General Principles and Processes of Isolation of Elements • Topic 2 of 3

Extraction of Metals

After concentration, the ore is converted into the metal oxide, because oxides are the easiest to reduce. Two thermal processes do this.

Calcination — heating the ore in the absence (or limited supply) of air to drive off water of hydration and $CO_2$ from carbonate and hydrated ores: $ZnCO_3 \xrightarrow{\Delta} ZnO + CO_2$; $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$.

Roasting — heating sulphide ores in excess air to give the oxide with evolution of $SO_2$: $2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2$; $2PbS + 3O_2 \rightarrow 2PbO + 2SO_2$.

Reduction of the oxide to the metal is the central step, and the reducing agent is chosen on thermodynamic grounds. A reduction $MO + C \rightarrow M + CO$ proceeds only if the overall Gibbs energy change is negative, $\Delta G < 0$. Since $\Delta G^0 = -RT\ln K$, a more negative $\Delta G^0$ means a larger equilibrium constant and a more complete reaction. The decomposition of $MO$ alone is unfavourable, so it is coupled with the strongly exergonic oxidation of the reducing agent; the sum is favourable if that agent has a more negative $\Delta G^0$ of oxidation than the metal.

The Ellingham diagram plots $\Delta G^0$ of oxide formation (per mole of $O_2$) against temperature $T$. Key features:

  • Most lines slope upward because forming a solid oxide consumes gaseous $O_2$, so $\Delta S < 0$ and $\Delta G = \Delta H - T\Delta S$ rises with $T$.
  • The carbon line ($2C + O_2 \rightarrow 2CO$) slopes downward — gas is produced, $\Delta S > 0$ — so it cuts below the metal lines at high $T$, making carbon a cheap high-temperature reducer.
  • A metal whose oxide line lies lower can reduce the oxide of any metal whose line lies above it; where two lines cross, the roles reverse.

Reduction by carbon / CO (smelting): $ZnO + C \xrightarrow{\Delta} Zn + CO$; $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$. Below the crossing temperature $CO$ is the reducer; above it carbon itself works.

Reduction by another metal (metallothermy): very stable oxides such as $Cr_2O_3$ are reduced by aluminium because the $Al/Al_2O_3$ line lies very low. The thermite reaction $2Al + Fe_2O_3 \rightarrow 2Fe + Al_2O_3$ is so exothermic that the iron is produced molten — used to weld rails.

Self-reduction: for less reactive metals (copper, lead), partial roasting gives a mix of sulphide and oxide which react together, the sulphide reducing the oxide with no external agent: $2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$; $2PbO + PbS \rightarrow 3Pb + SO_2$.

Simplified Ellingham diagram: standard Gibbs energy of oxide formation versus temperature, with the carbon line sloping down and crossing the metal linesΔG° (kJ/mol O2)Temperature (T) →0Fe -> FeOZn -> ZnOAl -> Al2O32C + O2 -> 2COcrossing T
Solved Examples
1
Worked Example
Differentiate calcination and roasting with one equation each.
Solution
  1. Calcination heats the ore in absence/limited air to remove water and $CO_2$ from carbonates and hydrated ores.
  2. Example: $ZnCO_3 \xrightarrow{\Delta} ZnO + CO_2$.
  3. Roasting heats sulphide ores in excess air to give the oxide plus $SO_2$.
  4. Example: $2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2$.

Answer: Calcination (limited air) is for carbonates/hydrates; roasting (excess air) is for sulphides and evolves $SO_2$.

2
Worked Example
Using $\Delta G^0 = -RT\ln K$, explain why a more negative $\Delta G^0$ favours reduction.
Solution
  1. For a reduction at equilibrium, $\Delta G^0 = -RT\ln K$.
  2. If $\Delta G^0$ is large and negative, then $\ln K$ is large and positive.
  3. Hence $K \gg 1$, so the equilibrium lies far towards the products (the metal).
  4. A more negative $\Delta G^0$ therefore means a more complete, more feasible reduction.

Answer: A more negative $\Delta G^0$ gives a larger $K$, so the reduction proceeds nearly to completion.

3
Worked Example
Why does the carbon line in the Ellingham diagram slope downwards while metal-oxide lines slope upwards?
Solution
  1. For $2C + O_2 \rightarrow 2CO$, two moles of gas are formed from one, so $\Delta S > 0$.
  2. From $\Delta G = \Delta H - T\Delta S$, a positive $\Delta S$ makes $\Delta G$ fall as $T$ rises — the line slopes down.
  3. For $2M + O_2 \rightarrow 2MO$, gaseous $O_2$ is consumed, so $\Delta S < 0$ and $\Delta G$ rises with $T$ — the line slopes up.
  4. Therefore the carbon line eventually cuts below the metal lines at high temperature.

Answer: The carbon-to-CO reaction increases gas moles ($\Delta S>0$) so its $\Delta G^0$ falls with $T$, unlike the metal lines whose $\Delta S<0$.

4
Worked Example
Write the thermite reaction and state two of its uses.
Solution
  1. Aluminium reduces ferric oxide: $2Al + Fe_2O_3 \rightarrow 2Fe + Al_2O_3$.
  2. The reaction is highly exothermic, so the iron is liberated in the molten state.
  3. The molten iron is used to weld broken railway tracks and machine parts (the Goldschmidt thermite process).
  4. The same metallothermic principle reduces $Cr_2O_3$ and $Mn_3O_4$ to their metals.

Answer: $2Al + Fe_2O_3 \rightarrow 2Fe + Al_2O_3$; used for welding rails and for extracting Cr and Mn from their oxides.

5
Worked Example
Explain self-reduction in the extraction of copper with equations.
Solution
  1. Copper matte ($Cu_2S$ with some FeS) is partially roasted in air.
  2. Part of the sulphide is converted to oxide: $2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$.
  3. On cutting off the air, the remaining sulphide reduces the oxide: $2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$.
  4. No external reducing agent is needed — the ore reduces itself.

Answer: Partial roasting makes some $Cu_2O$, which is then reduced by the leftover $Cu_2S$: $2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$.

6
Worked Example
From the Ellingham diagram, why can aluminium reduce $Cr_2O_3$ but carbon is preferred for $Fe_2O_3$?
Solution
  1. The $Al \rightarrow Al_2O_3$ line lies below the $Cr \rightarrow Cr_2O_3$ line at all ordinary temperatures.
  2. So aluminium can take oxygen from $Cr_2O_3$: $2Al + Cr_2O_3 \rightarrow 2Cr + Al_2O_3$ ($\Delta G < 0$).
  3. For iron, the carbon (C $\rightarrow$ CO) line drops below the $Fe \rightarrow FeO$ line at attainable furnace temperatures.
  4. Carbon is far cheaper than aluminium, so coke is used for iron while costly Al is reserved for very stable oxides.

Answer: Al's oxide line lies below Cr's, so Al reduces $Cr_2O_3$; for $Fe_2O_3$ the cheaper carbon line drops below it at furnace temperatures.

Key Points

  • Calcination heats an ore in limited air (carbonates/hydrates); roasting heats sulphides in excess air to oxides + $SO_2$.
  • Reduction is feasible only when overall $\Delta G$ is negative; $\Delta G^0 = -RT\ln K$ links it to the equilibrium constant.
  • The Ellingham diagram plots $\Delta G^0$ of oxide formation versus $T$; lower lines reduce higher ones.
  • The carbon-to-CO line slopes down ($\Delta S>0$) and cuts below metal lines at high $T$, making carbon a cheap reducer.
  • Very stable oxides are reduced by Al (thermite, $2Al + Fe_2O_3 \rightarrow 2Fe + Al_2O_3$); some metals reduce themselves (self-reduction).
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.Roasting differs from calcination chiefly in that roasting is carried out:
Explanation: Roasting heats sulphide ores in excess air to form oxides and $SO_2$, whereas calcination uses little or no air.
Q2.On the Ellingham diagram, the line for $2C + O_2 \rightarrow 2CO$:
Explanation: Forming CO increases the number of gas moles ($\Delta S>0$), so $\Delta G^0$ decreases with temperature — the line slopes down.
Q3.The thermite reaction is represented by:
Explanation: In the thermite (Goldschmidt) reaction aluminium reduces ferric oxide, liberating molten iron used for welding.
Q4.Self-reduction is used in the extraction of:
Explanation: Copper is extracted by self-reduction: $2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$, with no external reducing agent.
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