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Why is the reduction of a metal oxide ea...

Why is the reduction of a metal oxide easier, if the metal formed is in liquid state at the temperature of reduction?

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In the liquid state, the entropy of the system is higher than that in the solid state. The value of `DeltaG^(@)` becomes more negative, (i.e. `DeltaG= DeltaH-TDeltaS`) and the reduction becomes more spontaneous.
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