Question
Question: Point out the correct statement: A) Below \( {710^ \circ }C \) , C is better reducing agent than C...
Point out the correct statement:
A) Below 710∘C , C is better reducing agent than CO
B) Below 710∘C , CO is better reducing agent than C
C) Below 710∘C , CO is Oxidizing agent
D) Below 710∘C , CO2 is a reducing agent.
Solution
This question is based upon the Ellingham diagram. Ellingham diagram is a graph that shows the temperature dependence of the stability of the compounds. This analysis is used to evaluate the ease with which the oxides and sulphides can be reduced. This diagram was first given by Harold Ellingham in 1944.
Complete Step By Step Answer:
The Ellingham Diagram is a graph of ΔG v/s temperature for different reactions. The Ellingham diagram for reactions involving CO, C and CO2 can be given as follows.
From this graph we can conclude that the ΔG∘ values for reactions where all the lines meet are the same. Here the three lines meet at 983K. This is the crossover point.
On either side of the crossover point, the reaction on the lower side (i.e., the reaction with more negative value of ΔG∘ ) will be more spontaneous and will proceed in the forward direction as it has a negative ΔG∘ value. Whereas, the reaction on the upper side of the crossover point will have a more positive value of ΔG∘ and the reaction will proceed in the backward direction.
At the temperature below 710∘C , the reaction line of 2CO+O2→2CO2 has the most negative value and will be the most spontaneous. Hence, we can say that the line of (CO, CO2 ) lies below the (C, CO2 ) line. Hence below 710∘C , CO is a better reducing agent.
Therefore, the correct answer is Option (B).
Note:
If we consider the conditions above 710∘C , we can see that the line of (C, CO) lies below (CO, CO2 ). At higher temperatures C is a better reducing agent than CO. The data from the Ellingham diagram is used to determine a better oxidising/ reducing agent for a process.