When carbon acts as a reducing agent, it is either converted into $\mathrm{CO}$or $\mathrm{CO}_2$ or both. $2 \mathrm{C} + \mathrm{O}_2 \rightarrow 2 \mathrm{CO}$$\mathrm{C} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2$$\mathrm{CO}$ is oxidised to $\mathrm{CO}_2$ when it is used as a reducing agent.$2 \mathrm{CO} + \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2$From the Ellingham diagram (refer answer number 3), it is clear thatat the temperature 673 K, the ΔG° of the formation of $\mathrm{CO}_2$ from CO ismore negative than the formation of CO or $\mathrm{CO}_2$ from carbon. Hence, attemperature 673 K, CO is a better reducing agent than C.