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Redox equations!

  1. Feb 6, 2007 #1
    When hydroxide ion is added to Mn(II) in aqueous solution, an insoluble white precipitate of manganous hydroxide, Mn(OH)2, is formed. If there is oxygen present, this material reacts with it to form manganic hydroxide, Mn(OH)3


    Part one was easy because it wasnt a redox it just needed to be balanced so I got:

    +
    2OH + Mn --> Mn(OH)2

    The second part I am having trouble with....

    +2 -2+1 0 +3 -2+1
    Mn(OH)2 + O2 ---> Mn(OH)3

    +2 -2+1 +3 -2+1
    Oxidation - Mn(OH)2 ---> Mn(OH)3 + 1e-

    0 -2+1
    Reduction - O2 + 2e- ---> Mn(OH)3
    (I dont understand this bc theres no Mn on the left side)

    alright this I dont know if anyone can help me with because its really hard to type out and I dont think this is going to look like the way i set it up but if anyone can try to explain this to me through here it would be greatly appreciated.
     
  2. jcsd
  3. Feb 7, 2007 #2
    For starters, what do you have to do? Do you have to balance the redox rxn, or simply figure out the oxidation state?
     
  4. Feb 8, 2007 #3
    After a simple search i got this


    4Mn(OH)2 + O2 + 2H2O ==> 4Mn(OH)3
     
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