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Thread: Hall-Heroult process (electrolysis)

  1. #1 Hall-Heroult process (electrolysis) 
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    why is the anode reaction of the hall-heroult process (electrolysis of aluminum oxide)
    2 O[2-] → O2 + 4 e-

    instead of
    3/2 O2 → 3 O[2-] + 6 e-

    in other words, why do the oxygen ions form oxygen gas before they combine with the carbon to make CO2? I don't really understand the electrolysis process to begin with. An oxidation reaction produces two oxygen ions, and they bond together to make oxygen gas, but why does this happen still in the presence of the positive carbon which makes up anode?
    wouldn't it make more sense to have the oxygen with a total charge of 6-, since that is what it is in the compound Al2O3 ???


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  3. #2  
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    I'm not entirely sure of the complete chemical reaction you are asking about. Where is the carbon you speak of in the redox reaction? If carbon is solely the anode, then carbon does not react with oxygen. Rather, it is a medium that carries the electrons into the beaker (or wherever your solution may be). In addition, in Al2O3, oxygen has an oxidation state of -2, not -6. There are just 3 of them (3x2=6), and Al has an oxidation state of 3+.
    I don't think this answer is going to help at all, but write another post a little more detailed and I'll help if I can.


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  4. #3  
    Forum Isotope Zelos's Avatar
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    one reason is that on the left side of
    3/2 O2 → 3 O[2-] + 6 e-
    there is no charge
    while on the right there is -12 in charge, u get charge from nowhere
    and there is 18 electrons on the left side and on right side tehre is 30 electrons
    they just cant come from nowhere dude
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