The Gibbs energy vs T plot for the formation | vedclass

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(B) In an Ellingham diagram,a metal oxide can be reduced by a reducing agent if the line for the formation of the reducing agent's oxide lies below th

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At $600^{\circ} C$,$C$ can reduce $FeO$

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(B) In an Ellingham diagram,a metal oxide can be reduced by a reducing agent if the line for the formation of the reducing agent's oxide lies below the line for the formation of the metal oxide at that temperature.At $600^{\circ} C$:$1$. The line for $2C + O_2 \rightarrow 2CO$ is below the line for $2Fe + O_2 \rightarrow 2FeO$. Thus,$C$ can reduce $FeO$.$2$. The line for $2C + O_2 \rightarrow 2CO$ is above the line for $2Zn + O_2 \rightarrow 2ZnO$. Thus,$C$ cannot reduce $ZnO$.$3$. The line for $2CO + O_2 \rightarrow 2CO_2$ is below the line for $2Fe + O_2 \rightarrow 2FeO$. Thus,$CO$ can reduce $FeO$.$4$. The line for $2CO + O_2 \rightarrow 2CO_2$ is above the line for $2Zn + O_2 \rightarrow 2ZnO$. Thus,$CO$ cannot reduce $ZnO$.Therefore,the correct statement is that at $600^{\circ} C$,$C$ can reduce $FeO$.

The Gibbs energy vs $T$ plot for the formation of oxides is given below:
For the given diagram,the correct statement is-

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The Gibbs energy vs $T$ plot for the formation of oxides is given below:For the given diagram,the correct statement is-