(a) Name the method of refining which is (i) used to obtain semiconductor of high purity, (ii) used to obtain

Question

Q. (a) Name the method of refining which is

(i) used to obtain semiconductor of high purity,

(ii) used to obtain low boiling metal.

(b) Write chemical reactions taking place in the extraction of copper from

Cu_2S

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Concept understanding — Electrolytic Refining

Electrolytic Refining: From Intuition to Precision

Imagine you have a lump of copper that came straight out of a furnace. It's mostly copper, but it's full of impurities — bits of silver, gold, zinc, iron, maybe some nickel. You want pure copper, say 99.99% pure, for electrical wires. How do you get the copper atoms to leave the impurities behind and assemble themselves into a clean, solid sheet?

The trick is to turn the impure copper into an anode (positive electrode) in a bath of copper sulfate solution, and place a thin sheet of pure copper as the cathode (negative electrode). When you pass electricity through, the impure copper anode dissolves. But here's the key: only copper atoms leave the anode as copper ions ( $\text{Cu}^{2+}$ ). The impurities either fall off as sludge (if they are less reactive) or stay in solution (if they are more reactive). Meanwhile, those copper ions travel through the solution and deposit as pure copper atoms onto the cathode.

That is electrolytic refining in a nutshell: impure metal at the anode dissolves; pure metal deposits at the cathode; impurities are left behind.

The Precise Statement

\text{CuSO}_4

How It Works — Step by Step

Take copper refining as the classic example.

Setup: Electrolyte = acidified copper sulfate solution ( $\text{CuSO}_4 + \text{H}_2\text{SO}_4$ ). Anode = thick block of impure copper. Cathode = thin sheet of pure copper.
At the anode (oxidation): Copper atoms lose two electrons and go into solution as $\text{Cu}^{2+}$ ions.

\text{Cu (impure)} \rightarrow \text{Cu}^{2+} + 2e^-

Impurities like silver, gold, and platinum do not dissolve — they are less reactive than copper and simply fall off as anode mud (a valuable byproduct). Impurities like zinc and iron do dissolve, but their ions stay in the electrolyte because they are more difficult to reduce than copper.

At the cathode (reduction): Copper ions from the solution gain two electrons and deposit as pure copper metal.

\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu (pure)}

Net effect: Copper is transferred from the impure anode to the pure cathode. The anode gets thinner, the cathode gets thicker, and the impurities are left out.

Note

The voltage is kept low (around 0.2–0.4 V for copper) so that only copper ions are reduced at the cathode. A higher voltage would start depositing less noble metals like zinc, ruining the purity.

Why This Works — The Electrochemical Reason

The key is the standard reduction potential of the metal ions. Copper has a reduction potential of $+0.34\ \text{V}$ . Ions of more reactive metals (like $\text{Zn}^{2+}$ , $+0.76\ \text{V}$ ) are harder to reduce — they stay in solution. Ions of less reactive metals (like $\text{Ag}^+$ , $+0.80\ \text{V}$ ) would reduce more easily, but those metals do not even dissolve at the anode in the first place; they fall off as solid mud.

So the process naturally separates impurities by their electrochemical behaviour. …

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The refining methods are zone refining for high-purity semiconductors and distillation for low-boiling metals. Copper is extracted from $Cu_2S$ via self-reduction in a Bessemer converter, where the sulfide ore is partially roasted to $Cu_2O$ , which then reacts with remaining $Cu_2S$ to produce blister copper.

Concept and Intuition

Refining is the final step in metallurgy — removing trace impurities from a crude metal to achieve the desired purity. Different metals demand different techniques based on their physical and chemical properties.

For semiconductors like silicon or germanium, even parts-per-million impurities ruin electronic behaviour. Zone refining exploits the fact that impurities are more soluble in the molten metal than in the solid. A moving heater creates a narrow molten zone that sweeps impurities to one end, leaving ultra-pure crystal behind.

For low-boiling metals like zinc or mercury, distillation is simplest: the metal vaporises at a relatively low temperature, while higher-boiling impurities remain as residue. The vapour is then condensed back to pure metal.

Copper extraction from $Cu_2S$ (chalcocite) is a classic example of self-reduction — no external reducing agent like carbon is needed. The sulphide ore is first partially oxidised to oxide, then the oxide and remaining sulphide react to give copper metal and sulphur dioxide. This happens in a Bessemer converter, and the product is called blister copper (98–99% pure) because escaping $SO_2$ gas leaves a blistered surface.

Step-by-Step Solution

Part (a): Refining Methods

High-purity semiconductors — The method is zone refining.

A rod of impure metal is passed through a series of heating coils. Each coil creates a narrow molten zone. As the zone moves, impurities dissolve into the melt and are carried along, leaving a purer solid behind. After several passes, one end becomes highly enriched in impurities and is cut off. This yields semiconductor-grade purity (99.9999%+).
Low-boiling metals — The method is distillation.

Metals like zinc (boiling point $907^\circ\text{C}$ ) or mercury ( $357^\circ\text{C}$ ) are heated in a retort. The metal vaporises, leaving non-volatile impurities behind. The vapour is condensed in a separate chamber to collect the pure metal.

Tip

Zone refining is also called float-zone refining when the rod is vertical and the molten zone is held by surface tension — no crucible needed, avoiding contamination.

Part (b): Extraction of Copper from $Cu_2S$

The extraction involves two main chemical stages, both carried out in a Bessemer converter after the ore has been concentrated and roasted to remove some sulphur.

Step 1: Partial roasting of $Cu_2S$ to $Cu_2O$

Air is blown through molten $Cu_2S$ at around $1200^\circ\text{C}$ . Some of the sulphide is oxidised:

2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2 \uparrow

The $SO_2$ gas escapes and is used to make sulphuric acid. The key is that only part of the $Cu_2S$ is converted — we need both $Cu_2O$ and $Cu_2S$ present for the next step. …

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Q.(a) Name the method of refining which is

Concept and Intuition

Step-by-Step Solution

Part (a): Refining Methods

Part (b): Extraction of Copper from Cu2SCu_2SCu2​S

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Part (b): Extraction of Copper from $Cu_2S$

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