Metal Cu is not produces on :-

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The question asks which process does NOT produce copper (Cu).

A. Heating $C u_{2} S$ with $C u_{2} O$ : This is a self-reduction process (Bessemarization) where $C u_{2} S$ reacts with $C u_{2} O$ to produce $Cu$ .

$2 C u_{2} O + C u_{2} S \to 6 Cu + S O_{2}$

B. Heating $C u_{2} S$ with calcinated malachite: Malachite is $CuC O_{3} \cdot Cu (OH)_{2}$ . Calcination produces $CuO$ . $C u_{2} S$ reacts with $CuO$ to produce $Cu$ .

$C u_{2} S + 2 CuO \to 4 Cu + S O_{2}$

C. Heating copper matte in the absence of $O_{2}$ : Copper matte primarily consists of $C u_{2} S$ and $FeS$ . Heating in the absence of $O_{2}$ will not lead to the reduction of $C u_{2} S$ to $Cu$ . Oxidation (roasting) is required first to form $C u_{2} O$ , which then reacts with remaining $C u_{2} S$ for self-reduction. Without $O_{2}$ , no $C u_{2} O$ is formed, hence no $Cu$ is produced.

D. Electrolysis of $CuS O_{4} (aq.) + H_{2} S O_{4}$ by Pt electrode: This is electrolytic refining of copper. At the cathode, $C u^{2 +}$ ions are reduced to $Cu$ metal.

$C u^{2 +} + 2 e^{-} \to Cu$

Therefore, heating copper matte in the absence of $O_{2}$ does not produce $Cu$ .

The final answer is $C$

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