Balancing Oxidation Numbers

  • Thread starter MaxNumbers
  • Start date
  • #1
I'm not sure how to balance this equation with regards to the half reaction and electrons, as well as the elements:

Fe+2 + Cr2O-27--->Fe+3 + Cr+3

Right now, I've gotten this far, though I don't know if it's the right track:

Fe+2 + 2Cr+72O-27--->Fe+3 + 4Cr+3 + 7O-22

What should I do next?
 

Answers and Replies

  • #2
ShawnD
Science Advisor
668
1
My first post was totally wrong so here's a revamp.

Fe+2 + Cr2O-27--->Fe+3 + Cr+3
Ok well here's one part..

[tex]Fe^{2+} \rightarrow Fe^{3+} + e^-[/tex]

I can't figure out what happens with the Cr2O7. It usually reacts with acid, like this:

[tex]Cr_2O_7 ^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O[/tex]

But there's no acid so that ain't it.

Another thing is that I'm not sure your second equation works out.

Fe+2 + 2Cr+72O-27--->Fe+3 + 4Cr+3 + 7O-22
The charges are not the same on both sides.
 
Last edited:
  • #3
210
0
Maybe it should be 2CrO4(2-)? I noticed that on a website...Still can't figure it out though. It's pissing me off. :mad:
 
  • #4
ShawnD, that answered my question exactly. I wasn't sure how the notation worked, as far as just throwing the correct number of H+ and so forth.
Thanks. :wink:
 

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