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New answer posted
a year agoContributor-Level 10
7.58
Fluorine forms only one oxoacid i.e., HOF because of its high Electronegativity and small size.

New answer posted
a year agoContributor-Level 10
7.57
The general electronic configuration of halogens is np5, where n = 2-6. Thus, halogens need only one more electron to complete their octet and to attain the stable noble gas configuration.
Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an electron.
Whenever an atom accepts an electron from another atom, the atom accepting the electron is getting reduced and the atom donating the electron is oxidized.
But the electron accepting atom acts as an oxidizing agent while the electron donating atom acts as reducing agent. Hence,
New answer posted
a year agoContributor-Level 10
7.56
SO2 is a highly irritating gas and causes serious respiratory problems, and may cause a fit of coughing.
1. It reacts with water vapour present in the atmosphere to form sulphuric acid. This causes acid rain. Acid rain damages soil [soil become more acidic], plants, and buildings get corroded, especially those made of marble.
In the air, SO2 is oxidized to SO3 which is also an irritant.
2SO2 + O2 2SO3 SO3 + H2O H2SO4
2. Even in very low concentrations, SO2 causes irritation in the respiratory It causes throat and eye irritation and can also affect the larynx to cause breathlessness.
3. It is extremely harmful to plants. Plants exp
New answer posted
a year agoContributor-Level 10
Above 1350oC, the standard Gibbs free energy formation of Al2O3 from Al is less than that of MgO from Mg. Therefore, above 1350oC, Al can reduce MgO.
New answer posted
a year agoContributor-Level 10
(i) Zone refining Is the method based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.

(ii) Electrolytic refining works on the principle of refining impure metals by the use of electricity. In Electrolytic refining, the impure metal is made the anode and a strip of pure metal is made as the 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 a pure metal and the impure metal from the anode dissolves into the electrolyte in the form of ions. The impurities pre
New answer posted
a year agoContributor-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.

New answer posted
a year agoContributor-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.

New answer posted
a year agoContributor-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
New answer posted
a year agoContributor-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
New answer posted
a year agoContributor-Level 10

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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