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

Welcome to a fascinating journey into the thermodynamics of metallurgy! The Ellingham diagram might look like a simple collection of straight lines, but it holds the ultimate secret to extracting metals from their ores. Let's break down the two statements given in the question and see what the physics and chemistry behind them truly reveal.

Statement I introduces the core purpose of the Ellingham diagram. It states that the diagram is a plot of Gibbs free energy change ($\Delta G$) versus temperature ($T$), and it is used to choose a reducing agent.

This is absolutely true. The entire foundation of pyrometallurgy relies on this graph. If you want to extract a metal from its oxide, you need a reducing agent that has a stronger affinity for oxygen than the metal itself. On the Ellingham diagram, any element whose oxidation curve lies below the curve of the metal oxide you want to reduce can act as a successful reducing agent. Why? Because the net $\Delta G$ for the combined redox reaction will be negative, making the process thermodynamically spontaneous.

Now, let's tackle Statement II, which claims that the value of entropy change ($\Delta S$) increases from left to right in the diagram. To understand this, we need to look at the chemical reaction that is actually being plotted.

The standard reaction for an Ellingham curve is the oxidation of a metal:

Notice something critical here: we are starting with a solid metal and a gaseous oxygen molecule, and we are ending up with a solid metal oxide. A gas is being consumed!

In thermodynamics, gases have much higher entropy (randomness) than solids. Since a highly random gas is disappearing to form a highly ordered solid, the overall randomness of the system is decreasing. Therefore, the entropy change for this reaction, $\Delta S$, is negative.

How does this relate to the graph? Let's bring in the master equation of thermodynamics:

If we compare this to the equation of a straight line, $y=mx+c$, we can see that the $y$-axis is $\Delta G$, the $x$-axis is $T$, the $y$-intercept is $\Delta H$, and the slope ($m$) is $-\Delta S$.

Since $\Delta S$ is negative, the slope ($-\Delta S$) is positive. This is why almost all curves in the Ellingham diagram slope upwards!

But what happens as we move from left to right (increasing temperature)? Generally, $\Delta S$ remains fairly constant. However, if the temperature gets high enough for the solid metal to melt ($M(s)\to M(l)$), the entropy of the reactant metal increases.

Since $\Delta S_{\text{reaction}}=S_{\text{products}}-S_{\text{reactants}}$, an increase in $S_{\text{reactants}}$ makes the overall $\Delta S_{\text{reaction}}$ even more negative.

If $\Delta S$ becomes more negative (e.g., going from $-100\text{ J/K}$ to $-150\text{ J/K}$), its actual mathematical value is decreasing, not increasing.

Therefore, Statement II is false. The value of $\Delta S$ does not increase from left to right; it either remains constant or decreases (becomes more negative) when phase changes occur.

This leads us to our final conclusion: Statement I is true, but Statement II is false, making option (d) the correct choice.