Explain applications of Ellingham diagram in the selection of a reducing agent for the reduction of metal oxides.

Explanation

(1) `DeltaG` for the formation of `Al_(2)O_(3)` is `-1080 kJ "mol"^(-1)` Which is more negative hence `Al_(2)O_(3)` is more stable, while `DeltaG` for the formation of `Cr_(2)O_(3)` is `-675 kJ "mol"^(-1)` which is comparatively less stable, and lies above `Al_(2)O_(3)` in Ellingham diagram.
Hence `Al` is used to reduce `Cr_(2)O_(3)` to metal `Cr.`
`Cr_(2)O_(3) + 2Al rarr Al_(2)O_(3) + 2Cr DeltaG = - 421 kJ`
`DeltaG = -1080 - (-675) = - 405 kJ "mol"^(-1)`
Therefore any metal can reduce other metal oxide above it in the diagram.
(2) The metal oxides having low negative values of `DeltaG` for the formation are unstable and can be decomposed on heating at moderat temperatures. For example.
`2Ag_(2)O overset(600K)rarr4Ag + O_(2)`
(3) When carbon is used as a reducing agent, it favourably forms `CO_(2)` below `1000 K` while it forms CO above 1000 K, Which can be explained by Ellingham diagram.