Reduction to free Metal Questions in English

(D) The cyanide process (leaching) is primarily used for the extraction of $Gold$ $(Au)$ and $Silver$ $(Ag)$ from their ores.
In the froth flotation process for $Zinc$ sulfide $(ZnS)$,$NaCN$ or $KCN$ is used as a depressant to prevent $ZnS$ from forming froth.
However,$Copper$ $(Cu)$ extraction from its ores (like $CuFeS_2$) typically involves roasting and smelting,not a cyanide-based leaching process.

(D) During the extraction of copper from copper pyrites $(CuFeS_{2})$,the ore is roasted to remove sulfur as $SO_{2}$ and convert iron and copper into oxides.
In the smelting process,silica $(SiO_{2})$ is added as a flux to remove the iron oxide $(FeO)$ impurity.
$FeO$ reacts with $SiO_{2}$ to form iron silicate $(FeSiO_{3})$,which is removed as slag:
$FeO + SiO_{2} \rightarrow FeSiO_{3}$ (slag).
Therefore,only $FeO$ is removed as slag.

(A) The Gibbs free energy change is given by the equation $\Delta G = \Delta H - T \Delta S$.
Since the entropy of a liquid is higher than that of a solid,the entropy change $(\Delta S)$ for the formation of liquid metal is more positive compared to solid metal.
As $\Delta S$ increases,the term $-T \Delta S$ becomes more negative,which makes the overall $\Delta G^{\ominus}$ value more negative.
$A$ more negative $\Delta G^{\ominus}$ indicates a more spontaneous reaction,making the reduction of metal oxide easier.
Therefore,both Assertion $A$ and Reason $R$ are correct,and $R$ is the correct explanation of $A$.

(A) The correct option is $A$.
Extraction of silver is achieved by the initial complexation of the ore (argentite,$Ag_2S$) with cyanide ion $(CN^-)$ to form a soluble complex,followed by reduction with zinc $(Zn)$.
This process is known as the Mac-Arthur Forrest cyanide process.
The chemical reactions are:
$Ag_2S + 4CN^- \rightarrow 2[Ag(CN)_2]^- + S^{2-}$
$2[Ag(CN)_2]^- + Zn \rightarrow [Zn(CN)_4]^{2-} + 2Ag$
Here,$X = CN^-$ and $Y = Zn$.

(B) The metal is $Ag$ (Silver).
In the extraction of silver,the ore is treated with a dilute solution of $NaCN$ or $KCN$ in the presence of air $(O_2)$,which acts as an oxidizing agent to dissolve the metal as a complex:
$4Ag + 8CN^{-} + O_2 + 2H_2O \rightarrow 4[Ag(CN)_2]^{-} + 4OH^{-}$
Subsequently,the metal is recovered from the complex by displacement using a more electropositive metal like zinc $(Zn)$,which acts as a reducing agent:
$2[Ag(CN)_2]^{-} + Zn \rightarrow 2Ag \downarrow + [Zn(CN)_4]^{2-}$

(D) The copper ore contains iron as an impurity.
It is mixed with silica $(SiO_2)$ before heating in a reverberatory furnace.
$FeO$ acts as a basic impurity and reacts with acidic silica to form iron silicate slag $(FeSiO_3)$,which can be easily removed.
$FeO + SiO_2 \longrightarrow FeSiO_3$

(D) The reaction for the oxidation of carbon is $2C(s) + O_2(g) \rightarrow 2CO(g)$.
The entropy change $\Delta_{r}S^{\circ}$ is positive because the number of moles of gaseous products is greater than the number of moles of gaseous reactants.
According to the Gibbs-Helmholtz equation,$\Delta_{r}G^{\circ} = \Delta_{r}H^{\circ} - T\Delta_{r}S^{\circ}$. Since $\Delta_{r}S^{\circ} > 0$,the slope of the line in the Ellingham diagram (which is $-\Delta_{r}S^{\circ}$) is negative.
Therefore,Assertion $A$ is correct.
Carbon monoxide $(CO)$ is a very stable compound at high temperatures,and its tendency to decompose decreases as temperature increases. Thus,Reason $R$ is incorrect.

(A) In the Hall-Heroult process,the electrolytic reduction of $Al_2O_3$ is carried out in a steel vessel lined with carbon.
Graphite rods act as the anode and the carbon lining acts as the cathode.
During the process,$Al_2O_3$ is reduced to $Al$ at the cathode,while the graphite anode is consumed as it reacts with oxygen to form $CO$ and $CO_2$.

(B) In an Ellingham diagram,a metal oxide can be reduced by a reducing agent if the line for the formation of the reducing agent's oxide lies below the line for the formation of the metal oxide at that temperature.
At $600^{\circ} C$:
$1$. The line for $2C + O_2 \rightarrow 2CO$ is below the line for $2Fe + O_2 \rightarrow 2FeO$. Thus,$C$ can reduce $FeO$.
$2$. The line for $2C + O_2 \rightarrow 2CO$ is above the line for $2Zn + O_2 \rightarrow 2ZnO$. Thus,$C$ cannot reduce $ZnO$.
$3$. The line for $2CO + O_2 \rightarrow 2CO_2$ is below the line for $2Fe + O_2 \rightarrow 2FeO$. Thus,$CO$ can reduce $FeO$.
$4$. The line for $2CO + O_2 \rightarrow 2CO_2$ is above the line for $2Zn + O_2 \rightarrow 2ZnO$. Thus,$CO$ cannot reduce $ZnO$.
Therefore,the correct statement is that at $600^{\circ} C$,$C$ can reduce $FeO$.

(A) The reactions given are:
$(i) 2 Cu_2S + 3 O_2 \rightarrow 2 Cu_2O + 2 SO_2$
$(ii) 2 Cu_2O + Cu_2S \rightarrow 6 Cu + SO_2$
These reactions represent the self-reduction process in the extraction of copper.
Due to the evolution of $SO_2$ gas during the solidification of the molten copper,the surface of the metal develops bubbles or blisters.
Therefore,the copper obtained is referred to as blister copper.

(B) The Ellingham diagram plots $\Delta G^{\circ}$ versus $T$. The slope of the line is $-\Delta S$.
At the melting point of the metal $(Mg)$,the metal changes from solid to liquid state $(Mg_{(s)} \rightarrow Mg_{(l)})$.
This phase change involves a significant change in entropy $(\Delta S)$.
Since the entropy of the liquid state is higher than that of the solid state,the value of $\Delta S$ for the reaction $2Mg_{(l)} + O_2(g) \rightarrow 2MgO_{(s)}$ becomes more negative than for $2Mg_{(s)} + O_2(g) \rightarrow 2MgO_{(s)}$.
Consequently,the slope $(-\Delta S)$ increases sharply at the melting point $(\sim 1120^{\circ} C)$.
Thus,both Assertion $A$ and Reason $R$ are true,and $R$ is the correct explanation of $A$.

(B) . Self reduction: Used for metals like $Cu$ (from $Cu_2S$) and $Pb$ (from $PbS$). Thus,$A-P, R$.
$B$. Carbon reduction: Used for metals like $Pb$ (from $PbO$) and $Cu$ (from $CuO$). Thus,$B-P, R$.
$C$. Complex formation and displacement by metal: Used for $Ag$ (MacArthur-Forrest cyanide process). Thus,$C-Q$.
$D$. Decomposition of iodide: Used for refining of $B$ (Van Arkel method). Thus,$D-S$.
Matching: $A-P, R; B-P, R; C-Q; D-S$. Therefore,the correct option is $B$.

(B) Native silver metal $(Ag)$ reacts with a dilute aqueous solution of $NaCN$ in the presence of atmospheric oxygen $(O_2)$ to form a water-soluble complex,sodium dicyanoargentate$(I)$.
The chemical equation for this process is:
$4 Ag + 8 NaCN + 2 H_2O + O_2 \longrightarrow 4 Na[Ag(CN)_2] + 4 NaOH$

(C) $1.$ Partial roasting of chalcopyrite: $2CuFeS_2 + O_2 \rightarrow Cu_2S + 2FeS + SO_2 \uparrow$. Further roasting: $2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2 \uparrow$ and $2FeS + 3O_2 \rightarrow 2FeO + 2SO_2 \uparrow$. Thus,the products are $Cu_2O$ and $FeO$.
$2.$ Iron is removed as slag by adding silica $(SiO_2)$: $FeO + SiO_2 \rightarrow FeSiO_3$ (slag).
$3.$ In self-reduction: $2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2 \uparrow$. Here,the sulfur in $Cu_2S$ (as $S^{2-}$) is oxidized to $SO_2$ (where $S$ is $+4$),making $S^{2-}$ the reducing agent.

(A) The ore cassiterite is $SnO_{2}$.
$1$. The extraction involves the reduction of the oxide ore $SnO_{2}$ using carbon (smelting): $SnO_{2} + 2C \rightarrow Sn + 2CO$.
$2$. Cassiterite often contains iron as an impurity in the form of $FeWO_{4}$ (ferberite) or other iron oxides. This is removed by magnetic separation or by converting it into slag during the smelting process.
Therefore,the process involves carbon reduction of an oxide ore and the removal of iron impurity.

(D) The extraction of gold involves the following reactions:
$4Au_{(s)} + 8CN^{-}_{(aq)} + 2H_2O_{(l)} + O_{2(g)} \rightarrow 4[Au(CN)_2]^{-}_{(aq)} + 4OH^{-}_{(aq)}$
Here,$Q = O_2$ and $R = [Au(CN)_2]^{-}$.
The complex $R$ is then treated with zinc $(T = Zn)$ to displace gold:
$2[Au(CN)_2]^{-}_{(aq)} + Zn_{(s)} \rightarrow [Zn(CN)_4]^{2-}_{(aq)} + 2Au_{(s)}$
Here,$Z = [Zn(CN)_4]^{2-}$.
All statements $1, 2, 3,$ and $4$ are correct.

(B) The chemical equation for the reduction of cassiterite $(SnO_2)$ by coke $(C)$ is: $SnO_{2(s)} + C_{(s)} \longrightarrow Sn_{(s)} + CO_{2(g)}$
Calculate the standard enthalpy of reaction: $\Delta H^{\circ}_{rxn} = [\Delta_{f}H^{\circ}(Sn_{(s)}) + \Delta_{f}H^{\circ}(CO_{2(g)})] - [\Delta_{f}H^{\circ}(SnO_{2(s)}) + \Delta_{f}H^{\circ}(C_{(s)})]$
Since $\Delta_{f}H^{\circ}$ for elements in their standard state is $0$,$\Delta H^{\circ}_{rxn} = [-394.0] - [-581.0] = 187.0 \ kJ \ mol^{-1} = 187000 \ J \ mol^{-1}$
Calculate the standard entropy of reaction: $\Delta S^{\circ}_{rxn} = [S^{\circ}(Sn_{(s)}) + S^{\circ}(CO_{2(g)})] - [S^{\circ}(SnO_{2(s)}) + S^{\circ}(C_{(s)})]$
$\Delta S^{\circ}_{rxn} = [52.0 + 210.0] - [56.0 + 6.0] = 262.0 - 62.0 = 200.0 \ J \ K^{-1} \ mol^{-1}$
For the reaction to be spontaneous,$\Delta G^{\circ} < 0$. At equilibrium,$\Delta G^{\circ} = 0$,so $T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}$
$T = \frac{187000 \ J \ mol^{-1}}{200.0 \ J \ K^{-1} \ mol^{-1}} = 935 \ K$

(B) In the extraction of silver,argentite ore $(Ag_2S)$ is leached with a dilute solution of $NaCN$ in the presence of air $(O_2)$:
$Ag_2S + 4NaCN + \frac{1}{2}O_2 + H_2O \longrightarrow 2Na[Ag(CN)_2] + S + 2NaOH$
Here,$O_2$ acts as an oxidizing agent to facilitate the dissolution of silver.
Then,the silver is recovered from the complex by adding zinc dust,which acts as a reducing agent:
$2[Ag(CN)_2]^- + Zn \longrightarrow [Zn(CN)_4]^{2-} + 2Ag$
Thus,$O_2$ is the oxidizing agent and $Zn$ dust is the reducing agent.

(D) Carbon reduction is suitable for metals with moderate reactivity like $Sn$ and $Fe$.
$Al_2O_3$ (Aluminium oxide) and $Mg$ (Magnesium) are highly reactive metals.
Aluminium is extracted by the electrolytic reduction of molten $Al_2O_3$ in the presence of cryolite.
Magnesium is extracted by the electrolytic reduction of fused $MgCl_2$ or $MgO$.
Therefore,the carbon-based reduction method is $NOT$ used for $C$ and $D$.

(D) The process of self-reduction (or auto-reduction) is used in the extraction of copper from its sulfide ores.
$1. \ Cu_2S + 2 Cu_2O \longrightarrow 6 Cu + SO_2$
$2. \ Cu_2S + 2 CuO \longrightarrow 4 Cu + SO_2$
$3. \ Cu_2S + CuSO_4 \longrightarrow 3 Cu + 2 SO_2$
$CuFeS_2$ (chalcopyrite) is an ore that requires roasting and smelting to produce matte $(Cu_2S + FeS)$,but it does not directly produce copper metal upon simple heating with $Cu_2S$ in the same manner as the oxides or sulfate. Therefore,the reagents that produce copper metal are $CuO$,$Cu_2O$,and $CuSO_4$.

(B) Bessemerization is a process used in the metallurgy of $Copper$ to remove impurities like $FeS$ and $Cu_2S$ by oxidation. In this process,air is blown through the molten matte,which oxidizes the iron sulfide to iron oxide,which is then removed as slag.

(D) In the blast furnace,different temperature zones exist for different processes.
At $1500 \ K$,the formation of slag occurs where calcium oxide $(CaO)$ reacts with silica $(SiO_{2})$ impurity to form calcium silicate $(CaSiO_{3})$.
The reaction is: $CaO + SiO_{2} \rightarrow CaSiO_{3} \ (\text{slag})$.

(D) Copper is extracted from the ore copper pyrite $(CuFeS_{2})$ by smelting in a blast furnace.
In this process,$FeO$ is formed as an impurity (gangue).
To remove this impurity,silica $(SiO_{2})$ is added as an acidic flux.
It reacts with $FeO$ to form a fusible slag,$FeSiO_{3}$.
$FeO + SiO_{2} \rightarrow FeSiO_{3}$ (slag).

(A) Limestone $(CaCO_3)$ is used as a flux in the extraction of iron.
In the blast furnace,it decomposes at high temperatures to form calcium oxide $(CaO)$ and carbon dioxide $(CO_2)$.
$CaCO_3 \rightarrow CaO + CO_2$
The calcium oxide $(CaO)$ acts as a basic flux and reacts with the acidic impurity silica $(SiO_2)$ present in the ore to form fusible calcium silicate slag $(CaSiO_3)$.
$CaO + SiO_2 \rightarrow CaSiO_3$ (slag)

(C) In the Bessemer converter,air is blown through the molten matte. The remaining $FeS$ is oxidized to $FeO$,which reacts with silica to form slag $(FeSiO_3)$. Subsequently,a portion of $Cu_2S$ is oxidized to $Cu_2O$. This $Cu_2O$ then reacts with the remaining $Cu_2S$ to produce metallic copper through a process known as auto-reduction (or self-reduction): $2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2 \uparrow$.

(B) In the blast furnace,the extraction of iron involves different temperature zones.
At the zone of slag formation (approx. $1073-1273 \ K$),the limestone $(CaCO_3)$ decomposes to $CaO$.
This $CaO$ acts as a flux and reacts with the silica $(SiO_2)$ impurity present in the ore to form calcium silicate,which is the slag.
The reaction is: $CaO + SiO_2 \longrightarrow CaSiO_3 \text{ (slag)}$.

(C) In an Ellingham diagram,a sudden change in the slope of the $\Delta G^{\circ}$ vs $T$ plot indicates a phase change (melting or boiling) of the metal or the oxide.
Looking at the provided Ellingham diagram,the line corresponding to the formation of $Al_{2}O_{3}$ (represented by the reaction $\frac{4}{3} Al + O_{2} \rightarrow \frac{2}{3} Al_{2}O_{3}$) shows a distinct change in slope at a specific temperature,which corresponds to the melting point of aluminum.
Therefore,the correct option is $C$.

(B) In a blast furnace,the tuyeres are pipes located near the base of the furnace.
Their primary function is to blow a blast of preheated air into the furnace to facilitate the combustion of coke.
This combustion process produces $CO_2$ and generates a significant amount of heat,which is essential for the reduction of iron oxides.

(D) In the blast furnace,different temperature zones exist.
At the bottom of the furnace,the temperature is approximately $2000 \ K$.
In this high-temperature zone,carbon reacts with oxygen to form carbon monoxide: $2C + O_2 \longrightarrow 2CO$.
This reaction is exothermic and provides the heat required for the process.

(D) In an Ellingham diagram,the standard free energy change $(\Delta G^{\circ})$ is plotted against temperature for the oxidation of elements. The reactions are normalized such that they involve exactly $1 \text{ mole}$ of gaseous oxygen $(O_2)$ at $1 \text{ atm}$ pressure. For example,the reaction for the formation of magnesium oxide is written as $2Mg(s) + O_2(g) \rightarrow 2MgO(s)$,and for aluminum oxide,it is $\frac{4}{3}Al(s) + O_2(g) \rightarrow \frac{2}{3}Al_2O_3(s)$. Thus,$1 \text{ mole}$ of $O_2$ is used.

(D) The tendency to form an oxide is related to the standard Gibbs free energy of formation $({\Delta}G_f^{\circ})$ of the oxide. Elements with a less negative or positive ${\Delta}G_f^{\circ}$ have a lower tendency to form oxides. Among the given elements,$Hg$ (mercury) is a noble metal with a very low affinity for oxygen,making it the least likely to form a stable oxide under standard conditions.

(A) The stability of metal oxides is determined by their standard Gibbs free energy of formation $(\Delta G_f^\circ)$.
More negative values of $\Delta G_f^\circ$ indicate higher stability.
Based on the Ellingham diagram,the order of stability is:
$Fe_{2}O_{3} < Cr_{2}O_{3} < Al_{2}O_{3} < MgO$.

(B) In the Ellingham diagram,the line for the formation of $CO$ from $C$ and $O_{2}$ intersects the line for the formation of $CO_{2}$ at $983 \ K$.
Below $983 \ K$,the formation of $CO_{2}$ is more spontaneous (more negative $\Delta_{f}G^{\circ}$),meaning $CO_{2}$ is more stable than $CO$.
Above $983 \ K$,the formation of $CO$ becomes more spontaneous (more negative $\Delta_{f}G^{\circ}$),meaning $CO$ is more stable than $CO_{2}$.
Therefore,the statement that '$CO$ is less stable than $CO_{2}$ at more than $983 \ K$' is false.

(D) The extraction of gold involves the leaching of gold with a cyanide solution in the presence of air $(O_2)$:
$4Au + 8CN^{-} + 2H_2O + O_2 \longrightarrow 4[Au(CN)_2]^{-} + 4OH^{-}$
In this reaction,the gold is oxidized to the complex $[Au(CN)_2]^{-}$.
Next,the gold is recovered from the complex by displacement with zinc:
$2[Au(CN)_2]^{-} + Zn \longrightarrow 2Au + [Zn(CN)_4]^{2-}$
Comparing these with the given reactions,$X$ is $[Au(CN)_2]^{-}$ and $Y$ is $[Zn(CN)_4]^{2-}$.

(A) In the Ellingham diagram,the slope of the line for the formation of $CO$ is negative because the reaction $2C(s) + O_2(g) \rightarrow 2CO(g)$ involves an increase in the number of moles of gas $(\Delta n_g = 2 - 1 = 1)$.
According to the equation $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$,since $\Delta S^{\circ}$ is positive,the term $-T\Delta S^{\circ}$ becomes more negative as temperature $(T)$ increases.
Therefore,the formation of $CO$ becomes more thermodynamically stable and favorable at higher temperatures compared to metal oxides like $FeO$,$ZnO$,or $Cu_2O$,which show positive slopes.

(D) Copper is extracted from copper pyrites $(CuFeS_{2})$ by the auto-reduction process. The steps are as follows:
$2 CuFeS_{2} + O_{2} \longrightarrow Cu_{2}S + 2 FeS + SO_{2}$
$2 FeS + 3 O_{2} \longrightarrow 2 FeO + 2 SO_{2}$
$FeO + SiO_{2} \longrightarrow FeSiO_{3}$
$2 Cu_{2}S + 3 O_{2} \longrightarrow 2 Cu_{2}O + 2 SO_{2}$
$2 Cu_{2}O + Cu_{2}S \longrightarrow 6 Cu + SO_{2}$
In the final step,$Cu_{2}O$ reacts with $Cu_{2}S$ to produce metallic copper,which is known as auto-reduction.

(D) According to the Ellingham diagram,the line for the formation of $Fe_2O_3$ lies below the line for the formation of $CO$ at temperatures below $983 \ K$.
This indicates that the formation of $Fe_2O_3$ is more spontaneous (more negative $\Delta G^\circ$) than the formation of $CO$ in this temperature range.
Therefore,iron has a higher affinity for oxygen than carbon below $983 \ K$,making the reduction of $Fe_2O_3$ by carbon non-spontaneous.

(C) In the Bessemer converter,copper is extracted through a self-reduction process. The reactions involved are:
$2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$
$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$
These reactions occur because,at high temperatures,sulphur has a greater affinity for oxygen than copper does,allowing sulphur to remove oxygen from $Cu_2O$ to form $SO_2$ gas. Thus,copper has less affinity for oxygen than sulphur.

(D) In the blast furnace used for the extraction of iron,different zones have different temperature ranges.
The slag formation zone,where limestone $(CaCO_3)$ decomposes to $CaO$ and reacts with silica $(SiO_2)$ to form slag $(CaSiO_3)$,occurs at a temperature range of approximately $800-1000^{\circ} C$.
Therefore,the correct option is $D$.

(D) The incorrect statement about the Hall$-$Heroult process is given in option $(D)$.
In this process,the overall cell reaction is $2Al_2O_3 + 3C \longrightarrow 4Al + 3CO_2$.
In $Al_2O_3$,the oxidation state of oxygen is $-2$,and in $CO_2$,it is also $-2$.
Therefore,the oxidation state of oxygen does not change during the overall cell reaction.

(B) In the blast furnace,the extraction of iron from haematite $(Fe_2O_3)$ involves the following key processes:
$1$. Reduction of iron oxide: $Fe_2O_3$ is reduced by $CO$ to form metallic iron $(Fe)$ and $CO_2$. This is represented by reaction $i$: $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$.
$2$. Slag formation: The impurity $SiO_2$ (silica) is removed by adding limestone $(CaCO_3)$,which decomposes to $CaO$. $CaO$ reacts with $SiO_2$ to form calcium silicate slag $(CaSiO_3)$. This is represented by reaction $iv$: $CaO + SiO_2 \rightarrow CaSiO_3$.
Therefore,reactions $i$ and $iv$ are the main reactions occurring in the blast furnace.

(A) Copper matte is obtained during the extraction of copper from copper pyrites $(CuFeS_{2})$.
It is a molten mixture consisting mainly of copper$(I)$ sulphide $(Cu_{2}S)$ and iron$(II)$ sulphide $(FeS)$.

(C) Cryolite is $Na_{3}AlF_{6}$.
In the Hall-Heroult process,pure alumina $(Al_{2}O_{3})$ has a very high melting point $(2323 \ K)$ and is a poor conductor of electricity.
Cryolite $(Na_{3}AlF_{6})$ is added to the electrolyte to lower the melting point of the mixture to about $1240 \ K$ and to increase its electrical conductivity,making the electrolysis process more efficient.

(B) According to the Ellingham diagram,the line for the formation of $CO_2$ from $C$ and $O_2$ intersects the line for the formation of $Fe_2O_3$ from $Fe$ and $O_2$ at approximately $983 \ K$.
Above $983 \ K$,the Gibbs free energy change $(\Delta G)$ for the formation of $CO_2$ becomes more negative than the $\Delta G$ for the formation of $Fe_2O_3$.
This implies that at temperatures above $983 \ K$,carbon has a higher affinity for oxygen than iron does,allowing carbon to effectively reduce ferric oxide to metallic iron.

(C) In the blast furnace,the reduction of iron oxide is primarily carried out by $CO$ (carbon monoxide) rather than direct reduction by solid carbon as shown in option $C$.
The reaction $2Fe_2O_3 + 3C \longrightarrow 4Fe + 3CO_2$ is not the primary mechanism in the blast furnace.
The other reactions listed: $(A)$ decomposition of limestone,$(B)$ slag formation,and $(D)$ the Boudouard reaction,all occur within the furnace.

(A) The stability of an oxide is determined by the magnitude of its standard Gibbs free energy of formation,$\Delta G^{\circ}$.
$Al_{2}O_{3}$ is more stable than $Cr_{2}O_{3}$,which means the $\Delta G^{\circ}$ value for the formation of $Al_{2}O_{3}$ is more negative (lower) than that of $Cr_{2}O_{3}$.
Therefore,the $\Delta G^{\circ}$ for the formation of $Cr_{2}O_{3}$ is higher than that of $Al_{2}O_{3}$.

(C) Auto-reduction (or self-reduction) is a process used for metals like $Pb$,$Hg$,and $Cu$ where the sulfide ore is partially roasted to form oxide,which then reacts with the remaining sulfide to yield the metal.
For example,$2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$ followed by $2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$.
Titanium $(Ti)$ is a highly reactive metal and cannot be extracted by auto-reduction. It is typically extracted using the Kroll process,which involves the reduction of $TiCl_4$ with magnesium $(Mg)$ or sodium $(Na)$.

(A) The extraction of copper from copper glance $(Cu_2S)$ involves the following reactions:
$2 Cu_2S + 3 O_2 \rightarrow 2 Cu_2O + 2 SO_2$
$2 Cu_2O + Cu_2S \rightarrow 6 Cu + SO_2$
Here,the gas $\underline{X}$ evolved is sulfur dioxide $(SO_2)$.
In $SO_2$,the sulfur atom is $sp^2$ hybridized with one lone pair of electrons.
Due to the presence of one lone pair,the shape of the $SO_2$ molecule is angular or bent.

(D) The extraction of silver involves the formation of a dicyanoargentate$(I)$ complex followed by displacement with zinc.
$4 Ag_{(s)} + 8 CN^{-}_{(aq)} + 2 H_2O_{(l)} + O_{2(g)} \longrightarrow 4[Ag(CN)_2]^{-}_{(aq)} + 4 OH^{-}_{(aq)}$
Here,$[X]^{-}$ is $[Ag(CN)_2]^{-}$. The coordination number of $Ag$ is $2$.
$2[Ag(CN)_2]^{-}_{(aq)} + Zn_{(s)} \longrightarrow [Zn(CN)_4]^{2-}_{(aq)} + 2 Ag_{(s)}$
Here,$[Y]^{2-}$ is $[Zn(CN)_4]^{2-}$. The coordination number of $Zn$ is $4$.
Therefore,the coordination numbers are $2$ and $4$ respectively.

(C) In an Ellingham diagram,the metal whose oxide line is lower can reduce the oxide of the metal whose line is higher.
At point $A$ $(1673 \ K)$,the lines for the formation of $MgO$ and $Al_2O_3$ intersect,meaning $\Delta G^{\circ} = 0$ for the reaction $3MgO + 2Al \rightarrow Al_2O_3 + 3Mg$.
Below $1673 \ K$,the $MgO$ line is below the $Al_2O_3$ line,so $Mg$ can reduce $Al_2O_3$ to $Al$.
Above $1673 \ K$,the $Al_2O_3$ line is below the $MgO$ line,so $Al$ can reduce $MgO$ to $Mg$.
Therefore,the statement that $Al$ can reduce $MgO$ to $Mg$ below $1673 \ K$ is incorrect.