Question

According to the following diagram, A reduces $B{O}_2$ when the temperature is:

• A. $>1400{}^{\circ }C$
• B. $<1400{}^{\circ }C$
• C. $>1200{}^{\circ }C$ but $<1400{}^{\circ }C$
• D. $<1200{}^{\circ }C$

Answer 1

The correct option is A $>1400{}^{\circ }C$

In Ellingham’s diagram, the line of the element that lies below can reduce the oxide of the element which lies above it. Therefore, for A to reduce $B{O}_2$ , the temperature when the line for element A is below that of $B{O}_2$ according to the graph when $T>1400{}^{\circ }C,$
For $T>1400{}^{\circ }C,△G_r<0forA+B{O}_2\to B+A{O}_2$