Question

Reduction to free metal: Some of the methods commonly used to get free metal from the concentrates ore are below.
Smelting: The process of extracting a metal in the state of fusions is called smelting. In this process the ore is mixed with carbon and heated in a suitable furnace.
Electrolytic reduction process: This process is used in the extraction of the alkali and alkaline earth metals,zinc and aluminium.
Self reduction process: The sulphide ores of less electropositive metals like, Hg,Pb,Cu,etc. are heated in air so as to convert part of the ore into oxide or sulphate which then reacts with the remaining sulphide ore to give out the metal.
After Partial roasting, the sulphide of copper is reduced by:
A) Reduction by carbon
B) Electrolysis
C) Self-reduction
D) Cyanide process

Answer 1

The sulphide ores of less electropositive metals are usually dealt by the self reduction process.The sulphide ores of less electropositive metals like Hg, Pb, Cu etc. are heated in the air so as to convert part of the ore into oxide or sulphate which then reacts with the remaining sulphide ore to give the metal.

Complete step by step answer:
The extraction of pure metal from its ore is called metallurgy.
Reduction to free metal:some of the methods commonly used to get free metal from the concentrates ore are below.
Smelting: The process of extracting a metal in the state of fusions is called smelting. In this process the ore is mixed with carbon and heated in a suitable furnace.
Electrolytic reduction process:This process is used in the extraction of the alkali and alkaline earth metals,zinc and aluminium.
Self reduction process: The sulphide ores of less electropositive metals like, Hg,Pb,Cu,etc.are heated in air so as to convert part of the ore into oxide or sulphate which then reacts with the remaining sulphide ore to give out the metal.

Now, let us understand the partial roasting.
When $\mathrm{CuFeS}_{2}$ is heated with the presence of excess $\mathrm{O}_{2}$, it forms $\mathrm{Cu}_{2} \mathrm{S}$ and $\mathrm{FeS}$ and $\mathrm{SO}_{2}$. Then $\mathrm{Cu}_{2} \mathrm{S}$ again reacts with $\mathrm{O}_{2}$ to form $\mathrm{Cu}_{2} \mathrm{O}$ and $\mathrm{SO}_{2}$. Then $\mathrm{FeS}$ again reacts with $\mathrm{O}_{2}$ to form $\mathrm{FeO}$ and $\mathrm{SO}_{2} .$ But,here the amount of $\mathrm{Cu}_{2} \mathrm{O}$ is much less than $\mathrm{FeO}$ because Fe is much more reactive than Cu.
Now, the leftover $\mathrm{Cu}_{2} \mathrm{S}$ is reacted with $\mathrm{Cu}_{2} \mathrm{O}$ to form copper and $\mathrm{SO}_{2}$
Since, they reacted with each other, reducing their oxidation number from +2 to $0 .$ Thus, self reduction is the process here.
So, the correct answer is “Option C”.

Note: Self reduction is only possible here since all of the $\mathrm{Cu}_{2} \mathrm{S}$ could not convert itself into $\mathrm{Cu}_{2} \mathrm{O} .$ Thus this process is not valid for extraction of iron from $\mathrm{FeS}$ Copper Sulfide is a crystalline solid used as a semiconductor and in photo optic applications.