Posted by anonymous on Friday, November 18, 2011 at 8:54pm.
Here are the two half equations.
H2O2 ==>O2 + 2e + 2H^+
4H^+ + 2e + 2OCl^- ==> Cl2 + H2O
Add them and cancel anything common to each side.
where'd the sodium go
is sodium a spectator ion?
yes. I took the Na^+ out before I started.
i noticed that in other problems they do that too, why is that and what is the reason for ignoring it
Since it is a spectator ion it never enters into the reaction. That's what writing net ionic equations is all about; you include only those ions that are going about the reaction. Said another way, the H2O2 + OCl^- reaction is the same reaction with NaOCl, KOCl, LiOCl, etc. Of course those different ions would form different salts; e.g., Na would form NaCl, K would form KCl, Li would form LiCl, etc (if HCl were used as the acid) or Na2SO4, K2SO4 etc (if H2SO4 were used as the acid). H2O2 and OCl^- are what's reacting. I find it a little strange that you wondered where the Na^+ went but it didn't bother you that no anion was used for the H^+.
And is this the same case for if I am given a NIE and the problem says to complete the reaction by using Na(to make a compound)?
we learned that the H^+ cations were added to balance out the charge on both sides of the equation(for acids) and OH^- for bases
and also thanks a lot for explaining this to me
Yes, to complete a NIE, simply add cations of choice (or what someone give you) for the negative ions; the same with anions added to positive ions in the NIE.
thanks
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