Which respect to an ore, Ellingham diagram helps to predict the feasibility of its:

We first recall what an Ellingham diagram is. It is a plot of the standard Gibbs free-energy change $$\Delta G^\circ$$ for the formation of oxides (or sulphides, halides, etc.) versus temperature $$T$$ for a variety of metals. The basic thermodynamic relation used to draw every straight line on the diagram is the fundamental Gibbs-Helmholtz equation

$$ \Delta G^\circ \;=\; \Delta H^\circ \;-\; T\,\Delta S^\circ , $$

where $$\Delta H^\circ$$ is the standard enthalpy change and $$\Delta S^\circ$$ is the standard entropy change of the oxide-formation reaction

$$ \text{Metal} \;+\; \dfrac{1}{2}\,O_2 \;\longrightarrow\; \text{Metal oxide}. $$

The slope of each line equals $$-\Delta S^\circ$$ and the intercept equals $$\Delta H^\circ$$. Because most oxide-formation reactions involve a decrease in entropy (gas is consumed), $$\Delta S^\circ$$ is negative, so the slope $$-\,\Delta S^\circ$$ is positive. Thus, the lines usually rise with increasing temperature.

Now, the sign of $$\Delta G^\circ$$ decides spontaneity:

$$ \Delta G^\circ < 0 \quad\Longrightarrow\quad \text{reaction is spontaneous}; $$

$$ \Delta G^\circ > 0 \quad\Longrightarrow\quad \text{reaction is non-spontaneous}. $$

In metallurgical practice we use the diagram exactly the other way around. Consider the reduction of a metal oxide by carbon (or by another metal):

$$ \text{Metal oxide} \;+\; C \;\longrightarrow\; \text{Metal} \;+\; CO \quad\text{(or } CO_2\text{)}. $$

We can obtain the overall $$\Delta G^\circ$$ for this reduction by subtracting the $$\Delta G^\circ$$ line of the metal from that of the reducer (carbon forming CO/CO2). If the resulting $$\Delta G^\circ$$ is negative at a certain temperature, the reduction is feasible at that temperature.

Thus, the Ellingham diagram is primarily employed to judge whether an oxide can be thermally reduced by some reducing agent such as carbon, carbon monoxide, hydrogen, or another metal. This process—smelting in a blast furnace, electrolytic cell feed preparation, etc.—is broadly classified as thermal reduction or pyrometallurgical reduction.

Let us now examine the given options with this understanding:

• Electrolysis: The feasibility of electrolytic reduction depends on decomposition potential and over-potentials, not on Ellingham lines.
• Zone refining: This relies on the difference in solubility of impurities in solid and liquid phases; Ellingham diagram is not used here.
• Vapour phase refining: This depends on the volatility of metal halides (e.g., Mond process), again unrelated to Ellingham diagram.
• Thermal reduction: Exactly what the Ellingham diagram was devised for—to see if heat plus a reducing agent will convert the oxide to metal.

Therefore, only the fourth option matches.

Hence, the correct answer is Option D.