Answer

The reduction potentials of zinc and iron are lower than that of copper. In hydrometallurgy, zinc and iron can be used to displace copper from their solution.

$ \mathrm{Fe_{(s)}}+\mathrm{Cu_{(a q)}^{2+}} \longrightarrow \mathrm{Fe_{(a q)}^{2+}}+\mathrm{Cu_{(s)}} $

But to displace zinc, more reactive metals i.e., metals having lower reduction potentials than zinc such as $\mathrm{Mg}, \mathrm{Ca}, \mathrm{K}$, etc. are required. But all these metals react with water with the evolution of $\mathrm{H_2}$ gas.

$ 2 \mathrm{~K_{(s)}}+2 \mathrm{H_2} \mathrm{O_{(l)}} \longrightarrow 2 \mathrm{KOH_{(a q)}}+\mathrm{H_{2(g)}} $

As a result, these metals cannot be used in hydrometallurgy to extract zinc.

Hence, copper can be extracted by hydrometallurgy but not zinc.

Answer

In the froth floatation process, the role of the depressants is to separate two sulphide ores by selectively preventing one ore from forming froth. For example, to separate two sulphide ores ( $\mathrm{ZnS}$ and $\mathrm{Pbs}$ ), $\mathrm{NaCN}$ is used as a depressant which selectively allows PbS to come with froth, but prevents ZnS from coming to froth. This happens because $\mathrm{NaCN}$ reacts with $\mathrm{ZnS}$ to form $\mathrm{Na_2}\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]$.

$ 4 \mathrm{NaCN}+\mathrm{ZnS} \longrightarrow \mathrm{Na _2}\left[\mathrm{Zn}(\mathrm{CN}) _{4}\right]+\mathrm{Na _2} \mathrm{~S} $

Answer

The Gibbs free energy of formation $\left(\Delta_{1} \mathrm{G}\right)$ of $\mathrm{Cu_2} \mathrm{~S}$ is less than that of $\mathrm{H_2} \mathrm{~S}$ and $\mathrm{CS_2}$. Therefore, $\mathrm{H_2}$ and $\mathrm{C}$ cannot reduce $\mathrm{Cu_2} \mathrm{~S}$ to $\mathrm{Cu}$.

On the other hand, the Gibbs free energy of formation of $\mathrm{Cu_2} \mathrm{O}$ is greater than that of $\mathrm{CO}$. Hence, $\mathrm{C}$ can reduce $\mathrm{Cu_2} \mathrm{O}$ to $\mathrm{Cu}$.

$ \mathrm{C_{(s)}}+\mathrm{Cu_2} \mathrm{O_{(s)}} \longrightarrow 2 \mathrm{Cu_{(s)}}+\mathrm{CO_{(g)}} $

Hence, the extraction of copper from its pyrite ore is difficult than from its oxide ore through reduction.

Answer

(i) Zone refining:

This method is based on the principle that impurities are more soluble in the molten state of metal (the melt) than in the solid state. In the process of zone refining, a circular mobile heater is fixed at one end of a rod of impure metal. As the heater moves, the molten zone of the rod also moves with it. As a result, pure metal crystallizes out of the melt and the impurities pass onto the adjacent molten zone. This process is repeated several times, which leads to the segregation of impurities at one end of the rod. Then, the end with the impurities is cut off. Silicon, boron, gallium, indium etc. can be purified by this process.

(ii) Column chromatography:

Column chromatography is a technique used to separate different components of a mixture. It is a very useful technique used for the purification of elements available in minute quantities. It is also used to remove the impurities that are not very different in chemical properties from the element to be purified. Chromatography is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. In chromatography, there are two phases: mobile phase and stationary phase. The stationary phase is immobile and immiscible. $\mathrm{Al_2} \mathrm{O_3}$ column is usually used as the stationary phase in column chromatography. The mobile phase may be a gas, liquid, or supercritical fluid in which the sample extract is dissolved. Then, the mobile phase is forced to move through the stationary phase. The component that is more strongly adsorbed on the column takes a longer time to travel through it than the component that is weakly adsorbed. The adsorbed components are then removed (eluted) using a suitable solvent (eluant).

Answer

At $673 \mathrm{~K}$, the value of $\Delta \mathrm{G_{(\mathrm{CO}, \mathrm{CO_2})}}$ is less than that of $\Delta \mathrm{G_{(\mathrm{C}, \mathrm{CO})}}$. Therefore, $\mathrm{CO}$ can be oxidised more easily to $\mathrm{CO_2}$ than $\mathrm{C}$ to $\mathrm{CO}$. Hence, $\mathrm{CO}$ is a better reducing agent than $\mathrm{C}$ at $673 \mathrm{~K}$.

Answer

In electrolytic refining of copper, the common elements present in anode mud are selenium, tellurium, silver, gold, platinum, and antimony.

These elements are very less reactive and are not affected during the purification process. Hence, they settle down below the anode as anode mud.

Answer

In blast furnace, iron oxides are reduced at different temperature ranges. In the lower part of the blast furnace, the temperature is as high as $2200 \mathrm{~K}$. It is called combustion zone. At the top, the temperature is as low as $500-800 \mathrm{~K}$. It is called reduction zone. In the lower temperature range, carbon is the reducing agent and in the higher temperature range, $\mathrm{CO}$ is the reducing agent. In the reduction zone (500-800 $\mathrm{K}$ ), following reactions occur.

$ \begin{aligned} & 3 \mathrm{Fe}_2 \mathrm{O}_3+\mathrm{CO} \rightarrow 2 \mathrm{Fe}_3 \mathrm{O}_4+\mathrm{CO}_2 \\ & \mathrm{Fe}_3 \mathrm{O}_4+4 \mathrm{CO} \rightarrow 3 \mathrm{Fe}+4 \mathrm{CO}_2 \\ & \mathrm{Fe}_2 \mathrm{O}_3+\mathrm{CO} \rightarrow 2 \mathrm{FeO}+\mathrm{CO}_2 \end{aligned} $

In the temperature range $900-1500 \mathrm{~K}$, following reactions occur.

$ \begin{aligned} & \mathrm{C}+\mathrm{CO}_2 \rightarrow 2 \mathrm{CO} \\ & \mathrm{FeO}+\mathrm{CO} \rightarrow \mathrm{Fe}+\mathrm{CO}_2 \end{aligned} $

Around 1270 K (middle portion), decomposition of limestone gives lime (CaO) and $\mathrm{CO}_2$. Lime is a flux and combines with silicate impurity to form slag of calcium silicate.

$ \begin{aligned} & \mathrm{CaCO}_3 \rightarrow \mathrm{CaO}+\mathrm{CO}_2 \\ & \mathrm{CaO}+\mathrm{SiO}_2 \rightarrow \mathrm{CaSIO}_3 \end{aligned} $

Answer

The different steps involved in the extraction of zinc from zinc blende (ZnS) are given below:

(i) Concentration of ore

First, the gangue from zinc blende is removed by the froth floatation method.

(ii) Conversion to oxide (Roasting)

Sulphide ore is converted into oxide by the process of roasting. In this process, ZnS is heated in a regular supply of air in a furnace at a temperature, which is below the melting point of $\mathrm{Zn}$.

$ 2 \mathrm{ZnS}+3 \mathrm{O_2} \longrightarrow 2 \mathrm{ZnO}+2 \mathrm{SO_2} $

(iii) Extraction of zinc from zinc oxide (Reduction)

Zinc is extracted from zinc oxide by the process of reduction. The reduction of zinc oxide is carried out by mixing it with powdered coke and then, heating it at $673 \mathrm{~K}$.

$ \mathrm{ZnO}+\mathrm{C} \xrightarrow{\text { coke, } 673 \mathrm{~K}} \mathrm{Zn}+\mathrm{CO} $

(iv) Electrolytic Refining

Zinc can be refined by the process of electrolytic refining. In this process, impure zinc is made the anode while a pure copper strip is made the cathode. The electrolyte used is an acidified solution of zinc sulphate $\left(\mathrm{ZnSO_4}\right)$. Electrolysis results in the transfer of zinc in pure from the anode to the cathode.

Anode: $\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-}$

Cathode: $\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}$

Answer

During the roasting of pyrite ore, a mixture of $\mathrm{FeO}$ and $\mathrm{Cu_2} \mathrm{O}$ is obtained.

$2 \mathrm{CuFeS_2}+\mathrm{O_2} \xrightarrow{\Delta} \mathrm{Cu_2} \mathrm{~S}+2 \mathrm{FeS}+\mathrm{SO_2}$

$2 \mathrm{Cu_2} \mathrm{~S}+3 \mathrm{O_2} \xrightarrow{\Delta} 2 \mathrm{Cu_2} \mathrm{O}+2 \mathrm{SO_2}$

$2 \mathrm{FeS}+3 \mathrm{O_2} \xrightarrow{\Delta} 2 \mathrm{FeO}+2 \mathrm{SO_2}$

The role of silica in the metallurgy of copper is to remove the iron oxide obtained during the process of roasting as ‘slag’. If the sulphide ore of copper contains iron, then silica $\left(\mathrm{SiO_2}\right)$ is added as flux before roasting. Then, FeO combines with silica to form iron silicate, $\mathrm{FeSiO_3}$ (slag).

$\mathrm{FeO}+\mathrm{SiO_2} \xrightarrow{\Delta} \underset{\text {(Slag)}}{\mathrm{FeSiO_3}}$

Answer

Chromatography is a collective term used for a family of laboratory techniques for the separation of mixtures. The term is derived from Greek words ‘chroma’ meaning ‘colour’ and ‘graphein’ meaning ’to write’. Chromatographic techniques are based on the principle that different components are absorbed differently on an absorbent. There are several chromatographic techniques such as paper chromatography, column chromatography, gas chromatography, etc.

Answer

The stationary phase is selected in such a way that the components of the sample have different solubility’s in the phase. Hence, different components have different rates of movement through the stationary phase and as a result, can be separated from each other.

Answer

Nickel is refined by Mond’s process. In this process, nickel is heated in the presence of carbon monoxide to form nickel tetracarbonyl, which is a volatile complex.

$ \mathrm{Ni}+4 \mathrm{CO} \xrightarrow{330-350 \mathrm{~K}} \underset{\text{Nickel tetracarbonyl}}{\mathrm{Ni}(\mathrm{CO})_{4}} $

Then, the obtained nickel tetracarbonyl is decomposed by subjecting it to a higher temperature ( 450 - $470 \mathrm{~K}$ ) to obtain pure nickel metal.

$\mathrm{\underset{\text{Nickel tetracarbonyl}}{Ni(CO)_4} \xrightarrow{450-470K} \underset{\text{Nickel}}{Ni} +4CO}$

Answer

To separate alumina from silica in bauxite ore associated with silica, first the powdered ore is digested with a concentrated $\mathrm{NaOH}$ solution at 473 - $523 \mathrm{~K}$ and 35 - 36 bar pressure. This results in the leaching out of alumina $\left(\mathrm{Al_2} \mathrm{O_3}\right)$ as sodium aluminate and silica $\left(\mathrm{SiO_2}\right)$ as sodium silicate leaving the impurities behind.

$$ \underset{\text{Alumina}}{\mathrm{Al_2} \mathrm{O_{3(s)}}} +2 \mathrm{NaOH_{(a q)}}+3 \mathrm{H_2} \mathrm{O_{(l)}} \longrightarrow \underset{\text{Sodium aluminate}}{2 \mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_{4}\right] {(a q)}} $$

$\underset{\text{Silica}}{\mathrm{SiO_2}}+2 \mathrm{NaOH_{(a q)}} \longrightarrow \underset{\text{Sodium silicate}}{\mathrm{Na_2} \mathrm{SiO_{3(a q)}}}+\mathrm{H_2} \mathrm{O_{(l)}}$

Then, $\mathrm{CO_2}$ gas is passed through the resulting solution to neutralize the aluminate in the solution, which results in the precipitation of hydrated alumina. To induce precipitation, the solution is seeded with freshly prepared samples of hydrated alumina.

$\underset{\text{Sodium aluminate}}{\mathrm{2Na[Al(OH)_4]}(aq)} + \mathrm{CO_2(g)} \rightarrow \underset{\text{Hydrated alumina}}{\mathrm{Al_2O_3.xH_2O (s)}} + \underset{\text{Sodium hydrogen carbonate}}{\mathrm{2NaHCO_2 (aq)}}$

During this process, sodium silicate remains in the solution. The obtained hydrated alumina is filtered, dried, and heated to get back pure alumina.

$\underset{\text{Hydrated alumina}}{\mathrm{Al_2} \mathrm{O_3} \cdot x \mathrm{H_2} \mathrm{O_{(s)}}} \xrightarrow{1470 \mathrm{~K}} \underset{\text{Alumina}}{\mathrm{Al_2} \mathrm{O_{3(s)}}}+x \mathrm{H_2} \mathrm{O_{(g)}}$

Answer

Roasting is the process of converting sulphide ores to oxides by heating the ores in a regular supply of air at a temperature below the melting point of the metal. For example, sulphide ores of $\mathrm{Zn}, \mathrm{Pb}$, and $\mathrm{Cu}$ are converted to their respective oxides by this process.

$\underset{\text{Zinc blende}}{2 \mathrm{Zns}}+3 \mathrm{O_2} \xrightarrow{\Delta} 2 \mathrm{ZnO}+2 \mathrm{SO_2}$

$\underset{\text{Galena}}{2 \mathrm{PbS}}+3 \mathrm{O_2} \xrightarrow{\Delta} 2 \mathrm{PbO}+2 \mathrm{SO_2}$

$\underset{\text{Copper glance}}{2 \mathrm{Cu_2} \mathrm{~S}}+3 \mathrm{O_2} \xrightarrow{\Delta} 2 \mathrm{Cu_2} \mathrm{O}+2 \mathrm{SO_2}$

On the other hand, calcination is the process of converting hydroxide and carbonate ores to oxides by heating the ores either in the absence or in a limited supply of air at a temperature below the melting point of the metal. This process causes the escaping of volatile matter leaving behind the metal oxide. For example, hydroxide of Fe, carbonates of $\mathrm{Zn}, \mathrm{Ca}, \mathrm{Mg}$ are converted to their respective oxides by this process.

$\underset{\text{Limonite}}{\mathrm{Fe_2} \mathrm{O_3} \cdot 3 \mathrm{H_2} \mathrm{O}} \xrightarrow{\Delta} \mathrm{Fe_2} \mathrm{O_3}+3 \mathrm{H_2} \mathrm{O}$

$\underset{\text{Calamine}}{\mathrm{ZnCO_{3(s)}}} \xrightarrow{\Delta} \mathrm{ZnO_{(s)}}+\mathrm{CO_{2(g)}}$

$\underset{\text{Dolomite}}{\mathrm{CaMg (CO_3)2}} \xrightarrow{\Delta} \mathrm{CaO_{(s)}}+\mathrm{MgO_{(s)}}+2 \mathrm{CO}$

Answer

The iron obtained from blast furnaces is known as pig iron. It contains around $4 \%$ carbon and many impurities such as S, P, Si, Mn in smaller amounts.

Cast iron is obtained by melting pig iron and coke using a hot air blast. It contains a lower amount of carbon (3\%) than pig iron. Unlike pig iron, cast iron is extremely hard and brittle.

Answer

Minerals are naturally occurring chemical substances containing metals. They are found in the Earth’s crust and are obtained by mining.

Ores are rocks and minerals viable to be used as a source of metal.

For example, there are many minerals containing zinc, but zinc cannot be extracted profitably (conveniently and economically) from all these minerals.

Zinc can be obtained from zinc blende $(\mathrm{ZnS})$, calamine $\left(\mathrm{ZnCO_3}\right)$, Zincite $(\mathrm{ZnO})$ etc.

Thus, these minerals are called ores of zinc.

Answer

Copper matte contains $\mathrm{Cu_2} \mathrm{~S}$ and FeS. Copper matte is put in a silica-lined converter to remove the remaining FeO and $\mathrm{FeS}$ present in the matte as slag $\left(\mathrm{FeSiO_3}\right)$. Also, some silica is added to the silica-lined converter. Then, a hot air blast is blown. As a result, the remaining $\mathrm{FeS}$ and $\mathrm{FeO}$ are converted to iron silicate $\left(\mathrm{FeSiO_3}\right)$ and $\mathrm{Cu_2} \mathrm{~S}$ is converted into metallic copper.

$ \begin{aligned} & 2 \mathrm{FeS}+3 \mathrm{O_2} \longrightarrow 2 \mathrm{FeO}+2 \mathrm{SO_2} \\ & \mathrm{FeO}+\mathrm{SiO_2} \longrightarrow \mathrm{FeSiO_3} \\ & 2 \mathrm{Cu_2} \mathrm{~S}+3 \mathrm{O_2} \longrightarrow 2 \mathrm{Cu_2} \mathrm{O}+2 \mathrm{SO_2} \\ & 2 \mathrm{Cu_2} \mathrm{O}+\mathrm{Cu_2} \mathrm{~S} \longrightarrow 6 \mathrm{Cu}+\mathrm{SO_2} \end{aligned} $

Answer

Cryolite $\left(\mathrm{Na_3} \mathrm{AlF_6}\right)$ has two roles in the metallurgy of aluminium:

1. To decrease the melting point of the mixture from $2323 \mathrm{~K}$ to $1140 \mathrm{~K}$.
2. To increase the electrical conductivity of $\mathrm{Al_2} \mathrm{O_3}$.

Answer

In case of low grade copper ores, leaching is carried out using acid or bacteria in the presence of air. In this process, copper goes into the solution as $\mathrm{Cu}^{2+}$ ions.

$ \mathrm{Cu_{(s)}}+2 \mathrm{H_{(a q)}^{+}}+\frac{1}{2} \mathrm{O_{2(g)}} \longrightarrow \mathrm{Cu_{(a q)}^{2+}}+2 \mathrm{H_2} \mathrm{O_{(l)}} $

The resulting solution is treated with scrap iron or $\mathrm{H_2}$ to get metallic copper.

$ \mathrm{Cu_{(a q)}^{2+}}+\mathrm{H_{2(g)}} \longrightarrow \mathrm{Cu_{(s)}}+2 \mathrm{H_{(a q)}^{+}} $

Answer

The standard Gibbs free energy of formation of ZnO from Zn

is lower than that of $\mathrm{CO_2}$ from $\mathrm{CO}$. Therefore, $\mathrm{CO}$ cannot reduce $\mathrm{ZnO}$ to $\mathrm{Zn}$. Hence, $\mathrm{Zn}$ is not extracted from $\mathrm{ZnO}$ through reduction using $\mathrm{CO}$.

Answer

The value of $\Delta_{4} \mathrm{G}^{\ominus}$ for the formation of $\mathrm{Cr_2} \mathrm{O_3}$ from $\mathrm{Cr}\left(-540 \mathrm{kJmol}^{-1}\right)$ is higher than that of $\mathrm{Al_2} \mathrm{O_3}$ from $\mathrm{Al}\left(-827 \mathrm{kJmol}^{-1}\right)$. Therefore, $\mathrm{Al}$ can reduce $\mathrm{Cr_2} \mathrm{O_3}$ to $\mathrm{Cr}$. Hence, the reduction of $\mathrm{Cr_2} \mathrm{O_3}$ with $\mathrm{Al}$ is possible.

Alternatively,

$ \begin{array}{ll} 2 \mathrm{AI}+\frac{3}{2} \mathrm{O_2} \longrightarrow \mathrm{Al_2} \mathrm{O_3} & \Delta_{\mathrm{T}} \mathrm{G}^{\ominus}=-827 \mathrm{~kJmol}^{-1} \\ 2 \mathrm{Cr}+\frac{3}{2} \mathrm{O_2} \longrightarrow \mathrm{Cr_2} \mathrm{O_3} & \Delta_{\mathrm{r}} \mathrm{G}^{\ominus}=-540 \mathrm{~kJmol}^{-1} \end{array} $

Subtracting equation (ii) from (i), we have

$ \begin{aligned} & 2 \mathrm{Al}+\mathrm{Cr_2} \mathrm{O_3} \longrightarrow \mathrm{Al_2} \mathrm{O_3}+2 \mathrm{Cr} \\ & \Delta_{\mathrm{r}} \mathrm{G}^{\ominus}=-827-(-540) \\ & =-287 \mathrm{~kJmol}^{-1} \end{aligned} $

As $\Delta_{4} \mathrm{G}^{\theta}$ for the reduction reaction of $\mathrm{Cr_2} \mathrm{O_3}$ by $\mathrm{Al}$ is negative, this reaction is possible.

Answer

Temperature

Reduction of $\mathrm{ZnO}$ to $\mathrm{Zn}$ is usually carried out at 1673 K. From the above figure, it can be observed that above 1073 $\mathrm{K}$, the Gibbs free energy of formation of CO from C and above $1273 \mathrm{~K}$, the Gibbs free energy of formation of $\mathrm{CO_2}$ from $\mathrm{C}$ is lesser than the Gibbs free energy of formation of $\mathrm{ZnO}$. Therefore, $\mathrm{C}$ can easily reduce $\mathrm{ZnO}$ to $\mathrm{Zn}$.

On the other hand, the Gibbs free energy of formation of $\mathrm{CO_2}$ from $\mathrm{CO}$ is always higher than the Gibbs free energy of formation of $\mathrm{ZnO}$. Therefore, $\mathrm{CO}$ cannot reduce $\mathrm{ZnO}$. Hence, $\mathrm{C}$ is a better reducing agent than $\mathrm{CO}$ for reducing $\mathrm{ZnO}$.

Answer

Temperature

The above figure is a plot of Gibbs energy $\left(\Delta G^{9}\right)$ vs. T for formation of some oxides.

It can be observed from the above graph that a metal can reduce the oxide of other metals, if the standard free energy of formation $\left(\Delta_\mathrm{f} \mathrm{G}^{\mathrm{O}}\right)$ of the oxide of the former is more negative than the latter. For example, since $\Delta_\mathrm{f} \mathrm{G_{(\mathrm{Al}, \mathrm{Al_2}, \mathrm{O_3})}^{\ominus}}$ is more negative than $\Delta_{\mathrm{f}} \mathrm{G_{(\mathrm{Cu}, \mathrm{Cu_2}, \mathrm{O})}^{\ominus}}, \mathrm{Al}$ can reduce $\mathrm{Cu_2} \mathrm{O}$ to $\mathrm{Cu}$, but $\mathrm{Cu}$ cannot reduce $\mathrm{Al_2} \mathrm{O_3}$. Similarly, Mg can reduce $\mathrm{ZnO}$ to $\mathrm{Zn}$, but $\mathrm{Zn}$ cannot reduce $\mathrm{MgO}$ because $ { \Delta_{\mathrm{f}} \mathrm{G_{(\mathrm{Mg}, \mathrm{MgO})}^{\ominus}}}$ is more negative than $\Delta_{f} \mathrm{G_{(\mathrm{Zn}, \mathrm{ZnO})}^{\ominus}}$.

Answer

In the electrolysis of molten $\mathrm{NaCl}, \mathrm{Cl_2}$ is obtained at the anode as a by product.

$\mathrm{NaCl_\text {(melt) }} \longrightarrow \mathrm{Na_\text {(melt) }}^{+}+\mathrm{Cl}^{-}{ _\text {(melt) }}$

At cathode: $\mathrm{Na_{\text {(melt) }}^{+}}+\mathrm{e}^{-} \longrightarrow \mathrm{Na_{(s)}}$

At anode: $\mathrm{Cl_\text {(melt) }^{-}} \longrightarrow \mathrm{Cl_{(g)}}+\mathrm{e}^{-}$

$2 \mathrm{Cl_{(g)}} \longrightarrow \mathrm{Cl_{2(g)}}$

The overall reaction is as follows:

$\mathrm{NaCl_\text {(melt) }} \xrightarrow{\text { Electrolysis }} \mathrm{Na_{(s)}}+\frac{1}{2} \mathrm{Cl_{2(g)}}$

If an aqueous solution of $\mathrm{NaCl}$ is electrolyzed, $\mathrm{Cl_2}$ will be obtained at the anode but at the cathode, $\mathrm{H_2}$ will be obtained (instead of $\mathrm{Na})$. This is because the standard reduction potential of $\mathrm{Na}\left(\mathrm{E}^{\circ}=-2.71 \mathrm{~V}\right)$ is more negative than that of $\mathrm{H_2} \mathrm{O}\left(E^{\circ}=-0.83 \mathrm{~V}\right)$. Hence, $\mathrm{H_2} \mathrm{O}$ will get preference to get reduced at the cathode and as a result, $\mathrm{H_2}$ is evolved.

$\mathrm{NaCl_{(a q)}} \longrightarrow \mathrm{Na_{(a q)}}^{+}+\mathrm{Cl_{(a q)}^{-}}$

At cathode: $2 \mathrm{H_2} \mathrm{O_{(l)}}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H_{2(g)}}+2 \mathrm{OH_{(a q)}^{-}}$

At anode: $\mathrm{Cl_\text {(melt) }}^{-} \longrightarrow \mathrm{Cl_{(g)}}+\mathrm{e}^{-}$

$2 \mathrm{Cl_{(g)}} \longrightarrow \mathrm{Cl_{2(g)}}$

Answer

In the electrometallurgy of aluminium, a fused mixture of purified alumina $\left(\mathrm{Al_2} \mathrm{O_3}\right)$, cryolite $\left(\mathrm{Na_3} \mathrm{AlF_6}\right)$ and fluorspar $\left(\mathrm{CaF_2}\right)$ is electrolysed. In this electrolysis, graphite is used as the anode and graphite-lined iron is used as the cathode. During the electrolysis, $\mathrm{Al}$ is liberated at the cathode, while $\mathrm{CO}$ and $\mathrm{CO_2}$ are liberated at the anode, according to the following equation.

Cathode : $\mathrm{Al_{\text {(melt) }}^{3+}}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al_{(l)}}$

Anode: $\mathrm{C_{(s)} + O_{\text{(melt)}}^{2-} \longrightarrow CO_{(g)} + 2e^-}$

$ \mathrm{C_{(s)}}+2 \mathrm{O_{(\text {melt}) }^{2-}} \longrightarrow \mathrm{CO_{2(g)}}+4 \mathrm{e}^{-} $

If a metal is used instead of graphite as the anode, then $\mathrm{O_2}$ will be liberated. This will not only oxidise the metal of the electrode, but also convert some of the $\mathrm{Al}$ liberated at the cathode back into $\mathrm{Al_2} \mathrm{O_3}$. Hence, graphite is used for preventing the formation of $\mathrm{O_2}$ at the anode. Moreover, graphite is cheaper than other metals.

Answer

(i) Zone refining:

This method is based on the principle that impurities are more soluble in the molten state of metal (the melt) than in the solid state. In the process of zone refining, a circular mobile heater is fixed at one end of a rod of impure metal.

As the heater moves, the molten zone of the rod also moves along with it. As a result, pure metal crystallizes out of the melt and the impurities pass to the adjacent molten zone. This process is repeated several times, which leads to the segregation of impurities at one end of the rod. Then, the end with the impurities is cut off. Silicon, boron, gallium, indium etc. can be purified by this process.

(ii) Electrolytic refining;

Electrolytic refining is the process of refining impure metals by using electricity. In this process, impure metal is made the anode and a strip of pure metal is made the cathode. A solution of a soluble salt of the same metal is taken as the electrolyte. When an electric current is passed, metal ions from the electrolyte are deposited at the cathode as pure metal and the impure metal from the anode dissolves into the electrolyte in the form of ions. The impurities present in the impure metal gets collected below the anode. This is known as anode mud.

(iii) Vapour phase refining

Vapour phase refining is the process of refining metal by converting it into its volatile compound and then, decomposing it to obtain a pure metal. To carry out this process,

(i) the metal should form a volatile compound with an available reagent, and

(ii) the volatile compound should be easily decomposable so that the metal can be easily recovered.

Nickel, zirconium, and titanium are refined using this method.

Answer

Above $1350^{\circ} \mathrm{C}$, the standard Gibbs free energy of formation of $\mathrm{Al_2} \mathrm{O_3}$ from $\mathrm{Al}$ is less than that of $\mathrm{MgO}$ from $\mathrm{Mg}$. Therefore, above $1350^{\circ} \mathrm{C}$, $\mathrm{Al}$ can reduce $\mathrm{MgO}$.