Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 6.12:

Describe a method for refining nickel.

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.

Then, the obtained nickel tetracarbonyl is decomposed by subjecting it to a higher temperature (450 − 470 K) to obtain pure nickel metal.

Question 6.13:

How can you separate alumina from silica in bauxite ore associated with

silica? Give equations, if any.

Answer:

To separate alumina from silica in bauxite ore associated with silica, first the powdered ore 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.

Then, CO₂ 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.

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

Question 6.14:

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

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

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 Zn, Ca, Mg are converted to their respective oxides by this process.

Question 6.15:

How is ‘cast iron’ different from ‘pig iron”?

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.

Question 6.16:

Differentiate between “minerals” and “ores”.

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 (ZnS), calamine (ZnCO₃), Zincite (ZnO) etc.

Thus, these minerals are called ores of zinc.

Question 6.17:

Why copper matte is put in silica lined converter?

Answer:

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.

Question 6.18:

What is the role of cryolite in the metallurgy of aluminium?

Answer:

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.

Question 6.19:

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

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 Cu²⁺ ions.

The resulting solution is treated with scrap iron or H₂ to get metallic copper.

Question 6.20:

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

Answer:

The standard Gibbs free energy of formation of ZnO from Zn

is lower than that of CO₂ from CO. Therefore, CO cannot reduce ZnO to Zn. Hence, Zn is not extracted from ZnO through reduction using CO.

Question 6.21:

The value of   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?

Answer:

The value of  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,

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

As   for the reduction reaction of Cr₂O₃ by Al is negative, this reaction is possible.

Question 6.22:

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

Answer:

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.

Question 6.23:

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.

Answer:

The above figure is a plot of Gibbs energy   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   of the oxide of the former is more negative than the latter. For example, since   is more negative than  , Al can reduce Cu₂O to Cu, but Cu cannot reduce Al₂O₃. Similarly, Mg can reduce ZnO to Zn, but Zn cannot reduce MgO because   is more negative than  .

Question 6.24:

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?

Answer:

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

At cathode: 

At anode: 

The overall reaction is as follows:

If an aqueous solution of NaCl is electrolyzed, Cl₂ will be obtained at the anode but at the cathode, H₂ will be obtained (instead of Na). This is because the standard reduction potential of Na (E°= − 2.71 V) is more negative than that of H₂O (E° = − 0.83 V). Hence, H₂O will get preference to get reduced at the cathode and as a result, H₂ is evolved.

At cathode: 

At anode: 

Question 6.25:

What is the role of graphite rod in the electrometallurgy of aluminium?

Answer:

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.

If a metal is used instead of graphite as the anode, then O₂ will be liberated. This will not only oxidise the metal of the electrode, but also convert some of the Al liberated at the cathode back into Al₂O₃. Hence, graphite is used for preventing the formation of O₂ at the anode. Moreover, graphite is cheaper than other metals.

Question 6.27:

Outline the principles of refining of metals by the following methods:

(i) Zone refining

(ii) Electrolytic refining

(iii) Vapour phase refining

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.

Question 6.28:

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

Answer:

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.