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oxidation reduction question

Discussion in 'DAT Discussions' started by ATLATLATL, May 1, 2007.

  1. ATLATLATL

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    guys how do i do this....
    wat is the sum of the coefficients in the following redox rxn...

    MnO4- + I- + I2 + Mn2+

    i suck that these for some reason...poo=(
     
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  3. domonas

    domonas Junior Member

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    I get 5

    MnO4(-) + 2I(-) + 8H(+) + 5e(-) --> I2 + Mn(2+) + 4H2O + 2e-

    unless you add the coefficients of the H+, H2O, and e-
    in which case, the answer is = 23
     
  4. ATLATLATL

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    how did you come up with 8H...i'm just stumped on how to arrive at the answer
     
  5. Lonely Sol

    Lonely Sol cowgoesmoo fan!

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    This is how I would do it

    2I- ---->I2 + 2e-
    8H+ + 5e- + MnO4- --->Mn2+ + 4H20 (you get 8H+ by finding the balancing the equation)

    Now, you need to get equal number of electrons on each side

    so,
    (2I- ---->I2 + 2e-)*5
    (8H+ + 5e- + MnO4- --->Mn2+ + 4H20)*2

    Now, we get:

    16H+ + 10e- + 2MnO4- + 10I- -----> 5I2 + 10e- + 2Mn2+ + 8H20

    electrons cancel out
    So you are left with:
    16H+ + 2MnO4- + 10I- -----> 5I2 + 2Mn2+ 8H20

    So, sum of co-efficent should be : 16+2+10+5+2+8 = 46

    *Plz correct me if I am wrong (I am not sure if you are suppose to include H20 and H+ in the sum)
     
  6. domonas

    domonas Junior Member

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    OMG, i totally forgot to make e- on both sides equal. i do believe the above is correct. Thank you for correcting my error Lonely Sol.

    edit: actually, you'd end up with 8 H2O on the right, not 4.
     

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