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(B) In an Ellingham diagram,a metal $A$ can reduce the oxide of another metal $B$ $(BO_{2})$ if the line for the formation of $AO_{2}$ lies below the

Vedclass · Accepted Answer

$> 1400^{\circ} C$

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(B) In an Ellingham diagram,a metal $A$ can reduce the oxide of another metal $B$ $(BO_{2})$ if the line for the formation of $AO_{2}$ lies below the line for the formation of $BO_{2}$.From the given diagram,the line for $A + O_{2} \rightarrow AO_{2}$ intersects the line for $B + O_{2} \rightarrow BO_{2}$ at $1400^{\circ} C$.Above $1400^{\circ} C$,the line for $AO_{2}$ formation is below the line for $BO_{2}$ formation,meaning the $\Delta G^{\circ}$ for the reaction $A + BO_{2} \rightarrow AO_{2} + B$ becomes negative.Therefore,$A$ reduces $BO_{2}$ at temperatures $> 1400^{\circ} C$.

According to the following Ellingham diagram,$A$ reduces $BO_{2}$ when the temperature is

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