Given below are two statements: One is labelled as Assertion A and...

The entropy is higher if the metal is in liquid state than when it is in solid state. The value of entropy change $( ∆ S )$ of the reduction process is more on positive side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus the value of ${∆}_{f}G$ becomes more on negative side and the reduction becomes easier.

$\Delta G = \Delta H - T \Delta S$

$∵$ Entropy of liquid is more than solid

$∴$ on melting the entropy increases and $\Delta G$ becomes more negative and hence it becomes easier to reduce metal.