Reduction to free Metal Questions in English

(C) The Hall-Heroult process is the electrolytic reduction of alumina $(Al_2O_3)$ dissolved in molten cryolite $(Na_3AlF_6)$.
The overall reaction is: $2Al_2O_3 + 3C \longrightarrow 4Al + 3CO_2$.
In this process,the reactions occurring are:
Dissociation: $2Al_2O_3 \rightleftharpoons 4Al^{3+} + 6O^{2-}$.
At Cathode: $4Al^{3+} + 12e^- \longrightarrow 4Al$.
At Anode: $6O^{2-} \longrightarrow 3O_2 + 12e^-$.
The oxygen gas produced at the anode reacts with the carbon anode: $2C + 3O_2 \longrightarrow 3CO_2$.

(B) In the $Hall-Heroult$ process,the electrolytic cell consists of a steel vessel lined with carbon,which acts as the cathode.
Therefore,the cathode is made out of carbon.

(B) The Ellingham diagram,which represents the plots of $\Delta G$ versus $T$,helps in predicting the feasibility of the thermal reduction of metal oxides using a suitable reducing agent.

(B) The Goldschmidt reaction,also known as the thermite process,involves the reduction of metal oxides (such as $Fe_2O_3$ or $Cr_2O_3$) using metallic aluminum as a reducing agent.
The reaction is highly exothermic and is represented by the general equation: $M_xO_y + yAl \rightarrow xM + y/2 Al_2O_3 + \text{Heat}$.
Because aluminum has a very high affinity for oxygen,it effectively reduces these oxides to their respective metals.

(C) An acidic flux is used to remove basic impurities from an ore.
$SiO_2$ is an acidic flux because it reacts with basic impurities like $CaO$ or $FeO$ to form a fusible slag.
For example: $CaO + SiO_2 \rightarrow CaSiO_3$ (slag).
$CaO$,$MgO$,and $Na_2O$ are basic oxides and act as basic fluxes.

(B) In the extraction of iron from its ore (hematite,$Fe_2O_3$),limestone $(CaCO_3)$ is added as a flux.
This flux reacts with the silica $(SiO_2)$ impurity present in the ore to form calcium silicate $(CaSiO_3)$,which is known as slag.
$CaO + SiO_2 \rightarrow CaSiO_3$ (Slag).
Slag is a molten byproduct that floats on top of the molten iron and is removed separately.

(A) The extraction of gold and silver involves the formation of a soluble complex with $CN^{-}$ ions in the presence of air $(O_2)$.
$4Au(s) + 8CN^{-}(aq) + 2H_2O(aq) + O_2(g) \rightarrow 4[Au(CN)_2]^{-}(aq) + 4OH^{-}(aq)$
$2Ag(s) + 4CN^{-}(aq) + 2H_2O(aq) + O_2(g) \rightarrow 2[Ag(CN)_2]^{-}(aq) + 4OH^{-}(aq)$
The metal is then recovered from the complex by displacement using a more electropositive metal like zinc $(Zn)$:
$2[Au(CN)_2]^{-}(aq) + Zn(s) \rightarrow [Zn(CN)_4]^{2-}(aq) + 2Au(s)$
$2[Ag(CN)_2]^{-}(aq) + Zn(s) \rightarrow [Zn(CN)_4]^{2-}(aq) + 2Ag(s)$
Thus,the correct method is the displacement of the metal by some other metal from the complex ion.

(A) The Mac-Arthur Forrest process is used for the extraction of silver and gold from their ores.
In the case of silver,the finely powdered ore is treated with a dilute solution of sodium cyanide $(NaCN)$ or potassium cyanide $(KCN)$ in the presence of air.
The chemical reaction is: $4Ag + 8CN^- + 2H_2O + O_2 \rightarrow 4[Ag(CN)_2]^- + 4OH^-$.
The silver is dissolved in the form of the soluble dicyanoargentate$(I)$ complex,$[Ag(CN)_2]^-$.
Therefore,the correct option is $A$.

(B) Highly electropositive metals (like $Na, K, Mg, Ca, Al$) have a very high affinity for oxygen.
Their oxides are extremely stable,meaning the $\Delta G_f$ of these metal oxides is highly negative.
Carbon reduction is not feasible because these metals react with carbon at high temperatures to form stable ionic carbides rather than being reduced to the free metal.
Therefore,these metals are typically extracted by electrolytic reduction of their molten salts.

(B) To determine if $C$ can reduce $ZnO$,we consider the reaction: $ZnO_{(s)} + C_{(s)} \rightarrow Zn_{(s)} + CO_{(g)}$.
This reaction is the difference between reaction $(II)$ and reaction $(I)$:
$\Delta G^o_{reaction} = \Delta G^o_{(II)} - \Delta G^o_{(I)}$
$\Delta G^o_{reaction} = -460\,kJ\,mol^{-1} - (-360\,kJ\,mol^{-1}) = -100\,kJ\,mol^{-1}$.
Since the overall $\Delta G^o$ is negative,the reaction is spontaneous at $1000\,^{\circ}C$.
Therefore,$ZnO$ can be reduced to $Zn$ by $C$.

(A) The Ellingham diagram is a graphical representation that helps in predicting the feasibility of the thermal reduction of an ore.
It consists of plots of the Gibbs free energy change $(\Delta G)$ versus temperature $(T)$ for the formation of oxides of common metals and reducing agents.
The reaction typically represented is: $2xM_{(s)} + O_{2(g)} \rightarrow 2M_xO_{(s)}$.
Since $\Delta G = \Delta H - T\Delta S$,the slope of the line in the Ellingham diagram is equal to $-\Delta S$.

(A) The process described is known as $hydrometallurgy$.
In this process,the ore is treated with a suitable chemical reagent (like an aqueous solution of $NaCN$ or $KCN$ in the case of gold or silver) to dissolve the metal as a complex.
This is followed by the recovery of the metal from the solution by adding a more electropositive metal (like $Zn$),which acts as a reducing agent to precipitate the noble metal.

(D) In the alumino-thermite process,$Al$ acts as a reducing agent.
It reduces metal oxides (like $Fe_2O_3$) to their respective metals while getting oxidized to $Al_2O_3$.

(D) The Goldschmidt alumino-thermite process is a reduction process where metal oxides (like $Fe_2O_3$) are reduced to their respective metals using aluminum powder as a reducing agent.
This reaction is highly exothermic,producing enough heat to keep the metal in a molten state.
The chemical equation is: $2Al + Fe_2O_3 \to Al_2O_3 + 2Fe$.

(B) The extraction of zinc from zinc blende $(ZnS)$ is carried out by first roasting the ore to convert it into zinc oxide $(ZnO)$,followed by reduction with carbon $(C)$.
Step $1$: Roasting: $2 ZnS + 3 O_2 \rightarrow 2 ZnO + 2 SO_2$
Step $2$: Reduction with Carbon: $ZnO + C \xrightarrow{> 1270 \ K} Zn + CO$

(A) Refractory materials are substances that can withstand high temperatures without undergoing chemical or physical changes.
$CaO$ and $MgO$ are basic refractory materials.
$SiO_2$ is an acidic refractory material.
Graphite is a neutral refractory material used in furnaces.

(C) In the extraction of copper from its sulphide ore,the metal is formed by the self-reduction of $Cu_2O$ with $Cu_2S$.
The sulphide ore of copper is heated in air until a part is converted to oxide $(Cu_2O)$.
Then,upon further heating in the absence of air,the oxide reacts with the remaining unchanged sulphide $(Cu_2S)$ to produce metallic copper.
The chemical reactions involved are:
$2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$
$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$

(C) The given reactions represent the process of self-reduction (auto-reduction) of sulphide ores.
Metals like $Hg$ ($Cinnabar$,$HgS$),$Pb$ ($Galena$,$PbS$),and $Cu$ ($Copper \, glance$,$Cu_2S$) are extracted from their sulphide ores using this method.
$Zn$ is extracted from its sulphide ore ($Zinc \, blende$,$ZnS$) by roasting to form $ZnO$,followed by reduction with carbon $(C)$ or carbon monoxide $(CO)$,not by self-reduction.
Therefore,$X$ cannot be $Zn$.

(B) Metal oxides having very high melting points,e.g.,oxides of $Cr$,$Mn$,$Ti$,$Mo$,$Fe$,etc.,are difficult to reduce by carbon reduction. These can be reduced by thermite reduction using $Al$ powder as a reducing agent.
Alumino-thermite process: Certain oxides of $Mn$ and $Cr$ do not reduce by carbon due to their very high melting points. They are reduced by $Al$.
$Cr_2O_3 + 2 Al \rightarrow 2 Cr + Al_2O_3$
$3 Mn_3O_4 + 8 Al \rightarrow 9 Mn + 4 Al_2O_3$
In these reactions,$Al$ undergoes oxidation while reducing the metal oxide to metal. Thus,$Al$ acts as a reducing agent.

(C) Aluminium is a highly reactive metal,and its oxide $(Al_2O_3)$ cannot be reduced by common reducing agents like carbon or hydrogen because the affinity of aluminium for oxygen is much higher than that of carbon or hydrogen.
Therefore,the extraction of aluminium from bauxite ore is carried out by the electrolytic reduction of molten alumina $(Al_2O_3)$ dissolved in molten cryolite $(Na_3AlF_6)$ and fluorspar $(CaF_2)$,which lowers the melting point and increases the electrical conductivity.
This process is known as the Hall-$H$éroult process.

(C) Chromium is extracted from chromite ore $(FeCr_2O_4)$ by the aluminothermic process.
In this process,the concentrated ore is reduced using aluminium powder as the reducing agent.
The chemical reaction is: $FeCr_2O_4 + 2Al \rightarrow 2Cr + Fe + Al_2O_3$.

(C) The extraction of $Aluminium$ $(Al)$ is carried out by the electrolytic reduction of fused alumina $(Al_2O_3)$ mixed with cryolite $(Na_3AlF_6)$ and fluorspar $(CaF_2)$,which is known as the $Hall-Heroult$ process. Therefore,$Aluminium$ is the element recovered from an electrolytic process.

(C) Copper is extracted from its sulphide ore $(Cu_2S)$ primarily by the auto-reduction process.
In this process,the ore is heated in a limited supply of air,where a portion of the sulphide ore is converted to oxide,which then reacts with the remaining sulphide ore to produce metallic copper.
The chemical reaction is: $2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$
Followed by: $2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$
This method is known as auto-reduction because no external reducing agent like carbon or carbon monoxide is required.

(A) In the extraction of copper,the metal is formed in the Bessemer converter due to a process known as self-reduction (or auto-reduction).
In this process,copper$(I)$ sulfide reacts with copper$(I)$ oxide to produce metallic copper and sulfur dioxide gas.
The balanced chemical equation is: $Cu_2S + 2Cu_2O \to 6Cu + SO_2$.

(A) Calcium is extracted by the electrolysis of a fused mixture of $CaCl_2$ and $CaF_2$.
$CaCl_2$ has a high melting point $(772 \ ^\circ C)$,which makes the process energy-intensive.
Adding $CaF_2$ (fluorspar) lowers the melting point of the mixture to approximately $650 \ ^\circ C$,which facilitates the electrolysis process and improves the conductivity of the melt.

(B) Lead is mainly extracted from its ore,galena $(PbS)$,by the self-reduction method (also known as the air reduction process).
In this process,the ore is partially roasted in air to form lead oxide $(PbO)$ and lead sulfate $(PbSO_4)$,which then react with the remaining $PbS$ to produce metallic lead $(Pb)$.

(A) In the metallurgy of mercury from cinnabar $(HgS)$,the ore is heated in air to form $HgO$,which then undergoes self-reduction at higher temperatures to produce mercury metal.
$2HgS + 3O_2 \rightarrow 2HgO + 2SO_2$
$2HgO \rightarrow 2Hg + O_2$
Thus,no external reducing agent is required.

(A) Aluminium is a strong reducing agent and is used in the thermite process to reduce metal oxides that are less reactive than aluminium.
Specifically,it is used to reduce $Cr_2O_3$ to chromium metal according to the reaction: $Cr_2O_3 + 2Al \rightarrow 2Cr + Al_2O_3$.

(B) In the Hall-Heroult process,pure alumina $(Al_2O_3)$ has a very high melting point $(2050 \ ^\circ C)$ and is a poor conductor of electricity.
To overcome these issues,it is dissolved in molten cryolite $(Na_3AlF_6)$.
Fluorspar $(CaF_2)$ is added to the mixture to lower the melting point of the electrolyte further and to increase the electrical conductivity of the melt,which facilitates the electrolytic reduction process.

(B) During the extraction of copper from copper pyrites $(CuFeS_2)$,the ore is roasted to remove sulfur and then smelted.
Iron present in the ore is oxidized to $FeO$,which then reacts with the added flux,silica $(SiO_2)$,to form a fusible slag,iron $(II)$ silicate $(FeSiO_3)$.
The chemical reaction is:
$2CuFeS_2 + 2SiO_2 + 3O_2 \rightarrow Cu_2S + 2FeSiO_3 + 2SO_2$
Thus,iron is removed as $FeSiO_3$.

(B) In the extraction of iron from its ore,haematite $(Fe_2O_3)$,the ore is mixed with coke and limestone before being charged into the blast furnace.
Coke acts as a reducing agent,reducing $Fe_2O_3$ to molten iron.
Limestone $(CaCO_3)$ acts as a flux,which decomposes to $CaO$ and $CO_2$. The $CaO$ reacts with silica $(SiO_2)$ impurities present in the ore to form calcium silicate $(CaSiO_3)$ slag,which is easily removed.

(C) In the blast furnace,the combustion of coke occurs at the bottom near the tuyeres,where hot air is blasted in.
The reactions are:
$2C + O_2 \rightarrow 2CO + \text{heat}$
$C + O_2 \rightarrow CO_2 + \text{heat}$
Both of these combustion reactions are highly exothermic,which generates the maximum temperature in the combustion zone (near the tuyeres).

(A) The iron obtained from the blast furnace contains about $4 \%$ carbon and many impurities in smaller amounts (e.g.,$S$,$P$,$Si$,$Mn$).
This form of iron is known as pig iron and is cast into a variety of shapes.
Hence,the correct option is $(A)$.

(C) When $AgCl$ is fused with $Na_2CO_3$,it initially forms silver carbonate:
$2AgCl + Na_2CO_3 \to Ag_2CO_3 + 2NaCl$
Silver carbonate is thermally unstable and decomposes upon heating to form metallic silver:
$Ag_2CO_3 \xrightarrow{\Delta} 2Ag + \frac{1}{2} O_2 + CO_2 \uparrow$
Thus,the final product obtained is metallic silver $(Ag)$.

(D) Railway wagon axles are made by heating rods of iron embedded in charcoal powder. The process is known as case hardening.
Case hardening is the process of hardening the surface of wrought iron or low-carbon steel by diffusing carbon into the surface layer to form a thin layer of high-carbon steel.
This is typically achieved by heating the iron in contact with carbonaceous materials like charcoal or potassium ferrocyanide,followed by quenching in a suitable medium like oil or water.

(D) Pyrometallurgy involves the extraction of metals by heating ores at high temperatures,often with a reducing agent like carbon.
$Ag_2S$,$ZnS$,$Cu_2S$,$HgS$,$MnO_2$,and $SnO_2$ are all ores that can be reduced to their respective metals using pyrometallurgical processes (e.g.,roasting followed by reduction with carbon or self-reduction).
Since all the given pairs of ores can be converted into their corresponding metals by pyrometallurgy,the correct answer is that none of the pairs are excluded.

(D) The reduction of metal oxides by carbon is thermodynamically feasible only if the Gibbs free energy change $(\Delta G)$ for the reaction is negative.
$Al_2O_3$ has a very high negative enthalpy of formation,meaning it is extremely stable.
Carbon cannot reduce $Al_2O_3$ to $Al$ because the affinity of $Al$ for oxygen is much higher than that of $C$ for oxygen at the temperatures typically used in metallurgical processes.
Therefore,the heat of formation of $Al_2O_3$ is too high for carbon to act as an effective reducing agent.

(D) When alumina $(Al_2O_3)$ is heated with carbon $(C)$ in a nitrogen $(N_2)$ atmosphere,the reaction produces aluminum nitride $(AlN)$ and carbon monoxide $(CO)$.
The balanced chemical equation is:
$Al_2O_3 + 3C + N_2 \rightarrow 2AlN + 3CO$

(C) Tin $(Sn)$ and zinc $(Zn)$ are commercially extracted from their respective oxide ores by the carbon reduction method.
For example,$SnO_2 + 2C \rightarrow Sn + 2CO$ and $ZnO + C \rightarrow Zn + CO$.
Metals like $Al$ are extracted by electrolytic reduction,and $Cu$ is often extracted by self-reduction or roasting followed by reduction.

(B) In the commercial extraction of copper from copper glance $(Cu_2S)$,the ore is partially roasted to form copper$(I)$ oxide $(Cu_2O)$:
$2Cu_2S + 3O_2 \to 2Cu_2O + 2SO_2$
This is followed by self-reduction (autoreduction) where the remaining $Cu_2S$ reacts with the $Cu_2O$ formed to produce metallic copper:
$2Cu_2O + Cu_2S \to 6Cu + SO_2$
Copper is not reduced by carbon or carbon monoxide in this process because copper has a lower affinity for oxygen compared to carbon at these temperatures.

(D) The Parke's process is a metallurgical process for removing silver from lead.
Electrolytic refining is a standard method for the purification of metals like $Cu$,$Ni$,$Pb$,$Au$,and $Ag$,where silver is often recovered as anode mud.
The extraction of silver from silver sulphide $(Ag_2S)$ involves the following reactions:
$Ag_2S + 4NaCN \rightarrow 2Na[Ag(CN)_2] + Na_2S$ (reversible reaction)
$2Na[Ag(CN)_2] + Zn \rightarrow Na_2[Zn(CN)_4] + 2Ag$
However,heating $Na[Ag(CN)_2]$ does not yield metallic silver $(Ag)$ directly as it decomposes into complex products rather than simple metallic silver.
Therefore,option $D$ is the correct answer.

(B) Step $I$: Chromite ore $(FeCr_2O_4)$ is fused with $NaOH$ in the presence of air to form sodium chromate $(Na_2CrO_4)$.
$4FeCr_2O_4 + 8NaOH + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8H_2O$.
Step $II$: Sodium chromate is treated with carbon to reduce it to chromium$(III)$ oxide $(Cr_2O_3)$.
$2Na_2CrO_4 + 2C \rightarrow Cr_2O_3 + Na_2CO_3 + CO$.
Step $III$: Chromium$(III)$ oxide is reduced to metallic chromium using the aluminothermic process ($Al$ as a reducing agent).
$Cr_2O_3 + 2Al \rightarrow 2Cr + Al_2O_3$.

(A) The electrolysis of pure alumina is not feasible because it is a bad conductor of electricity and its fusion temperature is high.
Cryolite $(Na_3AlF_6)$ and fluorspar $(CaF_2)$ are added to alumina $(Al_2O_3)$,which lowers the fusion temperature to approximately $1140 \ K$ and makes the mixture a good electrical conductor.
Then,the electrolysis of this fused mixture is carried out.

(B) The reduction of metal oxides by aluminium powder is known as the Goldschmidt's aluminothermite process.
In this process,a metal oxide (like $Cr_2O_3$) is reduced by aluminium powder to obtain the metal in its molten state.
The reaction is highly exothermic:
$Cr_2O_3 + 2Al \rightarrow 2Cr + Al_2O_3 + \text{Heat}$

(C) In the Hall-Heroult process,$Al_2O_3$ is dissolved in molten cryolite $(Na_3AlF_6)$ to lower its melting point and increase electrical conductivity. $Al$ is deposited at the cathode,and $CO_2$ is released at the carbon anode. $CaF_2$ (fluorspar) is added to lower the melting point further and improve fluidity. $Li_2CO_3$ cannot be used as a substitute for cryolite because it does not provide the necessary electrolyte properties for the process. Therefore,option $C$ is the incorrect statement.

(B) In the blast furnace,limestone $(CaCO_3)$ is added as a flux.
First,it decomposes to provide $CaO$ and $CO_2$: $CaCO_3 \rightarrow CaO + CO_2$.
Then,$CaO$ acts as a flux to remove acidic impurities like silica $(SiO_2)$ present in the ore,forming slag $(CaSiO_3)$: $CaO + SiO_2 \rightarrow CaSiO_3$.
Thus,it provides a source of $CaO$ $(I)$ and removes impurities $(II)$. The $CO_2$ produced is a byproduct,not the primary purpose of adding limestone. Therefore,the correct statements are $I$ and $II$.

(C) In the Bessemer converter,the extraction of copper involves the process of self-reduction. The molten copper matte,which consists of $Cu_2S$ and $FeS$,is oxidized. After the removal of iron as slag,the remaining $Cu_2S$ reacts with $Cu_2O$ (formed by partial oxidation of $Cu_2S$) to produce metallic copper. The reaction is: $Cu_2S + 2Cu_2O \longrightarrow 6Cu + SO_2$.

(A) If air is used for the oxidation of pig iron,the molten metal will take up a small amount of nitrogen from the air.
This nitrogen makes the steel brittle.
Hence,pure $O_2$ is used instead of air to avoid the presence of nitrogen.

(D) In the extraction of silver from argentite ore $(Ag_2S)$,the ore is treated with a dilute solution of $NaCN$ or $KCN$ in the presence of air to form a soluble complex,$Na[Ag(CN)_2]$.
$Ag_2S + 4NaCN \rightarrow 2Na[Ag(CN)_2] + Na_2S$.
$KNO_3$ is added as a flux to oxidise the $Na_2S$ (or $S^{2-}$ ions) formed during the reaction to $SO_2$ or $SO_4^{2-}$,preventing the reverse reaction of the complex formation.
Specifically,it oxidises the sulphur impurities present in the ore to $SO_2$ which escapes,thereby shifting the equilibrium forward.

(D) Some metal oxides cannot be reduced satisfactorily by carbon.
For such oxides,$Al$ (aluminium),which is a more reactive metal,is used.
This process is known as the thermite process or alumino-thermic process.
In this reaction,$Al$ removes oxygen from the metal oxide,meaning it undergoes oxidation itself and reduces the metal oxide.
Therefore,$Al$ acts as a reducing agent.