A metal (M) exists in oxide form. Its oxide c | vedclass

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(B) According to the Ellingham diagram principle,a metal oxide can be reduced by a reducing agent if the line of the reducing agent lies below the lin

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(B) According to the Ellingham diagram principle,a metal oxide can be reduced by a reducing agent if the line of the reducing agent lies below the line of the metal oxide at that temperature.$1$. Since the oxide of $M$ cannot be reduced by carbon up to $800\, ^\circ C$,the line for $M$ must lie below the lines for $C \rightarrow CO_2$ and $C \rightarrow CO$ up to $800\, ^\circ C$.$2$. Since the oxide of $M$ can be reduced by $Al$ below $800\, ^\circ C$,the line for $Al$ must lie below the line for $M$ at temperatures below $800\, ^\circ C$.$3$. Combining these,the line for $M$ must be above the line for $Al$ but below the lines for carbon oxidation at temperatures below $800\, ^\circ C$. This corresponds to the diagram in option $B$.

$A$ metal $(M)$ exists in oxide form. Its oxide cannot be reduced by carbon up to $800\, ^\circ C$,but it can be reduced by $Al$ even below $800\, ^\circ C$. Select the correct Ellingham diagram for metal $(M)$.

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