StreakPeaked· Practice

ExamsJEE AdvancedChemistry

Which of the following statements about the metallurgical extraction of lead from galena (PbS) are correct? (A) Roasted ore (PbO) can be reduced using fresh galena (PbS) directly. (B) Copper impurities in molten lead can be removed by liquation. (C) Alkali metal cyanide selectively prevents galena from entering the froth during froth flotation of mixed sulfide ores. (D) Silver impurities in molten lead can be removed by oxidation with oxygen (Pattinson's process uses a different mechanism, but cupellation involves O2).

  1. Reduction of roasted ore with fresh galena
  2. Copper impurities can be remove by liquation
  3. Alkali metal cyanide selectively prevent galena coming to the froth
  4. Oxidation of silver impurity in molten metal by O2

Correct answer: Reduction of roasted ore with fresh galena

Solution

A: The self-reduction reaction PbS + 2PbO -> 3Pb + SO2 is a real step in lead metallurgy — correct. B: Liquation separates metals of different melting points; copper has a higher melting point than lead, so liquation (which drains the lower-MP metal) would remove lead, not copper — incorrect. C: Alkali cyanide (KCN) is used to depress ZnS (sphalerite) and pyrite in froth flotation of complex sulfide ores, not to depress galena — incorrect as stated. D: Silver in lead is removed by the Pattinson process (crystallization) or cupellation; in cupellation lead is oxidised to litharge (PbO) by blowing air, but silver is NOT oxidised — incorrect.

Related JEE Advanced Chemistry questions

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →