Statement-1: The reduction of a metal oxide is easier if the metal formed is in liquid state at the temperature of reduction.

Statement-2: The value of entropy change $\mathrm{\Delta }S$ of the reduction process is more on +ve side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus. the value of $\mathrm{\Delta G}$ becomes more on negative side.

If the metal generated is liquid at the temperature of reduction, reducing a metal oxide will be simpler. Because when the metal created is in a liquid state and the metal oxide being reduced is in a solid state, the value of entropy change S of the reduction process is greater on the positive side. As a result, the value of G shifts further toward the negative.

Hence, option A is correct.