Question

Given below are two statements .

Statement I : The choice of reducing agents for metals extraction can be made by using Ellingham diagram , a plot of $\Delta G^o$ 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 :

Answer 1

In a number of processes , one metal is used to reduce the oxide of another metal. Any metal will reduce the oxide of other metals which lie above it in the Ellingham diagram because the free energy will become more negative by an amount equal to the difference between the two graphs at that particular temperature .
From left to right the general reaction taking place is

$M\left(s\right)+\frac{1}{2}{O}_2\left(g\right)\to MO\left(s\right)$

In this reaction, the gaseous amount (hence molecular randomness) is decreasing from left to right due to consumption of gases leading to a negative value of $\Delta S$ i.e. entropy decreases from left to right in Ellingham diagram