Reduction to free Metal Questions in English

(B) Silica $(SiO_2)$ is added during the smelting process to remove iron oxide $(FeO)$ impurities present in the roasted copper ore.
During roasting,iron sulfides present in the ore react with oxygen to form iron oxides $(FeO)$.
$2FeS + 3O_2 \rightarrow 2FeO + 2SO_2$
Silica acts as an acidic flux and reacts with the basic impurity $FeO$ to form a fusible slag,iron silicate ($FeSiO_3$ or $FeO \cdot SiO_2$).
$FeO + SiO_2 \rightarrow FeSiO_3 \text{ (slag)}$
This slag is less dense than the molten matte and can be easily removed.

(C) In the Hall-$H$éroult process for the extraction of aluminium,pure alumina $(Al_2O_3)$ has a very high melting point and is a poor conductor of electricity.
To lower the melting point and increase electrical conductivity,it is dissolved in a molten mixture of cryolite $(Na_3AlF_6)$ and fluorspar $(CaF_2)$.

(B) Oxides of strong electropositive metals such as $K$,$Ca$,$Na$,$Al$,and $Mg$ are very stable.
It is difficult to reduce them into their metallic state by carbon reduction because their affinity for oxygen is much higher than that of carbon.
Such metals are extracted by passing electricity (electrolysis process) through their fused chlorides,oxides,or hydroxides.
Hence,the most suitable method for the extraction of a metal which has a high concentration in the Earth's crust and whose oxides cannot be reduced by carbon is the electrolysis process.

(B) catalyst is a substance that lowers the activation energy of a reaction by providing an alternate pathway. Generally,a catalyst increases the speed of a reaction.
In the thermite process,the reaction $2 Al + Fe_2O_3 \rightarrow Al_2O_3 + 2 Fe$ (or similar metal oxide reductions) is highly exothermic and proceeds rapidly without the need for a catalyst.
In contrast,the Contact process uses $V_2O_5$,the Ostwald process uses $Pt/Rh$ gauze,and the Haber process uses $Fe$ with $Mo$ as a promoter.

(B) Refractory materials are substances that have a very high melting point and can withstand extremely high temperatures without melting,softening,or undergoing chemical changes. Therefore,they are used to line the interior of furnaces.

(B) The most modern and widely used method for steel production is the $L.D.$ (Linz-Donawitz) process,which involves blowing oxygen into molten pig iron.

(A) In the Hall-Heroult process for the electrolytic reduction of alumina $(Al_2O_3)$,the melting point of pure alumina is very high $(2323 \ K)$.
To lower the melting point to about $1140 \ K$ and to increase the electrical conductivity of the melt,two auxiliary electrolytes are added: Cryolite $(Na_3AlF_6)$ and Fluorspar $(CaF_2)$.
Thus,$X$ and $Y$ are cryolite and fluorspar.

(A) During the extraction of metals like $Cu$ or $Fe$,slag is formed by the reaction of flux with gangue impurities.
Slag is generally lighter than the molten metal and has a lower melting point.
Due to its lower density,it floats on the surface of the molten metal,allowing for easy separation.

(A) The reduction of metal oxides by carbon $(C)$ depends on the thermodynamic stability of the oxide and the affinity of the metal for oxygen compared to carbon.
Highly reactive metals such as alkali metals ($K$,$Na$) and alkaline earth metals ($Ca$,$Mg$) form very stable oxides that cannot be reduced by carbon at standard metallurgical temperatures.
In the given options,$CaO$ and $K_2O$ are oxides of highly reactive metals ($Ca$ and $K$),which have a very high affinity for oxygen.
Therefore,they cannot be reduced by carbon to their respective metals.

(C) In the alumino thermite process,metal oxides like $Fe_2O_3$ are reduced to their respective metals using aluminum powder as the reducing agent. The reaction is highly exothermic: $Fe_2O_3(s) + 2Al(s) \rightarrow 2Fe(l) + Al_2O_3(s) + \text{Heat}$. Here,$Al$ undergoes oxidation (loses electrons) and reduces the metal oxide,hence it acts as a reducing agent.

(C) In the extraction of copper from its sulphide ore,the metal is formed by the self-reduction of $Cu_2O$ with $Cu_2S$.
First,the sulphide ore is roasted in air to convert a portion of it into copper$(I)$ oxide: $2 Cu_2S + 3 O_2 \rightarrow 2 Cu_2O + 2 SO_2$.
Then,the remaining copper$(I)$ sulphide reacts with the copper$(I)$ oxide produced to form metallic copper: $2 Cu_2O + Cu_2S \rightarrow 6 Cu + SO_2$.
This process is known as auto-reduction or self-reduction.

(B) The given reactions represent the MacArthur-Forrest cyanide process used for the extraction of noble metals like silver $(Ag)$ or gold $(Au)$.
The balanced chemical equations for silver are:
$4Ag + 8CN^{-} + 2H_2O + O_2 \rightarrow 4[Ag(CN)_2]^{-} + 4OH^{-}$
$2[Ag(CN)_2]^{-} + Zn \rightarrow [Zn(CN)_4]^{2-} + 2Ag$
Thus,the metal $M$ is silver $(Ag)$.

(A) The extraction of gold involves the leaching of gold ore with a dilute solution of $CN^-$ in the presence of air (source of $O_2$),which acts as an oxidizing agent.
The chemical reaction is:
$4Au(s) + 8CN^-(aq) + 2H_2O(aq) + O_2(g) \rightarrow 4[Au(CN)_2]^-(aq) + 4OH^-(aq)$
Here,$[x] = [Au(CN)_2]^-$.
Gold is then recovered from the complex by displacement using zinc,which is a more reactive metal:
$2[Au(CN)_2]^-(aq) + Zn(s) \rightarrow [Zn(CN)_4]^{2-}(aq) + 2Au(s)$
Here,$[y] = [Zn(CN)_4]^{2-}$.

(B) In the commercial extraction of copper from copper glance $(Cu_2S)$,the process is known as self-reduction or auto-reduction.
First,the sulphide ore is partially roasted in air to form copper$(I)$ oxide:
$Cu_2S + \frac{3}{2}O_2 \to Cu_2O + SO_2$
Then,the remaining copper$(I)$ sulphide reacts with the copper$(I)$ oxide formed to produce metallic copper:
$2Cu_2O + Cu_2S \to 6Cu + SO_2$
This is the major reaction in the Bessemer converter.

(C) In the Hall-Heroult process,the graphite anode is susceptible to oxidation by the oxygen gas liberated at the anode,which leads to its corrosion. To prevent this,$Coke$ powder is sprinkled on the surface of the molten electrolyte. This layer of $Coke$ powder prevents the oxygen from coming into contact with the graphite anode,thereby protecting it from oxidation.

(A) The extraction of $Zn$ from its ore (zinc blende,$ZnS$) involves roasting to form $ZnO$.
$ZnO$ is then reduced to $Zn$ metal using carbon (coke) at high temperatures: $ZnO + C \rightarrow Zn + CO$.
Since $Zn$ has a low boiling point $(907 \ ^\circ C)$,it is refined by the distillation method,where the metal is vaporized and then condensed to obtain pure metal.

(C) For the reaction $C_{(s)} + 1/2 O_{2(g)} \to CO_{(g)}$,the number of moles of gas increases $(\Delta n_g = +0.5)$,so the entropy change $\Delta S$ is positive.
For the reaction $C_{(s)} + O_{2(g)} \to CO_{2(g)}$,the number of moles of gas remains the same $(\Delta n_g = 0)$,so $\Delta S$ is approximately zero.
Therefore,$\Delta S(C \to CO) > \Delta S(C \to CO_2)$.
Thus,statement $(C)$ is incorrect.

(A) During the extraction of copper from copper pyrites $(CuFeS_2)$,the ore is roasted to convert iron sulphide $(FeS)$ into iron oxide $(FeO)$.
$2FeS + 3O_2 \rightarrow 2FeO + 2SO_2$
To remove the impurity of $FeO$,silica $(SiO_2)$ is added as a flux.
$FeO + SiO_2 \rightarrow FeSiO_3$ (slag).
Thus,$SiO_2$ is added to form the slag.

(D) Self-reduction,also known as auto-reduction,is a process where the metal sulfide ore is partially roasted to form its oxide,which then reacts with the remaining sulfide to produce the metal without the need for an external reducing agent.
This process is commonly used for metals like $Cu$,$Pb$,and $Hg$.
For example,in the extraction of lead $(Pb)$ from galena $(PbS)$:
$2PbS + 3O_2 \rightarrow 2PbO + 2SO_2$
$2PbO + PbS \rightarrow 3Pb + SO_2$
Therefore,the correct option is $D$.

(C) In the Mac-Arthur Forest cyanide process for the extraction of silver,silver sulfide $(Ag_2S)$ ore is treated with a dilute solution of sodium cyanide $(NaCN)$ or potassium cyanide $(KCN)$.
The chemical reaction is as follows:
$Ag_2S + 4KCN \rightarrow 2K[Ag(CN)_2] + K_2S$
This process involves the formation of a soluble complex,$K[Ag(CN)_2]$,which allows the silver to be separated from the ore.

(C) The process $(iii)$ is used for obtaining sodium metal via the electrolysis of molten $NaCl$ mixed with $CaCl_2$ (Down's process). The $CaCl_2$ lowers the melting point of $NaCl$.
The process $(i)$ is the Mond process,used for the refining of nickel. Impure nickel reacts with $CO$ to form volatile nickel tetracarbonyl,$Ni(CO)_4$,which is then decomposed at higher temperatures to yield pure nickel.
The process $(ii)$ is self-reduction (or auto-reduction),used for the extraction of copper from copper glance $(Cu_2S)$. The reactions are:
$2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$
$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$
Therefore,the processes for sodium,nickel,and copper are $(iii), (i)$ and $(ii)$ respectively.

(A) $1$. Metals like $Cu$ (from $Cu_2S$) and $Hg$ (from $HgS$) are extracted by self-reduction (roasting).
$2$. Metals that are less reactive than hydrogen in the electrochemical series do not evolve $H_2$ gas when reacted with dilute acids like $H_2SO_4$.
$3$. $Cu$ and $Hg$ are below hydrogen in the reactivity series,hence they do not displace $H_2$ from dilute $H_2SO_4$.
$4$. $Zn$,$Pb$,$Mn$,and $Al$ are more reactive than hydrogen and will evolve $H_2$ gas.

(C) $1$. The reactions describe the extraction of Zinc $(Zn)$.
$2$. Ore '$X$' is Zinc blende $(ZnS)$,which on roasting gives Zinc oxide $(Y = ZnO)$ and $SO_2$.
$3$. Ore '$Z$' is Calamine $(ZnCO_3)$,which on calcination gives Zinc oxide $(Y = ZnO)$ and $CO_2$.
$4$. The reduction of Zinc oxide $(ZnO)$ with Carbon $(C)$ at $1673 \ K$ is given by: $ZnO + C \xrightarrow{1673 \ K} Zn(g) + CO(g)$.
$5$. Here,'$A$' is Zinc vapour $(Zn)$ and '$B$' is Carbon monoxide $(CO)$.

(D) The correct option is $D$.
Carbon reduction is suitable for metals with moderate electropositivity.
$MgO$ (Magnesium oxide) has a very high negative Gibbs free energy of formation $(\Delta G_f^{\circ})$,meaning it is highly stable.
Carbon cannot reduce $MgO$ to $Mg$ at commercially feasible temperatures because the required temperature is extremely high,making electrolytic reduction the preferred method for $Mg$.
In contrast,$PbO$,$SnO_2$,and $ZnO$ are easily reduced by carbon.

(B) The extraction of $Cu$ from low-grade ores is done by hydrometallurgy,where the ore is leached with acid or bacteria.
The resulting solution contains $Cu^{2+}$ ions.
To recover $Cu$ metal,we use a more reactive metal to displace $Cu$ from the solution.
The reaction is: $Cu^{2+}(aq) + M(s) \rightarrow Cu(s) + M^{2+}(aq)$.
Both $Zn$ and $Fe$ can displace $Cu$ because they are more reactive than $Cu$.
However,$Fe$ is much cheaper and more abundantly available as scrap compared to $Zn$.
Therefore,$Fe$ scrap is more economically suitable for the reduction of leached $Cu$ ore.

(A) The reaction between copper$(I)$ oxide and copper$(I)$ sulfide is a self-reduction process used in the extraction of copper metal.
The balanced chemical equation is:
$2Cu_2O + Cu_2S \to 6Cu + SO_2$

(A) In an Ellingham diagram,the metal whose oxidation curve lies lower is a better reducing agent for the oxide of the metal whose curve lies higher.
Below $1673 \ K$,the curve for $2Mg + O_2 \rightarrow 2MgO$ lies below the curve for $4/3Al + O_2 \rightarrow 2/3Al_2O_3$,which means $\Delta G^o_{MgO} < \Delta G^o_{Al_2O_3}$. Thus,$Mg$ can reduce $Al_2O_3$ to $Al$ below $1673 \ K$.
Above $1673 \ K$,the curve for $Al$ oxidation lies below the curve for $Mg$ oxidation,meaning $Al$ can reduce $MgO$ to $Mg$. Therefore,$Al_2O_3$ is not a reducing agent,but $Al$ is.
Thus,only statement $(A)$ is correct.

(D) The extraction of silver $(Ag)$ from its ore involves the following steps:
$1$. The ore is treated with a dilute solution of $NaCN$ in the presence of air $(O_2)$ to form a soluble complex:
$4Ag + 8NaCN + 2H_2O + O_2 \to 4Na[Ag(CN)_2] + 4NaOH$
Here,$A = Na[Ag(CN)_2]$ (Sodium dicyanoargentate$(I)$).
$2$. The silver is then recovered from the complex by adding zinc $(Zn)$,which acts as a reducing agent:
$2Na[Ag(CN)_2] + Zn \to Na_2[Zn(CN)_4] + 2Ag$
Here,$B = Na_2[Zn(CN)_4]$ (Sodium tetracyanozincate$(II)$).
$3$. This process is a displacement reaction involving oxidation and reduction (redox process),where $Zn$ is oxidized and $Ag^+$ is reduced to $Ag$.
Therefore,all the given options are correct.

(A) Self-reduction (or auto-reduction) is a process where the metal sulfide is partially roasted to form oxide,which then reacts with the remaining sulfide to produce the metal.
$1. PbS$ (Galena) undergoes self-reduction: $2PbS + 3O_2 \to 2PbO + 2SO_2$ followed by $PbS + 2PbO \to 3Pb + SO_2$.
$2. Cu_2S$ (Copper glance) undergoes self-reduction: $2Cu_2S + 3O_2 \to 2Cu_2O + 2SO_2$ followed by $2Cu_2O + Cu_2S \to 6Cu + SO_2$.
$3. HgS$ (Cinnabar) undergoes self-reduction: $2HgS + 3O_2 \to 2HgO + 2SO_2$ followed by $2HgO + HgS \to 3Hg + SO_2$.
$4. ZnS$ (Zinc blende) cannot be reduced by self-reduction because zinc oxide $(ZnO)$ is not easily reduced by $ZnS$. Instead,$ZnO$ is reduced by carbon $(C)$ in a blast furnace.
Therefore,self-reduction is not possible for $ZnS \to Zn$.

(B) In the leaching of $Ag_2S$ (silver sulphide) with aqueous $NaCN$,the reaction is reversible:
$Ag_2S + 4 NaCN \rightleftharpoons 2 Na[Ag(CN)_2] + Na_2S$
To drive the reaction forward,the product $Na_2S$ must be removed.
$A$ stream of air (oxygen) is passed through the solution to oxidise $Na_2S$ into $Na_2SO_4$ (sodium sulphate) and sulphur,preventing the reverse reaction.
The overall reaction is: $2 Na_2S + 2 O_2 + H_2O \rightarrow Na_2SO_4 + 2 NaOH + S$ (or similar oxidation products like $Na_2SO_4$).

(D) In the blast furnace,the combustion zone is the region at the bottom where coke burns in the presence of a hot air blast to produce $CO_2$ and heat.
The primary reaction occurring in this zone is $C + O_2 \to CO_2$ $(\Delta H < 0)$.
Since this reaction is not among the given options,the correct answer is $D$.

(A) The electrolysis of pure alumina faces some difficulties. Pure alumina is a bad conductor of electricity.
The fusion temperature of pure alumina is about $2000^{\circ} C$ and at this temperature,the metal formed vaporizes,as the boiling point of aluminum is $1800^{\circ} C$.
These difficulties are overcome by using a mixture containing alumina $(Al_2O_3)$,cryolite $(Na_3AlF_6)$,and fluorspar $(CaF_2)$,which lowers the melting point to about $900^{\circ} C$ and increases electrical conductivity.

(C) The given equation represents the leaching process of noble metals like $Au$ (gold) or $Ag$ (silver) using a dilute solution of sodium cyanide $(NaCN)$ in the presence of air $(O_2)$ as an oxidizing agent.
This is a key step in the MacArthur-Forrest cyanide process for the extraction of gold and silver from their ores.
Therefore,the metal $M$ is $Au$ (gold) or $Ag$ (silver).

(D) In the Ellingham diagram, the reduction of $FeO$ by carbon to produce $CO$ is represented by the reaction: $2FeO + 2C \rightarrow 2Fe + 2CO$.
This reaction is spontaneous when the Gibbs free energy change $(\Delta G^\circ)$ for the coupled reaction is negative.
Looking at the diagram, the line for the oxidation of carbon to $CO$ $(2C + O_2 \rightarrow 2CO)$ intersects the line for the oxidation of iron to $FeO$ $(2Fe + O_2 \rightarrow 2FeO)$ at point $A$.
At temperatures above point $A$, the line for $2C + O_2 \rightarrow 2CO$ lies below the line for $2Fe + O_2 \rightarrow 2FeO$, meaning the reduction of $FeO$ by $C$ becomes thermodynamically favorable $(\Delta G^\circ < 0)$.
Therefore, the reduction occurs at temperatures above point $A$.

(B) In the cyanide process for the extraction of silver,the ore is treated with a dilute solution of $NaCN$ in the presence of air to form a soluble complex: $4Ag + 8CN^- + 2H_2O + O_2 \rightarrow 4[Ag(CN)_2]^- + 4OH^-$.
Zinc is then added to this solution to displace silver from the complex: $2[Ag(CN)_2]^- + Zn \rightarrow [Zn(CN)_4]^{2-} + 2Ag$.
In this reaction,$Zn$ is oxidized to $Zn^{2+}$ and $Ag^+$ is reduced to $Ag$. Therefore,zinc acts as a reducing agent.

(B) Copper matte is obtained during the extraction of copper from copper pyrites $(CuFeS_2)$.
It is a molten mixture consisting mainly of cuprous sulphide $(Cu_2S)$ and a small amount of ferrous sulphide $(FeS)$.
Therefore,it contains a major amount of $Cu_2S$.

(C) In the extraction of copper,the roasted ore is mixed with silica $(SiO_2)$ and heated in a reverberatory furnace.
Iron oxide $(FeO)$ reacts with silica to form iron silicate slag $(FeSiO_3)$,which is removed.
The remaining mixture,known as copper matte,primarily consists of cuprous sulphide $(Cu_2S)$ and ferrous sulphide $(FeS)$.
Thus,copper matte contains sulphides of copper $(I)$ and iron $(II)$.

(C) Carbon reduction is generally used for the extraction of metals like $Fe$,$Zn$,etc.
However,for metals like $Cr$ and $Mn$,carbon reduction is not preferred because these metals have a high affinity for carbon at high temperatures.
At the high temperatures required for the reduction process,$Cr$ and $Mn$ react with carbon to form stable interstitial carbides (e.g.,$Cr_3C_2$,$Mn_7C_3$).
These carbides contaminate the metal and make the extraction process inefficient and the product impure.

(A) In the Down's process,sodium is obtained by the electrolysis of a fused mixture of sodium chloride $(NaCl)$ and calcium chloride $(CaCl_2)$.
The mixture typically consists of $40\, \% \ NaCl$ and $60\, \% \ CaCl_2$.
The primary function of adding $CaCl_2$ is to lower the melting point of the electrolyte mixture from $801\, ^{\circ}C$ to approximately $600\, ^{\circ}C$,which reduces the operating temperature and prevents the volatilization of sodium metal.
At the cathode: $Na^{+} + e^{-} \longrightarrow Na$
At the anode: $2Cl^{-} \longrightarrow Cl_2 + 2e^{-}$

(B) The extraction of silver from its ore $Ag_2S$ (argentite) involves the following steps:
$1$. Leaching: $Ag_2S + 4NaCN \to 2Na[Ag(CN)_2] + Na_2S$. Here,$(A)$ is $Na[Ag(CN)_2]$.
$2$. Displacement: $2Na[Ag(CN)_2] + Zn \to Na_2[Zn(CN)_4] + 2Ag$. Here,$(B)$ is $Ag$ (silver metal).
Therefore,$(A)$ is $Na[Ag(CN)_2]$ and $(B)$ is $Ag$.

(C) The carbon reduction process is generally used for metals with moderate electropositivity,such as $Zn, Fe,$ and $Sn$.
Metals like $Al$ and $Mg$ are highly electropositive and have a very high affinity for oxygen.
Their oxides ($Al_2O_3$ and $MgO$) are extremely stable and cannot be reduced by carbon at standard commercial temperatures.
While $MgO$ can be reduced by carbon at extremely high temperatures (around $2000^{\circ} C$),it is not considered a commercially viable process due to the high energy requirements and technical difficulties.
Therefore,carbon reduction is not commercially applicable for $Al_2O_3$ and $MgO$.

(C) In the extraction of copper from copper glance $(Cu_2S)$,the ore is partially roasted to form cuprous oxide $(Cu_2O)$.
The remaining $Cu_2S$ then reacts with the formed $Cu_2O$ in a process known as auto-reduction or self-reduction.
The chemical equation is: $Cu_2S + 2Cu_2O \to 6Cu + SO_2$.
Thus,$Cu_2S$ is oxidized by $Cu_2O$ to produce copper metal.

(D) The relationship between Gibbs free energy change and equilibrium constant is $\Delta G^\circ = -RT \ln K_p$,which can be rewritten as $-\ln K_p = \frac{\Delta G^\circ}{RT}$.
For a reaction to be spontaneous,$\Delta G^\circ < 0$,which implies $-\ln K_p < 0$.
From the Ellingham diagram,at $T > 1200 \ K$,the value of $-\ln K_p$ for the reaction $C_{(s)} + \frac{1}{2} O_{2(g)} \to CO_{(g)}$ is lower than that for $M_{(s)} + \frac{1}{2} O_{2(g)} \to MO_{(s)}$.
This indicates that the formation of $CO$ is more thermodynamically stable than $MO$ at $T > 1200 \ K$.
Therefore,carbon can reduce $MO_{(s)}$ to $M_{(s)}$ at temperatures above $1200 \ K$ according to the reaction $MO_{(s)} + C_{(s)} \to M_{(s)} + CO_{(g)}$.

(C) In the extraction of copper,$FeO$ is present as a basic gangue impurity.
$SiO_2$ is added as an acidic flux to remove this impurity.
They react to form a fusible slag,$FeSiO_3$.
The reaction is: $\mathop {FeO}\limits_{\text{Basic impurity}} + \mathop {SiO_2}\limits_{\text{Acidic flux}}$ $\longrightarrow \mathop {FeSiO_3}\limits_{\text{Slag}}$

(C) The iron obtained from the blast furnace is known as $Pig \ Iron$. It contains about $4 \%$ carbon and many impurities like $S, P, Si, Mn$ in smaller amounts.

(C) The reduction of metal oxides using powdered aluminium is known as the Goldschmidt aluminothermic process.
This process is specifically employed for the extraction of metals that have very high melting points $(m.p.)$ and are obtained from their respective oxides.

(B) Carbon is used as a reducing agent in metallurgy to reduce metal oxides to their respective metals.
However,carbon cannot reduce oxides of highly reactive metals like $Ca$,$K$,$Na$,and $Mg$ because these metals have a much higher affinity for oxygen than carbon does.
In the given options,$CaO$ and $K_2O$ are oxides of highly reactive alkaline earth and alkali metals,respectively.
Therefore,they cannot be reduced by carbon.

(A) Fused alumina $(Al_2O_3)$ is a bad conductor of electricity and has a very high melting point. Cryolite $(Na_3AlF_6)$ is added to purified alumina to lower its melting point to about $1140 \ K$ and to increase its electrical conductivity.

(C) Extraction of $Zn$ from zinc blende $(ZnS)$ is achieved by roasting followed by reduction with carbon (coke).
Step $1$: Roasting of zinc blende in the presence of excess air:
$2ZnS + 3O_2 \xrightarrow{\Delta} 2ZnO + 2SO_2$
Step $2$: Reduction of zinc oxide with carbon (coke) at high temperature:
$ZnO + C \xrightarrow{\Delta} Zn + CO$

(A) In an Ellingham diagram,a metal can reduce the oxide of another metal if the line for the formation of its own oxide lies below the line for the formation of the other metal's oxide.
$1$. At $1400\,^\circ C,$ the line for the formation of $Al_2O_3$ is below the line for the formation of $ZnO.$ Therefore,$Al$ can reduce $ZnO$ to $Zn$.
$2$. At $500\,^\circ C,$ the line for the formation of $ZnO$ is below the line for the formation of $CO,$ so coke cannot reduce $ZnO$.
$3$. The line for the formation of $CO$ is below the line for the formation of $Cu_2O$ at all temperatures,so coke can reduce $Cu_2O$.
$4$. At $800\,^\circ C,$ the line for the formation of $ZnO$ is below the line for the formation of $Cu_2O,$ so $Cu$ cannot reduce $ZnO$.
Thus,option $A$ is correct.