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

(N/A) The entropy of a substance is higher in the liquid state compared to the solid state.
When the metal formed is in the liquid state,the entropy change $(\Delta S)$ for the reduction reaction becomes more positive.
According to the Gibbs free energy equation,${\Delta _r}{G^\Theta } = \Delta H - T\Delta S$.
As $\Delta S$ increases,the value of ${\Delta _r}{G^\Theta }$ becomes more negative,which makes the reduction process thermodynamically more favorable and easier.