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The minimum voltage required to electrolyse alumina in the Hall-Heroul process is [Given, `DeltaG^(@)_(f)(A1_(2)O_(3)) =- 1520kJ//mol` and `DeltaG^(@)_(f)(CO_(2)) = 394kJ//mol]`A. `1.60V`B. `1.575V`C. `1.312V`D. `-2.62V` |
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Answer» Correct Answer - A In hall-Heroult process the following reactions occur `3C+2A1_(2)O_(3)rarr4A1+3CO_(2)` `DeltaG^(@) = 3Delta_(f)G^(@)(CO_(2))-2Delta_(f)G^(@)(A1_(2)O_(2))` `=3 (-394)-2(-1520) = 1858kJ` `DeltaG^(@) = nFE^(@)=` or `=-E^(@)` `=(DeltaG^(@))/(nF)= = (1858xx1000)/(12xx96500) = 1.60V` |
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