JEE Advance - Chemistry (2010 - Paper 1 Offline - No. 24)
In self-reduction, the reducing species is
S
O$$^{2-}$$
S$$^{2-}$$
SO$$_2$$
Explanation
In the self-reduction step,
Cu$$_2$$S + $$\frac{3}{2}$$O$$_2$$ $$\to$$ Cu$$_2$$O + SO$$_2$$
Cu$$_2$$S + 2Cu$$_2$$O $$\to$$ $$\mathrm{\mathop {6Cu(l)}\limits_{Blister\,copper} }$$ + SO$$_2$$
In the reaction, the oxidation number of sulphur (S$$^{2-}$$) changes to (S$$^{4+}$$) so, S$$^{2-}$$ is the reducing agent.
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