cathode, while CO and CO are liberated at the anode, according to the following equation.
2

Cathode : Al
3+ −
+ 3e ⟶ Al(l)
(meth)

Anode : C
2− −
(s) + O ⟶ CO(g) + 2e
(meth)

2− −
C(s) + 2O ⟶ CO2(g) + 4e
(melt)

If a metal is used instead of graphite as the anode, then O will be liberated. This will not only
2

oxidise the metal Of the electrode, but also convert some Of the A1 liberated at the cathode back
into Al O ,. Hence, graphite is used for preventing the formation of O at the anode. Moreover,
2 3 2

graphite is cheaper than other metals.

Page : 169 , Block Name : Exercise

Q6.26 Outline the principles of re ning of metals by the following methods:
(i) Zone re ning
(ii) Electrolytic re ning
(iii) Vapour phase re ning

Answer. (i) Zone re ning:
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 re ning, a circular mobile heater is
xed 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 puri ed by this process.

(ii) Electrolytic re ning;
Electrolytic re ning iS the process of re ning 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 re ning:
Vapour phase re ning is the process of re ning 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 re ned using this method.

Page : 169 , Block Name : Exercise

Q6.27 Predict conditions under which Al might be expected to reduce MgO.

Answer. Above 1350 C, the standard Gibbs free energy of formation of Al O from Al is less than
∘
2 3

that of MgO from Mg. Therefore, above 13500C, Al can reduce MgO.

Page : 169 , Block Name : Exercise