Balance the following electrochemical equation in basic media OCN- + OCl- ==> CO32- + N2 + Cl-

Balance the following electrochemical equation in basic media OCN- + OCl- ==> CO32- + N2 + Cl-
what is the # of moles of H20 in the final balanced equation?

Here's my update:

This problem only needs two half-reactions. This is because the oxidation number on the C does not change and, at first glance, I thought it did. The C is +4 in OCN- and also +4 in the carbonate.

Here are the two half-reactions:

OCN- ---> CO3^2- + N2
OCl- ---> Cl-

Balance them as if in acid solution:

4H2O + 2OCN- ---> 2CO3^2- + N2 + 8H+ + 6e-
2e- + 2H+ + OCl- ---> Cl- + H2O

Change over to basic:

8OH- + 2OCN- ---> 2CO3^2- + N2 + 4H2O + 6e-
2e- + H2O + OCl- ---> Cl- + 2OH-

Multiply second half-reaction by 3, add them and delete duplicates.

Email me if you get stuck on finishing the problem.

Original message: This looks like it needs three half-reactions to balance. I'm going to go work on it and update my answer where done. Perhaps, in the interim, someone else will answer it.

I have a file of equations that required three half-reactions here:

http://www.chemteam.info/Redox/Redox-Thr…