Explanation
In some sulphide ores of copper, zinc and lead, the concentration is brought by the froth flotation process. In this method, the advantage is taken of the preferential wetting of the ore by an oil. The finely ground ore is taken in a tank containing water and 1% of pine oil or turpentine oil.
Collectors (e.g., pine oils, fatty acids, xanthates, etc.) enhance the non-wettability of the mineral particles and froth stabilisers (e. g., cresols, aniline) stabilise the froth.
Sometimes, it is possible to separate two sulphide ores by adjusting the proportion of oil to water or by using ‘depressants’. For example, in the case of an ore containing ZnS and PbS, the depressant used is NaCN. It selectively prevents ZnS from coming to the froth but allows PbS to come with the froth.
Option D is correct.
Si of high purity to be used in semiconductor can be prepared by following methods :
I. $$SiO_2\, +\, 2C\, \rightarrow\, Si\, +\, 2CO$$
II. $$Si\, +\, 2Cl_2\, \rightarrow\, SiCl_4$$
$$SiCl_4\, +\, 2Mg\, \rightarrow\, Si\, +\, 2MgCl_2$$
Use the relationship $$\Delta \,G^{\circ}\,=\,- nFE^{\circ}_{cell}$$ to estimate the minimum voltage required to electrolyse $$Al_2O_3$$ in the Hall-Heroult process.
$$\Delta G^{\circ}_{f}(Al_2O_3)\,=\,-1520\,kJ\,mol^{-1}$$
$$\Delta G^{\circ}_{f}(CO_2)\,=\,-394\,kJ\,mol^{-1}$$
The oxidation of the graphite anode to $$CO_2$$ permits the electrolysis to occur at a lower voltage than if the electrolysis reactions were :
Net reaction in Hall-Heroult process is
$$3C\,+\,2Al_2O_3\,\rightarrow\,4Al\,+\,3CO_2$$
Or $$4Al^{3+}\,+\,12e^{-}\,\rightarrow \,4Al,$$
Number of electrons (n) = 12
$$\Delta\,G^{\circ }\,=\,3 \Delta\, G^{\circ }_f (CO_2)\,-\,2 \Delta\,G^{\circ }_f(Al_2O_3)$$
$$=-3\,\times\,394\,-\,2(-1520)$$
$$\Delta G^{\circ}\,=\, -nFE^{\circ}_{cell}$$
$$-E^{\circ}_{cell}\,=\, \displaystyle \frac {\Delta G^{\circ}}{nF}\, =\, \displaystyle\frac {1858\,\times\, 1000}{12\, \times\, 96500}$$
$$= 1.60\ V$$
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