General Principles and Processes of Isolation of Elements
MCQs
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
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
Explain Zone refining.
Answer: 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.
Answer: 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.
Explain Column chromatography.
Answer: 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.
Answer: 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.
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.
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.
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.
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.
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 → 2FeO + 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
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 → 2FeO + 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
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.
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.
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.
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.
What is the role of cryolite in the metallurgy of aluminium ?
Answer: Cryolite (Na3AlF6) has two roles in the metallurgy of aluminium. They are-
to decrease the melting point of the mixture from 2323 K to 1140 K.
to increase the electrical conductivity of Al2O3.
Answer: Cryolite (Na3AlF6) has two roles in the metallurgy of aluminium. They are-
to decrease the melting point of the mixture from 2323 K to 1140 K.
to increase the electrical conductivity of Al2O3.
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)
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)
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.
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.
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) → CO(g) + 2e-
C(s) + 2O2-(melt) → CO2(g) + 4e-
At cathode:-
Al3+ (melt) + 3e- → Al(l)
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) → CO(g) + 2e-
C(s) + 2O2-(melt) → CO2(g) + 4e-
At cathode:-
Al3+ (melt) + 3e- → Al(l)
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.
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.
