Given below are two statements :
Statement I : According to Ellingham diagram, any metal oxide with higher $\Delta$G$^\circ$ is more stable than the one with lower $\Delta$G$^\circ$.
Statement II : The metal involved in the formation of oxide placed lower in the Ellingham diagram can reduce the oxide of a metal placed higher in the diagram.
In the light of the above statements, choose the most appropriate answer from the options given below :
Solution
Ellingham diagram is plot of $\Delta$G vs T.
The criterion for the feasibility of a thermal
reduction is that at a given temperature Gibbs
energy change of a reaction must be negative. The
change in Gibbs energy, $\Delta$G for any process at any
specified temperature is given by the equation<br/><br/>
$\Delta$G = $\Delta$H - T$\Delta$S<br/><br/>
where $\Delta$H = enthalpy change and $\Delta$S = entropy change<br/><br/>
According to the ellingham diagram, any metal
oxide with higher $\Delta$G° has a tendency of getting
reduced by the metal whose metal oxide has lower
value of $\Delta$G°.
About this question
Subject: Chemistry · Chapter: p-Block Elements · Topic: Group 13: Boron Family
This question is part of PrepWiser's free JEE Main question bank. 300 more solved questions on p-Block Elements are available — start with the harder ones if your accuracy is >70%.