Metallurgy Important Questions

CHAPTER: PRINCIPLES AND PROCESSES OF EXTRACTION OF METALS

Ellingham diagram:

·                     Ellingham plotted the experimentally determined standard free energy of formation, ∆G_f ⁰ of            

            various oxides using one mole of oxygen with temperature.

(2x/y) M(s) + O ₂ (g) → (2/y) M _x O _y (s) ; ∆G ⁰ =∆H ⁰ -T∆S ⁰

• Ellingham pointed out that, the standard enthalpy and entropy of formation of a compound do not change significantly with the temperature as long as there is no change of state of product or reactant. ∆G ⁰  =∆H ⁰  -T∆S ⁰
• Thus, the general forms of ∆G ⁰ –T ∆S ⁰ relationships could be approximated to straight lines over temperature ranges .The slope of free energy vs temperature plot would thus be (-∆S ⁰ ) entropy change of the reaction and the intercept is the enthalpy value.

• It provides a sound idea about selecting a reducing agent in the reduction of oxides. Such diagrams helps in predicting the feasibility of a thermal reduction of an ore. ∆G must be negative at a given temperature for a reaction to be feasible.

·         2x M(s) + O ₂ (g) → 2M _x O(s) In this reaction ∆S=-ve because solid and gaseous reactants are changing to metal oxide which is solid, hence entropy decreases. ∆G will be +ve if temperature is increased and T∆S>∆H. This results in +ve slope in the curve for the most of reactions shown in the diagram for the formation of metallic oxides.

Characteristics of Ellingham diagrams:

1. Each plot is a straight line when solid changes to liquid or liquid changes to gas i.e., change in phase occurs. The temperature at which change occurs is indicated by increase in slope on +ve side. The abrupt increase in Zn to ZnO plot shows the melting point of Zn metal.

      ii.            Curves in the Ellingham diagrams for the formation of metallic oxides are straight lines with a positive slope.

3. There is a point in the curve below which ∆G is –ve .All metals with negative value of ΔG may be oxidized spontaneously so metal oxide is stable. Above this point the oxides with positive ΔG are not stable and are easily decomposed to elemental metals.The lower the position of a metal in the Ellingham diagram more is the stability of its oxide. For example, the Ellingham diagram for Al is found to be below Fe ₂ O ₃ .
4. The value of ∆ _f G ^o for the formation of oxides at different temperatures helps in predicting the temperature and reducing agent suitable for particular oxide.

5. A metal found in the Ellingham diagram can act as a reducing agent for a metallic oxide found above it i.e., A given metal can reduce the oxides of other metals whose lines lie above theirs on the diagram. Thus, Al can act as a reducing agent to Cr which is above it in the diagram.
6. A substance whose formation enthalpy is lower (ΔG line lower on diagram) at given temperature, will reduce one whose formation enthalpy is higher on the diagram. Hence metallic aluminum can reduce iron from iron oxide into metallic iron, aluminum itself oxidizing into aluminum oxide.

  vii.            The lower the position of a metal in the Ellingham diagram, the greater is the stability of its oxide. For example, the Ellingham diagram for Al is found to be below Fe2O3.

8. The greater the gap between any two lines, the greater the efficiency of the reducing agent.
9. Stability of metallic oxides decrease with increase in temperature. Highly unstable oxides like Ag ₂ O and HgO easily undergo thermal decomposition.
10. The intersection of two lines imply the equilibrium of oxidation and reduction reaction between two substances. Reduction with using a certain reductant is possible at the intersection point and higher temperatures where the ΔG line of the reductant is lower on diagram than the metallic oxide to be reduced. At the point of intersection the Gibbs energy is 0(zero), below this point the Gibbs energy is <0 and the oxides are stable,while above the point of intersection the Gibbs energy is >0 and so, the oxides are unstable.
11. The formation enthalpy of carbon dioxide (CO ₂ ) is almost a temperature-independent constant, while that of carbon monoxide (CO) has negative slope. The line for CO intersects the lines of most of the metals. . Carbon monoxide is the dominant compound in higher temperatures, and the higher the temperature, the more efficient reductant carbon monoxide also is. This explains why carbon is considered as an energetic reducing agent at high temperature.

Very Short Answer Questions (VSA) 1 mark each

1. The reaction: Cr₂0₃ + 2 Al →  Al₂O₃ + 2Cr; ∆G^o = - 421 kJ is thermodynamically feasible

as is apparent from Gibbs energy value. Why does it not take place at room temperature?

Ans. The reaction is thermodynamically favourable. As we know, even thermodynamic,

favourable reactions require some activation energy to proceed, therefore, this reaction will

not occur at room temperature. Heating is required for this reaction to start.

Q.2. Why cannot Al be reduced by carbon?

Ans. Aluminium is stronger reducing agent than carbon and therefore, cannot be reduced by it.

Q.3. Which is better reducing agent at 983 K, C or CO?

Ans. Carbon.

Q.4. What is the role of a stabilizer in froth :floatation process? Give examples also.

Ans. In froth floatation process stabilizer stabilizes the froth. For example, aniline, cresol,etc

5. Is it true that under certain conditions magnesium can reduce Si02 and silica can reduce MgO. What are those conditions?

Ans. Below 1623 K (or 1350°C), magnesium can reduce Si02 but above 1623 K iron and

silicon can reduce MgO.

Q.6. Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?

Ans. The anode mud in the electrolytic refining of copper contains antimony, selenium tellurium, silver, gold and platinum. These are present as impurities in blister copper. They are less reactive and are not affected by CuSO4- H2SO4solution and hence settle down near anode as anode mud.

7. Which is the cheapest and most abundant reducing agent which is used in, the extraction of metals?

Ans. Carbon in the form of coke.

8. What is the significance of leaching in the extraction of aluminium ?

Ans. The principal ore of aluminium is bauxite (Al203). It contains Si02, iron oxides, titanium oxide as impurities. The significance of leaching in the extraction of aluminium from bauxite is to remove the impurities from the ore.

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

Ans. The graphite rod is useful in the electrometallurgy of aluminium for reduction of alumina to aluminium. ;   2Al203+ 3C → 4Al + 3C02

Short Answer Questions (SAQ-I) 2 marks each

Q.1.Why is the reduction of a metal oxide easier if the metal is formed in liquid state at the temperature of reduction?

 Ans. The entropy of a metal is higher in its liquid state than in its solid state. Therefore,

entropy change, ∆S of the reduction process is more on the positive side when the metal formed

in liquid state and metal oxide being reduced is in the solid state. As a result, the value of ∆G

becomes more on negative side and therefore, reduction becomes easier.

Q.2. At a site, low grade copper ores are available and zinc and iron scraps are also available. Which of the two scraps would be more suitable for reducing the leached copper ore and why?

Ans. Since zinc lies above iron in electrochemical series, it is more reactive than iron. As result, if zinc scraps are used, the reduction will be fast. However, zinc is a costlier metal than on. Therefore, it will be advisable and advantageous to use iron scraps.

Q.3. How is copper extracted from low grade ores or scraps?

Ans. Copper is extracted by hydrometallurgy from low grade ores: It is leached out using acid or bacteria. The solution containing copper ions (Cu2+) is treated with scrap iron or H₂ as:

Cu2+ (aq) + H2(g) → Cu(s) + 2H+ (aq)

In this way, copper is obtained.

Q.4. Although thermodynamically feasible, in practice magnesium metal is not used for the reduction of alumina in the metallurgy of aluminium. Why?

Ans. Inspection of Ellingham diagram shows that ∆G vs T curves for Al203 and MgO

intersect at a point corresponding to very high temperature of the order of 2000 K. This means

above this temperature, ∆G for the reaction:

Al203 + 3Mg →2Al + 3MgO

would become negative and hence reduction will be feasible. However, this temperature is very

high so that the process is uneconomical and technologically difficult.

Q.5. Copper can be extracted by hydrometallurgy but not zinc. Explain.

Ans. Zinc is more electropositive (E^O= - 0.76V) and therefore, is highly reactive metal. Hence, it cannot be easily displaced from its solution of ZnSO4. On the other hand, copper is less electropositive (E^O= + 0.34 V) and can be readily displaced from its solution by some more

active metal such as zinc.

Q.6. Why is the extraction of copper from pyrite difficult than that from its oxide through reduction?

Ans. The graph of ∆G^o vs T in Ellingham diagram for the formation of oxides shows

the copper-copper oxide line is almost at the top. Therefore, it is very easy to reduce oxide of copper directly to metal by heating with coke. This is because, ∆G^o vs T lines for CO has negative slope at higher temperature and therefore, can easily reduce Cu2O to copper. However the Gibbs energies of formation of most sulphides are greater than that for CS2.

Q.7. What is the role of depressant in froth floatation process?

Ans. The depressants are used to prevent certain types of particles from forming the froth with bubbles in froth floatation process. This helps to separate two sulphides ores. For example:

in case of an ore containing zinc sulphide (ZnS) and lead sulphide (PbS), sodium cyanide is also

is used as a depressant. It forms a layer of zinc complex Na2[Zn(CN)4] with ZnS on the surface as slag of ZnS and therefore, prevents it from forming the froth. Therefore, it acts as a depressant.

ZnS + 4 NaCN →    Na2[Zn(CN4)] +     Na2S

Sodium tetracyanozincate (II)

However, NaCN does not prevent PbS from forming the froth and allows it to come with the froth.

Q.8. Describe a method for refining nickel.

Ans. For refining nickel, nickel is heated in a stream of carbon monoxide forming volatile complex (nickel tetracarbonyl).

Ni + 4CO →   Ni (CO) 4

The carbonyl is subjected to high temperature so that the complex decomposes to give the pure metal                                                        Ni (CO)4→ Ni + 4CO

This process is called Mond’s process.

Q.9. Giving examples, differentiate between roasting and calcination.

Ans.Calcination

1. In calcination, the ore is heated in the absence of air.

Fe203.xH2O(s)  + Heat → Fe203  + x H2O(g)

ZnCO3 + Heat→ ZnO(s) + CO2(g)

CaC03.MgC03(s) +  Heat→ CaO(s) + MgO(s) + 2C02

2. It is used for carbonate and oxide ores.

3. Moisture and organic impurities are removed.

Roasting

1.      In roasting the ore is heated in regular supply of air below the melting point of metal.

2ZnS + 3O2 → 2ZnO + 2SO2

2Cu2S + 3O2 → 2Cu2O + 2SO2

2PbS + 3O2 → 2PbO + 2SO2

2..It is used for sulphide ores.

3. Volatile impurities are removed as oxides: S02,  P205,  As205 etc.

Q.10. Predict conditions under which Al might be expected to reduce MgO.

Ans. The two equations are: 2Mg + O2 → 2MgO

                                                 3Al + O2 → 3Al203

It is clear from Ellingham diagram that ∆G^o vs T plot for Mg, MgO is below ∆G^o vs T plot

Therefore, Al cannot reduce MgO to Mg. However, at the point of intersection corresponding to temperature 1600K,  the ∆G^o for the combined reaction becomes zero. After this point (1600 K), Mg, MgO curve, is higher than that for Al, Al203 curve. Thus Al can reduce Mg0 at temperature higher than 1600 K. But there are practical difficulties to attain higher temperatures.

Q.ll. Why copper matte is put in silica lined converter?

Ans. The copper matte containing Cu2S and FeS is put in silica lined converter. Some silica is also added and hot air blast is blown to convert remaining FeS to FeO, which is removed as slag with silica.

2FeS + O2 → 2FeO + 2S02

FeO + Si02 → FeSi03

Cu2S,or CuO gets converted to copper.

2Cu2S + 3O2 → 2Cu20 + 2S02

2Cu20 + Cu2S → 6 Cu + S02

Short Answer questions (SAQ-II) : 3 marks each

Q.1. Discuss the following methods of purification of metals:

(i) Electrolytic refining

(ii) Zone refining

(iii) Vapour phase refining

Ans. The process of purifying the crude metal is called refining.

(i) Electrolytic refining. This is most common method for the refining of metals and based upon the phenomenon of electrolysis. In this method, the impure metal is made to act as anode. A strip of

the same metal in pure form is used as cathode. Both anode and cathode are placed in a suitable electrolytic bath containing soluble salt of the same metal. On passing the current, metal ions from

the electrolyte are deposited at the cathode in the form of pure metal while equivalent amount of metal dissolves from the anode into the electrolyte in the form of metal . The impurities fall down below the anode as anode mud. The reaction occurring at the electrodes are :

At cathode: Mn+ + ne^-  →M

At anode: M → Mn+ + ne^-

Copper is refined using an electrolytic method as shown in figure. In this method crude copper is made anode, a thin sheet of pure copper is made cathode and acidified solution copper sulphate is used as an electrolyte. On passing the electric current, metal ions from the electrolyte are deposited at the cathode in the form of pure metal. On the other hand, an equivalent amount of metal dissolves from the anode into the electrolyte in the form of metal ions. The reactions occurring at electrodes are

At cathode: Cu²⁺ + 2e^- → Cu

At anode: Cu →Cu²⁺ + 2e^-

 (ii) Zone refining. This method is used for metals  which are required in very high purity,

For example, extremely pure silicon, germanium, boron, gallium and indium are refined by this

method. This method is based on the principle that the impurities are more soluble in the melt

than in the solid state of the metal. Therefore, an impure metal on solidification will deposit

crystals of pure metal and the impurities will remain behind in the molten part of the metal

In this method, the impure metal is cast into a thin bar. A circular mobile heater is fixed at

one end of the rod of impure metal. One  zone of the bar is melted by a circular mobile heater

in the atmosphere of a noble gas like argon. At the heated zone, the metal melts . As the heater moves slowly, the impurities also move into the adjacent molten part.

In this way, the impurities are made to move into one end which is finally cut off and discarded,

The molten metal present at the colder region solidifies in the  mean time since it is away from

the heater. Thus, we get completely pure metal by this method. This method is specially useful

producing semiconductors of very high purity. Germanium

(iii) Vapour phase refining. This method is based on the fact that certain metals are converted to their volatile compounds while the impurities are not affected during compound formation. The compound formed decomposes on heating to give pure metal.

For example, nickel is refined by this technique and the method is known as Mond’s process. In this method, nickel is heated in a steam of carbon monoxide to form volatile nickel carbonyl Ni(CO)4

The carbonyl vapours when subjected to still higher temperature (450-470 K) undergo thermal decomposition giving pure nickel.

Ni + 4CO          →Ni(CO)4               →Ni + 4CO

                                    Impure                     Nickelcarbonyl            Pure

Q.2.Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.

Ans. In blast furnace, reduction of iron oxides takes place in different temperature ranges.

The lower part of the blast furnace has high temperature of the order of 2200 K (called combustion zone) and the top of the furnace has low temperature of the order of 500-800 K (called reduction zone). The reduction occurring in the lower temperature range (upper part) , by carbon and in the higher temperature range (lower part) is by carbon monoxide.

At lower temperature range (500to 800 K) in upper part of furnace the reactions occurring are:

3Fe203 + CO      → 2Fe3O 4 + CO2

Fe3O 4 + 4 CO     →3 Fe + 4 CO2

Fe203 + CO          →2 FeO + CO2

At higher temperature range (900-1500 K) in lower part of furnace the reactions occurring ire:

C + C02 → 2CO

FeO + CO →Fe + CO2

In the middle portion (at about 1270 K, limestone decomposes to give lime (CaO) and CO2

Lime acts as a flux and combines with silicate impurity to form slag.

CaC03 → CaO + CO2

CaO        +       Si02        →          CaSiO3

                                              Lime                 Impurity              Calcium silicate (slag)

Slag is in the molten state and separates out from iron.

NOTE: Pig iron: The iron obtained from blast furnace is called pig iron. It is impure form of iron contains 4% carbon and small amount of S, P, Si and Mn. It can be casted into variety of shapes.

Cast iron: It is made by melting pig iron with scrap iron and contains 3% of carbon content. It is hard and brittle.

Malleable iron: It is the purest form of iron. It is also called malleable iron. It is prepared by oxidative refining of pig iron in reverberatory furnace lined with haematite which oxidizes carbon to carbon dioxide.

Fe203 + 3C → 2Fe + 3CO

Q.3. Write chemical reactions taking place in the extraction of zinc from zinc blende.

Ans. (i) The concentrated zinc blende ore (ZnS) is roasted in the presence of excess air about

1200  to convert it to zinc oxide.

2ZnS +3O2    →    2ZnO + 2S02

Zinc blende      Zinc oxide

(ii) Zinc oxide is reduced to zinc by heating with crushed coke at 1673 K

ZnO + C →Zn + CO

(iii) The impure copper is refined by electro refining method. In this method, the impure zinc is made anode and a plate of pure zinc is made cathode in an electrolytic bath containing zinc sulphate and a small amount of dilute H2SO4. On passing current, the following reaction occur:

At anode: Zn →  Zn²⁺ + 2e-

At cathode: Zn²⁺ + 2e- →Zn

The zinc gets deposited on cathode and is collected.

Q.4. How can you separate alumina from bauxite ore associated with silica? Give equations.

Ans. Bauxite, a principal ore of aluminium contains silica, iron oxide and titanium oxide

as impurities. The ore is digested with a concentrated solution of NaOH at 473- 523 K and

35-36 bar pressure. Al2O3 is leached out as sodium aluminate and silica as sodium silicate while.

impurities  are left behind.

Al2O3 (s) + 2NaOH(aq) + 3H20 → 2Na[AI(OH)4](aq)

The aluminate in the solution is neutralized by passing CO2 gas and hydrated Al203 is precipitated.

2Na[Al(OH)4] (aq) + CO2(g)   →   Al203.xH20 + 2NaHCO3(aq)

The sodium silicate remains in the solution while hydrated alumina is filtered and dried,

The hydrated alumina thus precipitated is filtered, dried and heated at 1473 K to give back pure alumina.

Al203.xH20 →  Al2O3(s) + xH20

                                                   Hydrated alumina     Alumina

Q.5. (a) What is the role of cryolite in the metallurgy of aluminium ?

(b) How is leaching carried out in case of low-grade copper ores?

(c) Why is zinc not extracted from zinc oxide -through reduction using CO?

Ans.(a) Cryolite is added to bauxite ore before electrolysis because of the following reasons:

(i) It acts as a solvent.

(ii) It lowers the melting point of alumina to about 1173 K.

(iii) Addition of cryolite to alumina increases the electrical conductivity.

(b) Copper is leached out from low grade copper by using acid in the presence of air when copper goes into the solution as Cu2+ ions.

Cu(s) + 2H⁺(aq) + 2O2(g) → Cu²⁺(aq) + H2O(l)

The solution containing Cu²⁺ ions is treated with scrap iron or H2.

Cu ²⁺(aq) + H2(g) → Cu(s) + 2H⁺(aq)

(c) In Ellingham diagram, the ∆G; vs T plot representing CO, CO2 lies above ∆G^o vs T plot of Zn, ZnO. Therefore, CO cannot act as reducing agent for the reduction of ZnO.

Q.7. What is hydrometallurgy? Explain with an example.

Ans. The process of extraction of metals by dissolving the ore in a suitable chemical reagent

and the precipitation of the metal by more electropositive metal is called hydrometallurgy.

For example, concentrated argentite, Ag2S is first treated with a dilute solution of NaCN to

form the soluble complex, sodium dicyanoargentate (I)       2Na[Ag(CN)2]

The solution is decanted off and made alkaline by adding NaOH and then treated with zinc

or aluminium to precipitate silver.

2Na[Ag(CN)2] + Zn →2Ag + Na[Zn(CN)4]

 Silver    Sodium tetracyanozincate(II)

Na[Zn(CN)4] + 4NaOH → NaZn02 + 4NaCN

Gold is also precipitated from its complex salt solution in a similar way.

2K[Au(CN)2] + Zn   →   K2[Zn(CN)4] + 2Au

Q.8. What is leaching process? How is it used for the concentration of silver and gold ores?

Ans. This is a chemical method of concentration and is useful in case the ore is soluble in suitable solvent. In this method, the powdered ore is treated with certain reagents in which the ore is soluble but the impurities are not soluble. The impurities left undissolved are removed by filtration.

This method is used to concentrate silver and gold ores. The ore containing native metal (silver or gold) is treated with a dilute solution (0.5%) of NaCN or KCN in the presence of atmospheric oxygen. The metal dissolves in the solution as a complex.

For example, For Ag2S (argentite), an ore of silver the reactions are :

Ag2S + 4NaCN        →     2Na[Ag(CN)2]         + Na2S

                                                                     Sodium dicyanoargentate (I)

The above solution after filtration and removing insoluble impurities is heated with zinc

to get silver.

2Na[Ag(CN)2]    +    ss Zn → Na2[Zn(CN)4]    +    2Ag

HIGHER ORDER THINKING SKILL QUESTIONS (HOTS)

1. Why do ore particles float in froth floatation process?

2. Why is the reduction of metal oxide difficult if metal formed is in solid state at the temperature    of reduction?

3. What is flux used in extraction of copper and why?

4. A metal is used in making bodies of aero planes. It is extracted by electrolytic reduction of its oxide ore. It does not occur in free state. Identify the metal.

5. What happens when cinnabar is roasted? Give reason for the products formed.

6. What are depressants? How would you separate zinc sulphide (ZnS) from PbS ore?

7. How is zirconium purified? Name the process.

8. Why is chalcosite (CuzS) roasted and not calcined in recovery of copper?

9. When the ore haematite is burnt in air with coke at 2000°Calong with lime, the process not only

produces steel but also produces silicate slag that is useful in making building materials like cement

Discuss it through balanced equations.

10.                                'X'     Roasting.   Y+ Z  + S02(g)

                             powedered coke

                            +Flux (Si02)

                                smelting .

                                               Molten matte           ^{Bassemerisatton} .             Crude metal

                                                   + slag   

Identify 'X', 'Y', 'Z' and crude metal formed and write chemical equations involved.

UNSOLVED QUESTIONS

VSA TYPE QUESTIONS (1 MARK)

1. Name three metals which occur in native state in nature.

2. What are collectors in froth flotation process? Give one example.

3. Give the names and formulae of three ores which are concentrated by froth floatation process.

4. Among Fe, Cu, Al and Pb, which metal (s) can not be obtained by smelting.

5. What is the thermodynamic criteria for the feasibility of a reaction?

6. Why CO is a better reducing agent than C at 673 K?

7. Indicate the temperature at which carbon can be used as a reducing agent for FeO.

[Ans. : Refer Ellingham diagram T > 1123 K]

8. Why aluminium cannot be reduced by carbon?

[Hint : Al is stronger reducing agent than carbon]

9. Name the most important form of iron. Mention its one use.

10. Name the impurities present in bauxite ore.

11. What is the composition of Copper matte?

12. Which from of copper is called blister copper?

13. What are froth stabilizers? Give two examples.

14. A sample of galena is contaminated with zinc blende. Name one chemical which can be used    

      to concentrate galena selectively by froth floatation method. [Ans. : NaCN]

15. What does a steep increase in the slope of a line on Ellingham diagram indicates.

16. What are the constituents of German silver?

17. Why is froth floatation process selected for concentration of the sulphide ore?

[Ans. : Sulphide ore particles are welted by oil (pire oil) and gangue particles by water]

18. Which form of iron is used in making anchors, chains and agricultural implements?

[Ans. : Wrought Iron]

19. Write the reaction involved in the extraction of copper from low grade ores.

[Ans. : First Step is leaching of ore wrong acind or bacteria then

Cu2+ (aq) + H2 (g) Cu(s) + 2H+ (g)]

20. Although aluminium is above hydrogen in the electrochemical series, it is stable in air and water. Why?

21. Zinc is used but not copper for the recovery of metallic silver from the complex [Ag(CN)₂]–,

although electrode potentials of both zinc and copper are less than that of Ag. Explain why?

SA (I) QUESTIONS (2 MARKS)

22. What is hydrometallurgy? Give one example where it is used for metal extraction.

23. Name the process for the benefaction/concentration of (i) an ore having impurities lighter  

      than it (ii) Sulphide ores.

24. What is cryolite? Mention its use in the extraction of aluminium.

25. What is the role of following :

(a) SiO2 in the metallurgy of Cu.

(b) CaCO3 in the metallurgy of Fe.

26. Extraction of copper directly from sulphide ore is less favourable than from its oxide through

reduction. Explain. ∆G value is more –ve in second case as compared with first case.

27. The graphite electrodes in the extraction of ‘Al’ by Hall-Heroult process need to be changed

frequently. Why?

28. Write the chemical formulae of the following ores (a) Haematite (b) Magnetite (c) Limonite

(d) Siderite.

29. Give equations for the industrial extraction of zinc from calamine.

30. Name the elements contained in anode mud during refining of copper. Why does it contain such elements?

31. What kind of elements are suitable for purification by Chromotography?

32. Write the Chemical reactions taking place in different zones in the blast furnace for the extraction of iron from its ore.

33. How are impurities separated from bauxite ore to get pure alumina?

34. Why is the reduction of a metal oxide easier if metal formed is in liquid state at the temperature of radiation?

35. Name the alloying element added to iron for making

(i) steel used in cutting tools and crushers.

(ii) steel used in making cables, measuring tapes and aeroplane parts.

36. What is pyrometallurgy? Explain with one example.

37. Write the method to produce Copper matte from copper pyrites.

38. Copper can be extracted by hydrometallurgy but not zinc. Explain why?

39. Free energies of formation Gf of MgO(s) and CO(g) at 1273K and 2273 K are given below:

∆Gf [MgO(s)] = –941 KJ mol–1 at 1273 K.

∆Gf [CO(g)] = –439 KJ mol–1 at 1273 K.

∆Gf [MgO(s)] = –314 KJ mol–1 at 2273 K.

∆Gf [CO(g)] = –628 KJ mol–1 at 2273 K.

On the basis of above data, predict the temperature at which carbon can be used as a reducing

agent for MgO(s).

SA (II) TYPE QUESTIONS (3 MARKS)

40. State the principles of refining of metal by the following methods.

(a) Zone refining (b) Electrolytic refining (c) Vapour phase refining.

41. How is pure copper obtained from its principle ore? Write the chemical reactions occurring during the extraction.

42. Name the method of refining of the following metals – (a) Hg (b) Sn (c) Cu (d) Ge (e) Ni (f) Zr

43. Suggest a condition under which : (i) Mg can reduce alumina (Al2O3)

 (ii) Al can reduce MgO.

*44. The native silver forms a water soluble compound (B) with dilute aqueous solution of NaCN in the

presence of a gas (A). The silver metal is obtained by the addition of a metal (C) to (B) and

complex (D) is formed as a byproduct. Write the structures of (C) and (D) and identify (A) and

(B) in the following sequence –

Ag + NaCN + [A] + H2O [B] + OH– + Na+.

[C] + [B] [D] + Ag.

[Ans. : [A] = O2

[B] = Na [Ag(CN)2]

[C] = Zn

[D] = Na2 [Zn (CN)4] ].