Saved Bookmarks
| 1. |
Explain applications of Ellingham diagram in the selection of a reducing agent for the reductionof metal oxides. |
|
Answer» SOLUTION :(1) `DeltaG` for the formation of `Al_(2)O_(3)` is `-1080 kJ "mol"^(-1)` Which is more negativehence `Al_(2)O_(3)` is more stable, while `DeltaG` for the formation of `Cr_(2)O_(3)` is `-675 kJ "mol"^(-1)` WHICHIS 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. |
|