Answer

(a) Following reaction takes place in the extraction of chlorine by electrolysis

$$ 2 H_2 O(l)+2 Cl^{-}(aq) \rightarrow H_2(g)+Cl_2(g)+2 OH^{-}(aq) $$

From the above reaction it is very clear that the oxidation of $Cl^{-}$ion to chlorine gas occurs. Thus, option (a) is the correct answer.

$\Delta^{\circ} G$ for this reaction is $+422 kJ$. Therefore, thermodynamically reaction is not feasible. We know that

$$ \Delta^{\circ} G=-n F E^{\circ} $$

By using this equation value of $E^{\circ}$ can be calculated. Thus,

$$ E^{\circ}=-\frac{\Delta G^{\circ}}{n F}=-2.2 V $$

Therefore, for the occurrence of this reaction, external potential (emf) greater than $2.2 V$ is required. Due to this fact (c) is not the correct option.

• (b) Reduction of $Cl^{-}$ ion to chlorine gas does not occur. Instead, oxidation of $Cl^{-}$ ion to chlorine gas occurs, as shown in the reaction: $$ 2 H_2 O(l)+2 Cl^{-}(aq) \rightarrow H_2(g)+Cl_2(g)+2 OH^{-}(aq) $$

• (c) The overall reaction $\Delta G^{\circ}$ has a positive value ($+422 kJ$), indicating that the reaction is not thermodynamically feasible without external potential. Therefore, $\Delta G^{\circ}$ does not have a negative value.

• (d) A displacement reaction does not take place in the electrolysis of brine. The process involves the oxidation of $Cl^{-}$ ions to chlorine gas and the reduction of water to hydrogen gas, not a displacement reaction.

Answer

(c) When copper ore is mixed with sillica in a reverberatory furnace copper matte is produced. The copper matte contains sulphide of copper (I) and iron (II).

Copper matte $\rightarrow Cu_2 S$ and $FeS$

• (a) sulphides of copper (II) and iron (II): This option is incorrect because in the copper matte, copper is present as copper (I) sulfide ($Cu_2S$) and not as copper (II) sulfide ($CuS$). The iron is correctly identified as iron (II) sulfide ($FeS$).

• (b) sulphides of copper (II) and iron (III): This option is incorrect because in the copper matte, copper is present as copper (I) sulfide ($Cu_2S$) and not as copper (II) sulfide ($CuS$). Additionally, iron is present as iron (II) sulfide ($FeS$) and not as iron (III) sulfide ($Fe_2S_3$).

• (d) sulphides of copper (I) and iron (III): This option is incorrect because while copper is correctly identified as copper (I) sulfide ($Cu_2S$), iron is present as iron (II) sulfide ($FeS$) in the copper matte, not as iron (III) sulfide ($Fe_2S_3$).

Answer

(d) $Cu_2 O+\frac{1}{2} Cu_2 S \longrightarrow 3 Cu+\frac{1}{2} SO_2$

This reaction includes reduction of copper (I) oxide by copper (I) sulphide. In this process, copper is reduced by itself hence this process is known as autoreduction and the solidified copper. So, obtained is known as blister copper.

• (a) $Fe_3 O_4+4 CO \longrightarrow 3 Fe+4 CO_2$: This reaction involves the reduction of iron oxide by carbon monoxide, not by itself. Therefore, it is not an example of autoreduction.

• (b) $Cu_2 O+C \longrightarrow 2 Cu+CO$: This reaction involves the reduction of copper(I) oxide by carbon, not by itself. Hence, it is not an example of autoreduction.

• (c) $Cu^{2+}(aq)+Fe(s) \longrightarrow Cu(s)+Fe^{2+}(aq)$: This reaction involves the reduction of copper ions by iron, not by itself. Therefore, it is not an example of autoreduction.

Answer

(a) Among a number of elements which are available in earth crust, the most abundant elements are aluminium and iron. Aluminium is third most abundant element in earth crust. i.e., $8.3 $% by weight while iron present in earth crust with $4.2 $% by weight. Copper and silver are also found in earth crust but their abundance percentage is low.

• Option (b) is incorrect because copper (Cu) is not among the most abundant elements in the earth’s crust. While aluminum (Al) is indeed abundant, copper’s abundance percentage is much lower compared to iron (Fe).

• Option (c) is incorrect because copper (Cu) is not among the most abundant elements in the earth’s crust. Iron (Fe) is abundant, but copper’s abundance percentage is significantly lower compared to aluminum (Al).

• Option (d) is incorrect because neither copper (Cu) nor silver (Ag) are among the most abundant elements in the earth’s crust. Both elements have much lower abundance percentages compared to aluminum (Al) and iron (Fe).

Answer

(b) Zone refining is based on the principle that the impurities are more soluble in molten state than in solid state of the metal. A circular mobile heater fixed at one end of impure metal rod. The molten zone moves along with heater which is moved forward.

As the heater moves forward, the pure metal crystallises out of the melt and the impurities pass on into adjacent molten zone. The process is repeated several times and the heater is moved in the same direction.

At one end, impurities get concentrated. This end is cut off. e.g., germanium, silicon, gallium etc., are refined by this method.

• (a) Impurities of low boiling metals can be separated by distillation: This option is incorrect because distillation is a process used to separate components based on differences in their boiling points, not on their solubility in molten versus solid states.

• (c) Different components of a mixture are differently adsorbed on an adsorbent: This option is incorrect because adsorption is a surface phenomenon where different components adhere to the surface of an adsorbent material, which is not related to the solubility of impurities in molten and solid states.

• (d) Vapours of volatile compound can be decomposed in pure metal: This option is incorrect because it describes a process involving the decomposition of volatile compounds, which is not related to the principle of solubility differences between molten and solid states in zone refining.

Answer

(c) In the extraction of copper from its sulphide ore, the metal is formed by the reduction of $Cu_2 O$ with $Cu_2 S$. This reaction completes with the process of autoreduction.

Chemical reaction occurring in this reaction is as follows

$$ Cu_2 O+\frac{1}{2} Cu_2 S \longrightarrow 3 Cu+\frac{1}{2} SO_2 $$

In this process, copper appears as blister copper.

• (a) FeS: Iron sulfide (FeS) is not used in the reduction of $Cu_2 O$ to form copper. FeS does not participate in the autoreduction process that is specific to copper extraction.

• (b) $CO$: Carbon monoxide (CO) is a reducing agent used in various metallurgical processes, but it is not used in the reduction of $Cu_2 O$ in the extraction of copper from its sulphide ore. The specific autoreduction process involves $Cu_2 S$.

• (d) $SO_2$: Sulfur dioxide (SO_2) is a byproduct of the reduction reaction and not a reducing agent. It does not reduce $Cu_2 O$ to copper.

Answer

(a) Brine is electrolysed by using inert electrodes. The possible reactions occurring at anode are

$$ \begin{matrix} Cl^{-}(aq) \longrightarrow \frac{1}{2} Cl_2(g)+e^{-} ; & E_Cell^{s}=1.36 V \\ 2 H_2 O(l) \longrightarrow O_2(g)+4 H^{+}+4 e^{-} ; & E_Cell^{s}=1.23 V \end{matrix} $$

The reaction at anode with lower value of $E^{\circ}$ is preferred and therefore water should get oxidised in preference to $Cl^{-}(a q)$. However, $Cl_2$ is produced instead of $O_2$. This unexpected result is explained on the basis of the fact that water needs greater voltage for oxidation to $O_2$ (as it is kinetically slow process) than that needed for oxidation of $Cl^{-}$ions to $Cl_2$.

• Option (b): Although the standard electrode potential for the oxidation of water to oxygen is lower ($E_{\text{Cell}}^{s} = 1.23 , \text{V}$) compared to the oxidation of chloride ions to chlorine ($E_{\text{Cell}}^{s} = 1.36 , \text{V}$), the oxidation of water is kinetically slow and requires a higher overpotential. Therefore, in practice, chloride ions are oxidized preferentially.

• Option (c): This option describes the reduction of sodium ions to sodium metal, which occurs at the cathode, not the anode. The question specifically asks for the reaction at the anode.

• Option (d): This option describes the reduction of hydrogen ions to hydrogen gas, which also occurs at the cathode, not the anode. The question specifically asks for the reaction at the anode.

Answer

(b) In the metallurgy of aluminium, electrolysis is performed in a steel vessel with lining of carbon acts as cathode and graphite acts as anode. During this process graphite anode is oxidised to $CO$ and $CO_2$.

Chemical reaction occurring in this process is as follows

$$ 2 Al_2 O_3+3 C \longrightarrow 4 Al+3 CO_2 $$

This process is known as Hall - Heroult process. The electrolytic reactions are

At cathode $Al^{3+}($ melt $)+3 e^{-} \longrightarrow Al(l)$

At anode $C(s)+O^{2-}($ melt $) \longrightarrow CO(g)+2 e^{-}$

$C(s)+2 O^{2-}$ (melt) $\longrightarrow CO_2(g)+4 e^{-}$

• (a) $Al^{3+}$ is oxidised to $Al(s)$: This statement is incorrect because in the electrolysis process, $Al^{3+}$ ions are reduced, not oxidized. The reduction reaction at the cathode is $Al^{3+} + 3e^{-} \rightarrow Al(s)$.

• (c) oxidation state of oxygen changes in the reaction at anode: This statement is incorrect because the oxidation state of oxygen does not change in the reaction at the anode. The anode reactions involve the oxidation of carbon to form carbon monoxide ($CO$) and carbon dioxide ($CO_2$), but the oxidation state of oxygen remains -2 in both products.

• (d) oxidation state of oxygen changes in the overall reaction involved in the process: This statement is incorrect because the oxidation state of oxygen remains -2 throughout the overall reaction. The overall reaction is $2 Al_2 O_3 + 3 C \rightarrow 4 Al + 3 CO_2$, where oxygen is in the -2 oxidation state in both $Al_2 O_3$ and $CO_2$.

Answer

(a) Copper and zinc are two metals which are generally purified by using electrolyte refining. In this process, impure metal is used as anode and pure metal is used as a cathode. Impurities from the blister copper or impure zinc deposit as anode mud.

• (b) Ge and Si: Germanium (Ge) and Silicon (Si) are typically purified using methods such as zone refining or chemical vapor deposition, not electrolytic refining.

• (c) Zr and Ti: Zirconium (Zr) and Titanium (Ti) are usually purified through processes like the Kroll process or the Van Arkel process, rather than electrolytic refining.

• (d) Zn and Hg: While zinc (Zn) can be purified by electrolytic refining, mercury (Hg) is not typically purified using this method. Mercury is usually purified through distillation.

Answer

(a) Extraction of gold and silver involves leaching the metal with $CN^{-}$ion. The metal is recovered by displacement of metal by some other metal from the complex ion.

This is an oxidation reaction.

$$ 4 Au(s)+8 CN^{-}(aq)+2 H_2 O(aq)+O_2(g) \longrightarrow 4[Au(CN)_2]^{-}(aq)+4 OH^{-}(aq) $$

$$ 4[Au(CN_2)] (aq)+Zn(s) \longrightarrow 2Au(s) + [Zn(CN)_4]^{2-} (aq) $$

Here, $Zn$ acts as a reducing agent.

Direction (Q. Nos. 11-13) Answer the questions on the basis of figure

• (b) Roasting of metal complex: Roasting is a process that involves heating the ore in the presence of oxygen. This method is typically used to convert sulfide ores into oxides, which can then be reduced to extract the metal. However, in the case of gold and silver extraction, the metal is already in a complex ion form with cyanide, and roasting is not applicable for recovering the metal from this complex.

• (c) Calcination followed by roasting: Calcination is the process of heating the ore in the absence of air to remove volatile impurities, and roasting follows to convert the ore into an oxide. This method is generally used for carbonate and sulfide ores. Since gold and silver are leached with cyanide to form a complex ion, calcination and roasting are not relevant steps for recovering the metal from the cyanide complex.

• (d) Thermal decomposition of metal complex: Thermal decomposition involves breaking down a compound by heating it. This method is not suitable for recovering gold and silver from their cyanide complexes because the complex ions are stable and do not decompose easily with heat. Instead, a displacement reaction with another metal, such as zinc, is used to recover the gold and silver.

Thinking Process

This problem is based on concept of Ellingham diagram which relates Gibbs free energy of reaction and temperature. The point above which greater the negative value of formation of compound is observed, compound will form at that temperature. At a particular temperature compound having higher negative value of Gibbs free energy will be formed first.

Answer

(d) In the figure shown above point $A$

$$ \Delta_{f} G_{(C, CO)}^{\circ}<\Delta_{f} G_{(Fe, FeO)}^{\circ} $$

Therefore, above point $A, C$ reduces $FeO$ to $Fe$ forming carbon monoxide. Hence, reduction of $FeO$ by carbon occurs above point $A$ only.

• Option (a) Below temperature at point A: Below point A, the Gibbs free energy change for the formation of CO from C is not sufficiently negative compared to the Gibbs free energy change for the formation of FeO from Fe. Therefore, carbon cannot effectively reduce FeO to Fe below this temperature.

• Option (b) Approximately at the temperature corresponding to point A: At the temperature corresponding to point A, the Gibbs free energy changes for the formation of CO from C and FeO from Fe are approximately equal. This means that the reduction reaction is not favored or is at equilibrium, so carbon cannot effectively reduce FeO to Fe at this temperature.

• Option (c) Above temperature at point A but below temperature at point D: This option is partially correct but not specific enough. While it is true that carbon can reduce FeO to Fe above point A, the critical information is that the reduction starts just above point A. The range between point A and point D is too broad and does not precisely indicate the onset of the reduction reaction.

Answer

(a) Below point $A$ Gibbs free energy change for the formation of $CO_2$ from $CO(\Delta G_{CO_2, CO_2}^{\circ})$ has lower value (more negative value) than Gibbs free energy change for the formation of $FeO(\Delta G_{Fe, FeO})$. Hence, $FeO$ will be reduced by $CO$ only below point $A$.

• (b) FeO cannot be reduced by both carbon monoxide and carbon below point A because the Gibbs free energy change for the formation of CO₂ from CO is more negative than that for the formation of FeO, indicating that CO is a more effective reducing agent than carbon in this context.

• (c) FeO cannot be reduced by carbon only below point A because the Gibbs free energy change for the formation of CO₂ from CO is more negative than that for the formation of FeO, making CO the preferred reducing agent over carbon.

• (d) FeO can be reduced by carbon monoxide below point A, as indicated by the more negative Gibbs free energy change for the formation of CO₂ from CO compared to the formation of FeO. Therefore, it is incorrect to say that FeO cannot be reduced by both carbon and carbon monoxide.

Answer

(a) At point $D, \Delta G$ curve for formation of $CO_2$ from $CO$ and $FeO$ from $Fe$ intersect each other therefore, overall reduction of $FeO$ with $CO$ is zero.

Hence, (a) is the correct choice.

• (b) $\Delta G$ value for the overall reduction reaction with a mixture of $1 mol$ carbon and 1 mol oxygen is positive.

  The statement is incorrect because the mixture of 1 mol carbon and 1 mol oxygen would form carbon dioxide ($CO_2$), which is a highly exothermic reaction. This would result in a negative $\Delta G$ value, not a positive one.

• (c) $\Delta G$ value for the overall reduction reaction with a mixture of $2 mol$ carbon and 1 mol oxygen will be positive.

  The statement is incorrect because the mixture of 2 mol carbon and 1 mol oxygen would form carbon monoxide ($CO$), which is also an exothermic reaction. This would result in a negative $\Delta G$ value, not a positive one.

• (d) $\Delta G$ value for the overall reduction reaction with carbon monoxide is negative.

  The statement is incorrect because at point $D$, the $\Delta G$ value for the overall reduction reaction with carbon monoxide is zero, not negative. This is due to the intersection of the $\Delta G$ curves for the formation of $CO_2$ from $CO$ and $FeO$ from $Fe$.

Answer

$(b, d)$

Below point $B$ and $E, FeO$ will be reduced to $Fe$ by all the three reactions shown above in the question. $\Delta G_{(C, CO_2)}^{\circ}, \Delta G_{(C, CO)}^{\circ}, \Delta G_{(CO_2, CO_2)}^{\circ}$ lie below $\Delta f G_{(Fe, FeO)}^{\circ}$ curve at point $B$ and $E$. Therefore, $FeO$ will be reduced by all three reactions.

Hence, options (b) and (d) are correct choice.

• Option (a) Point A: At point A, the Gibbs free energy changes for the reactions involving carbon and carbon monoxide do not lie below the Gibbs free energy change for the reduction of FeO to Fe. Therefore, FeO will not be reduced by all three reactions at this point.

• Option (c) Point D: At point D, the Gibbs free energy changes for the reactions involving carbon and carbon monoxide do not lie below the Gibbs free energy change for the reduction of FeO to Fe. Therefore, FeO will not be reduced by all three reactions at this point.

Thinking Process

This process is based on concept of purification techniques and formation of cast iron.

Answer

$(a, d)$

Correct statements are

(a) Cast iron is obtained by remelting pig iron with scarp iron and coke using hot air blast.

(d) $Zr$ and $Ti$ are purified by van Arkel method as

$$ \underset{\text { Impure }}{Zr}+2 I_2 \longrightarrow \underset{\text { Volatile }}{ZrI_4} \stackrel{\Delta}{\longrightarrow} \underset{\text { Pure }}{Zr+2 I_2} $$

(b) and (c) can be correctly stated as

(b)In extraction of silver, silver is extracted as anionic complex $[Ag(CN)_2]^{-}$

(c)Nickel is purified by vapour phase refining method.

$$ Ni+4 CO \rightarrow Ni(CO)_4 \xrightarrow{450-470 K} Ni+4 CO $$

• (b) In extraction of silver, silver is extracted as anionic complex $[Ag(CN)_2]^{-}$: The statement is incorrect because silver is not extracted as a cationic complex but as an anionic complex, specifically $[Ag(CN)_2]^{-}$.

• (c) Nickel is purified by zone refining: The statement is incorrect because nickel is not purified by zone refining. Instead, it is purified by the vapour phase refining method, where nickel reacts with carbon monoxide to form a volatile complex, nickel tetracarbonyl, which decomposes upon heating to yield pure nickel.

Thinking Process

This problem is based on extraction of aluminium using Hall-Heroult process.

Answer

$(a, b)$

In the metallurgy of aluminium $Al_2 O_3$ is mixed with $Na_3 AlF_6$ and $CaF_2$ which causes following affects.

(i) Lower the melting point of $Al_2 O_3$ (ii) Increase the conductivity of molten mixture

• (c) $CaF_2$ does not reduce $Al^{3+}$ into $Al(s)$. The reduction of $Al^{3+}$ to $Al(s)$ is achieved through electrolysis, where an electric current is passed through the molten mixture to facilitate the reduction process.
• (d) $CaF_2$ does not act as a catalyst in the Hall-Heroult process. Its primary roles are to lower the melting point of $Al_2 O_3$ and increase the conductivity of the molten mixture.

Thinking Process

This problem is based on the method involved in froth floatation method and function of collector and depressants.

Answer

$(a, c)$

Froth floatation process is used to extract metal from sulphide ore. This method utilises collectors and depressants whose functions are as follows

(i) Collectors enhance the non-wettability of the mineral particles.

(ii) By using depressants in the process two sulphide ores can be separated. e.g., sodium cyanide is used as a depressant to separate lead sulphide ore from zinc sulphide ore.

• Collectors do not enhance the wettability of gangue particles; they enhance the non-wettability of mineral particles, making option (b) incorrect.
• Froth stabilisers do not decrease the wettability of gangue; they help in stabilizing the froth formed during the process, making option (d) incorrect.

Answer

$(b, c)$

Froth floatation method is used to extract metal from sulphide ore. ZnS and PbS can be separated by using depressant and adjusting the proportion of oil to water. Depressant used for this purpose is $NaCN$. It selectively prevents $ZnS$ from coming to the froth.

Hence, (b) and (c) are correct choices.

• (a) using collectors: Collectors are reagents that increase the non-wettability of the mineral particles, making them attach to air bubbles and rise to the froth. However, collectors do not specifically help in separating zinc sulphide (ZnS) from lead sulphide (PbS) as they generally enhance the floatability of both types of sulphide minerals.

• (d) using froth stabilisers: Froth stabilisers are used to maintain the froth formed during the floatation process, ensuring that it is stable enough to carry the mineral particles to the surface. They do not play a role in selectively separating ZnS from PbS.

Answer

$(c, d)$

Bauxite is an ore of aluminium which contain $Fe_2 O_3$ and $SiO_2$ as common impurities.

• (a) $CuO$: Copper oxide ($CuO$) is not a common impurity in bauxite. Bauxite primarily contains impurities such as iron oxides and silica, not copper compounds.

• (b) $ZnO$: Zinc oxide ($ZnO$) is not typically found in bauxite. The common impurities in bauxite are iron oxides and silica, rather than zinc compounds.

Answer

$(b, c)$

Haematite $(Fe_2 O_3)$ and magnetite $(Fe_3 O_4)$ are oxide ores while galena ( $.PbS)$ and copper pyrites $(CuFeS_2)$ are sulphide ores. As we know sulphide ores are extracted by using froth floatation method. Hence, (b) and (c) are the correct choices.

• Haematite $(Fe_2O_3)$ and magnetite $(Fe_3O_4)$ are oxide ores, and froth flotation is not typically used for concentrating oxide ores.

Thinking Process

This process is based on concept of calcination.

Answer

$(a, c)$

Calcination involves heating of the ore below its melting point in the absence of air or in limited supply of air. Oxygen containing ores like oxide, hydroxides and carbonates are calcined. Thus, the following reactions occur during calcination.

$$ \begin{aligned} & CaCO_3 \stackrel{\Delta}{\longrightarrow} CaO+CO_2 \\ & Al_2 O_3 \cdot x H_2 O \stackrel{\Delta}{\longrightarrow} Al_2 O_3+x H_2 O \end{aligned} $$

• Option (b): The reaction $2 FeS_2+\frac{11}{2} O_2 \longrightarrow Fe_2 O_3+4 SO_2$ involves the oxidation of iron sulfide (FeS₂) in the presence of oxygen, which is a roasting process, not calcination. Calcination occurs in the absence or limited supply of air, whereas roasting requires a significant amount of air or oxygen.

• Option (d): The reaction $ZnS+\frac{3}{2} O_2 \longrightarrow ZnO+SO_2$ also involves the oxidation of zinc sulfide (ZnS) in the presence of oxygen, which is another example of roasting. As with option (b), this process requires a significant amount of air or oxygen, which is not characteristic of calcination.

Answer

$(a, b)$

In the metallurgical process, the oxide ores are reduced by carbon. Sulphide ore cannot be reduced by carbon. Here, haematite $(Fe_2 O_3)$ and calamine $(ZnO)$ are oxide ores of iron and zinc respectively while iron pyrites $(FeS_2)$ and sphalerite $(ZnS)$ are sulphide ores of iron and zinc respectively.

Therefore, haematite and calamine can be reduced by carbon.

• Iron pyrites (FeS₂) is a sulphide ore and cannot be reduced by carbon.
• Sphalerite (ZnS) is a sulphide ore and cannot be reduced by carbon.

Answer

$(a, d)$

In extraction of iron from haematite ore following reactions take place.

(i) $Fe_2 O_3+3 CO \longrightarrow 2 Fe+3 CO_2$

This reaction represents reduction of $Fe_2 O_3$ to $Fe$.

(ii) $\underset{\text { Formation of slag }}{CaO}+{SiO_2} \longrightarrow \underset{Slag}{CaSiO_3}$

• Option (b) $FeO + SiO_2 \longrightarrow FeSiO_3$ is incorrect because this reaction does not occur in the blast furnace. Instead, $FeO$ is reduced to $Fe$ by carbon monoxide ($CO$) or carbon ($C$), and $SiO_2$ reacts with $CaO$ to form slag ($CaSiO_3$).

• Option (c) $Fe_2 O_3 + 3 C \longrightarrow 2 Fe + 3 CO$ is incorrect because, in the blast furnace, carbon monoxide ($CO$) is the primary reducing agent for $Fe_2 O_3$, not carbon ($C$). The reaction involving carbon would require much higher temperatures and is not the main pathway in the blast furnace process.

Thinking Process

This problem is based on purification of metal using vapour phase refining method.

Answer

( $a, b$ )

Vapour phase refining method includes

(i) Heating of metal with stream of $CO$

$Ni+4 CO \rightarrow Ni(CO)_4 {\xrightarrow{450-470 K}} Ni+4 CO$ (Mond’s process)

(ii) Heating with iodine

$$ Zr+2 I_2 \xrightarrow{870 K} ZrI_4 \xrightarrow[\substack{\text { Tungsten } \\ \text { filament }}]{2075 K} Zr+2 I_2 \text { (van Arkel method) } $$

• Liquation: This method involves melting the metal and allowing it to flow away from impurities which have higher melting points. It does not involve the conversion of the metal to a volatile compound.

• Distillation: This method involves heating the metal to its boiling point and then condensing the vapor to obtain the pure metal. It does not involve the formation of a volatile compound that decomposes to give the pure metal.

Answer

$(a, c)$

Correct statements are

(i) A depressant prevents certain type of particle to come to the froth. e.g., $NaCN$ is added as a depressant during separation of $PbS$ and $ZnS$.

(ii) The solidified copper obtained from reverberatory furnace has blistered appearance due to evolution of $SO_2$ during the extraction.

(b) and (d) are incorrect statements, and can be correctly stated as

(iii) Copper matte contains $Cu_2 S$ and $FeS$.

(iv) Zinc can be extracted by reduction of $ZnO$ with carbon.

• (b) Copper matte contains (Cu_2S) and (FeS), not (ZnS).
• (d) Zinc is extracted by reduction of (ZnO) with carbon, not by self-reduction.

Answer

$(b, c)$

Electrolysis of brine solution is used to extract chlorine. Overall chemical reaction occurring in this process and value of $\Delta G^{\circ}$ can be shown as

$$ 2 H_2 O(l)+2 Cl^{-}(aq) \longrightarrow H_2(g)+Cl_2(g)+2 OH^{-}(aq) $$

For the given reaction, value of $\Delta G^{\circ}$ is $422 kJ$

Using $\Delta G^{\circ}=-n F E^{\circ}$, the value of $E^{\circ}=E^{\circ}-2.2 V$.

Therefore, $\Delta G^{\circ}$ for the overall reaction is positive and $E^{\circ}$ has negative value.

• Option (a) is incorrect because the value of $\Delta G^{\circ}$ for the overall reaction is positive, not negative. A positive $\Delta G^{\circ}$ indicates that the reaction is non-spontaneous under standard conditions.

• Option (d) is incorrect because the value of $E^{\circ}$ for the overall reaction is negative, not positive. A negative $E^{\circ}$ indicates that the reaction is non-spontaneous under standard conditions.

Answer

$2 Cl^{-}(aq)+2 H_2 O(l) \longrightarrow 2 OH^{-}(aq)+H_2(g)+Cl_2(g)$

For the given reaction, value of $\Delta G^{\circ}$ is $+422 kJ$.

Using $\Delta G^{\circ}=-n F E^{\circ}$, the value of $E^{\circ}=-2.2 V$.

Therefore, an external emf greater than $2.2 V$ is required for the extraction of $Cl_2$ from brine.

Answer

Using Ellingham diagram, we observe that at temperature greater than $1073 K$; $\Delta G_{(C, CO)}<\Delta G_{(Fe, FeO)}$. We know that according to Ellingham diagram, compound having lower $\Delta_{t} G^{s}$ undergo its formation.

Hence, coke can reduce $FeO$ to $Fe$.

Answer

(a)

$\underset{\substack{\text { Haematite } \\ \text { lining }}}{Fe_2 O_3}+\underset{\begin{matrix} \text { Impurities } \\ \text { (present in cast iron) }\end{matrix} }{3 C} \longrightarrow \underset{\begin{matrix} \text { Wrought } \\ \text { iron }\end{matrix} }{2 Fe}+3 CO$

This reaction takes place in reverberatory furnace lined with haematite.

(b) The haematite oxidises $S$ to $SO_2$, $Si$ to $SiO_2$ and $P$ to $P_4 O_{10}$. Sometimes limestone is added as flux. Impurities of $S, Si$ and $P$ oxidise and pass into slag. The metal is removed and freed from slag by passing through rollers.

Answer

Copper is extracted from low grade copper ore using hydrometallurgy method. For this purpose, ore is leached out using bacteria. The solution containing $Cu^{2+}$ is treated with scrap iron and $H_2$.

$$ Cu^{2+}(aq)+H_2(g) \longrightarrow Cu(s)+2 H^{+}(aq) $$

Answer

Two basic requirements are

(i) Metal should form volatile compound with available reagent.

(ii) The volatile compound should be easily decomposable so that it can be recovered easily.

(a) Mond’s process includes conversion of $Ni$ to $Ni(CO)_4$ and then decomposition of $Ni(CO)_4$ to $Ni$.

$$ \begin{gathered} Ni+4 CO \longrightarrow Ni(CO)_4 \\ Ni(CO)_4 {\xrightarrow{450-470 K}} Ni+4 CO \end{gathered} $$

(b) van Arkel method includes conversion of $Zr$ to volatile $ZrI_4$ and then decomposition of $ZrI_4$ to $Zr$ and $I_2$.

$$ Zr+2 I_2 \stackrel{870 K}{\longrightarrow} ZrI_4 \xrightarrow{2075 K} Zr+2 I_2 $$

Answer

Because at high temperatures, carbon and hydrogen react with metals to form carbides and hydrides respectively. So, carbon and hydrogen are not better reducing agents to reduce metallic oxide.

Answer

Two sulphide ores can be separated by adjusting the proportion of oil and water or by using depressants, e.g., if an ore contains $ZnS$ and PbS both, a depressant $NaCN$ is added during froth floatation. It forms a complex with $ZnS$ and prevents it from coming into froth. PbS comes into the froth and thus separated.

Answer

(a) Haematite $(Fe_2 O_3)$ ore is used to line the furnace.

$$ \underset{\substack{\text { Haematite } \\ \text { lining }}}{\text { (b) } Fe_2 O_3}+\underset{\substack{ \text{impurites}\\ \text { (present in cast iron) }}}{3 C} \rightarrow \underset{\substack{\text { Wrought } \\ \text { iron }}}{2 Fe}+3 CO $$

Answer

Compounds which are adsorbed to the more extent comes out later while the another one which absorbed to the less extent comes out readily. Since, compound ’ $A$ ’ comes out before compound ’ $B$ ‘, the compound ’ $B$ ’ is more readily adsorbed on the column.

Answer

Silica (flux) being an acidic flux removes the impurities of iron oxide the basic impurity, by reacting with it. Thus, iron silicate (slag) is formed.

$$ \underset{\text { Gangue }}{FeO}+\underset{\text { Flux }}{SiO_2} \longrightarrow \underset{\text { Slag }}{FeSiO_3} $$

Answer

This is because oxides are reduced to metals easily while sulphides are not. So, sulphide ores are converted to oxide before reduction.

Answer

$Zr$ and $Ti$ are refined by van Arkel method. It involves 2 steps

(a) Formation of iodide e.g., zirconium

$$ Zr+2 I_2 \longrightarrow ZrI_4 $$

(b) Decomposition of iodide

$$ ZrI_4 \xrightarrow{1800 K} \underset{\text { Pure }}{Zr}+2 I_2 $$

Answer

During extraction of metals by electrochemical method the following two points must be considered

(i) Reactivity of metals If the metals are quite reactive and are expected to react with water then the metals should be extracted by the electrolysis of their purified molten ore rather than their aqueous solution.

(ii) Suitability of electrodes The electrods selected should not react with the product of electrolysis. If at all they react, then the electrodes must be made up of a material which is quite cheap since their periodic replacement should not increase the cost of the process.

Answer

Role of flux in metallurgical process

(i) Flux is used to remove the gangue by combining with it. Thus, slag formation takes place.

(ii) It makes the molten mass more conducting.

Answer

Metals of high grade like germanium, silicon etc., used as semiconductors are refined by zone refining method.

Principle This method is based upon the principle that impurities are more soluble in molten state of metal than in solid state.

Answer

Chemical reactions occurring in blast furnace related to the metallurgy of iron in the temperature range $500-800 K$ are

(i) $3 Fe_2 O_3+CO \longrightarrow 2 Fe_3 O_4+CO_2$

(ii) $Fe_3 O_4+4 CO \longrightarrow 3 Fe+4 CO_2$

(iii) $Fe_2 O_3+CO \longrightarrow 2 FeO+CO_2$

Answer

Two requirements for vapour phase refining are

(i) The metal should form a volatile compound with an available reagent.

(ii) The volatile compound should be easily recovered by decomposition.

Answer

(i) $4 Au(s)+8 CN^{-}(aq)+2 H_2 O(aq)+O_2(g) \longrightarrow 4[Au(CN)_2]^{-}(aq)+4 OH^{-}(aq)$

(ii) $2[Au(CN)_2]^{-}(aq)+Zn(s) \longrightarrow 2 Au(s)+[Zn(CN)_4]^{2-}(aq)$

Zinc acts as a reducing agent in this reaction.

Answer

(b) A. $\rightarrow$ (2)

B. $\rightarrow(4)$

C. $\rightarrow(5)$

D. $\rightarrow(3)$

A. Pendulum is made up of nickel steel.

B. Molecular formula of malachite is $Cu CO_3 \cdot Cu(OH)_2$.

C. Molecular formula of calamine is $ZnCO_3$.

D. Molecular formula of cryolite is $Na_3 AlF_6$.

Answer

(b) A. $\rightarrow$ (4)

B. $\rightarrow$ (3)

C. $\rightarrow(1)$

D. $\rightarrow(2)$

A. Coloured bands are observed in chromatography.

B. Impure metal is converted to volatile complex by using Mond’s process.

C. Purification of $Ge$ ans $Si$ are purified by zone refining method.

D. Purification of mercury is done by fractional distillation.

Answer

(a) A. $\rightarrow$ (4) $\quad$

B. $\rightarrow$ (2) $\quad$

C. $\rightarrow$ (3) $\quad$

D. $\rightarrow$ (1)

A. Cyanide process is used for extraction of Au through formation of anionic complex $.[AuCN)_2]^{-}$.

B. Froth floatation process is used for dressing of ZnS.

C. Electrolytic reduction method is used for extraction of aluminium. Graphite electrode is used for this purpose.

D. Zone-refining is used for purification of Ge.

Answer

(a) A. $\rightarrow$ (3)

B. $\rightarrow(4)$

C. $\rightarrow(2)$

D. $\rightarrow(1)$

A. Sapphire is a gemstone which contain Co.

B. Molecular formula of sphalerite is $ZnS$. C. $NaCN$ is used as a depressant in froth floatation method.

D. Molecular formula of corundum is $Al_2 O_3$.

Answer

(a) A. $\rightarrow$ (2)

B. $\rightarrow(3)$

C. $\rightarrow(4)$

D. $\rightarrow(1)$

A. Blisterred $Cu$ can be prepared by means of following chemical reaction

$$ 2 Cu_2 O+Cu_2 S \longrightarrow 6 Cu+SO_2 $$

B. Iron is extracted by using blast furnace.

C. In reverberatory furnace formation of slag occurs as

$$ FeO+SiO_2 \longrightarrow \underset{Slag}{FeSiO_3} $$

D. Hall-Heroult process is used for extraction of aluminium.

Answer

(a) Both assertion and reason are true and reason is the correct explanation of assertion. Nickel can be purified by Mond’s process in which formation of a volatile compound $Ni(CO)_4$ takes place which further decomposes to $Ni$ at $460 K$.

Answer

(a) Both assertion and reason are true and reason is the correct explanation of assertion. Zirconium can be purified by van Arkel method which include formation of volatile $ZrI_4$ which decomposes at $1800 K$ to $Zr$.

Answer

(b) Both assertion and reason are true but reason is not the correct explanation of assertion.

Sulphide ores are concentrated by froth floatation method. Sulphide ore particles are preferentially wetted by oil, become lighter and rise to the surface along with the froth while gangue particles are preferentially 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 main reason for extraction of metal. Metal ore comes out along with froth.

Answer

(b) Both assertion and reason are true but reason is not the correct explanation of assertion.

Zone refining method is very useful for producing semiconductors of high purity as in this process pure metal crystallises while impurities pass on into adjacent molten zone when impure metal rod is heated.

Answer

(b) Assertion and reason both are correct but reason is not the correct explanation of assertion.

Hydrometallurgy involves dissolving the ore in suitable reagent followed by precipitation with the help of more electropositive metal in which pure metal is get replaced by more electropositive metal.

Thinking Process

This problem is based on concept of Ellingham diagrams and purification of metal.

Answer

(a) As shown in Ellingham diagram which relates Gibbs free energy and temperature at below $710 K$.

$\Delta G_{(C, CO_2)}<\Delta G_{(C, CO)}$ So, $CO_2$ is a better reducing agent than $CO$ while above $710 K$ it becomes a very good reducing agent.

(b) Generally, sulphide ores are converted to oxides before reduction as reduction of oxides can easily be done using $C$ or $CO$ depending upon metal ore and temperature.

(c) Silica is a flux added to the sulphide ore of copper in the reverberatory furnace leading to the formation of slag

$$ FeO+SiO_2 \rightarrow \underset{Slag}{FeSiO_3} $$

(d) Carbon and hydrogen are not used as reducing agents at high temperature. At high temperature carbon and hydrogen readily form their carbides and hydrides respectively.

(e) Vapour phase refining method is used for the purification of $Ti$ as

$$ Ti+2 I_2 {\xrightarrow{523 K}} TiI_4 {\xrightarrow{1700 K}} Ti+2 I_2 $$