Given below are two statements.
Statement I: The choice of reducing agents for metals extraction can be made by using the Ellingham diagram, a plot of $$\Delta G$$ vs temperature.
Statement II: The value of $$\Delta S$$ increases from left to right in the Ellingham diagram.
In the light of the above statements, choose the most appropriate answer from the options given below:

We recall the basic thermodynamic relation for any chemical reaction

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

If we plot $$\Delta G^\circ$$ on the vertical axis and temperature $$T$$ on the horizontal axis, the above linear equation tells us that

• the intercept on the $$\Delta G^\circ$$ axis is $$\Delta H^\circ$$,
• the slope of the straight line is $$-\Delta S^\circ$$.

Such a plot for the oxidation reactions of different metals is known as the Ellingham diagram. It contains a family of straight lines, one for each metal, and is extensively used to decide which metal can reduce the oxide of another metal. The criterion is very simple: at a particular temperature the metal whose line lies lower (has more negative $$\Delta G^\circ$$) will serve as an effective reducing agent for the oxide represented by the higher line. Hence the choice of reducing agent during metal extraction can indeed be made by consulting the Ellingham diagram.

Thus, Statement I – “The choice of reducing agents for metals extraction can be made by using the Ellingham diagram, a plot of $$\Delta G$$ vs temperature” – is true.

Now we analyse Statement II. For a given oxidation reaction such as

$$\mathrm{M\,(s) + \dfrac{1}{2}O_2\,(g) \longrightarrow MO\,(s)}$$

the change in entropy $$\Delta S^\circ$$ is practically independent of temperature over a wide range and has a negative value because one mole of gaseous $$O_2$$ is consumed and only solids remain. Hence each individual line in the Ellingham diagram possesses a fixed negative $$\Delta S^\circ$$, and consequently a fixed positive slope $$\bigl(-\Delta S^\circ\bigr)$$. As we move to the right (towards higher temperatures) along any line, the value of $$\Delta S^\circ$$ itself does not change; only the product $$T\Delta S^\circ$$ changes. Therefore it is incorrect to say that “the value of $$\Delta S$$ increases from left to right in the Ellingham diagram.”

So, Statement II is false.

Combining the conclusions, Statement I is true while Statement II is false. This matches Option C, the third option in the list.

Hence, the correct answer is Option C.