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

For the reduction $MO(s) + \text{reducing agent} \rightarrow M + \text{oxide of reducing agent}$, if M is produced as a liquid, the system gains extra randomness (entropy) compared to M staying solid. Since $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, a more positive $\Delta S^\circ$ makes $\Delta G^\circ$ more negative as T increases, so the reduction becomes more spontaneous/easier once the metal is molten at the temperature of reduction. …

Step 1 – Write the general reduction reaction

Step 2 – Recall the Gibbs–Helmholtz relation

A reaction becomes more spontaneous (more negative $\Delta G^\circ$) as T increases whenever $\Delta S^\circ$ (of the overall reaction) is positive — a larger positive $\Delta S^\circ$ means a steeper favourable slope on the Ellingham-type ($\Delta G^\circ$ vs T) plot.

Step 3 – Effect of the metal being liquid

If the metal produced, M, is in the liquid state at the reduction temperature (rather than remaining solid), the products side of the reaction has extra disorder/randomness (a liquid has higher entropy than the corresponding solid). This raises the overall $\Delta S^\circ$ of the reduction step compared to the case where M stays solid.

Step 4 – Consequence …

Apply $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$: a liquid metal product raises the entropy of the products side, so $\Delta S^\circ$ is larger and posi …

• Thinking only about enthalpy ($\Delta H^\circ$) of melting and ignoring the entropy argument, which is the actual reason invoked in the Ellingham-diagram treatment. …