Ncert Solutions Chemistry Class 12th

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6 months ago

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Vishal Baghel

Contributor-Level 10

Graphite rod acts as anode and graphite lined iron acts as a cathode in the electrometallurgy of aluminium. Carbon reacts with oxygen liberated at anode producing CO and CO2 otherwise oxygen liberated at the anode may oxidize some of the liberated aluminium back to Al2O3.

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Vishal Baghel

Contributor-Level 10

In the electrolysis of NaCl by Down's process, chlorine is obtained as a by-product. This process involves the electrolysis of a fused mixture of NaCl and CaCl2 at 873K. during electrolysis, sodium is liberated at the cathode and Cl2 is liberated at the anode.

If an aqueous solution of NaCl is electrolysed, H2 is evolved at the cathode and Cl2 is obtained at the anode, the reason being that E0 of Na+/Na redox couple is much lower (E0 = - 2.71 V) than that of H2O (EH2O/H20    = - 0.83V ) and hence water is reduced to H in presence of Na+ ions. However, NaOH is obtained in the solution.

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alok kumar singh

Contributor-Level 10

7.55

Theory: This process involves the catalytic oxidation of sulphur dioxide into sulphur trioxide by atmospheric air.

2SO2 + O2  2 SO3; AH = - 196.6 kJ

The reaction is reversible, exothermic and involves a decrease in the number of moles. Therefore, according to the Le-Chatelier's principle, the favourable conditions for the maximum yield of sulphur trioxide are as follows.

[i]. Low temperature: A decrease in temperature would favour the forward reaction. The optimum temperature is experimentally found to be 670-720 K.

[ii]. High pressure: An increase in pressure should favour the forward reaction because the reaction involves a de

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Vishal Baghel

Contributor-Level 10

Thermodynamic factors help us in choosing a suitable reducing agent for the reduction of a particular metal state as described below.

From Ellingham diagram, it is evident that metals for which the standard free energy of formation of their oxides is more negative can reduce those metal oxides for which the standard free energy of formation of their respective oxides is less negative. In other words, any metal will reduce the oxides of other metals which lie above in the Ellingham Diagram because the standard free energy change of the combined redox reaction will be negative by any amount of equal to the difference in Δ fG0 of the two

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Vishal Baghel

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The free energy formation ( ΔfGo ) of CO from C becomes lower at temperatures above 1120K whereas that of CO2 from C becomes lower above 1323K than ΔfG0 of ZnO. However, Δ fG0 of CO2 from CO is always higher than that of ZnO. Therefore, C can reduce ZnO to Zn but not CO. therefore, out of C and CO, C is a better reducing agent than CO for ZnO.

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Vishal Baghel

Contributor-Level 10

The value of Δ fG0 for the formation of Cr2O3 is – 540 kJ mol-1 which are higher than that of Al2O3 is – 827 Kjmol-1. Thus, Al can reduce Cr2O3 to Cr. Hence, the reduction of Cr2O3 with Al is possible.

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alok kumar singh

Contributor-Level 10

7.54

Freons or chlorofluorocarbons [CFCs] are aerosols that accelerate the depletion of ozone. In presence of ultraviolet radiation [UV radiation], these chlorofluorocarbons break down to give chlorine ions which combine with the ozone atoms in the atmosphere to give oxygen atoms.

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Vishal Baghel

Contributor-Level 10

The standard Gibbs free energy formation of ZnO from Zn is lower than that of CO2 from CO. That is why CO cannot reduce ZnO to Zn. thus, Zn is not extracted from ZnO through reduction using CO.

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Vishal Baghel

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For the group of low-grade copper ores, leaching is carried out using acid or bacteria in the presence of air or oxygen. In this process, copper goes into the solution as Cu2+ ions.

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

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6 months ago

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alok kumar singh

Contributor-Level 10

7.53

Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable it will be. Lattice energy is directly proportional to the charge carried by an ion.

When a metal combines with oxygen, the lattice energy of the oxide involving O2- ion is much more than the oxide involving O- ion. Hence, the oxide having O2-ions are more stable than oxides having O-. Hence, we can say that formation of O2- is energetically more favorable that formation of O-.

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