NCERT Solutions for Class 12 Chemistry Chapter 6 

Q.1 Copper can be extracted by hydrometallurgy but not zinc. Explain.

Ans.

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.

Fe(s) + Cu²⁺ (aq) → Fe²⁺(aq) + Cu(s)

But to displace zinc, more reactive metals i.e., metals having lower reduction potentials than zinc such as Mg, Ca, K, etc. are required. But all these metals react with water with the evolution of H₂ gas.

2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g)

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

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

Q.2 What is the role of depressant in froth floatation process?

Ans.

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 (ZnS and Pbs), NaCN is used as depressant which selectively allows PbS to come with froth, but prevents ZnS from coming to froth. This happens because NaCN reacts with ZnS to form Na₂[Zn(CN)₄].

4NaCN + Zns → Na₂ [Zn(CN)₄] + Na₂S

Q.3 Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?

Ans.

The Gibbs free energy of formation (Δ_fG) of CS₂ and H₂S is higher than that of Cu₂S. Therefore, C and H₂ cannot reduce Cu₂S to Cu. On the other hand, the Gibbs free energy of formation (Δ_fG) of Cu₂O is higher than that of CO and thus C can reduce Cu₂O to Cu.

Q.4 Explain: (i) Zone refining (ii) Column chromatography.

Ans.

(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. Al₂O₃ 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).

Q.5 Out of C and CO, which is a better reducing agent at 673 K?

Ans.

At 673 K, the value of

$\mathrm{\Delta }{\text{G}}_{(\text{CO},{\text{CO}}_{\text{2}})}$

is less than that of ∆G_(C,CO) . Therefore, CO can be reduced more easily to CO₂ than C to CO. Hence, CO is a better reducing agent than C at 673 K.

Q.6 Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?

Ans.

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.

Q.7 Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.

Ans.

During the extraction of iron, the reduction of iron oxides takes place in the blast furnace. In this process, hot air is blown from the bottom of the furnace and coke is burnt to raise the temperature up to 2200 K in the lower portion itself. The temperature is lower in the upper part. Thus, it is the lower part where the reduction of iron oxides (Fe₂O₃ and Fe₃O₄) takes place. The reactions taking place in the lower temperature range (500 – 800 K) in the blast furnace are:

3Fe₂O₃ + CO → 2Fe₃O₄ + CO₂

Fe₃O₄ + 4CO → 3FeO + 4CO₂

Fe₂O₃ + CO → 2Fe₃ + CO₂

The reactions taking place in the higher temperature range (900 – 1500 K) in the blast furnace are:

C + CO₂ → 2CO

FeO + CO → Fe + CO₂

The silicate impurity of the one is removed as slag by calcium oxide (CaO), which is formed by the decomposition of limestone (CaCO₃).

CaCO₃ → CaO + CO₂

CaO + SiO₂ → CaSiO₃

Calcium silicate

(Slag)

Q.8 Which method of refining may be more suitable if element is obtained in minute quantity?

Ans.

Column chromatograhy

Q.9 Write chemical reactions taking place in the extraction of zinc from zinc blende.

Ans.

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 one 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 Zn.

2ZnS + 3O₂ → 2ZnO + 2SO₂

(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 K.

$\text{ZnO}+\mathrm{C}\stackrel{\mathrm{coke},673\mathrm{k}}{\to }\mathrm{Zn}+\mathrm{CO}$

(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 (ZnSO₄). Electrolysis results in the transfer of zinc In pure from the anode to the cathode.

Anode: Zn → Zn²⁺ + 2e^–

Cathode: Zn²⁺ + 2e^– → Zn

Q.10 State the role of silica in the metallurgy of copper.

Ans.

During the roasting of pyrite one, a mixture of FeO and Cu₂O is obtained.

$\begin{array}{l}2{\mathrm{CuFeS}}_2+{\mathrm{O}}_2\stackrel{\mathrm{\Delta }}{\to }{\mathrm{Cu}}_2\mathrm{S}+2\mathrm{FeS}+{\mathrm{SO}}_2\\ 2{\mathrm{Cu}}_2\mathrm{S}+3{\mathrm{O}}_2\stackrel{\mathrm{\Delta }}{\to }2{\mathrm{Cu}}_2\mathrm{O}+2{\mathrm{SO}}_2\\ 2\mathrm{FeS}+3{\mathrm{O}}_2\stackrel{\mathrm{\Delta }}{\to }2\mathrm{FeO}+2{\mathrm{SO}}_2\\ \begin{array}{l}\text{The role of silica in the metallurgy of copper is to remove the iron oxide obtained during the process of roasting as}‘\text{slag}’.\text{}\\ \text{If the sulphide ore of copper contains iron},\text{then silica}\left({\text{SiO}}_{\text{2}}\right)\text{is added as flux before roasting}.\end{array}\\ \text{Then},\text{FeO combines with silica to form iron silicate},{\text{FeSiO}}_{\text{3}}\left(\text{slag}\right).\\ \mathrm{FeO}+\mathrm{Si}{\mathrm{O}}_2\stackrel{}{\to }{\mathrm{FeSiO}}_3\end{array}$

Q.11 Which method of refining will you suggest for an element in which impurities present have chemical properties close to the properties of that element?

Ans.

Zone refining

Q.12 Describe a method for refining nickel.

Ans.

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.

$$\begin{gathered} \begin{array}{l}\mathrm{Ni}+4\mathrm{CO}\stackrel{330-350\mathrm{K}}{\to }\underset{\mathrm{Nickel}\mathrm{tera}\mathrm{carbonyl}}{\mathrm{Ni}{\left(\mathrm{CI}\right)}_4}\\ \text{Then},\text{the obtained nickel tetracarbonyl is decomposed by subjecting it to a higher temperature} \\ (\text{45}0–\text{47}0\text{K})\text{\hspace{0.17em}to obtain pure nickel metal}.\\ \underset{\mathrm{Nickel}\mathrm{tera}\mathrm{carbonyl}}{\mathrm{Ni}{\left(\mathrm{CI}\right)}_4}\stackrel{450-470\mathrm{K}}{\to }\mathrm{Ni}+4\mathrm{CO}\end{array} \end{gathered}$$

Q.13 How can you separate alumina from silica in bauxite are associated with silica? Give equations, if any?

Ans.

To separate alumina from silica in bauxite are associated with silica, first the powdered one is digested with a concentrated NaOH solution at 473 – 523 K and 35 – 36 bar pressure. This results in the leaching out of alumina (Al₂O₃) as sodium aluminate and silica (SiO₂) as sodium silicate leaving the impurities behind.

$\begin{array}{l}\begin{array}{l}\underset{\text{Alumina}}{{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}\left(\text{8}\right)}}{\text{+2NaOH}}_{\left(\text{aq}\right)}{\text{+3H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to \underset{\text{Sodium\hspace{0.17em}aluminate}}{\text{2Na}{\left[\text{Al}{\left(\text{OH}\right)}_{\text{4}}\right]}_{\left(\text{aq}\right)}}\\ \underset{\text{Silica}}{{\text{SiO}}_{\text{2}}}{\text{+2NaOH}}_{\left(\text{aq}\right)}\to \underset{\text{Sodium\hspace{0.17em}silicate}}{{\text{Na}}_{\text{2}}{\text{SiO}}_{\text{3}\left(\text{aq}\right)}}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ \begin{array}{l}\text{Then},{\text{CO}}_{\text{2}}\text{gas is passed through the resulting solution to neutralize the aluminate in the}\\ \text{solution},\text{which resultsi n the precipitation of hydrated alumina}.\text{To induce precipitation},\\ \text{the solution is seeded with freshly prepared samples of hydrated alumina}.\end{array}\end{array}\\ \underset{\text{Sodium aluminate Hydrated alumina}}{\text{2Na}{\left[\text{Al}{\left(\text{OH}\right)}_{\text{4}}\right]}_{\left(\text{aq}\right)}+{\text{CO}}_{\text{2}\left(\text{g}\right)}}\to \underset{\text{Sodiumhydrogen carbonate}}{{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{.}_{\text{x}}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{s}\right)}+{\text{2NaHCO}}_{\text{3}\left(\text{aq}\right)}}\\ \text{During this process},\text{sodium silicate remains in the solution}.\text{The obtained hydrated alumina}\\ \text{is filtered, dried, and heated to get back pure alumina}\\ \underset{\text{Hydrated\hspace{0.17em}alumina}}{{\text{Al}}_2{\mathrm{O}}_3{.}_{\mathrm{x}}{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{s}\right)}}\stackrel{1470\mathrm{K}}{\to }\underset{\text{Alumina}}{{\text{Al}}_{\text{2}}{{\text{O}}_{\text{3}}}_{\left(\text{s}\right)}}{+}_{\mathrm{x}}{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}\end{array}$

Q.14 Giving examples, differentiate between ‘roasting’ and ‘calcination’.

Ans.

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 Zn, Pb, and Cu are converted to their respective oxides by this process.

$\begin{array}{l}\begin{array}{l}\underset{\mathrm{Zinc}\text{blende}}{2\mathrm{Zns}}+3{\text{O}}_2\stackrel{\text{Δ}}{\to }2\mathrm{ZnO}+2{\mathrm{SO}}_2\\ \underset{\mathrm{Galena}}{2\mathrm{PbS}}+3{\text{O}}_2\stackrel{\text{Δ}}{\to }2\mathrm{PbO}+2{\mathrm{SO}}_2\\ \underset{\mathrm{Galena}}{2{\mathrm{Cu}}_2\text{S}}+3{\text{O}}_2\stackrel{\text{Δ}}{\to }2{\mathrm{Cu}}_2\text{O}+2{\mathrm{SO}}_2\end{array}\\ \begin{array}{l}\begin{array}{l}\text{On the other hand},\text{calcination is the process of converting hydroxide and carbonate}\\ \text{ones to oxides by heating the ones either in the absence or in a limited supply of air at}\\ \text{a temperature below the melting point of the metal}.\end{array}\\ \begin{array}{l}\text{This process causes the escaping of volatile matter leaving behind the metal oxide}.\\ \text{For example},\text{hydroxide of Fe},\text{carbonates of Zn},\text{Ca},\text{Mg are converted to their}\\ \text{respective oxides by this process}.\end{array}\end{array}\\ \begin{array}{l}\underset{\mathrm{Limonite}}{{\mathrm{Fe}}_2{\text{O}}_{3}3{\text{H}}_2\text{O}}\stackrel{\text{Δ}}{\to }{\mathrm{Fe}}_2{\text{O}}_3+3{\text{H}}_2\text{O}\\ \underset{\text{Calamine}}{{\mathrm{ZnCO}}_{3\left(\text{s}\right)}}\stackrel{\text{Δ}}{\to }{\mathrm{ZnO}}_{\left(\text{s}\right)}+{{\mathrm{CO}}_2}_{\left(\text{g}\right)}\\ \underset{\mathrm{Dolomite}}{\mathrm{CaMg}{\left({\mathrm{CO}}_3\right)}_2}\stackrel{\text{Δ}}{\to }{\mathrm{CaO}}_{\left(\text{s}\right)}+{\mathrm{MgO}}_{\left(\text{s}\right)}2\mathrm{CO}\end{array}\end{array}$

Q.15 How is ‘cast iron’ different from ‘pig iron”?

Ans.

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.

Q.16 Differentiate between “minerals” and “ores”.

Ans.

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 (ZnS), calamine (ZnCO₃), Zincite (ZnO) etc. Thus, these minerals are called ores of zinc.

Q.17 Why copper matte is put in silica lined converter?

Ans.

Copper matte contains Cu₂S and FeS. Copper matte is put in a silica-lined converter to remove the remaining FeO and FeS present in the matte as slag (FeSiO₃). Also, some silica is added to the silica-lined converter. Then, a hot air blast is blown. As a result, the remaining FeS and FeO are converted to iron silicate (FeSiO₃) and Cu₂S is converted into metallic copper.

$\begin{array}{l}2\mathrm{FeS}+3{\mathrm{O}}_2\to 2\mathrm{FeO}+2{\mathrm{SO}}_2\\ \mathrm{FeO}+{\mathrm{SiO}}_2\to {\mathrm{FeSiO}}_3\\ 2{\mathrm{Cu}}_2\mathrm{S}+3{\mathrm{O}}_2\to 2{\mathrm{Cu}}_2\mathrm{O}+2{\mathrm{SO}}_2\\ 2{\mathrm{Cu}}_2\mathrm{O}+{\mathrm{Cu}}_2\mathrm{S}\to \mathrm{}6\mathrm{Cu}+{\mathrm{SO}}_2\end{array}$

Q.18 What is the role of cryolite in the metallurgy of aluminium?

Ans.

Cryolite (Na₃AlF₆) has two roles in the metallurgy of aluminium:

1. To decrease the melting point of the mixture from 2323 K to 1140 K.

2. To increase the electrical conductivity of Al₂O₃.

Q.19 How is leaching carried out in case of low grade copper ores?

Ans.

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 Cu²⁺ ions.

$\begin{array}{l}{\mathrm{Cu}}_{\left(\mathrm{s}\right)}+2{\mathrm{H}}_{\left(\mathrm{aq}\right)}^{+}+\frac{1}{2}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to {\mathrm{Cu}}_{\left(\mathrm{aq}\right)}^{2+}+2{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}\\ {\text{The resulting solution Is treated with scrap iron or H}}_{\text{2}}\text{to get metallic copper}..\\ {\mathrm{Cu}}_{\left(\mathrm{aq}\right)}^{2+}+{\mathrm{H}}_{2\left(\mathrm{g}\right)}\to {\mathrm{Cu}}_{\left(\mathrm{s}\right)}+2{\mathrm{H}}_{\left(\mathrm{aq}\right)}^{+}\end{array}$

Q.20 Why is zinc not extracted from zinc oxide through reduction using CO?

Ans.

The standard Gibbs free energy of formation of zinc oxide from zinc is lower than that of CO₂ from CO. Therefore, CO cannot reduce zinc oxide to zinc. Hence, zinc is not extracted from zinc oxide through reduction using CO.

Q.21 The value of ∆_fG^Φ for formation of Cr₂O₃ is − 540 kJmol^-1 and that of Al₂ O₃ is – 827 kJmol^-1. Is the reduction of Cr₂O₃ possible with Al?

Ans.

The value of ∆_fG^Φ for the formation of Cr₂O₃ from Cr (−540 kJmol^-1) is higher than that of Al₂O₃ from Al (−827 kJmol^-1). Therefore, Al can reduce Cr₂O₃ to Cr. Hence, the reduction of Cr₂O₃ with Al is possible.

Alternatively,

$\begin{array}{l}\mathrm{Al}+\frac{3}{2}{\mathrm{O}}_2\to {\mathrm{Al}}_2{\mathrm{O}}_3{\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{\varnothing }=827\mathrm{}{\mathrm{kJmol}}^{-1}\\ 2\mathrm{Cr}+\frac{3}{2}{\mathrm{O}}_2\to {\mathrm{Cr}}_2{\mathrm{O}}_3\mathrm{}{\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{\varnothing }=540\mathrm{}{\mathrm{kJmol}}^{-1}\\ \text{Subtracting equation}\left(\text{ii}\right)\text{from}\left(\text{i}\right),\text{we have}\\ 2\mathrm{Al}+{\mathrm{Cr}}_2{\mathrm{O}}_3\to {\mathrm{Al}}_2{\mathrm{O}}_3+2\mathrm{Cr}\\ {\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{\varnothing }=-827-(-540)\\ =-287\mathrm{}{\mathrm{kJmol}}^{-1}\\ \end{array}$

As ∆_fG^Φ for the reduction reaction of Cr₂O₃ by Al is negative, this reaction is possible.

Q.22 Out of C and CO, which is a better reducing agent for ZnO?

Ans.

Reduction of ZnO to Zn is usually carried out at 1673 K. From the above figure, it can be observed that above 1073 K, the Gibbs free energy of formation of CO from C and above 1273 K, the Gibbs free energy of formation of CO₂ from C is lesser than the Gibbs free energy of formation of ZnO. Therefore, C can easily reduce ZnO to Zn. On the other hand, the Gibbs free energy of formation of CO₂ from CO is always higher than the Gibbs free energy of formation of ZnO. Therefore, CO cannot reduce ZnO. Hence, C is a better reducing agent than CO for reducing ZnO.

Q.23 The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.

Ans.

The above figure is a plot of Gibbs energy (∆G^Φ) 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 (∆_fG^Φ) of the oxide of the former is more negative than the latter. For example, since ∆_fG^Φ(Al, Al₂,O₃) is more negative than, ∆_fG^Φ(Cu, Cu₂,O), Al can reduce Cu₂O to Cu, but Cu cannot reduce Al₂O₃. Similarly, Mg can reduce ZnO to Zn, but Zn cannot reduce MgO because ∆_fG^Φ(Mg, MgO) is more negative than ∆_fG^Φ(Zn, ZnO)

Q.24 Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous solution of NaCl is subjected to electrolysis?

Ans.

In the electrolysis of molten NaCl, Cl₂ is obtained at the anode as a by product.

$\begin{array}{l}\begin{array}{l}{\mathrm{NaCl}}_{\left(\mathrm{melt}\right)}\to {\mathrm{Na}}_{\left(\mathrm{melt}\right)}^{+}+{\mathrm{Cl}}_{\left(\mathrm{melt}\right)}^{-}\\ \mathrm{At}\mathrm{cathode}:{\mathrm{Na}}_{\left(\mathrm{melt}\right)}^{+}+{\text{e}}^{-}\to {\mathrm{Na}}_{\left(\text{s}\right)}\\ \text{At}\text{anode}:{\mathrm{Cl}}_{\left(\mathrm{melt}\right)}^{-}+\to 1/2\mathrm{Cl}{2}_{\left(\text{g}\right)}+{\text{e}}^{-}\\ 2{\mathrm{Cl}}_{\left(\text{g}\right)}\to {\mathrm{Cl}}_{2\left(\text{g}\right)}\\ \text{The overall reaction is as follows}:\\ {\mathrm{NaCl}}_{\left(\mathrm{melt}\right)}\stackrel{\mathrm{Electrolysis}}{\to }{\mathrm{Na}}_{\left(\text{s}\right)}+\frac{1}{2}{\mathrm{Cl}}_{2\left(\text{g}\right)}\end{array}\\ \mathrm{If}{\text{an aqueous solution of NaCl is electrolyzed, Cl}}_2\text{will be obtained at the anode but at the cathode,}\\ {\text{H}}_2\text{will be obtained}\left(\mathrm{instead}\text{of Na}\right).\text{This is because the standard reduction potential of Na}\left({\text{E}}^{\text{°}}=-2.71\text{V}\right)\\ {\text{is more negative than of H}}_2\text{O}\left({\text{E}}^{\text{°}}=-0.83\text{V}\right).{\text{Hence, H}}_2\text{O}\\ {\text{will get preference to get reduced at the cathode and as a result, H}}_2\text{is evolved.}\\ \begin{array}{l}{\mathrm{NaCl}}_{\left(\mathrm{aq}\right)}\to {\mathrm{Na}}_{\left(\mathrm{aq}\right)}^{+}+{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}\\ \text{At cathode}:\\ 2{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}+2_{\text{e}}^{-}\to {\text{H}}_{2\left(\text{g}\right)}+2{\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-}\end{array}\\ \begin{array}{l}\text{At anode}:\\ {\mathrm{Cl}}_{\left(\mathrm{melt}\right)}^{-}\to {\mathrm{Cl}}_{\left(\text{g}\right)}+{\text{e}}^{-}\end{array}\\ 2{\mathrm{Cl}}_{\left(\text{g}\right)}\to {\mathrm{Cl}}_{2\left(\text{g}\right)}\end{array}$

Q.25 What is the role of graphite rod in the electrometallurgy of aluminium?

Ans.

In the electrometallurgy of aluminium, a fused mixture of purified alumina (Al₂O₃), cryolite (Na₃AlF₆) and fluorspar (CaF₂) is electrolysed. In this electrolysis, graphite is used as the anode and graphite-lined iron is used as the cathode. During the electrolysis, Al is liberated at the cathode, while CO and CO₂ are liberated at the anode, according to the following equation.

Cathode: Al³⁺(melt) + 3e^–→ Al(l)

Anode: C(s) + O^2- (melt) → CO(g) + 2e^–

C(s) + 2O^2- (melt) → CO₂(g) + 4e^–

Q.26 Outline the principles of refining of metals by the following methods:
(i) Zone refining
(ii) Electrolytic refining
(iii) Vapour phase refining

Ans.

(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.

$\begin{array}{l}\text{Anode}:\mathrm{M}\to {\mathrm{M}}^{\mathrm{n}+}+{\mathrm{ne}}^{-}\\ \text{Cathode}:{\mathrm{M}}^{\mathrm{n}+}+{\mathrm{ne}}^{-}\to \mathrm{M}\end{array}$

(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.

Q.27 Predict conditions under which Al might be expected to reduce MgO.

Ans.

Above 1350°C, the standard Gibbs free energy of formation of Al₂O₃ from Al is less than that of MgO from Mg. Therefore, above 1350°C, Al can reduce MgO.