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

Intext Exercise -1

1. Which of the Ores Mentioned Can be Concentrated by Magnetic Separation Method?

Ans: Ores which are magnetic in nature can be separated from non-magnetic gangue particles by magnetic separation method.

For Example: ores of iron such as hematite ($F{{e}_{2}}{{O}_{3}}$ ), magnetite ($F{{e}_{3}}{{O}_{4}}$), siderite ($FeC{{O}_{3}}$) and iron pyrites ($Fe{{S}_{2}}$ ) being magnetic can be separated from non-magnetic silica and other impurities can be separated by magnetic separation method. Ores of copper such as Copper pyrites ($CuFe{{S}_{2}}$ ) are also separated by this process.

2. What is the Significance of Leaching in the Extraction of Aluminium?

Ans: Aluminium contains silica ($Si{{O}_{2}}$), iron oxide ($F{{e}_{2}}{{O}_{3}}$ ) and titanium oxide ($Ti{{O}_{4}}$ ) as impurities. These impurities can be removed by the process of leaching. During leaching, the powdered bauxite ore is heated with a concentrated (45% ) solution of $NaOH$ at 473-523 K, where alumina dissolves as sodium meta aluminate and silica as sodium silicate leaving $F{{e}_{2}}{{O}_{3}}$ , $Ti{{O}_{2}}$  and other impurities behind:

\[A{{l}_{2}}{{O}_{3}}(s)+2NaOH(aq)\overset{473-523K}{\mathop{\to}}\,\underset{sodium\text{}meta-aluminate}{\mathop{2Na[Al{{(OH)}_{4}}](aq)}}\,\]

\[Si{{O}_{2}}(l)+2NaOH(aq)\overset{473-523K}{\mathop{\to}}\,\underset{sodium\text{}silicate}{\mathop{N{{a}_{2}}Si{{O}_{3}}(aq)}}\,+{{H}_{2}}O(l)\] 

The impurities are filtered off and the solution of sodium meta-aluminate is neutralized by passing $C{{O}_{2}}$  when hydrated alumina separates out while sodium silicate remains in solution. The hydrated alumina thus obtained is filtered, dried and heated to give back pure alumina.

\[A{{l}_{2}}{{O}_{3}}.x{{H}_{2}}O\overset{1473K}{\mathop{\to}}\,A{{l}_{2}}{{O}_{3}}(s)+x{{H}_{2}}O(g)\] 

Thus, by leaching, pure alumina can be obtained from bauxite ore.

3. The reaction $C{{r}_{2}}{{O}_{3}}+2Al\to A{{l}_{2}}{{O}_{3}}+2Cr(\Delta {{G}^{\theta }}=-421kJ)$ is Thermodynamically Feasible as is Apparent from the Gibbs Energy Value. Why Does it not Take Place at Room Temperature?

Ans: This is explained on the basis of ${{K}_{eq}}$ , the equilibrium constant. In the given redox reaction, all reactants and products are solids at room temperature, so, there is no equilibrium between the reactants and products and hence the reactions do not occur at $RT$ . At high temperature, Cr melts and values of $T\Delta S$  increases. As a result, the value of ${{\Delta }_{r}}{{G}^{\theta }}$  becomes more negative and hence the reaction proceeds rapidly.

4. Is It True that Under Certain Conditions, Magnesium Can Reduce $A{{l}_{2}}{{O}_{3}}$  and Aluminum Can Reduce $MgO$ ? What are those Conditions?

Ans: Yes. When we studied Ellingham Diagrams indicates that,   below ${{1350}^{o}}C$ , Magnesium can reduce $A{{l}_{2}}{{O}_{3}}$ and above ${{1350}^{0}}C$ , Aluminum can reduce$MgO$ . This can be inferred from $\Delta {{G}^{0}}$ Vs T plots. 

NCERT Exercise:

1. Copper Can be Extracted by Hydrometallurgy but not Zinc. Explain.

Ans: Copper can be extracted by hydrometallurgy but not zinc, this is because ${{E}^{0}}_{Z{{n}^{+2}}/Zn}=0.76V$ is lower than that of ${{E}^{0}}_{C{{u}^{+2}}/Cu}=0.34V$ .  Hence, Zinc can replace copper from solutions of $C{{u}^{+2}}$ ions.

\[Zn(s)+C{{u}^{+2}}(aq)\to Z{{n}^{+2}}(aq)+Cu(s)\] 

In order to displace zinc from zinc solution, a more reactive metal is   required, such as:

\[Al({{E}^{0}}_{A{{l}^{+3}}/Al}=-1.66V),Mg({{E}^{0}}_{M{{g}^{+2}}/Mg}=-2.37V)] 

\[Ca({{E}^{0}}_{C{{a}^{+2}}/Ca}=-2.87V),K({{E}^{0}}_{{{K}^{+}}/K}=-2.93V)\] 

But with water, these metals (Al, Mg, Ca and K) form their corresponding ions with the evolution of ${{H}_{2}}$  gas. Thus, Al, Mg, Ca, K, etc., cannot be used to displace zinc from zinc solution, and only copper can be extracted by hydrometallurgy but not the zinc.

2. What is the Role of Depressant in the Froth-Floatation Process?

Ans: The role of depressant is to prevent one type of sulphide ore particles from forming the froth with air bubbles. Sodium cyanide ($NaCN$ ) is used as a depressant to separate lead sulphide ($PbS$ ) ore from zinc sulphide ($ZnS$ ) ore. Sodium cyanide ($NaCN$ ) forms a zinc complex, $N{{a}_{2}}[Zn{{(CN)}_{4}}]$ on the surface of Zinc sulphide , thereby preventing it from the formation of the froth.

\[4NaCN+ZnS\to\underset{Sodium\text{}tetracyno-zincate(II)}{\mathop{N{{a}_{2}}[Zn{{(CN)}_{4}}]}}\,+N{{a}_{2}}S\] 

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: ${{\Delta }_{f}}{{G}^{o}}$ of $C{{u}_{2}}S$ is more negative than ${{\Delta }_{f}}{{G}^{o}}$ of $C{{S}_{2}}{{H}_{2}}S$ . So, $C{{u}_{2}}S$ cannot be reduced by carbon or hydrogen.  ${{\Delta }_{f}}{{G}^{o}}$ of $C{{O}_{2}}$ is more negative than ${{\Delta }_{f}}{{G}^{o}}$ of $C{{u}_{2}}O$. So, $C{{u}_{2}}O$ can be reduced by carbon. So, pyrites are first converted to oxide before reduction to copper.

\[C{{u}_{2}}S(s)+\frac{3}{2}{{O}_{2}}(g)\to C{{u}_{2}}O(s)+S{{O}_{2}}(g)\] 

\[C{{u}_{2}}O(s)+C(s)\to 2Cu(s)+CO(s)\] 

4. Explain: 

1. Zone Refining

Ans: 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 a 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 crystallizes 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. 

2. Chromatography

Ans: 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    ($A{{l}_{2}}{{O}_{3}}$ ) 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 673K?

Ans: This can be explained thermodynamically, taking entropy and free energy changes into account

(a) $C(s)+{{O}_{2}}(g)\to C{{O}_{2}}(g)$ 

(b) $2C(s)+{{O}_{2}}(g)\to 2CO(g)$

Case (i): Volume of $C{{O}_{2}}$  produced = Volume of   

 ${{O}_{2}}$  Used.

Therefore,

$\Delta S$ is very small and $\Delta G$ does not change with temperature. 

Therefore, Plot of $\Delta G$ Vs T is almost horizontal.

Case (ii): Volume of $CO$  produced = 2 × volume of ${{O}_{2}}$  used.

Therefore, 

$\Delta S$is positive and hence $\Delta G$ becomes increasingly.

Negative as the temperature increases.

Therefore, Plot of ${{\Delta }^{o}}G$ VsT slopes downwards.

Volume of CO produced

As can be seen from ${{\Delta }^{o}}G$ VsT  plot (Ellingham diagram), 

lines for the reactions, 

$C\to C{{O}_{2}}$ and $C\to CO$ cross at 983 K. Below 983 K, the  

 reaction.

(a) is energetically more favourable but above 673 K, reaction. 

(b) is favourable and preferred. Thus, below 673 K both C and $CO$ can act as a reducing agent but since $CO$ can be more easily oxidised to $C{{O}_{2}}$ than C to$C{{O}_{2}}$ , therefore, below 673K, $CO$ is more effective reducing agent than carbon.

6. Name the Common Elements Present in the Anode Mud in Electrolytic Refining of Copper. Why are They So Present?

Ans: The common elements present in the anode mud are antimony, selenium, tellurium, silver, gold and platinum. These elements settle down under anode as anode mud because they are less reactive and are not affected by $CuS{{O}_{4}}-{{H}_{2}}S{{O}_{4}}$ solution.

7. Write Down the Reactions Taking Place in Different Zones in the Blast Furnace During the Extraction of Iron.

Ans: In the blast furnace reduction of iron oxides take place at different temperature ranges as shown below.

• At 500-800K: 

\[3F{{e}_{2}}{{O}_{3}}+CO\to 2F{{e}_{3}}{{O}_{4}}+C{{O}_{2}}\]

\[F{{e}_{2}}{{O}_{3}}+4CO\to 3Fe+4C{{O}_{2}}\]

\[F{{e}_{2}}{{O}_{3}}+CO\to 2Fe+C{{O}_{2}}\] 

• At 900-1500K

\[C+C{{O}_{2}}\to 2CO\] 

\[FeO+CO\to Fe+C{{O}_{2}}\] 

\[C+{{O}_{2}}\to C{{O}_{2}}\] 

• Above 1570K

\[FeO+C\to Fe+CO\]

\[CaC{{O}_{3}}\overset{\Delta }{\mathop{\to }}\,CaO+C{{O}_{2}}\] 

\[CaO+Si{{O}_{2}}\to CaSi{{O}_{3}}(slag)\] 

The Following Reactions Can Occur in the Blast Furnace:

(a) In Zone of Combination,

\[C+{{O}_{2}}\to C{{O}_{2}},\Delta H=-393.3kJ\] 

(b) In Zone of Heat Absorption,

\[C{{O}_{2}}+C\to 2CO,\Delta H=+163.2kJ\] 

(c) In Zone of Slag Formation,

\[CaC{{O}_{3}}\to CaO+C{{O}_{2}}\]

\[CaO+Si{{O}_{2}}\to\underset{\text{CalciumSilicate(Slag)}}{\mathop{Cai{{O}_{3}}}}\,\] 

(d) In Zone of Reduction,

\[F{{e}_{2}}{{O}_{3}}+CO\overset{823K}{\mathop{\to }}\,2FeO+C{{O}_{2}}\] 

\[F{{e}_{3}}{{O}_{4}}+CO\overset{823K}{\mathop{\to }}\,3FeO+C{{O}_{2}}\]

\[F{{e}_{2}}{{O}_{3}}+3C\overset{>1123K}{\mathop{\to }}\,2Fe+3CO\] 

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 air at about 1200 K as a result zinc oxide is formed.

\[\underset{\text{Zinc blende}}{\mathop{2ZnS}}\,+3{{O}_{2}}\overset{\Delta}{\mathop{\to }}\,\underset{\text{Zinc oxide}}{\mathop{2ZnO}}\,+2S{{O}_{2}}\] 

(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.

\[ZnO+C\overset{1673K}{\mathop{\to }}\,Zn+CO\] 

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. $ZnS{{O}_{4}}$ Solution acidified with dil. ${{H}_{2}}S{{O}_{4}}$ 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: During roasting, copper pyrites are converted into a mixture of $FeO$ and $C{{u}_{2}}O$ . Thus, acidic flux silica is added during smelting to remove FeO (basic).  Ferrous oxide ($FeO$) combines with Silica ($Si{{O}_{2}}$ ) to form ferrous silicate ($FeSi{{O}_{3}}$ ) slag which floats over molten matte.

10. What is Meant by the Term “Chromatography”?

Ans: The technique which is used for separation, purification, identification and characterization of the components of a mixture whether colored or colorless is known as Chromatography. The term chromatography was originally derived from the Greek word ‘chroma’ meaning color and ‘graphy' for writing because the method was first used for the separation of colored substances (plant pigments) into individual components.

11. What Criterion is Followed for the Selection of the Stationary Phase in Chromatography?

Ans: The stationary phase is selected in such a way that the impurities are more strongly adsorbed or are more soluble in this phase than component to be purified. Thus, when the column is extracted, the impurities will be retained by the stationary phase while the pure component is easily eluted.

12. Describe a Method for Refining Nickel.

Ans: When impure nickel is heated in presence of $CO$ at 330-350 K, it forms volatile nickel tetra carbonyl leaving behind the impurities. The nickel tetracarbonyl thus obtained is then heated to higher temperature (450-470K), and then it undergoes thermal decomposition to give pure nickel.

\[\underset{(\text{impure})}{\mathop{Ni}}\,+4CO\overset{330-350K}{\mathop{\to }}\,\underset{\text{Nickel tetracarbo}nyl}{\mathop{Ni{{(CO)}_{4}}}}\,\] 

\[Ni{{(CO)}_{4}}\overset{450-470K}{\mathop{\to}}\,\underset{Nickel}{\mathop{Ni}}\,+4CO\] 

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 Baeyer’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.

\[A{{l}_{2}}{{O}_{3}}(s)+2NaOH(aq)+3{{H}_{2}}O(l)\to \underset{\text{Sodium meta-aluminate}}{\mathop{2Na[Al{{(OH)}_{4}}](aq)}}\,\] 

\[Si{{O}_{2}}(s)+2NaOH(aq)\overset{473-523K}{\mathop{\to }}\,\underset{\text{Sodium silicate}}{\mathop{N{{i}_{2}}Si{{O}_{3}}(aq)}}\,+{{H}_{2}}O(l)\] 

The resulting solution is filtered to remove the undissolved impurities, sodium meta-aluminate can be precipitated as hydrated aluminium oxide by passing $C{{O}_{2}}$  vapours. The sodium silicate formed cannot be precipitated and can be filtered off..

\[2Na[Al{{(OH)}_{4}}](aq)+C{{O}_{2}}(g)\to\underset{\begin{smallmatrix} \text{Hydrated alumina} \\ \text{oxide ppt} 

\end{smallmatrix}}{\mathop{A{{l}_{2}}{{O}_{3}}.x{{H}_{2}}O(s)}}\,+2NaHC{{O}_{3}}(aq)\] 

The hydrated alumina thus precipitated is filtered, dried and heated to give back pure

\[\underset{Hydrated\text{alumina}}{\mathop{A{{l}_{2}}{{O}_{3}}.x{{H}_{2}}O(s)}}\,\overset{1473K}{\mathop{\to }}\,\underset{\text{pure alumina}}{\mathop{A{{l}_{2}}{{O}_{3}}(s)}}\,+x{{H}_{2}}O\] 

14. Giving Examples Differentiate Between ‘Roasting’ and ‘Calcination’.

Ans: Calcination is a process of converting carbonates and hydroxide ores of metals to their respective oxides by heating them, strongly below their melting points either in absence or limited supply of air.

\[F{{e}_{2}}{{O}_{3}}.3{{H}_{2}}O\overset{\Delta}{\mathop{\to }}\,F{{e}_{2}}{{O}_{3}}+3{{H}_{2}}O\] 

\[CaC{{O}_{3}}.MgC{{O}_{3}}\overset{\Delta}{\mathop{\to }}\,CaO+MgO+2C{{O}_{2}}\] 

\[CuC{{O}_{3}}.Cu{{(OH)}_{2}}\overset{\Delta}{\mathop{\to }}\,2CuO+{{H}_{2}}O+C{{O}_{2}}\] 

\[ZnC{{O}_{3}}\overset{\Delta }{\mathop{\to }}\,ZnO+C{{O}_{2}}\] 

Roasting is a process of converting sulphide ores into its metallic oxides by heating strongly below its melting point in excess of air.

\[2ZnS+3{{O}_{3}}\to 2ZnO+2S{{O}_{2}}\uparrow \] 

\[2PbS+3{{O}_{2}}\to PbO+2S{{O}_{2}}\uparrow \] 

15. How is ‘Cast Iron’ Different from ‘Pig Iron’?

Ans: The iron obtained from blast furnaces is called pig iron. It contains about $4%$  carbon and many other impurities in smaller amounts (eg., S. P, Si and Mn). Cast iron is made by melting pig iron with scrap iron and coke using hot air blast. It has slightly lower carbon content (about $3%$ ) and is extremely hard and brittle.

16. Differentiate Between Minerals’ and Ores’.

Ans: Minerals: The natural substances in which the metals or their compounds occur in the earth is called minerais. 

Ores: The minerals from which the metals can be conveniently and economically extracted are called ores. 

Note: All ores are minerals but all minerals are not ores.

17. Why is Copper Matte Put in a Silica Lined Converter?

Ans: Copper matte consists of copper sulphide ($C{{u}_{2}}S$ ) along with some unchanged Ferrous sulphide ($FeS$ ). When a blast of hot air is passed through molten matte placed in silica lined converter, $FeS$ present in matte is oxidised to Ferrous oxide, which combines with silica ($Si{{O}_{2}}$ ) to form slag.

\[2FeS+3{{O}_{2}}\to 2FeO+2S{{O}_{2}}\] 

\[FeO+\underset{silica}{\mathop{Si{{O}_{2}}}}\,\to \underset{slag}{\mathop{FeSi{{O}_{3}}}}\,\] 

When whole of iron has been removed as slag, some of the  $C{{u}_{2}}S$ undergoes oxidation to form $C{{u}_{2}}O$ which then reacts with more $C{{u}_{2}}S$to form copper metal.

\[2C{{u}_{2}}S+3{{O}_{2}}\to 2C{{u}_{2}}O+2S{{O}_{2}}\] 

\[2C{{u}_{2}}O+3{{O}_{2}}\to 6Cu+S{{O}_{2}}\] 

Thus, copper matte is heated in a silica lined converter to remove $FeS$ present in matte as $FeSi{{O}_{3}}$  slag.

18. What is the Role of Cryolite in the Metallurgy of Aluminum?

Ans: 

1. It lowers the fusion (melting) point of the bath from 2323 K to about 1140 K., 

2. It makes alumina a good conductor of electricity.

19. How Is Leaching Carried Out in Case of Low Grade Copper Ores?

Ans: Leaching in case of low grade copper ores is carried out with acids in presence of air. In this process, copper is oxidised to $C{{u}^{+2}}$ ions which pass into the solution.

\[2Cu(s)+2{{H}_{2}}S{{O}_{4}}(aq)+{{O}_{2}}(g)\to 2CuS{{O}_{4}}(aq)+2{{H}_{2}}O(l)\] 

\[Cu(s)+2{{H}^{+}}(aq)+\frac{1}{2}{{O}_{2}}(g)\to C{{u}^{+2}}(aq)+{{H}_{2}}O(l)\] 

20. Why is Zinc not Extracted from Zinc Oxide Through Reduction Using $CO$ ?

Ans: This is because the standard free energy of formation of $C{{O}_{2}}$  from $CO$ is higher than that of standard free energy of formation of Zinc oxide   ($ZnO$) from Zinc.

21. The value of ${{\Delta }_{f}}{{G}^{o}}$ for the Formation of $C{{r}_{2}}{{O}_{3}}$ is $-540kJmo{{l}^{-1}}$ and that of $A{{l}_{2}}{{O}_{3}}$ is $-827kJmo{{l}^{-1}}$. Is the Reduction of $C{{r}_{2}}{{O}_{3}}$ Possible with Aluminum?

Ans: Chemical equation for the formation $C{{r}_{2}}{{O}_{3}}$of and are $A{{l}_{2}}{{O}_{3}}$ as follows:

(a)$\frac{4}{3}Cr(s)+\frac{3}{2}{{O}_{2}}(g)\to \frac{2}{3}C{{r}_{2}}{{O}_{3}}(s);{{\Delta }_{f}}{{G}^{o}}=-540kJmo{{l}^{-1}}$ 

(b)$\frac{4}{3}Al(s)+\frac{3}{2}{{O}_{2}}(g)\to \frac{2}{3}A{{l}_{2}}{{O}_{3}}(s);{{\Delta }_{f}}{{G}^{o}}=-827kJmo{{l}^{-1}}$ 

Subtracting equation (a) from equation (b), we get

\[\frac{4}{3}Al(s)+\frac{2}{3}C{{r}_{2}}{{O}_{3}}\to \frac{2}{3}A{{l}_{2}}{{O}_{3}}(s)+\frac{4}{3}Cr(s),\Delta {{G}^{0}}=-287kJmo{{l}^{-1}}\] 

As can be seen ${{\Delta }_{r}}{{G}^{o}}$ is negative, thus, reduction of by $C{{r}_{2}}{{O}_{3}}$ by Aluminum is possible.

22. Out of C and $CO$ , Which is a better Reducing Agent for $ZnO$ ?

Ans: The two reduction reactions are

(i) $ZnO(s)+C(s)\to Zn(s)+CO(g)$ 

(ii) $ZnO(s)+CO(s)\to Zn(s)+C{{O}_{2}}(g)$ 

In the first case, there is an increase in the magnitude of change in the entropy ($\Delta {{S}^{0}}$ ), while in the second case it almost remains the same.

In other words, a change in the standard Gibbs free energy ($\Delta {{G}^{0}}$ ) will have more negative value in the first case when Carbon is used as reducing agent than in the second case carbon monoxide 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 Factors. How Far Do You Agree With This Statement? Support Your Opinion With Two Examples.

Ans: We can study the choice of a reducing agent in a particular case using the Ellingham diagram. It is evident from the diagram that metals for which the standard free energy of formation of their oxides is more negative can reduce those metal oxides for which the standard free energy of formation of their respective oxides is less negative it means that any metal will reduce the oxides of other metals which lie above it in the Ellingham diagram. This is because the standard free energy change (${{\Delta }_{r}}{{G}^{0}}$ ) of the combined redox reaction will be negative by an amount equal to the difference in ${{\Delta }_{f}}{{G}^{0}}$of the two metal oxides. Thus both Aluminum and Zinc can reduce $FeO$  to Iron but Fe cannot reduce $A{{l}_{2}}{{O}_{3}}$ to Aluminum and $ZnO$  to Zinc. 

Note: Only that reagent will be preferred as a reducing agent which will lead to decrease in free energy value $\Delta {{G}^{0}}$at a certain specific temperature.

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 $CaC{{l}_{2}}$  at 873 K. Sodium is discharged at the cathode while $C{{l}_{2}}$  is obtained at the anode as a by-product.

\[NaCl(l)\overset{Electrolysis}{\mathop{\to }}\,N{{a}^{+}}(melt)+C{{l}^{-}}(melt)\] 

At Cathode: $N{{a}^{+}}+{{e}^{-}}\to Na$ 

At Anode:  $C{{l}^{-}}\to  

 \frac{1}{2}C{{l}_{2}}+{{e}^{-}}$ 

If, an aqueous solution of $NaCl$is electrolyzed, ${{H}_{2}}$  is evolved at the cathode while $C{{l}_{2}}$ is obtained at the anode.

25. What is the Role of Graphite Rod in the Electrometallurgy of Aluminium?

Ans: In the electrometallurgy of aluminium, oxygen gas is evolved at anode. ${{O}_{2}}$ reacts with graphite or carbon (graphite electrodes) to form carbon monoxide and carbon dioxide. In case of some other metal carbon (graphite electrodes) to form carbon monoxide and carbon dioxide. In case of some other metal aluminium oxide ($A{{l}_{2}}{{O}_{3}}$ ) which will pass into the reaction mixture resulting into wastage of Aluminium. Since graphite is cheaper than aluminium, its wastage can be tolerated.

26. Outline the Principles of Refining of Metals by the Following Methods.

1. Zone Refining: 

Ans: 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 a 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. 

2. Electrolytic Refining: 

Ans: Many metals, such as Cu, Ag, Au, Al, Pb, etc., are purified by this method. The impure metal is made the anode while a thin sheet of pure metal acts as a cathode. The electrolytic solution consists of a salt or a complex salt solution of the metal. On passing the current, the pure metal is deposited on the cathode while the impurities fall down as anode mud. 

3. Vapour-Phase Refining: 

Ans: The crude metal is freed from impurities by first converting it into a suitable volatile compound by heating it with a specific reagent at a lower temperature and then decomposing the volatile compound at some higher temperature to give the pure metal. 

(a) Mond’s Process: When impure nickel is heated with a current of $CO$ at 330-350 K, it forms a volatile nickel tetracarbonyl complex leaving behind the impurities. The complex is then heated to a higher temperature (450.470K) when it undergoes thermal decomposition giving pure nickel.

\[\underset{(impure)}{\mathop{Ni}}\,+4CO\overset{330-350K}{\mathop{\to }}\,Ni{{(CO)}_{4}}\] 

\[Ni{{(CO)}_{4}}\overset{450-470K}{\mathop{\to }}\,\underset{(pure)}{\mathop{Ni}}\,+4CO\] 

(b) Van Arkel Method: This method is Used for preparing ultra-pure metals by removing all the oxygen and nitrogen present as impurities in metals like zirconium and titanium (which are used in space technology).Crude Zirconium is heated in an evacuated vessel with iodine at 870 K. Zirconium tetraiodide thus formed is separated. It is then decomposed by heating over a tungsten filament at 1800–2075 K to give pure Zirconium.

\[\underset{(impure)}{\mathop{Zr(s)}}\,+2{{I}_{2}}\to Zr{{I}_{4}}\] 

\[Zr{{I}_{4}}\underset{Tungstenfilament}{\overset{1800-2075K}{\mathop{\to }}}\,\underset{pure}{\mathop{Zr(s)}}\,+2I(g)\] 

27. Predict Conditions Under Which Al Might be Expected to Reduce $MgO$.

Ans: The equations for the formation of the two oxides are

(i) $\frac{4}{3}Al(s)+{{O}_{2}}(s)\to \frac{2}{3}A{{l}_{2}}{{O}_{3}}(s)$ 

(ii) $2Mg(s)+{{O}_{2}}(s)\to 2MgO(s)$ 

If we look at the plots for the formation of the two oxides of the Ellingham diagram, we find that they intersect at certain points. The corresponding value of $\Delta {{G}^{0}}$  becomes zero for the reduction of $MgO$ by Al metal.

\[2MgO(s)+\frac{4}{3}Al(s)\rightleftharpoons 2Mg(s)+\frac{2}{3}A{{l}_{2}}{{O}_{3}}(s)\] 

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 $\Delta {{G}^{0}}$for $A{{l}_{2}}{{O}_{3}}$ is less as compared to that of $MgO$

\[3MgO(s)+2Al(s)\overset{(>1665K)}{\mathop{\to }}\,A{{l}_{2}}{{O}_{3}}(s)+3Mg(s)\] 

Important Topics Covered in NCERT Solutions for Class 12 Chemistry Chapter 6

The concepts of ores, extraction of metals, thermodynamic principles of metallurgy, etc. are discussed in this chapter. This chapter is designed to lay the foundation of the advanced concepts of metallurgy. 

The important topics covered in General Principles and Processes of Isolation of Elements are listed below.

• Occurrence of Metals

• Extraction of Crude Metal from Ores

• Concentration of Ores

• Oxidation-Reduction

• Thermodynamic Principles of Metallurgy

• Uses of Aluminium, Zinc, Copper, and Iron

• Electrochemical Principles of Metallurgy

• Refining

Students are suggested to learn and revise all the topics and subtopics covered in CBSE Class 12 Chemistry Chapter 6 thoroughly to answer all the questions asked in the 12th Board exam from this chapter.

NCERT Solutions For Class 12 Chemistry Chapters 6 PDF

Class 12 Chemistry Chapter 6 NCERT Solutions provide the answers to the in-text and exercise questions. The questions are mostly theoretical, and can be categorised primarily in two types:

• Long Answers

• Short Answers

Both these types of questions have been interspersed in the textbook. Students using NCERT Solutions for Class 12 Chemistry Chapter 6 PDF will find it useful for their daily study routine.

NCERT Solutions for Class 12 Chemistry PDF Download

• Chapter 1 - The Solid State

• Chapter 2 - Solutions

• Chapter 3 - Electrochemistry

• Chapter 4 - Chemical Kinetics

• Chapter 5 - Surface Chemistry

• Chapter 6 - General Principles and Processes of Isolation of Elements

• Chapter 7 - The p-Block Elements

• Chapter 8 - The d and f Block Elements

• Chapter 9 - Coordination Compounds

• Chapter 10 - Haloalkanes and Haloarenes

• Chapter 11 - Alcohols, Phenols and Ethers

• Chapter 12 - Aldehydes, Ketones and Carboxylic Acids

• Chapter 13 - Amines

• Chapter 14 - Biomolecules

• Chapter 15 - Polymers

• Chapter 16 - Chemistry in Everyday life

Benefits of NCERT Solutions for Class 12 Chemistry Chapter 6 

• The NCERT Solutions for General Principles and Processes of Isolation of Elements are prepared by our subject experts as per the latest guidelines for CBSE Class 12. Thus, students can learn the format for answering questions in their 12th Board exams by referring to these solutions.

• These NCERT Solutions provide a concise explanation of the topics covered in the chapter. So these solutions come in handy for a quick revision of the chapter.

• Students can refer to these solutions online or download them for their offline self-study purpose, all free of cost.

• These solutions are as per the latest syllabus of CBSE Class 12 Chemistry. Hence, students can rely upon these solutions for their exam preparation.

• Properly labelled diagrams, balanced equations, and logical explanations are provided wherever required. 

Why Vedantu?

We at Vedantu offer comprehensive study resources and customised online classes ensuring a seamless learning experience for all students. We provide NCERT Solutions, CBSE Class 12 Chemistry previous years’ question papers with solutions, sample papers, reference book solutions, topic-wise short notes, Class 12 Chemistry chapter-wise important questions, and revision notes for CBSE Class 12 Chemistry, NCERT Exemplar Solutions, and several other study resources to help students with their exam preparation for Class 12 Chemistry. Similar study materials are available on Vedantu for the other subjects of CBSE Class 12. All our study resources are available for free. 

 

Our experts guide students with their doubts and make it a point to encourage them to give their best in every exam. We also provide mock test series for JEE and NEET. So sign up on Vedantu and avail the best study resources for your exam preparation!