Identify the Reducing Species in Self Reduction Process

In the process of self-reduction (or auto-reduction), less electropositive metals such as copper ($\text{Cu}$), lead ($\text{Pb}$), and mercury ($\text{Hg}$) are extracted from their sulfide ores without adding an external reducing agent.

1. Chemical Process

A portion of the sulfide ore is first roasted in air to convert it into its oxide:
$2\text{Cu}_2\text{S} + 3\text{O}_2 \rightarrow 2\text{Cu}_2\text{O} + 2\text{SO}_2$

The unreacted sulfide ore then reacts with the formed metal oxide in the absence of air:
$2\text{Cu}_2\text{O} + \text{Cu}_2\text{S} \rightarrow 6\text{Cu} + \text{SO}_2$

2. Redox Analysis

To determine the reducing species, we analyze the oxidation states of the involved species in the reaction $2\text{Cu}_2\text{O} + \text{Cu}_2\text{S} \rightarrow 6\text{Cu} + \text{SO}_2$:

• Copper ($\text{Cu}$):

  • Oxidation state in $\text{Cu}_2\text{O}$ and $\text{Cu}_2\text{S}$: $+1$
  • Oxidation state in elemental $\text{Cu}$: $0$
  • $\text{Cu}^+$ gains electrons to form $\text{Cu}$ (it is reduced):
    $\text{Cu}^+ + e^- \rightarrow \text{Cu}^0$

• Sulfur ($\text{S}$):

  • Oxidation state in $\text{Cu}_2\text{S}$: $-2$ ($\text{S}^{2-}$)
  • Oxidation state in $\text{SO}_2$: $+4$
  • $\text{S}^{2-}$ loses electrons to form $\text{S}^{4+}$ (it is oxidized):
    $\text{S}^{2-} \rightarrow \text{S}^{4+} + 6e^-$

• Oxygen ($\text{O}$):

  • Oxidation state remains unchanged at $-2$ ($\text{O}^{2-}$ in both $\text{Cu}_2\text{O}$ and $\text{SO}_2$).

Conclusion

Since $\text{S}^{2-}$ loses electrons and reduces the metal cation ($\text{Cu}^+$) to elemental metal ($\text{Cu}$), the sulfide ion ($\text{S}^{2-}$) is the reducing species.

Correct Option: C) $\text{S}^{2-}$