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1. Copper can be extracted by hydrometallurgy but not zinc. Explain.
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Solution:
Hydro metallurgy method is based on the fact that more electreopositive metals displace less
electropositive metals from their salt solution. Copper is precipitated From copper sulphate solution
by adding iron.
CuSO4 + Fe → FeSO4 + Cu↓
But zinc is not precipitated from zinc sulphate solution by adding iron. Hence copper can be extracted
by hydro metallurgy but not zinc.
2. What is the role of depressant in froth floatation process?
•
Solution:
The role of depressant in froth floatation process is to prevent certain type of materials from forming
the froth with bubbles.
3. Why is the extraction of copper from pyrites more difficult than that
from its oxide ore through reduction?
•
Solution:
In the graph of ∆rG0 Vs T for the formation of oxides, the Cu2O line is almost at the top. So it is quite
easier to reduce oxide ores of copper directly to the metal by heating with coke, (both the lines of
C,CO and C,CO2 are at much lower positions in the graph particularly after 500-600K).
However sulphide ores are roasted/smelted to give oxides.
2Cu2S + 3O2 → 2Cu2O + 2SO2
The oxide can then be easily reduced to metallic copper using coke.
Cu2O + C → 2Cu + CO
4. Explain: (i) Zone refining (ii) Column chromatography.
•
in
Solution:
(i) Zone Refining
This method is employed for preparing highly pure metal (silicon, tellurium. germanium), which are
used as semiconductors. It is based on the principle that melting point of a substance is lowered by
the presence of impurities. Consequently, when an impure molten metal is cooled, crystals of the
pure metal are solidified, and the impurities remain cooled, crystals of the pure metal are solidified,
and the impurities remain behind the remaining metal.
1
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The process consists in casting the impure metal in the form of a bar. A circular heater fitted around
this bar is slowly moved longitudinally from one end to the other. At the heated zone, the bar melts,
and as the heater moves on, oure metal crystallizes, while the impurities are swept from one end of
the bar to the other. By repeating the process, ultra pure metal can be obtained.
(ii) Column Chromatography
Column chromatography in chemistry is the preparative application of chromatography. It is used to
obtain pure chemical compounds from a mixture of compounds on a scale from micrograms up to
kilograms using large industrial columns.
in
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Column chromatography on a large scale in the 1950s, the chemist uses a ladder to refill eluent, he
operates not one but 11 columns.
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The classical preparative chromatography column is a glass tube with a diameter from 5 to 50 mm
and a height of 50 cm to 1 m with a tap at the bottom. A slurry is prepared of the eluent with the
stationary phase powder and then carefully poured into the Column. Care must be taken to avoid air
bubbles. A solution of the organic material is pipetted on top of the stationary phase. This layer is
usually topped with a small layer of sand or with cotton or glass wool to protect the shape of the
organic layer from the velocity of newly added eluent. Eluent is slowly passed through the Column to
advance the organic material. Often a spherical eluent reservoir or an eluent-filled and stoppered
separating funnel is put on top of the Column.
The individual components are retained by the stationary phase differently and separate from each
other while they are running at different speeds through the Column with the eluent. At the end of
the Column they elute one at a time. During the entire chromatography process the eluent is
collected in a series of fractions. The composition of the eluent flow can be monitored and each
fraction is analyzed for dissolved compounds, e.g. by analytical chromatography, UV absorption, or
fluorescence. Coloured compounds (or fluorescent compounds with the aid of an UV lamp) can be
seen through the glass wall as moving bands.
Stationary Phase (Adsorbent)
The stationary phase or adsorbent in Column chromatography is a solid. The most common
stationary phase for Column chromatography is -C18H37, followed by alumina. Cellulose powder has
often been used in the past. The stationary phases are usually finely ground powders or gels and/or
are microporous for an increased surface.
Mobile Phase (Eluent)
The mobile phase or eluent is either a pure solvent or a mixture of different solvents. It is chosen so
that the retention factor value of the compound of interest is roughly around 0.75 in order to
minimize the time and the amount of eluent to run the chromatography. The eluent has also been
chosen so that the different compounds can be separated effectively.
A faster flow rate of the eluent minimizes the time required to run a Column and thereby minimizes
diffusion, resulting in a better separation. A simple laboratory Column runs by gravity flow.
Extending the fresh eluent filled Column above the top of the stationary phase or decreased by the
tap controls can increase the flow rate of such a Column. Better flow rates can be achieved by using
a pump or by using compressed gas (e.g. air, nitrogen, or argon) to push the solvent through the
Column (flash Column chromatography).
5. Out of C and CO, which is a better reducing agent at 673 K?
in
•
Solution:
CO acts as a better reducing agent between 500 – 800K and C acts as a better reducing agent
between 900 – 1500K.
6. Name the common elements present in the anode mud in
electrolyticrefining of copper. Why are they so present?
•
Solution:
Impurities from the blister copper deposit as anode mud which contains antimony, selenium,
tellurium, silver, gold and platinum; recovery of these elements may meet the cost of refining.They
are so present because they are less electropositive metals while more electropositive elements
remain in the solution.
3
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7. Write down the reactions taking place in different zones in the blast
furnace during the extraction of iron.
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Solution:
The following reactions takes place in the furnace during the extraction of iron:
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Combustion zone
C + O2 → CO2; ∆H = - 393.5 kJ
The heat produced raises the temperature to about 1775K.
Fusion zone
CO2 + C → 2CO; ∆H = + 163.2 kJ
This is an endothermic reaction and therefore, the temperature is lowered to 1475 - 1575 K. The iron
produced in the upper region melts here. Any Fe2O3 if present undergoes reduction by hot coke to
iron. This region is called fusion zone.
Fe2O3 + 3C → 2Fe + 3CO + Heat
Slag formation zone
The temperature is about 1075 to 1275K.
CaCO3 → CaO + CO2
(Limestone)
CaO + SiO2 → CaSiO3
(lime) (Calcium silicate - slag)
The molten slag forms a separate layer (being lighter) above the molten iron. This region is called
slag formation zone.
Reduction zone
The temperature near the top of the furnace is of the order of 875K. The oxides of iron are reduced
by carbon monoxide to iron.
Fe2O3 + 2CO → 2FeO + CO2
FeO + CO → Fe + CO2
This region of the furnace is called reduction zone.
8. Write chemical reactions taking place in the extraction of zinc from
zinc blende.
•
Solution:
Zinc can be extracted from zinc blende by the following steps:
in
1.Concentration
The ore is concentrated by froth floatation process.
2. Roasting
Concentrated ore is roasted in the presence of excess of air at about 1200K to convert zinc zulphide
into zinc oxide.
2ZnS + 3O2 → 2ZnO + 3SO2
Zinc blende Zinc oxide
3. Reduction
Zinc oxide is reduced to zinc by heating with crushed coke at 1673 K in vertical fire clay retorts.
ZnO + C → Zn + CO
The vapours of the zinc formed are collected and condensed.
4
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4. Refining
The impure metal is refined by fractional distillation or by electrolytic method.
9. State the role of silica in the metallurgy of copper.
Solution:
Silica is used as a flux in the metallurgy of copper.It is used to convert the formed Ferrous oxide to
Ferrous silicate (slag) during the extraction of copper from copper pyrites (CuFeS2).
The reaction is written as follows:
FeO + SiO2 → FeSiO3
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10. What is meant by the term "chromatography"?
•
Solution:
Chromatography (from Greek :chroma, colour and "grafein" to write) is the collective term for a
family of laboratory techniques for the separation of mixtures. It involves passing a mixture
dissolved in a "mobile phase" through a stationary phase, which separates the analyte to be
measured from other molecules in the mixture and allows it to be isolated.
Chromatography is a physical method of separation in which the components to be separated are
distributed between two phases, one of which is stationary (stationary phase) while the other (the
mobile phase) moves in a definite direction.
11. What criterion is followed for the selection of the stationary phase in
chromatography?
•
Solution:
Criteria for the selection of the stationary phase:
1. The stationary phase must be immobile and immiscible.
2. The stationary phase must dissolve or must have atleast little affinity towards any of the
components in the mixture to be analysed.
12. Describe a method for refining nickel.
•
in
Solution:
The carbonyl method is used for the refining o0f metals like nickel and iron. For example, in case of
nickel the impure metal is heated with CO. The nickel carbonyl thus formed is then decomposed
(after distilling off the impurities) to get pure nickel metal and Co. The process is known as Mond's
process. This is a best example for vapour phase refining.
Ni + 4CO → Ni (CO)4 → Ni + 4CO
It is based on the following facts:
Only nickel (and not Cu, Fe, etc) forms a volatile carbonyl Ni (CO)4, when CO is passed over it at
50° C.
The nickel carbonyl decomposes at 180° C to yield pure nickel.
13. How can you separate alumina from silica in a bauxite ore associated
with silica? Give equations, if any.
•
Solution:
5
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To separate silica impurity from bauxite,it is treated with a conc. Solution of NaOH at 473-520K and
35 bar pressure. The Al2O3 in bauxite dissolves forming sodium aluminate and silica is left behind. It
is separated by filtration.
Al2O3 +2NaOH+3H2O -------- 2Na[Al(OH)4]aq.
Alternatively,
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Bauxite ore which contains silica as one of the impurity is called white bauxite.
The white bauxite is purified by Serpeck's process.
In this process, the ore is powdered, mixed with coke and heated to 2073 K in an atmosphere of
nitrogen. Alumina changes to aluminium nitride while silica gets reduced to Silicon which being
volatile is removed as vapours.
Al2O3 + N2 + 3C → 2AlN + 3CO
Alumina
Aluminium nitride
SiO2 + 2C →€Si + 2CO
Silica
14. Giving examples differentiate between ‘roasting’ and ‘calcination’.
•
Solution:
Roasting
1. It is the method where the sulphide ore is
converted to its metallic oxide by subjecting it to
the action of heat in excess of air at a
temperature below its melting point.
Calcination
It is the method where the carbonate or hydroxide
ore is converted to its metallic oxide by subjecting
it to the action of heat in the absence of air at a
temperature below its melting point.
2. Eg. 2ZnS + 3O2 → 2ZnO + 3SO2€
CaCO3 → CaO + CO2
zinc blende
(Limestone)
2PbS + 3O2 → 2PbO + 3SO2
CuCO3.Cu(OH)2 → 2CuO + H2O + CO2
Galena
Malachite
15. How is ‘cast iron’ different from ‘pig iron"
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Solution:
o
o
The iron obtained from blast furnace is pig iron. It has about 4% of carbon content. It has other
impurities like S,P,Mn.
Cast iron is made by heating pig iron with scrap iron and coke in hot air.The carbon content is
reduced to 3%.. It is hard and brittle.
16. Differentiate between "minerals" and "ores".
•
Solution:
Minerals
Ores
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When a mineral contains sufficient amount of a
metal in the combined state, from which it can be
readily and profitably separated on commercial
scale, then the mineral is said to be an ore.
Aluminium can be profitably extracted only from
bauxite and not from the clay.
Hence bauxite is an ore of aluminium and not
clay.
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1. The natural material in which the metal or their
compounds occur in the earth is known as
mineral. A mineral can be a single compound or a
complex mixture of various compounds.
2. For example, clay (Al2O32SiO22H2O) and
bauxite (Al2O32H2O) are two minerals of
aluminium.
3. All minerals are not ores.
All ores are minerals.
17. Why copper matte is put in silica lined converter?
•
Solution:
A mixture containing sulphide of copper and iron is called matte. Since matte can form only basic
oxides like FeO, it is put in silica (acidic oxide) lined converter where silica can act as a flux for FeO
and forms slag.
FeO + SiO2 → FeSiO3
Ferrous oxide Silica Ferrous silicate (slag).
18. What is the role of cryolite in the metallurgy of aluminium?
•
Solution:
Aluminium is obtained by the electrolysis of fused aluminium oxide. Since fused aluminium oxide
(alumina) is not an electrolyte, it is made an electrolyte by dissolving in fused cryolite. A molten
cryolite bath is maintained at 950° C - 1000° CAnd alumina is dumped periodically into each cell to
keep a 3-5% Al2O3 concentration.
Cryolite reduces the melting point of alumina and makes it more conducting.
19. How is leaching carried out in case of low grade copper ores?
•
in
Solution:
Leaching is the process of extracting a substance from a solid by dissolving it in a suitable liquid.
Leaching of sulfides is a more complex process due to the refractory nature of sulfide ores. It often
involves the use of pressurized vessels, called autoclaves. A good example of the autoclave leach
process can be found in the metallurgy of copper ores. It is best described by the following chemical
reaction:
2CuS + O2 + 2H2SO4 → 2CuSO4 + 2H2O + 2S
This reaction proceeds at temperatures above the boiling point of water, thus creating a vapour
pressure inside the vessel. Oxygen is injected under pressure, making the total pressure in the
autoclave more than 0.6 MPa.
20. Why is zinc not extracted from zinc oxide through reduction using
CO?
•
Solution:
The standard free energy of formation of CO2 from CO is higher than that of Zn to ZnO. Hence Co
cannot be used as reducing agent.
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The process of extraction of zinc is carried out at a temperature of above 1200K where Carbon
serves its best purpose of being a reducing agent. Carbon monoxide is not preferable as a reducing
agent at this temperature.
21. The value of ∆fG0 for formation of Cr2O3 is – 540 kJmol-1and that of
Al2O3 is – 827 kJmol-1Is the reduction of Cr2O3 possible with Al?
Solution:
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2Al + 3/2 O2 --------- Al2O3. ∆f G0 =-827kJ/mol. (1)
2Cr +3/2 O2--------------------- Cr2O3. ∆f G0 =-540kJ/mol.(2)
(1) –(2) , gives, 2Al + Cr2O3 --- 2Cr +Al2O3 ∆r G0 =-287kJ.
Hence reduction of Cr2O3 is possible with Al.
22. Out of C and CO, which is a better reducing agent for ZnO?
•
Solution:
According to Ellingham diagram, the free energy of formation of CO from C becomes lower above
1000K while the free energy of formation of CO2from CO is always higher than that of Zn O from
Zn.So ,only C can be used as reducing agent and not CO.
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 example.
•
Solution:
The choice of a reducing agent in a particular case depends on thermodynamic factor. It can be
explained by having a following example.
For oxidation of the reducing agent Al, ∆fG0. is -827 kJmol-1.
For reduction of the metal oxide Cr2O3, ∆fG0 is -540 kJmol
The sum total of ∆fG0 for the two above mentioned processes is (-827+ 540) kJmol-1
= -287 kJmol-1
Since the sum total of ∆fG0 for the reduction of Cr2O3 to Cr and oxidation of the reducing agent Al to
Al2O3 is negative, Al can be used as a reducing agent for Cr2O3.
in
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?
•
Solution:
Chlorine is obtained as a byproduct in the following processes:
1. Down's process of extraction of sodium
2. Electrolysis of sea water.
8
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When an aqueous solution of sodium chloride is subjected to electrolysis, chlorine is formed as one of
the byproduct. The chemistry of the the electrolysis is explained as follows:
Water is reduced to hydrogen at the cathode. According to the electrochemical series, oxygen should
be produced at the anode As the oxygen over potential is greater than that of chlorine, the latter is
produced at the anode, particularly at high chloride concentrations.
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Cathode: 2 Na+ + 2 H2O + 2 e- → 2 Na+ + 2 OH- + H2
Anode:
2 Cl- → Cl2 + 2 e__________________________________________________
2 Na+ + 2 Cl- + 2 H2O → 2 Na+ + 2 OH- + H2 + Cl2
__________________________________________________
25. What is the role of graphite rod in the electrometallurgy of
aluminium?
•
Solution:
Graphite is used as anode in the electrometallurgy of aluminium.
When oxygen is liberated at anode, graphite reacts with it to form CO2.Otherwise the liberated
oxygen will react with aluminium to convert it back to Al2O3.Graphite rods are replaced periodically.
26. Outline the principles of refining of metals by the following methods:
(i) Electrolytic Refining
(ii) Vapour Phase Refining
•
Solution:
in
(i) Electrolytic Refining
This is one of the most convenient and important method of refining and gives a metal of high purity.
This method is applicable to many metals such as copper, silver, lead, gold, nickel, tin, zinc etc. The
blocks of impure metal form the anode and this sheets of metal form the cathode. A solution of a salt
of the metal is taken as an electrolyte. On passing an electric current through the solution pure metal
dissolves from the anode and deposits on the cathode. By this process, more metal ions undergo
reduction and pure metal is deposited at the cathode. The insoluble impurities either dissolve in the
electrolyte or fall at the bottom and collect as anode mud. For example, in the refining of copper,
impurities like Fe and Zn dissolve in the electrolyte, while Au, Ag and Pt are left behind as anode
mud.
(ii) Vapour Phase Refining
The carbonyl method is used for the refining o0f metals like nickel and iron. For example, in case of
nickel the impure metal is heated with CO. The nickel carbonyl thus formed is then decomposed
(after distilling off the impurities) to get pure nickel metal and Co. The process is known as Mond's
process. This is a best example for vapour phase refining.
Ni + 4CO →€Ni (CO)4 →€Ni + 4CO
It is based on the following facts:
Only nickel (and not Cu, Fe, etc) forms a volatile carbonyl Ni (CO)4, when CO is passed over it at
9
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50° C.
The nickel carbonyl decomposes at 180° C to yield pure nickel.
27. Predict conditions under which Al might be expected to reduce MgO.
Solution:
Below 1350°C Mg can reduce Al2O3 and above 1350°C, Al can reduce MgO.
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•
28. Which of the ores mentioned can be concentrated by magnetic
separation method?
•
Solution:
Ores in which one of the components (either the impurity or the actual ore) is magnetic can be
concentrated, e.g., ores containing iron (haematite, magnetite, siderite and iron pyrites).
29. What is the significance of leaching in the extraction of aluminium?
•
Solution:
Leaching is significant as it helps in removing the impurities like SiO2, Fe2O3,etc. from the bauxite
ore. When bauxite is leached with NaOH, only Al2O3 dissolves leaving behind, Fe2O3, which is filtered
and removed.
30. The reaction,
Cr2 O3 + 2 Al → Al2 O3 + 2 Cr (∆G0 = – 421 kJ) is thermodynamically feasible
as is apparent from the Gibbs energy value.
Why does it not take place at room temperature?
•
Solution:
Certain amount of activation energy is essential even for such reactions which are thermodynamically
feasible, therefore heating is required.
31. Is it true that under certain conditions, Mg can reduce SiO2 and Si
canreduce MgO? What are those conditions?
•
Yes, below 1350°C Mg can reduce SiO2 and above 1350°C, Al can reduce MgO
in
Solution:
Above this temperature, the free energy of formation curve of MgO lies above that of SiO2.Below this
temperature free energy of formation curve of SiO2 lies above that of MgO and so Mg can reduce
SiO2.
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