Posted by Laura on Tuesday, May 24, 2011 at 4:04pm.
I can't help you without doing all of the work. You need to learn how to do this yourself. What, specifically, do you not understand?
i don't understand how i'm supposed to use the reduction potentials table to blance it
To be completely honest, I don't understand that either since I've never been asked to use reduction potentials to balance a redox equation. However, I think it is done this way.
Look up the standard reduction potential of H2O2 ==> H2O and I see the equation as follows (see +1.77 volts in the table):
H2O2 + 2H^+ + 2e ==> 2H2O
Then look up the I2 ==> I^- couple (see +0.535 volts) and I see the following:
I2 + 2e ==> 2I^- but we want the reverse of that which is 2I^- ==>I2 + 2e, then we add the two together.
H2O2 + 2H^+ + 2e ==> 2H2O
2I^- ==> I2 + 2e
-----------------------------
H2O2 + 2H^+ + 2I^- +2e ==> 2H2O + I2 + 2e
Check that I didn't make a typo in getting all of this down. Apparently one positive for this method is that a look at the standard reduction tables gives you the half reaction ALREADY balanced, then you balance the electrons (they are already ok in this example) and add them.
For the oxidation number method, here is a very good site. Post with specific questions if you still have questions and I'll be happy to help you through them.
http://www.chemteam.info/Redox/Redox.html
wow thank you very much, that site is very helpful and you've been very helpful! Thank you!
how are we supposed to add them together when there is no I in the first equation?
You are adding the two halves. You add the H2O2 + 2H^+ + 2e ==> 2H2O which is the first half, then you place the second half below it (as I've done in my previous post) which is
2I^- ==> I2 + 2e and add those two halves together to obtain the final of
H2O2 + 2H^+ + 2I^- ==> 2H2O + I2