Given below are two statements :
Statement I : According to Ellingham diagram, any metal oxide with higher ΔG° is more stable than the one with lower ΔG°.
Statement II : The metal involved in the formation of oxide placed lower in the Ellingham diagram can reduce the oxide of a metal placed higher in the diagram.
In the light of the above statements, choose the most appropriate answer from the options given below :

Solution

Chemistry ma’am · 4 steps

Lower on the Ellingham diagram means more negative Δ G°, and that means a more stable oxide.

1. What the diagram plots. Each line shows Δ G° for forming an oxide, 2xM + O₂ → 2MₓO, against temperature. The more negative Δ G° is, the more readily the oxide forms and the harder it is to break apart.
2. Check Statement I. A higher Δ G° (less negative) means the oxide forms less readily, so it is less stable. Statement I has this backwards, so it is incorrect.
3. Check Statement II. For a reduction MₓO + N → xM + NO, add the reverse of the upper line to the lower line. The overall Δ G° is (lower line) - (upper line), which is negative, so the reaction is spontaneous. That is why Al, which lies below Cr, reduces Cr₂O₃ in the thermite process. Statement II is correct.
4. So. Statement I is incorrect and Statement II is correct, which is option (d).

The idea to keep. On an Ellingham diagram, lower line means a more negative Δ G° and a more stable oxide. A metal can reduce the oxide of any metal whose line lies above its own.

Answer from the NTA final answer key. Solution by PrepWiser.