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

Explanation

Entropy of a metal in liquid state is higher than its entropy in solid state. The entropy change `(DeltaS)` of the reduction is more `+ve` when the metal formed is in liquid state and the metal oxide which is reduced, is in solid state. Thus the `DeltaG^(@)` becomes more `-ve` making the reduction spontaneous.