Write the thermodynamic idea for reduction

Consider reduction of a metal oxide: $\text{MO} \rightarrow \text{M} + \frac{1}{2}\text{O}_2$

The feasibility is governed by: $\Delta G^\Theta = \Delta H^\Theta - T\Delta S^\Theta$

If $\Delta G^\Theta$ becomes more negative, reduction becomes easier.

Effect of metal being formed in liquid state

A liquid has higher entropy than the corresponding solid: $S_{\text{liquid}} > S_{\text{solid}}$

Therefore, if the product metal is formed as a liquid instead of a solid, the entropy change $\Delta S^\Theta$ for the reduction process increases.

Effect on Gibbs free energy

Since $\Delta G^\Theta = \Delta H^\Theta - T\Delta S^\Theta,$ a larger positive $\Delta S^\Theta$ makes the term $-T\Delta S^\Theta$ more negative.

Hence, $\Delta G^\Theta \text{ becomes more negative}$ and the reduction becomes thermodynamically easier.

Check Assertion A

Assertion A says: The reduction of a metal oxide is easier if the metal formed is in liquid state than solid state.

This is correct, because formation of liquid metal increases entropy and favors reduction.

Check Reason R

Reason R says: The value of $\Delta G^\Theta$ becomes more on negative side as entropy is higher in liquid state than solid state.

This is also correct, due to the relation: $\Delta G^\Theta = \Delta H^\Theta - T\Delta S^\Theta$

Does R explain A?

Yes. The assertion is true because higher entropy of liquid metal makes $\Delta G^\Theta$ more negative, making reduction easier.

Correct option

$\boxed{\text{A: Both A and R are correct and R is the correct explanation of A}}$