Home | | Chemistry 12th Std | Metallurgy: Answer the following questions

Book Back and Important Questions Answers | Chemistry - Metallurgy: Answer the following questions | 12th Chemistry : UNIT 1 : Metallurgy

Chapter: 12th Chemistry : UNIT 1 : Metallurgy

Metallurgy: Answer the following questions

Chemistry: Metallurgy: Book Back Important Questions, Answers, Solutions : Answer briefly, Answer in detail, Exercise Numerical Problems

Chemistry : Metallurgy

Answer the following questions:

1.        What are the differences between minerals and ores?


Mineral

• A naturally occurring substance which contains the metal in free state or in the form of compounds

• Low percentage of metal present

• All minerals are not ores

Ore

• Minerals that contain a high percentage of metal, from which it can be extracted conveniently and economically

• Large percentage of metal present

• All ores are minerals

 

2.     What are the various steps involved in extraction of pure metals from their ores?

(i) Concentration of the ore

(ii) Extraction of crude metal

(iii) Refining of crude metal


3.     What is the role of Limestone in the extraction of Iron from its oxide Fe2O3 ?

The silica gangue present in the iron ore is acidic in nature. Silica removed by using basic flux, limestone (CaO). The limestone combines with silica gangue to form calcium silicate (slag). The slag was removed.

CaCO3 ___Δ__→ CaO + CO2

CaO(s) [Flux] + SiO2(S) [Gangue] → CaSiO3(S) [Slag]

Flux + Gangue → Slag

 

4.     Which type of ores can be concentrated by froth floatation method? Give two examples for such ores.

Sulphide ores can be concentrated by froth flotation method.

Example: Zinc blende (ZnS), Silver glance (Ag2S)


5.     Out of coke and CO, which is better reducing agent for the reduction of ZnO? Why?

Coke (C) is a better reducing agent for the reduction of ZnO. From the Ellingham diagram, the free energy for the formation of CO from C is lower at above 1120 K while that of CO2 from carbon is lower at above 1320 K than free energy of formation of ZnO. However, the free energy of formation of CO2 from CO is always higher than that of ZnO.


6.     Describe a method for refining nickel.

The impure nickel is heated in a stream of carbon monoxide at around 350 K. The nickel reacts with the CO to form a highly volatile nickel tetracarbonyl. The solid impurities are left behind.

Ni (s) + 4 CO (g) → Ni(CO)4 (g)

The nickel tetracarbonyl heated around 460 K, the complex decomposes to give pure nickel.

Ni(CO)4 (g) → Ni (s) + 4 CO (g)

 

7.     Explain zone refining process with an example using the Ellingham diagram given below.

Zone refining process:

i) This method is based on the principle of fractional crystallisation. When an impure metal is melted and allowed to solidify, the impurities are more soluble in the melt than in the solid state metal.

ii) The impure metal is taken in the form of a rod. One end of the rod is heated using a mobile induction heater which results in melting of the metal on that portion of the rod.

iii) When the heater is slowly moved to the other end, the pure metal crystallises while the impurities will move on to the adjacent molten zone formed. As the heater moves further away, the impurities also moves along with it.

iv) The process is repeated several times to achieve the pure element.

v) This process is carried out in an inert gas atmosphere to prevent the oxidation of metals.

Example:

Elements such as germanium (Ge), silicon (Si) and galium (Ga) that are used as semiconductor are refined using this process.


8.     (A) Predict the conditions under which

(i) Aluminium might be expected to reduce magnesia.

In the Ellingham diagram, above 1623 K, the ∆G° value for Al2O3 is more negative than that of MgO. Thus ∆G° of the reaction is negative.

Therefore above 1623 K, Al can reduce MgO to Mg.

 3MgO + 2Al  ___1623 K__→  Al2O3 + 3Mg

(ii) Magnesium could reduce alumina.

In the Ellingham diagram, below 1623 K the ∆G° value of Al2O3, is less negative than that of MgO. Thus, ∆G°of the reaction is negative. Therefore, below 1623 K, Mg can reduce Al2O3 to Al.

Al2O3 + 3Mg ___below 1623__→ 3MgO + 2Al

(B) Carbon monoxide is more effective reducing agent than carbon below 983K but, above this temperature, the reverse is true –Explain.

The value of ∆G° for change of C to CO2 is less than the value of ∆G° for change of CO to CO2. Therefore, coke (C) is a better reducing agent than CO at 983 K or above this temperature. However below this temperature, CO is more effective reducing agent than C.

(C) it is possible to reduce Fe2 O3 by coke at a temperature around 1200K

Yes, it is possible to reduce Fe2O3 by coke at a temperature around 1200K . Ellingham diagram for the formation of FeO and CO intersects around 1000 K. Below this temperature, the carbon line lies above the iron line which indicates that FeO is more stable and the reduction is thermodynamically not feasible. Around 1200 K carbon line lies below the iron line and hence, coke can be used as reducing agent.

Fe2O3 + C → 2Fe + 3CO 


9.     Give the uses of zinc.

Metallic zinc is used in galvanising metals.

It is used to produce die-castings in the automobile, electrical and hardware industries.

Zinc oxide is used in the manufacture of many products such as paints, rubber, cosmetics, pharmaceuticals, plastics, inks, batteries, textiles and electrical equipment.

Zinc sulphide is used in making luminous paints, fluorescent lights and X-ray screens.

Brass an alloy of zinc is used in water valves and communication equipment as it is highly resistant to corrosion.


10. Explain the electrometallurgy of aluminium.

Hall-Heroult process:

Cathode: Iron tank lined with carbon

Anode: The carbon blocks immersed in the electrolyte

A 20% solution of alumina, obtained from the bauxite ore is mixed with molten cyrolite and is taken in the electrolysis chamber.

About 10% calcium chloride is also added to the solution, which lowers the melting point of the mixture.

The fused mixture is maintained at a temperature of above 1270 K.

Ionisation of alumina

Al2O3 → 2Al3+ + 3O2−

Reaction at cathode

Al3+ (melt) + 3e → Al (1)

Reaction at anode

2O2− (melt) → O2 + 4e

Carbon anode consumed slowly.

C(s) + O2− (melt) → CO + 2e

C(s) + 2O 2− (melt) → CO2 + 4e

The pure aluminium is formed at the cathode and settles at the bottom. The net electrolysis reaction is

4Al3+ (melt) + 6O2(melt) + 3C(s) → 4Al() + 3CO2(g)


11. Explain the following terms with suitable examples.

(i) Gangue (ii) slag

Gangue:

The ores are associated with nonmetallic impurities, rocky materials and siliceous matter which are collectively known as gangue.

Example:

The silica gangue present in the iron ore

Slag:

When gangue present in the roasted or calcined ore combined with the flux forms a fusible material called slag.

Example:

Silica gangue in the iron ore removed by using limestone (CaO). The slag was removed.

CaO(s) [Flux] + SiO2(s) [Gangue] → CaSiO3(s) [Slag]

Flux + Gangue → Slag 


12. Give the basic requirement for vapour phase refining.

The metal is treated with a suitable reagent it should form a volatile compound with the metal.

The volatile compound is easily decomposed to give the pure metal.


13. Describe the role of the following in the process mentioned.

(i) Silica in the extraction of copper.

(ii) Cryolite in the extraction of aluminium.

(iii) Iodine in the refining of Zirconium.

(iv) Sodium cyanide in froth floatation.

(i) Silica in the extraction of copper.

Silica is used as flux material in the extraction of copper.

The concentrated ore is heated with an acidic flux silica. The ferrous oxide formed due to melting is basic in nature and it combines with silica to form ferrous silicate (slag).

FeO + SiO2 → FeSiO3

(ii) Cryolite in the extraction of aluminium.

(a) The melting point of alumina is very high.

Hence it is mixed with cryolite (Na3AlF6) which lowers its melting point

(b) Cryolite increase the electrical conductivity of alumina.

(c) The function of cryolite is to lower the fusion temperature.

(iii) Iodine in the refining of Zirconium.

The impure zirconium metal is heated with iodine at a temperature of 550 K to form the volatile zirconium tetra-iodide. The impurities are not reacting with iodine. The volatile tetraiodide vapour is passed over a tungsten filament at high temperature and pure zirconium was obtained.

Zr(s) + 2I2 (s) __ 550 K_→ ZrI4 (vapour)

ZrI4 (vapour) ___1800 K_→ Zr (s) + 2I2 (s)

(iv) Sodium cyanide in froth floatation

Sodium cyanide (NaCN) is added to depress the floatation property.

When a sulphide ore of a metal contains other metal sulphides as impurities, depressing agent sodium cyanide is used to selectively prevent other metal sulphides from coming to the froth.


14. Explain the principle of electrolytic refining with an example.

The crude metal is refined by electrolysis.

Anode - Impure metal rod

Cathode - Thin strips of pure metal

The metal dissolves from the anode, pass into the solution while the same amount of metal ions from the solution will be deposited at the cathode. The insoluble impurities in the anode settles at the bottom of the anode as anode mud.

Example: Electrolytic refining of silver

Cathode : Pure silver

Anode : Impure silver rods

Electrolyte : Silver nitrate in HNO3

Reaction at cathode

Ag+ (aq) + e → Ag(s)

Reaction at anode

Ag(s) → Ag+ (aq) + e

During electrolysis, at the anode the silver atoms lose electrons and enter the solution. The positively charged silver cations migrate towards the cathode and get discharged by gaining electrons and deposited on the cathode. The impurities settle down at the bottom of anode as anode mud. 


15. The selection of reducing agent depends on the thermodynamic factor: Explain with an example.

The extraction of metals from their oxides can be carried out by using different reducing agents.

2/y MxOy(s) → 2x / y M(s) + O2(g)

The above reduction may be carried out with carbon. The reducing agent carbon may be oxidised to either CO or CO2.

C + O2 → CO2

2 C + O2 → 2 CO

If carbon monoxide is used as a reducing agent, it is oxidised to CO2 as follows

2CO + O2 → 2 CO2

A suitable reducing agent is selected based on the thermodynamic considerations. The change in free energy (∆G) should be negative, for a spontaneous reaction. Therefore, thermodynamically, the reduction of metal oxide with a given reducing agent can occur if the free energy change for the coupled reaction is negative. Hence, the reducing agent is selected in such a way that it provides a large negative ∆G value for the coupled reaction.


16. Give the limitations of Ellingham diagram.

i) Ellingham diagram is constructed based only on thermodynamic considerations. It gives information about the thermodynamic feasibility of a reaction. It does not tell anything about the rate of the reaction. Moreover, it does not give any idea about the possibility of other reactions that might be taking place.

ii) The interpretation of ∆G is based on the assumption that the reactants are in equilibrium with the product which is not always true.


17.   Write a short note on electrochemical principles of metallurgy.

The reduction of oxides of active metals by carbon is thermodynamically not feasible. Such metals are extracted by using electrochemical methods. The metal salts are taken in a fused form or in solution form. The metal ion present can be reduced by treating it with some suitable reducing agent or by electrolysis.

Gibbs free energy change for the electrolysis process is given by the following expression

∆G° = −nFE°

Where n is number of electrons involved in the reduction process

F is the Faraday

E° is the electrode potential of the redox couple.

If E° is positive then the ∆G is negative and the reduction is spontaneous. When a more reactive metal is added to the solution containing the relatively less reactive metal ions, the more reactive metal will go into the solution.

For example,

Cu (s) + 2Ag+ (s) → Cu2+ (aq) + 2Ag(s)

Cu+2 (aq) + Zn(s) → Cu (s) + Zn+2 (aq)


EVALUATE YOURSELF:

 

1. Write the equation for the extraction of silver by leaching with sodium cyanide and show that the leaching process is a redox reaction.

In the metallurgy of silver metal is leached with a dilute solution of NaCN in the presence of air (O2).

4Ag + 8CN+ 2H2O + O2 → 4[Ag(CN)2]+ 4OH

In this reaction, Ag → Ag+ oxidation number of Ag increases from 0 to +1, hence oxidation

O2 → OH (oxidation number of oxygen decreases from 0 to −2, hence reduction)

Hence Leaching of silver is a redox reaction.

 

2. Magnesite (Magnesium carbonate) is calcined to obtain magnesia, which is used to make refractory bricks. Write the decomposition reaction.

Magnesite (Magnesium carbonate) is heated in the absence of oxygen decomposes to form Magnesium oxide (Magnesia)

MgCO3 → MgO + CO2

 

3. Using Ellingham diagram indicate the lowest temperature at which ZnO can be reduced to Zinc metal by carbon. Write the overall reduction reaction at this temperature.

Ellingham diagram for the formation of ZnO and CO intersects around 1233 K below this temperature, Carbon line lies above Zinc line. Hence ZnO is more stable than CO so the reduction is thermodynamically not feasible at this temperature range.

However above 1233 K carbon line lies below the zinc line, hence carbon can be used as a reducing agent above 1233 K.

2Zn + O2 → 2ZnO ………….. (1)

2C + O2 → 2CO …………… (2)

Reversing (1) and adding with equation (2)

2ZnO → 2Zn + O2

2C +O2 → 2CO

2ZnO + 2C → 2Zn + 2CO

ZnO + C → Zn + CO

 

4. Metallic sodium is extracted by the electrolysis of brine (aq.NaCl). After electrolysis the electrolytic solution becomes basic in nature. Write the possible electrode reactions.

Sodium metal is prepared by Down's process. This involves the electrolysis of fused NaCl and CaCl2 at 873K. During electrolysis sodium is discharged at the cathode and Cl2 is obtained at the anode.

NaCl(l) → Na+(melt) + Cl(melt)

Cathode : Na+(melt) + e→ Na(s)

Anode : 2Cl(aq) → Cl2(g) + 2e

If an aqueous solution of NaCl is electrolysed, H2 is evolved at cathode and Cl2 is evolved at anode. NaOH is obtained in the solution.

NaCl(aq) ___Electrolysis­_→ Na+(aq) + Cl(aq)

Cathode : 2H2O(I) + 2e → H2(g) + 2OH(aq)

Anode : Cl(aq) → 1/2 Cl2(g) + 2e

Na+ and OH ions to form NaOH

Hence solution is basic in nature.

 

Tags : Book Back and Important Questions Answers | Chemistry , 12th Chemistry : UNIT 1 : Metallurgy
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
12th Chemistry : UNIT 1 : Metallurgy : Metallurgy: Answer the following questions | Book Back and Important Questions Answers | Chemistry


Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright © 2018-2024 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.