General principles and processes of isolation of elements practice · Mixed difficulty

Option A is correct because the negative Gibbs free energy change for the formation of Al2O3 is more negative than that of Cr2O3, making the reduction of Cr2O3 with aluminum thermodynamically feasible.

Option B (Hall’s process) is correct because Hall's process is used for the electrolytic reduction of alumina (Al2O3) to obtain aluminum.

Option D is correct because the cyanide process is used for the extraction of both gold (Au) and silver (Ag) from their ores.

Option B (Al) is correct because Hoope's process is an electrolytic refining method specifically used for the purification of aluminum.

Option A is correct because lead is extracted from its sulfide ore (PbS) by self-reduction, while tin is extracted from its oxide ore ($SnO_2$) by carbon reduction.

Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because at 673K, CO is a better reducing agent than coke, and this is because the change in Gibbs free energy for CO to CO2 is more negative than that of C to CO.

Option A is correct because the metal extracted by the cyanide process is primarily silver, although it is also used for gold.

Option B (Al) is correct because aluminum has a strong affinity for oxygen, making its oxide very stable and difficult to reduce with carbon.

Option B (${Ti}+2 {I}_2$) is correct because $\mathrm{TiI}_4$ is unstable at high temperatures and decomposes into titanium and iodine.

Option D is correct because both the assertion and the reason are false. Ellingham diagrams show that the $\Delta G^\circ$ for the formation of FeO is less negative than that of $Cu_2O$, meaning $Cu_2O$ is easier to reduce than FeO by carbon. Also, the given $\Delta G^\circ$ values are incorrect.