Redox

We have to write the balanced redox equation for this and this is the only one im stuck on. Incase theres another way to do it, we have to do it where you write all the oxidation numbers and then find out which one is reducaed which is oxidized, then write the half reactions, then the full ones. Anyways my problem is the oxidation numbers for htis one, I know your supposed to split up ionic compounds but this one is just not working for me based on what I know. The equation is:
CN- + IO3- --> I- +CNO- (In the IO3-, its not the charge that's 3-, its the O has a subscript 3 and the overal charge is -)


My one rule I have is I NEVER NEVER EVER change oxygen or hydrogen unless it is absolutely necessary. It just causes problems. Sometimes we must but only if we must.
I in IO3^- is +5. It changes to -1 in I^-. For the CN^- it really doesn't matter which one you want to change but be consistent. For example, if we say C will have an oxidation state of +4, and we know oxygen is -2, then N MUST be -5 (C = +4, N = -5, leaves a charge of -1 on CN^- and that is what we have.) Armed with that, lets go to the CNO^- on the other side. C is +4 (because we have already decided that), oxygen is -2 (because we NEVER NEVER EVER change that unless it is a must) so N must be -3. (C = +4, N = -3, O = -2, which leaves +4 and -5 or -1 which is the charge on the CNO^-). Now, see if it isn't much easier to balance. Post again and tell us what you are stuck with and perhaps we can help you through it. I hope this helps. I know this will help.


Thank you very much, I knew we never chnaged the O's so I was confused on how the rest of it worked. That helped alot, thanks.


If you want to do something just for the fun of it, on the CN^- try calling N +19 and be consistent. Then CN^- will have N as +19 and C will be -20 so that C = -20 and N = +19 leaves a -1 charge on CN^- which is what we want. On the other side of the equation, for CNO^-, we have N = +19, O = -2, so C = -18. That gives us +19-2-18 = +19-20 = -1 which is what we want for CNO^-. So C changes from -20 on the left to -18 on the right which means C lost 2 electrons. Isn't that what we had on the first one? N changed from -5 to -3 which is a loss of 2 electrons. What that means is that with the electron loss being the same, all of the other rules for writing the half equation remain the same and the equation comes out exactly has you wrote it the first time. Most students don't believe this when I start this exercise and everyone goes away in amazement. What this REALLY means is that if we remember the basics of H, O, group I, II, III, etc., then we can assign any number we wish to something like CN or SiC and the like. That is, we can allow EITHER one to change and use any number we wish to start. The trick to this is that the C was all in one place on the left (as CN^-) and all in one place on the right (CNO^-). If we had a CN^- on one side going to CO2 and CNO on the other we couldn't take such liberties. Clever, huh! Chemistry sure is interesting.

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