NCERT Solutions for Class 12 Science Chemistry Chapter 6 - General Principles And Processes Of Isolation Of Elements

Answer:

In the extraction of chlorine by electrolysis of brine, oxidation of Cl^–ion to chlorine gas occurs due to loss of electrons.
2Cl^–(aq) + 2H₂O(l) → 2OH^–(aq) + H₂(g) + Cl₂(g)
For overall reaction, ∆G^⊖ has positive value.

Hence, the correct answer is option A. 

Answer:

When copper ore is mixed with silica, in a reverberatory furnace, copper is produced in the form of copper matte which contains sulphide of copper (I), i.e., Cu₂S and sulphide of iron (II), i.e., FeS.

Hence, the correct answer is option C.

 

Answer:

Autoreduction is a process in which sulphide ores of less electropositive metals like Cu, Hg, Pb etc. are heated in presence of air to get crude metal. No external reducing agent is used in this process. Thus, the reaction that undergoes autoreduction is given below.
Cu₂O + $\frac{1}{2}$ Cu₂S → 3Cu +$\frac{1}{2}$ SO₂

Hence, the correct answer is option D. 
_​
 

Answer:

The most abundant elements available in earth's crust are Al and Fe.

Hence, the correct answer is option A. 
 

Answer:

Zone refining is based on the principle that impurities are more soluble in molten metal than in solid metal.
A mobile heater surrounding the rod of impure metal is fixed at its one end. The molten zone moves along with the heater which is moved forward. As the heater moves forward, the pure metal crystallises out of the melt left behind and the impurities pass on into the adjacent new molten zone created by movement of heaters. The process is repeated several times and the heater is moved in the same direction again and again. Impurities get concentrated at one end. This end is cut off.

Hence, the correct answer is option B. 

Answer:

In the extraction of copper from its sulphide ore, the metal is formed by the reduction of Cu₂O with Cu₂S. 

 2Cu₂O + Cu₂S → 6Cu+SO₂ ​

Hence, the correct answer is option C. 
​

Answer:

Although reduction potential of Cl⁻ ion is higher than that of H₂O, yet oxidation of Cl ⁻ ions occurs in preference to H₂O due to over voltage of O₂ .​ The following oxidation reactions are possible at anode when brine is electrolysed using inert electrodes:
$$\begin{gathered} 1) 2{\mathrm{Cl}}^{-}(\mathrm{aq}) \to {\mathrm{Cl}}_2(\mathrm{g}) + 2{\mathrm{e}}^{-}, {E^{\Theta }}_{cell}=1.36 \mathrm{V} \\ 2) 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \to {\mathrm{O}}_2\left(\mathrm{g}\right) + 4{\mathrm{H}}^{+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-}, {E^{\Theta }}_{cell}=1.23 \mathrm{V} \end{gathered}$$
The reaction at anode with lower value of $E_{\mathrm{cell}}^{\Theta }$ is preferred and therefore, water should get oxidised in preference to Cl^–(aq). However, on account of overpotential of oxygen, reaction 1) is preferred.
Cl^–(aq) → $\frac{1}{2}$Cl₂(g) + e^–;            $E_{\mathrm{cell}}^{\Theta }$= 1.36V

Hence, the correct answer is option A. 

Answer:

In the metallurgy of aluminium, graphite anode is oxidised to carbon monoxide and carbon dioxide by the oxygen liberated at anode.


Hence, the correct answer is option B. 
 

Answer:

In electrolytic refining, the impure metal is made to act as anode. A strip of the same metal in pure form is used as cathode. They are put in a suitable electrolytic bath containing soluble salt of the same metal. The more basic metal remains in the solution and the less basic ones go to the anode mud. Less reactive metals like Cu and Zn are purified by electrolytic refining.

Hence, the correct answer is option A.

Answer:

Extraction of gold (Au) and silver (Ag) involves leaching the metal with CN^–ion. The metal is recovered by displacement of metal by some other metal from the complex ion.


Hence, the correct answer is option A.

Answer:

One of the main steps in the reduction of iron oxide is

It is a combination of two simpler reactions.

Net Gibbs energy change

The result reaction will take place when  is negative.

In the given Ellingham diagram, the two lines representing formation of FeO and CO intersect each other at a point  A. Above the temperature at a point A, the value becomes more negative for the combined process making reduction of FeO to iron by carbon possible and in-turn leads to the production of CO. 

Answer:

Below point A, FeO can be reduced by carbon monoxide only as the curve of formation of CO₂ ​from CO lies below the oxidation curve of Fe.

Hence, the correct answer is option A. 

Answer:

At the point D, both the curves of oxidation of Fe and oxidation of CO intersect, therefore at this point, △G° value for the overall reduction of FeO to Fe with CO is zero.

Hence, the correct answer is option A. 
​

Answer:

At the temperature corresponding to point B and point E, FeO will be reduced to Fe by coupling the reaction 2FeO → 2Fe + O₂ with all of the given reactions.

Hence, the correct answers are options B and D. 
 

Answer:

The correct statements are as follows:

• Cast iron is obtained by remelting pig iron with scrap iron and coke using hot air blast.
• Zr and Ti are purified by van Arkel method.

Answer:

In the extraction of aluminium by Hall-Heroult process, purified Al₂O₃ is mixed with CaF₂ to lower the melting point of Al₂O₃  and  increase the conductivity of molten mixture.

Hence, the correct answers are option A and B. 
 

Answer:

Froth floatation process is used for removing gangue from sulphide ores. This method utilises collectors, froth stabilisers and depressants whose functions are as follows:

• Collectors enhance the non-wettability of the mineral particles.
• Depressants separate two sulphide ores by selectively preventing one sulphide ore from coming to the froth but allowing other ore to come with the froth.
• Froth stabilisers stabilise the froth.

Answer:

Froth floatation method is used for removing gangue from sulphide ores. In an ore containing ZnS and PbS, the depressant used is NaCN which selectively prevents ZnS from coming to the froth but allows PbS to come with the froth. They can also be separated by using adjusting the proportion of oil to water.

Hence, the correct answers are option B and C.

Answer:

Bauxite is the principal ore of aluminium. It usually contains silica (SiO₂), iron oxide (Fe₂O₃) and titanium oxide (TiO₂) as impurities.

Hence, the correct answers are options C and D.

Answer:

Froth floatation is used for removing gangue from sulphide ores. The ores that are concentrated by froth floatation are galena (PbS) and copper pyrites (CuFeS₂).

Hence, the correct answers are options B and C. 
 

Answer:

Calcination involves heating. It removes the volatile matter which escapes leaving behind the metal oxide. The reactions that occur during calcination are: 
CaCO₃ → CaO + CO₂
Al₂O₃.xH₂O → Al₂O₃ + xH₂O

Hence, the correct answers are options A and C.

 

Answer:

Calcinaton involves heating. It removes the volatile matter which escapes leaving behind the metal oxide. The haematite ore (Fe₂O₃.xH₂O) and calamine (ZnCO₃) can be calcinated to leave behind FeO and ZnO which can be reduced by carbon to Fe and Zn.

Hence, the correct answers are options A and B.
 

Answer:

The main reactions occurring in blast furnace during extraction of iron from haematite are as follows:

• Reduction of Fe₂O₃ with CO : Fe₂O₃+ 3CO → 2Fe + 3CO₂
• Slag Formation : CaO + SiO₂ → CaSiO₃

Answer:

The methods of purification in which metal is converted to its volatile compound which is decomposed to give pure metal are :

• Mond Process for Refining Nickel: In this process, nickel is heated in a stream of carbon monoxide forming a volatile complex named as nickel tetracarbonyl. This complex is decomposed at higher temperature to obtain pure metal.
• van Arkel Method for Refining Zirconium or Titanium: This method is very useful for removing all the oxygen and nitrogen present in the form of impurity in certain metals like Zr and Ti. The crude metal is heated in an evacuated vessel with iodine. The metal iodide being more covalent, volatilises. The metal iodide is decomposed on a tungsten filament, electrically heated to about 1800K. The pure metal deposits on the filament.

Answer:

The correct statements are:

• A depressant selectively prevents one type of ore from coming to the froth but allows other type of ore to come with the froth
• The solidified copper obtained from reverberatory furnace has blistered appearance due to evolution of SO₂ during the  extraction.

Answer:

The reaction for the extraction of chlorine from brine is a s follows:
2Cl⁻ + 2H₂O → 2OH⁻ + H₂ + Cl₂
 ∆G^⊖ for the reaction is +422 kJ.
Using ∆G^⊖ = −nFE^⊖
We get, E^⊖ = − 2.2V

Hence, the correct answers are options B and C.

Answer:

For the reaction, 2Cl⁻ (aq) + 2H₂O (l) → 2OH⁻(aq) + H₂(g) + Cl₂(g)
Value of ΔG^Θ is + 422kJ.
Using the equation ΔG^Θ = – nFE^Θ the value of E^Θ comes out to be –2.2V. Therefore, extraction of Cl₂ from brine will require an external emf of greater than 2.2V.
 

Answer:

As per Ellingham diagram, at temperature greater than 1073K , ∆G^o for the formation of FeO is less negative than ∆G^o for the formation of carbon monoxide from carbon. Thus, above 1073K , carbon can reduce FeO to Fe , i.e., ​∆G_r^o for the reaction is negative.
 

Answer:

The reaction used for the preparation of wrought iron from cast iron is given as:
Fe₂O₃ + 3C → 2Fe + 3CO
​Limestone is added as flux and the impurities of sulphur, silicon and phosphorus change to their oxides and pass into slag.

Answer:

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 follows:
Cu²⁺(aq) + Fe(s) → Cu(s) + Fe²⁺(aq)
​Cu²⁺(aq) + H₂(g) → Cu(s) + 2H⁺(aq)                               
 

Answer:

The basic requirements for both processes are :
(i) The metal should form a volatile compound with an available reagent.
(ii) The volatile compound should be easily decomposable so that recovery of metal is easy.

Answer:

Carbon and hydrogen are better reducing agents but they are not used to reduce metallic oxides at high temperatures because at high temperatures carbon and hydrogen react with metals to form carbides and hydrides respectively.

Answer:

Two sulphide ores can be separated by adjusting the proportion of oil to water or by using depressants. For example, in the ore containing ZnS and PbS, the depressant NaCN is used. It forms complex with ZnS and prevents it from coming with froth but PbS remains with the froth.

Answer:

Wrought iron or malleable iron is the purest form of commercial iron and is prepared from cast iron by oxidising impurities in a reverberatory furnace lined with haematite. The haematite oxidises carbon to carbon monoxide. Thus, the iron ore used to line the furnace is Haematite. The reaction is given as follows:
Fe₂O₃+ 3C → 2Fe + 3CO

Answer:

Compound that is adsorbed to a greater extent is eluted later while the one which is absorbed to a lesser extent is eluted readily. Since compound ‘A’ comes out before compound ‘B’, the compound ‘B’ is more readily adsorbed on the column.

Answer:

The sulphide ore of copper is heated in reverberatory furnace because iron oxide present as an impurity in sulphide ore of copper forms slag after mixing with silica which is iron silicate and copper is produced in the form of copper matte which contains Cu₂S and FeS.
FeO + SiO₂ → FeSiO₃

Answer:

The sulphide ores are converted to oxide before reduction because sulphides cannot be reduced directly whereas oxides can be reduced easily.

Answer:

The van Arkel Method is used for refining Zirconium or Titanium. This method is very useful for removing all the oxygen and nitrogen present in the form of impurity in certain metals like Zr and Ti. The crude metal is heated in an evacuated vessel with iodine. The metal iodide being more covalent, volatilises:
Zr + 2I₂ → ZrI₄
The metal iodide is decomposed on a tungsten filament, electrically heated to about 1800K. The pure metal deposits on the filament.
ZrI₄ → Zr + 2I₂

Answer:

The main considerations during the extraction of metals by the electrochemical method are as follows:

1. Reactivity of the metal produced.
2. Electrodes to be made of suitable material.

Answer:

The role of flux in metallurgical processes is to make the molten mass more conducting and combines with ‘gangue’ to form ‘slag’. Slag separates more easily from the ore than the gangue. 

Answer:

Semiconductors are refined by the method of Zone refining. This method is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal. 

Answer:

The reactions taking place in blast furnace related to the metallurgy of iron in the temperature range 500-800 K are as follows:
3Fe₂O₃ + CO → 2Fe₃O₄ + CO₂
Fe₃O₄ + 4CO → 3Fe + 4CO₂
Fe₂O₃ + CO → 2FeO + CO₂

Answer:

In the method of vapour phase refining, the metal is converted into its volatile compound which is collected and decomposed to give pure metal. So, the two requirements are:
(i) the metal should form a volatile compound with an available reagent
(ii) the volatile compound should be easily decomposable, so that the recovery is easy. 

Answer:

The chemical reactions involved in the extraction of gold by cyanide process is as follows:

$$\begin{gathered} \left(\mathrm{i}\right) 4\mathrm{Au} \left(\mathrm{s}\right)+8{\mathrm{CN}}^{-}\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{aq}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\stackrel{}{\to }4{\left[\mathrm{Au}{\left(\mathrm{CN}\right)}_2\right]}^{-}\left(\mathrm{aq}\right)+4{\mathrm{OH}}^{-}\left(\mathrm{aq}\right) \\ \left(\mathrm{ii}\right) 2{\left[\mathrm{Au}{\left(\mathrm{CN}\right)}_2\right]}^{-}\left(\mathrm{aq}\right)+\mathrm{Zn}\left(\mathrm{s}\right)\stackrel{}{\to }2\mathrm{Au}\left(\mathrm{s}\right)+{\left[\mathrm{Zn}{\left(\mathrm{CN}\right)}_4\right]}^{2-}\left(\mathrm{aq}\right) \end{gathered}$$

Here, zinc (Zn) acts as a reducing agent.

Answer:

(A) Pendulum is made up of nickel steel.
(B) Molecular formula of malachite is CuCO₃.Cu(OH)₂.
(C) Molecular formula of calamine is ZnCO₃.
(D) Molecular formula of cryolite is Na₃AlF₆.
So, the correct match is as follows:
A → 2, B → 4, C → 5, D → 3.

Hence, the correct answer is option (ii).

Answer:

(A) Coloured bands are observed in chromatography.
(B) Impure metal is converted to volatile complex by using Mond's process,
(C) Purification of Ge and Si is done by zone refining method.
(D) Purification of mercury can be done by fractional distillation method.
So, the correct match is as follows:
A → 4, B → 3, C → 1, D → 2

Hence, the correct answer is option (ii).

Answer:

(A) Extraction of Au can be done by using the cyanide process.
(B) Froth flotation process is used for dressing of ZnS.
(C) Extraction of Al can be done by using electrolytic reduction method.
(D) Zone refining is used for purification of Ge.
So, the correct match is as follows:
A → 4, B → 2, C → 3, D → 1.

Hence, the correct answer is option (i).

Answer:

(A) Sapphire is a gemstone. It contains cobalt.
(B) Molecular formula of sphalerite is ZnS.
(C) NaCN is used as depressant in froth floatation method.
(D) Molecular formula of corundum is Al₂O₃.
So, the correct match is as follows:
A → 3, B → 4, C → 2, D → 1.

​Hence, the correct answer is option (i).

Answer:

• Blistered Cu can be prepared by the following reaction:

• Iron is extracted from its oxides in blast furnace.
• Formation of slag occurs in reverberatory furnace by the following reaction: 

• Extraction of aluminium can be done by Hall-Heroult process.

Answer:

Nickel can be purified by Mond's process. This process involves the formation of a volatile compound Ni(CO)₄ which further decomposes to Ni at 460 K.


​Hence, the correct answer is option (i).

Answer:

Zirconium can be purified by Van Arkel method which include the formation of volatile compound ZrI4 that decomposes at 1800K to Zr.
Hence, the correct answer is option (i).

Answer:

Sulphide ores are concentrated by Froth Floatation method. The sulphide ores particles are wetted by oil, become lighter and rise to the surface along with the froth. The gangue particles, on the other hand, are wetted by water, become  heavier and thus settle down at the bottom of the tank and cresols stabilise the froth in Froth Floatation method.
Formation of froth is the main reason for extraction of metal. The metal ores separates out along with the froth.

​Hence, the correct answer is option (ii).

Answer:

Zone refining method is very useful for producing semi-conductors of high purity. In this method, when impure metal rod is heated, the pure metal is crystallised while the impurities pass on to the adjacent molten zone.

Hence, the correct answer is option (ii).

Answer:

Hydrometallurgy involves dissolving the ore in a suitable reagent followed by precipitation with the help of more electropositive metal. In this, pure metal gets replaced by more electropositive metal. Copper is extracted by hydrometallurgy from low grade ores. It is leached out using acid or bacteria. The solution containing Cu²⁺ is treated with scrap iron or H₂.

Hence, the correct answer is option (ii).

Answer:

(a) With reference to Ellingham diagram which relates Gibbs free energy and temperature, ΔG^Θ for formation of CO from C is more negative at temperatures below 710 K while at temperatures above 710 K, ΔG^Θ for formation of CO₂ from CO is more negative. Therefore, CO₂ is a better reducing agent below 710 K whereas CO is a better reducing agent above 710 K.
(b) Generally sulphide ores are converted into oxides before reduction. This is because reduction of oxides can easily be done using C or CO depending upon the metal ore and temperature.
(c) Silica is a flux added to the sulphide ore of copper in the reverberatory furnace. Sulphide ore of copper contains iron as impurity which is removed as iron slag. The impurities of iron present with copper ore are removed as iron silicate.
$\mathrm{FeO}+\mathrm{Si}{\mathrm{O}}_2\stackrel{}{\to }\underset{\mathrm{Slag}}{\mathrm{FeS}}{\mathrm{iO}}_3$
(d) At high temperatures, carbon and hydrogen readily form metal carbides and hydrides respectively, therefore, they are not used as reducing agents.
(e) The purification of Ti can be done by the vapour phase refining method.
The reactions involved are as follows:
$\mathrm{Ti}+2{\mathrm{I}}_2\stackrel{∆}{\to }\mathrm{Ti}{\mathrm{T}}_4\stackrel{1800\mathrm{K}}{\to }\mathrm{Ti}+2{\mathrm{I}}_2$