1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Gold Member

    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

    User Avatar
    Gold Member

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Balancing Redox Reaction
Loading...