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Worked Examples · Example 6.2

Q.Although thermodynamically feasible, in practice, magnesium metal is not used for the reduction of alumina in the metallurgy of aluminium. Why ?

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Step 1 – Note the distinction: thermodynamic feasibility vs practical/economic viability

A negative ΔrG∘\Delta_rG^\circ only tells us a reaction can proceed; it says nothing about whether it is the best industrial choice. Cost, availability of reagents, energy consumption, and ease of product separation all matter in practice.

Step 2 – Why Mg reduction is thermodynamically fine but practically poor

  • Mg metal itself is far more expensive than the value of the Al it would help produce — Mg is itself obtained only by electrolysis of fused MgCl2MgCl_2 (Dow process), so using it as a "reducing agent" simply shifts the expensive electrolysis step from Al to Mg, and then you still need to separate Al from the MgO/excess Mg byproduct.
  • The reaction 3Mg+Al2O3→3MgO+2Al3Mg + Al_2O_3 \rightarrow 3MgO + 2Al consumes 3 moles of Mg per mole of Al2_2O3_3 — a large mass of an expensive metal for a modest yield of aluminium.
  • Handling a violently exothermic solid-state reduction (thermite-type) at the huge tonnages aluminium is produced at is impractical and hard to control safely.

Step 3 – Why electrolysis (Hall–Héroult) wins instead

  • Purified alumina is dissolved in molten cryolite (Na3AlF6Na_3AlF_6), which lowers the melting point from ~2323 K to about 1140–1200 K, and electrolysed directly using cheap carbon (graphite) electrodes. …

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