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Isolation of Elements question

2010 · Shift 1 · Q24
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Isolation of Elements question

2010 · Shift 1 · Q24

JEE AdvancedChemistryIsolation of ElementsMCQ+3 / −1
Copper is the most noble of the first row transition metals and occurs in small deposits in several countries. Ores of copper inculcated chalcanthite (CuSO 4{}_44​ . 5H 2{}_22​ O), atacamite (Cu 2{}_22​ Cl(OH) 3{}_33​), cuprite (Cu 2{}_22​ O), copper glance (Cu 2{}_22​ S) and malacite (Cu 2{}_22​(OH) 2{}_22​ CO 3{}_33​). However, 80% of the world copper production comes from the ore of chalcopyrite (CuFeS 2{}_22​). The extraction of copper from chalcopynite involves partial roasting, removal of iron and self-reduction.In self-reduction, the reducing species is
  1. A
    S
  2. B
    O 2−{}^{2-}2−
  3. C
    S 2−{}^{2-}2−
  4. D
    SO 2{}_22​
View written solutionFree

Correct answer: C

The extraction of copper from chalcopyrite (CuFeS₂) involves several steps, including concentration, roasting, smelting, and bessemerisation. The question specifically asks about the self-reduction step, which primarily occurs during bessemerisation in a Bessemer converter.

  1. Roasting and Smelting: In the initial steps, the chalcopyrite ore is partially roasted to convert iron sulfides into iron oxides, which are then removed as slag (FeSiO₃) during smelting. This process yields a mixture called 'copper matte', which is predominantly cuprous sulfide (Cu₂S) with some remaining ferrous sulfide (FeS).

  2. Bessemerisation and Self-Reduction: The molten copper matte is transferred to a Bessemer converter, and a blast of hot air is passed through it. This stage involves the self-reduction process.

    • Step A: Partial Oxidation: A portion of the cuprous sulfide (Cu₂S) is oxidized by the hot air to form cuprous oxide (Cu₂O). 2Cu2S+3O2→2Cu2O+2SO22Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_22Cu2​S+3O2​→2Cu2​O+2SO2​

    • Step B: Self-Reduction: The cuprous oxide (Cu₂O) formed then reacts with the remaining cuprous sulfide (Cu₂S) to produce molten copper. This reaction is called self-reduction because one part of the ore (Cu₂S) reduces another part that has been oxidized (Cu₂O). 2Cu2O+Cu2S→6Cu+SO22Cu_2O + Cu_2S \rightarrow 6Cu + SO_22Cu2​O+Cu2​S→6Cu+SO2​

  3. Identifying the Reducing Species: To find the reducing species, we need to analyze the changes in oxidation states in the self-reduction reaction: 2Cu2+1O−2+Cu2+1S−2→6Cu0+S+4O2−22\stackrel{+1}{Cu_2}\stackrel{-2}{O} + \stackrel{+1}{Cu_2}\stackrel{-2}{S} \rightarrow 6\stackrel{0}{Cu} + \stackrel{+4}{S}\stackrel{-2}{O_2}2Cu2​+1​O−2​+Cu2​+1​S−2​→6Cu0+S+4​O2​−2​

    • Oxidation of Sulfur: The oxidation state of sulfur (S) increases from -2 in Cu₂S to +4 in SO₂. An increase in oxidation state signifies oxidation. The species that gets oxidized is the reducing agent.

    • Reduction of Copper: The oxidation state of copper (Cu) decreases from +1 in both Cu₂O and Cu₂S to 0 in elemental Cu. A decrease in oxidation state signifies reduction.

    Since the sulfide ion (S²⁻) in Cu₂S is oxidized, it acts as the reducing agent. It reduces the copper(I) ions (Cu⁺) from Cu₂O to elemental copper.

  4. Conclusion: The reducing species in the self-reduction step is the sulfide ion, S²⁻.

  5. Evaluating the Options:

    • A: S - Incorrect. Sulfur is present as the sulfide ion, S²⁻, not as elemental sulfur.
    • B: O²⁻ - Incorrect. The oxide ion's oxidation state remains -2 throughout the reaction.
    • C: S²⁻ - Correct. The sulfide ion is oxidized from -2 to +4, acting as the reducing agent.
    • D: SO₂ - Incorrect. SO₂ is a product of the reaction, not a reactant acting as a reducing agent.
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