Question

Observe the following statements Statement - I: The choice of reducing agent for the reduction of an oxide ore can be predicted by using Ellingham diagram, a plot of $\Delta G^\circ$ Vs T. Statement - II: According to Ellingham diagram, metal oxide with higher $\Delta G^\circ$ is more stable than the oxide with lower $\Delta G^\circ$. The correct answer is

• A. Both statements I and II are correct
• B. Statement I is correct, but statement II is not correct
• C. Statement I is not correct, but statement II is correct
• D. Both statements I and II are not correct

Hint:

In an Ellingham diagram: 1. Lower line means more stable oxide. 2. A metal can reduce an oxide of another metal whose line lies above it at a given temperature. 3. The slope of the lines is related to the change in entropy ($\Delta S$) for the formation reaction. Most lines slope upwards because the reactions consume gas (O$_2$), decreasing entropy.

Answer 1

The Correct Option is B

Let's analyze each statement regarding Ellingham diagrams.
Statement - I: The choice of reducing agent for the reduction of an oxide ore can be predicted by using Ellingham diagram, a plot of $\Delta G^\circ$ Vs T.
An Ellingham diagram plots the standard Gibbs free energy of formation ($\Delta_f G^\circ$) of oxides as a function of temperature.
For a reduction reaction like $M' + MO \rightarrow M'O + M$ to be spontaneous, the overall $\Delta G^\circ$ for the reaction must be negative.
In an Ellingham diagram, any element can reduce the oxide of another element whose line lies above it on the diagram.
Therefore, the diagram is used to predict the feasibility of reduction and choose a suitable reducing agent.
This statement is correct.
Statement - II: According to Ellingham diagram, metal oxide with higher $\Delta G^\circ$ is more stable than the oxide with lower $\Delta G^\circ$.
The $\Delta G^\circ$ values on the diagram are for the formation of the oxide from the metal and oxygen.
A more negative (i.e., lower) value of $\Delta G^\circ$ indicates a more spontaneous formation reaction and, consequently, a more thermodynamically stable oxide.
A higher (less negative or positive) $\Delta G^\circ$ indicates a less stable oxide.
This statement claims the opposite. Therefore, this statement is not correct.