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.

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 $FeS$

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.

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.

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.

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.

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 $Cl_2$.

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) $\longrightarrow CO_2(g)+4 e^{-}$

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.

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

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.

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 $A$.

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.

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.

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.

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

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

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.

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.

Answer

$(c, d)$

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

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.

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.

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

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.

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

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

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

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.

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.

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

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