NCERT Solutions for Class 12 Chemistry Chapter 6 - General Principles and Processes of Isolation of Elements
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1. Copper can be extracted by hydrometallurgy but not zinc. Explain.
Ans. In hydrometallurgy, zinc and iron can be used to replace copper from their solution because reduction potential of zinc and iron are lower than that of copper.
$$\text{Fe}_s + \text{Cu}^{2+} \text{(aq)}\xrightarrow{} \text{Fe}^{2+}\text{(aq) + C}$$
But to replace zinc, more reactive metal i.e., metals having lower reduction potential than zinc such as Mg, Ca, K etc are required. However, all these metals react with water with the evolution of hydrogen gas. Thus, these metals cannot be used in hydrometallurgy to extract zinc.
$$2\text{K}_s + 2\text{H}_2\text{Ol}\xrightarrow{} 2\text{KOH(aq) + H}_2\text{(g)}$$
Hence, copper can be extracted by hydrometallurgy but not zinc.
2. What is the role of depressant in froth-floatation process?
Ans. In the froth-floatation process, the role of depressant is to separate two sulphide by selectively preventing one ore from forming the froth with air bubbles. NaCN is used as a depressant to separate lead sulphide (PbS) ore from zinc sulphide (ZnS) ore. NaCN forms a zinc complex, Na2[Zn(CN)4] on the surface of ZnS thereby preventing it from the formation of the froth.
$$\text{4NaCN + ZnS}\xrightarrow{}\underset{\text{Sodium tetracy anozincate (II)}}{\text{Na}_2[\text{Zn(CN)}_4]+\text{Na}_2\text{S}}$$
In this condition, only lead sulphide forms froth and thus can be separated from zinc sulphide ore.
3. Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?
Ans. The standard free energy of formation (∆fG°) of Cu2S is less than that of CS2 and H2S. So, Cu2S cannot be reduced by carbon or hydrogen.
$$\text{2Cu}_2\text{S+ C}\xrightarrow{}4\text{Cu + CS}_2(\text{Not feasible)}\\\text{Cu}_2\text{S+ H}_2\xrightarrow{}2\text{Cu+ H}_2\text{S}(\text{Not feasible})$$
However, the standard free energy of formation of ∆fG° of copper oxide is greater than that of CO2. Therefore, the sulphide ore is first converted to oxide by roasting and then reduced.
$$2\text{Cu}_2\text{S}+3\text{O}_2\xrightarrow[\text{Heat}]{\text{Roasting}}2\text{Cu}_2\text{O}+2\text{SO}_2\\2\text{Cu}_2\text{O + C}\xrightarrow{}4\text{Cu + CO}_2\space\text{(feasible)}$$
4. Explain:
(i) Zone refining
(ii) Column chromatography.
Ans. (i) Zone refining: This method is used for production of semiconductors and other metals of very high purity, e.g., Ge, Si, B, Ca and In.
It is based on the principle that the impurities are more soluble in the molten state (melt) than in the solid state of the metal. The impure metal in the form of bar is heated at one end with a moving circular heater. As the heater is slowly moved along the length of the rod, the pure metal crystallises out of the melt whereas the impurities pass into the adjacent molten zone. This process is repeated several times till the impurities are completely driven to one end of the rod which is then cut off and discarded.
(ii) Chromatography: It is based on the principle that the different components of a mixture are adsorbed to different extents on an adsorbent. In column chromatography, an adsorbent, such as alumina (Al2O3) or silica gel is packed in a column. This forms the stationary phase. The mixture to be separated is dissolved in a suitable solvent (mobile phase) and applied to the top of the column. The adsorbed components are extracted (eluted) from the column with a suitable solvent (eluent). The component which is more strongly adsorbed on the column takes longer time to travel through the column than a component which is weakly adsorbed. Thus, the various components of the mixture are separated as they travel through absorbent (stationary phase).
5. Out of C and CO which is a better reducing agent at 673 K?
Ans. If we observe the ellingham diagram, the curve involving ∆G° vs T intersect for CO, CO2 is lower than that of C, CO line at 673K. Hence, CO can be used as a better reducing agent than C at 673K.
6. Name the common elements present in anode mud in the electro-refining of copper. Why are they so present?
Ans. In the electrolytic refining of copper, the anode mud contains metals like, Ag, Au, Pt, Se, which are less reactive than Cu. They are unaffected by CuSO4 — H2SO4 solution and hence, settle down under anode as anode mud.
$$\text{Cu}_{(s)}\xrightarrow{}\text{Cu}^{\normalsize2+}_{(aq)}+2e^{\normalsize-}$$
7. Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.
Ans. The following reactions occur in the blast furnace:
(a) In zone of combustion,
$$\text{C + O}_2\xrightarrow{}\text{CO}_2,\Delta\text{H}=-393.3\space\text{kJ}$$
(b) In zone of heat absorption,
$$\text{CO}_2+\text{C}\xrightarrow{}2\text{CO,}\space\Delta\text{H}=+163.2\space\text{kJ}$$
(c) In zone of slag formation,
$$\text{CaCO}_3\xrightarrow{}\text{CaCO + CO}_2\\\text{CaO + SiO}_2\xrightarrow{}\underset{\text{Calcium silicate (Slag)}}{\text{CaSiO}_3}\\\text{(d) In zone of reduction}\\\text{Fe}_2\text{O}_3+\text{CO}\xrightarrow{\text{823\text{K}}}2\text{FeO + CO}_2\\\text{Fe}_3\text{O}_4+\text{CO}\xrightarrow{\text{823 K}}\space 3\text{FeO + CO}_2\\\text{Fe}_2\text{O}_3+3\text{C}\xrightarrow{>1123\text{K}}2 \text{Fe + 3CO}$$
8. Write chemical reactions taking place in the extraction of zinc from zinc blende.
Ans. The following processes are involved in the extraction of zinc from zinc blende:
(i) Concentration: Zinc blende ore is crushed and the concentration done by froth- floatation process.
(ii) Roasting: The concentrated ore is then roasted in presence of excess of air at about 1200 K as a result zinc oxide is formed.
$$\underset{\text{Zinc blende}}{2\text{ZnS}}+ 3\text{O}_2\xrightarrow{\Delta}\underset{\text{Zinc oxide}}{2\text{ZnO + 2SO}_{2}}$$
(iii) Reduction: Zinc oxide obtained above is mixed with powdered coke and heated to 1673 K in a fire clay retort where it is reduced to zinc metal.
$$\text{ZnO}+\text{C}\xrightarrow{1673\text{K}}\text{Zn + CO}$$
At 1673 K, zinc metal being volatile (boiling point 1180 K), distills over and is condensed.
(iv) Electrolytic refining: Impure zinc is made the anode while pure zinc strip is made the cathode. ZnSO4 solution acidified with dil. H2SO4 is the electrolyte used. On passing electric current, pure zinc gets deposited on the cathode.
9. State the role of silica in the metallurgy of copper.
Ans. Silica (SiO2) acts as an acidic flux during the metallurgy of copper. In the metallurgy of copper, its role is to remove the impurities of iron and form slag.
$$\text{2FeS + 3O}_2\xrightarrow{}\text{2 FeO + 2SO}_2\\\underset{\text{Gangue}}{\text{FeO}}+\underset{\text{Flux}}{\text{SiO}_2}\xrightarrow{}\underset{\text{Iron silicate (Slag)}}{\text{FeSiO}_3}$$
10. Which method of refining may be more suitable if element is obtained in minute quantity?
Ans. Column chromatography is more suitable for an elements is obtained in minutes quantity. It is a technique used to separate different components of a mixture.
If is very useful technique used for the purification of elements available in minute quantities.
11. Which method of refining will you suggest for an elements in which impurities present have chemical properties close to the properties of that elements?
Ans. I suggest column chromatography for an elements in which impurities present have chemical properties close to the properties of that element. It is a technique used to separate different components of mixture. It is a very useful technique used for the purification of elements available in minute quantities in chemical properties from the element to purified. Chromatography is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. In chromatography, these are two phases, mobile phase and stationary phase.
12. Describe a method for refining nickel.
Ans. Mond’s process: Nickel is heated in a steam of carbon mono-oxide forming a volatile complex, nickel tetracarbonyl.
$$\text{Ni + 4CO}\xrightarrow{\text{330-350 K}}\text{Ni(CO)}_4\\\text{Nickel tetracarbonyl is strongly heated to decompose. Thus, pure nickel metal is obtained.}\\\underset{\text{Nickel tetracarbonyl}}{\text{Ni(CO)}_4}\xrightarrow{450-470\text{K}}\underset{\text{pure metal}}{\text{Ni + 4CO}}$$
13. How can you separate alumina from silica in a bauxite ore associated with silica? Give equations, if any.
Ans. Pure alumina can be separated from silica in bauxite by Bayer’s process. The bauxite ore associated with silica is heated with a concentrated solution of NaOH at 473-523 K and 35-36 bar pressure. Under these conditions, alumina dissolves as sodium meta-aluminate and silica as sodium silicate leaving behind the impurities.
$$\underset{\text{Alumina}}{\text{Al}_2\text{O}_3}+2\text{NaOH (aq)}+3\text{H}_2\text{O(l)}\xrightarrow{}\underset{\text{Sodium meta-aluminate}}{2\text{Na[Al(OH)}_{4}](aq)}\\\underset{\text{Silica}}{\text{SiO}_2(s)+2\text{NaOH}(aq)}\xrightarrow{473-523\text{K}}\underset{\text{Sodium silicate}}{\text{Ni}_{2}\text{SiO}_{3}\text{(aq)}+\text{H}_2\text{O(l)}}$$
The resulting solution is filtered to remove the undissolved impurities, sodium meta-aluminate can be precipitated as hydrated aluminium oxide by passing CO2 vapours. The sodium silicate formed cannot be precipitated and can be filtered off.
$$2\text{Na[Al(OH)}_{4}](aq)+ \text{CO}_2\text{(g)}\xrightarrow{}\underset{\text{Hydrated alumina}}{\text{Al}_2\text{O}_3.X\text{H}_2\text{O(s)}}+2\text{NaHCO}_3(aq)\\\text{The hydrated alumina thus precipitated is filtered, dired and heated to give back pure Al}_2\text{O}_3.\\\underset{\text{Hydrated alumina}}{\text{Al}_2\text{O}_3.x\text{H}_2\text{O}}\xrightarrow{1473 \text{K}}\underset{\text{Pure alumina}}{\text{Al}_{2}\text{O}_{3}(s)}+x\text{H}_{2}\text{O}$$
14. Giving examples, differentiate between ‘roasting’ and ‘calcination’.
Ans.
15. How is ‘cast-iron’ different from ‘pig iron’?
Ans.
| Cast iron | Pig iron |
|
It is obtained from blast furnace. It contains about 4% carbon and other impurities like S, P, Si, Mn etc. |
|
It has slightly higher melting point than cast iron. |
16. Differentiate between ‘minerals’ and ‘ores’.
Ans.
| Minerals | Ores |
|
Minerals which can be used to obtain the metal profitable are called ores. |
|
These definitely contain metals. |
|
All ores are minerals. Example: Bauxite, Haematite, Rock salt. |
| Roasting | Calcination |
|
The process of heating the concentrated ore in the presence of air to a high temperature so as not to melt it is called roasting. |
|
Usually sulphide ores are roasted. |
|
The impurities of P, As and S are removed as their oxides which being volatile, escape as gases. |
|
It ia also done in reverberatory furnace but the holes of the furnace are kept open to allow the entry of air into the furnace. $$\text{Example: 2ZnS + 3O}_2\xrightarrow{\Delta}2\text{Zno+2SO}_{2}\uparrow$$ |
17. Why copper matte is put in silica lined converter?
Ans. Copper matte consists of Cu2S along with some unchanged FeS. When a blast of hot air is passed through molten matte placed in silica lined converter, FeS present in matte is oxidised to FeO which combines with silica (SiO2) to form FeSiO3 slag.
$$2\text{2FeS + 3O}_2\xrightarrow{}2\text{FeO + 2SO}_{2}\\\text{FeO}+\underset{\text{Silica}}{\text{SiO}_{2}}\xrightarrow{}\underset{\text{Slag}}{\text{FeSiO}_{3}}\\\text{2S undergoes oxidation to form Cu}_2\text{O which then reacts with more Cu}_2\text{S to form copper metal.}\\2\text{Cu}_{2}\text{S + 3O}_{2}\xrightarrow{}2\text{Cu}_{2}\text{O + 2SO}_{2}\\2\text{Cu}_{2}\text{O + Cu}_{2}\text{S}\xrightarrow{}6\text{Cu + SO}_{2} $$
Thus, copper matte is heated in silica lined converter to remove FeS present in matte as FeSiO3 slag.
18. What is the role of cryolite in the metallurgy of aluminium?
Ans. Cryolite (Na3AlF3) has following rules in the metallurgy of aluminium:
(i) It acts as a solvent.
(ii) It lowers the melting point of the mixture from 2323 K to 1140 K.
(iii) It increase the electrical conductivity of Al2O3.
19. How is leaching carried out in case of low grade copper ores?
Ans. In case of low grade ores of copper, leaching is carried out with acids in the presence of air In this process, copper goes into the solution as Cu2+ ions..
$$\text{Cu(s)}+2\text{H}^{\normalsize+}\text{(aq)}+\frac{1}{2}\text{O}_{2}\text{(g)}\xrightarrow{}\text{Cu}^{2+}\text{(aq)}+2\text{H}_{2}\text{O(I)}\\\text{The resulting solution is then treated with scrap iron or hydrogen to get metallic copper}\\\text{Cu}^{\normalsize+2}\text{(aq) + H}_{2}\text{(g)}\xrightarrow{}\text{Cu(s) + 2 H}^{\normalsize+}\text{(aq)}$$
20. Why is zinc not extracted from zinc oxide through reduction using CO?
Ans. ∆G° for the conversion of Zn into ZnO is –650 kJ and for the conversion of CO into CO2 is –250 kJ.
$$\text{2Zn + O}_{2}\xrightarrow{}2\text{ZnO}\qquad \Delta \text{G}\degree=-650\space\text{kJ}\\2\text{CO+O}_{2}\xrightarrow{}2\text{CO}_{2}\space\Delta \text{G}\degree=-450\text{kJ}\\\text{For the reaction, \space 2ZnO + 2CO}\xrightarrow{}\text{2Zn + 2CO}_{2}\Delta \text{G}\degree=+200\text{kJ}$$
Positive value of ∆G° (+200 kJ) shows that reaction is not feasible. That is why CO cannot be used for the reduction of ZnO into Zn.
21. The value of ΔfG° for formation of Cr2O3 is – 540 kJ mol–1 and that of Al2O3 is – 827 kJ mol–1. Is the reduction of Cr2O3 possible with Al?
Ans. Chemical equation for the formation of Cr2O3 and Al2O3 are as follows:
$$\text{(a)}\space\frac{4}{3}\text{Cr(s)}+\frac{3}{2}\text{O}_{2}\text{(g)}\xrightarrow{}\frac{2}{3}\text{Cr}_{2}\text{O}_{3}(s);\\\Delta\text{G}\degree_{f}=-540\space\text{kJ mol}^{\normalsize-1}\\\text{(b)}\space\frac{4}{3}\text{Al(s)}+\frac{3}{2}\text{O}_{2}\text{(g)}\xrightarrow{}\frac{2}{3}\text{Al}_{2}\text{O}_{3}\text{(s)};\\\Delta\text{G}\degree_f=-827\space\text{kJ mol}^{\normalsize-1}\\\text{Subtracting equation (a) from equation (b), we get}\\\frac{4}{3}\text{Al(s)}+\frac{2}{3}\text{Cr}_{2}\text{O}_{3}\text{(s)}\xrightarrow{}\frac{2}{3}\text{Al}_{2}\text{O}_{3}(s)+\frac{4}{3}\text{Cr(s)}\\\Delta\text{G}\degree=-287\text{kJ mol}^{\normalsize-1}$$
As can be seen ∆rG° is negative, thus, reduction of Cr2O3 by Al is possible.
22. Out of C and CO, which is a better reducing agent for ZnO?
Ans. The two reduction reactions are :
$$\text{ZnO(s)+\text{C(s)}}\xrightarrow{}\text{Zn(s) + CO(g)}\qquad...\text{(i)}\\\text{ZnO(s) + CO(g)}\xrightarrow{}\text{Zn(s)+CO}_{2}(g)\qquad...\text{(ii)}$$
In the first case, there is increase in the magnitude of ΔS° while in the second case, it almost remains the same. In other words ΔG° will have more negative value in the first case when C(s) is used as the reducing agent than in the second case when CO(g) acts as the reducing agent. Therefore, C(s) is a better reducing agent.
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.
Ans. Thermodynamic factors have a major role in selecting the reducing agent for a particular reaction.
(i) The value of ∆G should be negative.
(ii) ∆S should be positive.
(iii) A metal oxide placed lower in the Ellingham diagram cannot be reduced by the metal involved in the formation of oxide placed higher in the diagram.
Example:
1. Al2O3 cannot be reduced by Cr present in Cr2O3. Since, the curve of Al2O3 is placed below that of Cr2O3 in the Ellingham diagram.
2. CO cannot reduce ZnO because there is hardly any change in standard free energy (∆G°) as a result of reaction.
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?
Ans. Down process is used for the preparation of sodium metal, where chlorine is obtained as a by- product. This process involves the electrolysis of a fused mixture of NaCl and CaCl2 at 873 K.Sodium is discharged at the cathode while Cl2 is obtained at the anode as a by-product.
$$\text{NaCl(l)}\xrightarrow{\text{Electrolysis}}\text{Na}^{\normalsize+}\text{(melt) + Cl}^{\normalsize-}\text{(melt)}\\\textbf{At cathode:}\space\text{Na}^{\normalsize+}+e^{\normalsize-}\xrightarrow{}\text{Na}\\\textbf{At anode:}\space\text{Cl}^{\normalsize-}\xrightarrow{}\frac{1}{2}\text{Cl}_{2}+e^{\normalsize-}$$
If, an aqueous solution of NaCl is electrolysed, H2 is evolved at the cathode while Cl2 is obtained at the anode.
25. What is the role of graphite rod in the electrometallurgy of aluminium?
Ans. Graphite rod is used as an anode and graphite line iron as cathode during the electrolysis of alumina. During the electrolysis aluminium is liberated at the cathode and oxygen at anode. This oxygen reacts with the graphite anode to produce CO and CO2 preventing the oxygen evolved from oxidising aluminium. Thus the role of graphite in electrometallurgy of aluminium is to prevent the liberation of oxygen so that the aluminium is not oxidised by oxygen.
26. Outline the principles of refining of metals by the following methods:
(i) Zone refining
(ii) Electrolytic refining
(iii) Vapour phase refining
Ans. (i) Zone refining: It is based on the principle that the impurities are more solute in molten state than in the solid state of metal.
(ii) Electrolytic refining: In this method, impure metal is taken as anode, while cathode is made up of pure strip of same metal. Electrolyte is the solution of a soluble salt of the metal. When an electric current is passed, metal ions from electrolyte deposited on cathode, while equal ions of metals dissolves from anode. As a results, pure metal is transfers from anode to cathode.
(iii) Vapour phase refining: In this method, the crude metal is freed from impurities by first converting into a suitable volatile compound by heating it with specific reagent at a lower temperature and then decomposing the volatile compound at higher temperature to give the pure metal.
27. Predict conditions under which Al might be expected to reduce MgO.
Ans. The equations for the formation of the two oxides are:
$$\frac{4}{3}\text{Al(s)+\text{O}}_{2}\text{(s)}\xrightarrow{}\frac{2}{3}\text{Al}_{2}\text{O}_{3}\text{(s)}\\2\text{MgO(s)\space+}\space\frac{4}{3}\text{Al(s)}\rightleftharpoons2\text{Mg(s)}+\frac{2}{3}\text{Al}_{2}\text{O}_{3}\text{(s)}$$
This means that the reduction of MgO by Al metal can occur below this temperature. Aluminium (Al) metal can reduce MgO to Mg above this temperature because Δ°G for Al2O3 is less as compared to that of MgO.
$$3\text{MgO(s)+2\text{AI}(s)}\xrightarrow{(>1665\text{K})}\text{Al}_{2}\text{O}_{3}\text{(s)}+3\text{Mg}\text{(s)}$$
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