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Homework Help: Balancing Redox Reaction

  1. Mar 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Cr(OH)3(s) + ClO3−(aq) ® CrO42−(aq) + Cl−(aq) (basic)

    I am up the point where I need to balance both half-reactions for electron charge and I'm confused as to why this half reaction: ClO3− + 6H+ + 6e− ® Cl− + 3H2O, has a total of 6 electrons instead of 5. Isn't the overall charge on that side +5, from the difference of 6H+ and 1 ClO3-?
     
  2. jcsd
  3. Mar 7, 2010 #2

    Char. Limit

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    Actually both sides of the half-reaction have charge -1. I looked and it looked balanced to me. If it's balanced, why mess with it?
     
  4. Mar 7, 2010 #3
    Really they are both -1? I thought that 6H+ (6e) + ClO- (-1e) = 5 electrons. Wouldn't all the hydrogen atoms carry a 6+ charge on that side? And on the other side, I see why Cl- + 3H20 = -1. H20 carries no charge and Cl- is obviously -1.
     
  5. Mar 7, 2010 #4

    Char. Limit

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    ClO- counts as an extra electron, not a deficiency.
     
  6. Mar 7, 2010 #5
    What about the 6H+? It doesn't carry a charge?
     
  7. Mar 7, 2010 #6
    Actually, I just realized the 6 electrons were not for total balance charge but rather the change in Cl by itself, so the -1 charge does make sense. Thank you!
     
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