General Principles and Processes of Isolation of Elements • Topic 3 of 3

Refining & Thermodynamics

The crude metal obtained after reduction still contains impurities; refining raises it to commercial purity, the method depending on the metal and its impurities.

1. Distillation — for low-boiling metals (zinc, mercury): the metal is vaporised and condensed pure, leaving non-volatile impurities.

2. Liquation — for low-melting metals (tin, lead): the metal melts and flows down a sloping hearth while higher-melting impurities stay behind.

3. Electrolytic refining — the chief method for copper, silver, gold and aluminium. The impure metal is the anode, a thin pure sheet the cathode, and a metal salt the electrolyte. The anode dissolves ($M \rightarrow M^{n+} + ne^-$) and pure metal deposits on the cathode ($M^{n+} + ne^- \rightarrow M$). Noble impurities (Ag, Au) settle as anode mud; reactive ones stay in solution.

4. Zone refining — for ultra-pure semiconductor metals (Si, Ge). It uses the fact that impurities are more soluble in the melt than in the solid: a mobile heater melts a narrow zone and is moved along the rod, sweeping impurities to one end, which is cut off.

5. Vapour-phase refining — the metal is made into a volatile compound, then decomposed. Mond (Ni): $Ni + 4CO \xrightarrow{330\,K} Ni(CO)_4 \xrightarrow{450\,K} Ni + 4CO$. van Arkel (Ti, Zr): $Ti + 2I_2 \rightarrow TiI_4 \xrightarrow{1700\,K} Ti + 2I_2$ on a hot filament, giving ductile titanium.

6. Chromatography — for trace impurities; components separate by their different rates of adsorption on a stationary phase.

Worked extractions.

Iron (Blast furnace). Ore, coke and limestone are charged at the top; hot air enters the base. Coke burns to $CO$ ($C + O_2 \rightarrow CO_2$; $CO_2 + C \rightarrow 2CO$). In the cooler upper zone $CO$ reduces the ore ($Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$); in the hot lower zone carbon reduces $FeO$ directly. Limestone gives $CaO$, which fluxes silica to a fusible slag ($CaO + SiO_2 \rightarrow CaSiO_3$) that floats on the molten iron.

Aluminium (Hall-Heroult). Since the $Al/Al_2O_3$ line is so low, carbon cannot reduce alumina; it is reduced electrolytically. $Al_2O_3$ is dissolved in molten cryolite ($Na_3AlF_6$) with fluorspar, lowering the temperature to ~950 $^\circ C$ and raising conductivity. Cathode: $Al^{3+} + 3e^- \rightarrow Al$; the $O_2$ at the carbon anode burns it to $CO/CO_2$.

Copper. Blister copper from self-reduction is purified by electrolytic refining ($CuSO_4/H_2SO_4$), with Ag and Au as anode mud. Zinc: $ZnO$ is reduced by coke ($ZnO + C \rightarrow Zn + CO$) above the crossing temperature, then distilled — each choice tracing back to the Ellingham diagram.

Blast furnace zones: charge of ore, coke and limestone enters the top; hot air enters the base; molten iron and slag are tapped at the bottomCharge: ore + coke + limestone~500 K: drying, CO reduces Fe2O3~1100 K: CaCO3 -> CaO + CO2~1500 K: C reduces FeO; slag forms~2000 K: combustion zonehot airblastmolten ironslag
Solved Examples
1
Worked Example
Explain the principle of zone refining and name two metals purified by it.
Solution
  1. Impurities are more soluble in the molten state of a metal than in its solid state.
  2. A circular heater melts a narrow zone of the metal rod and is moved slowly along its length.
  3. As the molten zone moves, impurities are dragged along with it to one end.
  4. That impure end is finally cut off, leaving an ultra-pure rod.

Answer: Zone refining relies on impurities preferring the melt; it purifies semiconductor metals such as silicon and germanium.

2
Worked Example
Write the reactions of the Mond process for refining nickel.
Solution
  1. Impure nickel is warmed with CO at about 330 K to form volatile nickel tetracarbonyl: $Ni + 4CO \rightarrow Ni(CO)_4$.
  2. The vapour is led away from the non-volatile impurities.
  3. It is then heated to about 450 K, where it decomposes: $Ni(CO)_4 \rightarrow Ni + 4CO$.
  4. Pure nickel is deposited and the CO is recycled.

Answer: $Ni + 4CO \xrightarrow{330\,K} Ni(CO)_4 \xrightarrow{450\,K} Ni + 4CO$ — refining by a volatile carbonyl.

3
Worked Example
In the electrolytic refining of copper, what happens at each electrode and what is anode mud?
Solution
  1. The impure copper is the anode and a thin pure copper sheet is the cathode in $CuSO_4/H_2SO_4$.
  2. At the anode copper dissolves: $Cu \rightarrow Cu^{2+} + 2e^-$.
  3. At the cathode pure copper is deposited: $Cu^{2+} + 2e^- \rightarrow Cu$.
  4. Noble impurities (Ag, Au, Pt) do not dissolve and collect below the anode as anode mud.

Answer: Anode dissolves, pure Cu deposits on the cathode, and the valuable Ag/Au settle as anode mud.

4
Worked Example
Why is alumina dissolved in molten cryolite in the Hall-Heroult process?
Solution
  1. Pure alumina melts only at about 2050 $^\circ C$, which is impractical.
  2. Dissolving it in molten cryolite ($Na_3AlF_6$) with some fluorspar lowers the working temperature to about 950 $^\circ C$.
  3. The molten mixture also conducts electricity well, allowing electrolysis.
  4. At the cathode $Al^{3+} + 3e^- \rightarrow Al$ and oxygen liberated at the carbon anode burns it away.

Answer: Cryolite lowers the melting point of alumina and raises conductivity, making electrolytic reduction feasible at ~950 $^\circ C$.

5
Worked Example
Describe the role of limestone in the blast furnace extraction of iron.
Solution
  1. Limestone decomposes in the furnace: $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$.
  2. The lime acts as a basic flux toward the acidic silica gangue.
  3. It combines with silica to form a fusible slag: $CaO + SiO_2 \rightarrow CaSiO_3$.
  4. The molten slag floats on the denser pig iron and is tapped off separately.

Answer: Limestone gives CaO, which fluxes the silica gangue to fusible $CaSiO_3$ slag that is removed from the molten iron.

6
Worked Example
Why is aluminium extracted electrolytically rather than by reduction with carbon?
Solution
  1. On the Ellingham diagram the $Al \rightarrow Al_2O_3$ line lies below the carbon (C $\rightarrow$ CO) line up to very high temperatures.
  2. Hence carbon cannot make $\Delta G$ negative for reducing $Al_2O_3$ at attainable temperatures.
  3. So a stronger, electrical method is required.
  4. Electrolysis of molten $Al_2O_3$ in cryolite forces the reduction: $Al^{3+} + 3e^- \rightarrow Al$.

Answer: Because Al's oxide line lies below carbon's, carbon cannot reduce alumina, so electrolytic (Hall-Heroult) reduction is used.

Key Points

  • Distillation purifies low-boiling metals (Zn, Hg); liquation purifies low-melting metals (Sn, Pb).
  • In electrolytic refining the impure metal is the anode, pure metal the cathode, and noble impurities form anode mud.
  • Zone refining (Si, Ge) uses the greater solubility of impurities in the melt; Mond (Ni) and van Arkel (Ti, Zr) use volatile compounds.
  • Iron is extracted in a blast furnace where CO/C reduce the ore and CaO slags off silica as $CaSiO_3$.
  • Aluminium is obtained by Hall-Heroult electrolysis of $Al_2O_3$ in molten cryolite because carbon cannot reduce alumina.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.Which metals are purified by zone refining?
Explanation: Zone refining gives the ultra-high purity needed for semiconductor metals such as silicon and germanium.
Q2.In the van Arkel method, titanium is purified through the volatile compound:
Explanation: $Ti + 2I_2 \rightarrow TiI_4$, which decomposes on a hot filament: $TiI_4 \rightarrow Ti + 2I_2$, giving pure ductile titanium.
Q3.The substance added to lower the fusion temperature of alumina in the Hall-Heroult process is:
Explanation: Cryolite ($Na_3AlF_6$), with some fluorspar, lowers the melting point of $Al_2O_3$ to about 950 $^\circ C$ and improves conductivity.
Q4.In the blast furnace, the slag formed is:
Explanation: Lime fluxes the silica gangue: $CaO + SiO_2 \rightarrow CaSiO_3$, a fusible slag that floats on the molten iron.
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