Question

Using the Ellingham diagram, how to determine that in between $C$ and $CO$ which is a better reducing agent?

Answer 1

Hint : It depends on what you are reducing and the temperature. Ellingham diagram is a graph that is plotted against temperature and Gibbs free energy ( $\vartriangle G$ ) which shows the reduction of elements or compounds such as metals or oxides.

Complete Step By Step Answer:
Lets us understand what Ellingham diagram is;
Ellingham graphs show the temperature of the stability of compounds. It shows the ease at reduction of sulphides and oxides.
An Ellingham diagram is a graph of $\vartriangle G$ v/s temperature for the given reaction where $\vartriangle G$ is Gibbs free energy change.
At this point $\vartriangle G$ is the same for all the reactions.
On the other side of the crossover point, the reaction is represented by the lower line; the one which has the most negative value of $\vartriangle G$ in forward direction will be spontaneous, while the ones represented by the upper line will be spontaneous in reverse direction.
Hence, it is possible to predict the temperature above which, for example carbon or carbon dioxide will reduce any metal oxide like $FeO$
Above $800K$ , reduction is done by converting coke to carbon dioxide is spontaneous
Above $900K$ , reduction is done by converting coke to carbon monoxide is spontaneous.
Below $600K$ , only $CO$ reduces $FeO$
Hence, for reducing $FeO$ , carbon monoxide is a better reducing agent below $600K$ , but carbon is better reducing above $800K$ .

Note :
This graph form is basically of principle that the thermodynamic feasibility of reaction that depends on $\vartriangle G$ or Gibbs free energy change, that is equal to $\vartriangle H - T\vartriangle S$ , where $\vartriangle S$ is entropy change and $\vartriangle H$ is the enthalpy.