In the extraction of copper from its sulphide ore, metal is finally obtained by the oxidation of cuprous sulphide with:

In the metallurgical process of copper extraction, the concentrated sulphide ore is first roasted and then smelted. After these operations we obtain a molten “matte”, which is chiefly a mixture of cuprous sulphide and ferrous sulphide:

$$\text{Matte} = \text{Cu}_2\text{S} + \text{FeS}$$

This matte is transferred to a Bessemer-type converter where a blast of hot air is blown through it. Let us examine, one by one, the chemical changes that occur in the converter.

1. First, the ferrous sulphide is oxidised to ferrous oxide:

$$\text{2FeS} + 3O_2 \;\longrightarrow\; 2\text{FeO} + 2\text{SO}_2 \uparrow$$

2. Silica is added, and ferrous oxide combines with it to form a slag which floats away. The reaction is:

$$\text{FeO} + \text{SiO}_2 \;\longrightarrow\; \text{FeSiO}_3 \;(\text{slag})$$

3. Simultaneously, part of the cuprous sulphide present in the matte is oxidised by the same air blast to cuprous oxide:

$$\text{Cu}_2\text{S} + O_2 \;\longrightarrow\; \text{Cu}_2\text{O} + \text{SO}_2 \uparrow$$

Now comes the pivotal step that actually produces metallic copper. The freshly formed cuprous oxide reacts with the remaining cuprous sulphide. We first state the general redox principle involved:

When an oxide of a metal reacts with a sulphide of the same metal, the oxide is reduced to the metal, while the sulphide is oxidised to sulphur dioxide.

Applying this principle, the reaction between $$\text{Cu}_2\text{O}$$ and $$\text{Cu}_2\text{S}$$ is:

$$\text{Cu}_2\text{O} + \text{Cu}_2\text{S} \;\longrightarrow\; 4\text{Cu} + \text{SO}_2 \uparrow$$

Algebraically balancing it in the more familiar six-copper form, we also write:

$$2\text{Cu}_2\text{O} + \text{Cu}_2\text{S} \;\longrightarrow\; 6\text{Cu} + \text{SO}_2 \uparrow$$

Observe the redox roles clearly:

• $$\text{Cu}_2\text{O}$$ is reduced to $$\text{Cu}$$ (its oxidation state goes from $$+1$$ in the oxide to $$0$$ in the metal).

• $$\text{Cu}_2\text{S}$$ is oxidised; the sulphur in it goes from $$-2$$ in the sulphide to $$+4$$ in $$\text{SO}_2$$.

Thus, in the very last step copper metal appears because cuprous sulphide is oxidised by cuprous oxide. Therefore, the reagent responsible for the final oxidation of $$\text{Cu}_2\text{S}$$, and consequently for the production of metallic copper, is $$\text{Cu}_2\text{O}$$.

The option list is:

A. $$\text{SO}_2$$     B. $$\text{Fe}_2\text{O}_3$$     C. $$\text{Cu}_2\text{O}$$     D. CO

We have just shown that the correct reagent is $$\text{Cu}_2\text{O}$$, which corresponds to Option C (Option 3 in the given numbering).

Hence, the correct answer is Option C.