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    Aqueous sodium ferrate (VI) will oxidise ammonia into substance X.
    What is X?
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    (Original post by ILoveUSA)
    Aqueous sodium ferrate (VI) will oxidise ammonia into substance X.
    What is X?
    Ammonia has nitrogen in the 3- oxidation state. Oxidation would increase the oxidation state of the nitrogen.

    However, nitrogen also exists in the 0, +1, +2, +3, +4 and +5 oxidation states.

    There is no way to find out the product apart from looking it up!
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    (Original post by ILoveUSA)
    Aqueous sodium ferrate (VI) will oxidise ammonia into substance X.
    What is X?

    (Original post by charco)
    Ammonia has nitrogen in the 3- oxidation state. Oxidation would increase the oxidation state of the nitrogen.

    However, nitrogen also exists in the 0, +1, +2, +3, +4 and +5 oxidation states.

    There is no way to find out the product apart from looking it up!
    I suppose it will depend upon the decomposition pathway of the ferrate. If it is a series of one electron reductions then presumably it will oxidise the ammonia to N2 as it is the most stable oxidation state for N (might want to check the frost diagram).
    If it a concerted 4e- oxidation then you may get the +1 form.
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    (Original post by JMaydom)
    I suppose it will depend upon the decomposition pathway of the ferrate. If it is a series of one electron reductions then presumably it will oxidise the ammonia to N2 as it is the most stable oxidation state for N (might want to check the frost diagram).
    If it a concerted 4e- oxidation then you may get the +1 form.
    Ammonia oxidises to many different products right up to the nitrate ion. There is no way of knowing the likely products unless you have all of the electrode potential data.
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    (Original post by charco)
    Ammonia oxidises to many different products right up to the nitrate ion. There is no way of knowing the likely products unless you have all of the electrode potential data.
    It was just an educated guess.
 
 
 
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