Class 12 - Chemistry - Isolation of Elements

Question6.1.

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.

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.

 

Question 6.2.

What is the role of depressant in froth floatation process?

Answer:

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 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 Na2 [Zn (CN) 4].

 4 NaCN + ZnS → Na2 [Zn (CN) 4] + Na2S

 

Question 6.3.

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

Answer:

The Gibbs free energy of formation (ΔfG) of Cu2S is less than that of H2S and CS2. Therefore, H2 and C cannot reduce Cu2S to Cu.

On the other hand, the Gibbs free energy of formation of Cu2O is greater than that of CO. Hence, C can reduce Cu2O to Cu.

C(s) +Cu2O(s)   →   2Cu(s) + CO (g)

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

 

Question 6.4.

Explain: (i) Zone refining (ii) Column chromatography.

Answer:

  • Zone refining

Zone refining method is based on the principle that impurities are more soluble in the molten state of metal (the melt) than in the solid state.

The impure metal is heated with the help of a circular mobile heater at one end.

This results in the formation of the molten zone or melts.

As the heater is removed along with the length of the rod, the pure metal crystallizes out

of the melt and impurities pass into the adjacent molten zone.

This process is repeated several times till the impurities are completely driven to the end of the rod which is then cut off and discarded.

The method is very useful for semiconductor and other metals of very high purity, e.g., silicon, germanium, boron, and gallium.

  • Column chromatography

It is based on the principle that different components of a mixture are differently adsorbed on an adsorbent.

The mixture to be separated is put in a liquid or gaseous medium which is moved through the adsorbent.

Different components are adsorbed at different levels on the column.

Later, the adsorbed components are removed (eluted) by using suitable solvents (eluents).

There are several chromatographic techniques such as paper chromatography, column chromatography, gas chromatography, etc.

chromatography is used for the purification of elements which are available in minute quantities and the impurities

are not very different in chemical properties from the element to be purified.

Question 6.5.

Out of C and CO, which is a better reducing agent at 673 K?

Answer:

At 673 K, the value of Gibbs free energy (CO,CO2) is less than that of Gibbs free energy of (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.

 

Question 6.6.

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

Answer:

The common elements present in anode mud in electrolytic refining are antimony, selenium, tellurium, silver, gold, and platinum.

These elements being less reactive are not affected by CuSO4 + H2SO4 solution and hence settle down under anode 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:

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:

3Fe3O3 + CO --> 2Fe3O4 + CO2

Fe3O4 + 4CO --> 3Fe + 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 ore is removed as slag by calcium oxide (CaO), which is formed by the decomposition of limestone (CaCO3).

CaCO3 --> CaO + CO2

CaO + SiO2 -->CaSiO3

                    (Calcium silicate) (Slag)

Class_12_Chemistry_Isolation_Of_Elements_Blast_Furance

                                                                                                                                                                                   

Question 6.8.

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

Answer:

The steps involved in the extraction of zinc from zinc blende (ZnS) are as followed:-

1) Concentration: the ore is crushed and then concentrated by froth floatation process.

2) Roasting: the concentrated ore is heated in the presence of an excess of air at about 1200K to form zinc oxide.

ZnS + 3O2→ 2 ZnO + 2 SO2

3) Reduction: ZnO obtained above is mixed with powdered coke and heated to 1673K in a fireclay resort.

ZnO + C --> Zn + CO

4) Electrolytic Refining: Zinc is refined by the process of electrolytic refining.

In this process, impure zinc is made the anode and a pure copper strip is made the cathode.

Electrode used in an acidified solution of zinc sulphate (ZnSO4).

Electrolysis results in the transfer of zinc in pure form from the anode to the cathode.

Anode: Zn--> Zn2+ + 2e-

Cathode: Zn2+ + 2e- --> Zn

 

Question 6.9.

State the role of silica in the metallurgy of copper?

Answer:

During the roasting of pyrite ore, a mixture of FeO and Cu2O is obtained.

2 CuFeS2 + O2→ Cu2S + 2 FeS + SO2

2 Cu2S + 3 O2 → 2 Cu2O + 2 SO2

2 FeS + 3 O2 → 2 FeO + 2 SO2

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 iron, then silica (SiO2) is added as a flux before roasting.

Then, FeO combines with silica to form silicate, FeSiO3 (slag).

FeO + SiO2→ FeSiO3 (slag)

 

Question 6.10.

What is meant by the term “chromatography”?

Answer:

Chromatography is a term used for numerous laboratory techniques for the separation of mixtures.

The term is derived from Greek word ‘chroma’ meaning ‘color’ and ‘graphein’ meaning ‘to write’.

Chromatographic techniques are based on the principles that different components are adsorbed differently on an adsorbent.

There are various chromatographic techniques used such as paper chromatography, column chromatography, gas chromatography, etc.

 

Question 6.11.

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 solubility's 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.

 

Question 6.12.

Describe a method for refining nickel?

Answer:

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.

Ni + 4CO    -->       Ni (CO) 4

           (330-350 K)   Nickel tetra Carbonyl

 

Question 6.13.

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

Answer:

 To separate alumina from silica in bauxite ore associated with silica, firstly, the powdered ore is digested

with a concentrated NaOH solution at a temperature of 473 – 523K 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.

Al2O3 (s) + 2NaOH (aq) + 3H2O (l) → 2 Na [Al (OH) 4] (aq)

Alumina

SiO2 + 2NaOH (aq) → Na2SiO3 (aq) + H2O (l)

Silica

After this, CO2 gas is passed through the resultant solution to neutralize the alumina present in the solution,

this results in the precipitation of hydrated samples of alumina.

To include precipitation, the solution is seeded with freshly prepared samples of hydrated alumina.

2Na [Al (OH) 4] (aq) + CO2 (g) --> Al2O3x2H2O (s) + 2NaHCO3 (aq)

Sodium aluminate                      Hydrated Alumina Sodiumhydrogencarbonate

During this process, sodium silicate remains in the solution.

The hydrated alumina obtained is then filtered, dried, and heated to get back pure alumina.

Al2O3x2H2O (s) à   Al2O3(s) + xH2O (g)

Hydrated Alumina   Alumina

 

Question 6.14.

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.

2ZnS + 3O2 --> 2ZnO + 2SO2

Zinc Blend

2PbS + 3O2 --> 2PbO + 2SO2

Galena

2Cu2S + 3O2 --> 2Cu2O + 2 SO2

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.

Fe2O3. 3H2O --> Fe2O3 + 3H2O

Limonite

ZnCO3(s) --> ZnO(s) + CO2 (g)

Calamine

CaMg (CO3)2 --> CaO(s) + MgO(s) + 2CO

Dolomite

  

Question 6.15.

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 and 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; cast iron is extremely hard and brittle.

 

Question 6.16.

Differentiate between “minerals” and “ores”.

Answer:

Minerals

Ores

All the naturally occurring substances of metals which are present in the earth’s crust are known as Minerals.

Ores are usually used to extract metals economically. A large amount of metals are present.

All Minerals are not ores.

All ores are minerals.

Minerals are native form in which metals exist.

Ores are mineral deposits.

 

Question 6.17.

Why copper matte is put in silica lined converter?

Answer:

Copper matte contains Cu2S and FeS. When a blast of hot air is passed through molten matte took in a silica lined convertor,

FeS present in matte is oxidized to FeO which combines with Silica (SiO2) to form FeSiO3, slag.

2FeS + 3O2 --> 2FeO + 2SO2 (gas)

FeO + SiO2 --> FeSiO3

          Silica    Slag

When the whole of iron has been removed as slag, some of the Cu2S undergoes oxidation to form Cu2O

which then reacts with more Cu2O to form copper metal.

2Cu2S + 3O2 --> 2Cu2O + SO2 (gas)

2Cu2O + Cu2S --> 6Cu + SO2 (gas)

 

Question 6.18.

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

Answer:

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.

Question 6.19.

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

 Answer:

For the group of low-grade copper ores, leaching is carried out using acid or bacteria in the presence of air or oxygen.

In this process, copper goes into the solution as Cu2+ ions.

Cu(s) +2H+ (aq) + (1/2) O2 (g) --> Cu2+ (aq) +2H2O (l)

The resulting solution is then treated with scrap iron or H2 to obtain metallic copper.

Cu2+ (aq) + H2 (g) --> Cu(s) + 2H+ (aq)

 

Question 6.20.

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 CO2 from CO.

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

 

Question 6.21.

The value of ΔfG0 for formation of Cr2O3 is – 540 kJmol−1and that of Al2O3 is –827 kJmol−1. Is the reduction of Cr2 3 possible with Al?

Answer:

The value of Δ fG0 for the formation of Cr2O3 is –540kJ mol-1 which is higher than that of Al2O3 is – 827 Kjmol-1.

Therefore Al can reduce Cr2O3 to Cr. Hence, the reduction of Cr2O3 with Al is possible.

 

Question 6.22.

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

Answer:

Class_12_Chemistry_Isolation_Of_Elements_Graph

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.

 

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:

Class_12_Chemistry_Isolation_Of_Elements_Graph_1

Thermodynamic factors help us in choosing a suitable reducing agent for the reduction of a particular metal state as described below.

From Ellingham diagram, it is evident that metals for which the standard free energy of formation of their

oxides is more negative can reduce those metal oxides for which the standard free energy of formation of

their respective oxides is less negative.

In other words, any metal will reduce the oxides of other metals which lie above in the Ellingham Diagram

because the standard free energy changes of the combined redox reaction will be

negative by any amount of equal to the difference in Δ fG0 of the two metal oxides.

Hence, both Al, Zn can reduce FeO to Fe, but Fe cannot reduce Al2O3 to Al and ZnO to Zn.

 

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 the electrolysis of molten NaCl, Cl2 is obtained at the anode as a by-product.

NaCl (melt)   → 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:

By doing electrolysis:-

NaCl (melt) --> Na(s) + (1/2) Cl2 (g)

If an aqueous solution of NaCl is electrolyzed, Cl2 will be obtained at the anode but at the cathode, H2 will be obtained (instead of Na).

This is because the standard reduction potential of Na (E°= - 2.71 V) is more negative than that of H2O (E° = - 0.83 V).

Hence, H2O will get preference to get reduced at the cathode and as a result, H2 is evolved.

NaCl (aq) → Na+ (aq) + Cl-(aq)

At cathode:  2 H2O (l) + 2e-   →   H2 (g) + 2OH-(aq)  

At anode: Cl-(melt)    → Cl (g) + e-

2Cl (g) → Cl2 (g)

 

Question 6.25.

What is the role of graphite rod in the electrometallurgy of aluminium?

Answer:

Graphite rod acts as anode and graphite lined iron acts as a cathode in the electrometallurgy of aluminium.

Carbon reacts with oxygen liberated at anode producing CO and CO2 otherwise oxygen

liberated at the anode may oxidize some of the liberated aluminium back to Al2O3.

At Anode:-

C(s) + O2-(melt) à--> CO (g) + 2e-

C(s) + 2O2-(melt) --> CO2 (g) + 4e-

At cathode: - Al3+ (melt) + 3e- -->Al (l)

 

Question 6.26.

Outline the principles of refining of metals by the following methods:

(i) Zone refining

(ii) Electrolytic refining

(iii) Vapour phase refining?

Answer:

  • Zone refining: Zone refining is based on the principle that impurities are more soluble in the melt than in the solid state of the metal.
  • The impure metal bar is heated with the help of circular coils fitted around this bar and slowly moved longitudinally from one end to another of impure metal.
  • At the heated zone the bar melts and as the heater mover on pure metal crystallizes while the impurities pass into the adjacent molten portion.
  • In this way the impurities are swept from one end of the bar to another. As the molten bar is removed from the heating coil, it cools
  • Pure metal crystallizes out, leaving the impurities in the molten metal portion.
  • When the molten metal portion carrying impurities with increased concentration reaches the end of the rod, it is allowed to cool and is cut off.
  • The process is repeated several times to get pure metal of 99.99% purity.
  • Class_12_Chemistry_Isolation_Of_Elements_Zone_Refining
  • 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 (or sludge).

            Class_12_Chemistry_Isolation_Of_Elements_Electrolytic_Refining  

  • 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.

 

Question 6.27.

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

Answer:

Above 1350oC, the standard Gibbs free energy formation of Al2O3 from Al is less than that of MgO from Mg.

Therefore, above 1350oC, Al can reduce MgO.

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