I don't see any OH^- added.
Perhaps you mean CN^- ==> CNO^- in basic solution?
1. CN^- ==> CNO^-
C is +2 on left, +4 on right so we must add 2e to the right.
CN^- ==> CNO^- + 2e
2. Count up the charge on each side. I see -1 on the left and -3 on the right. Add OH^- to the appropriate side.
2OH^- + CN^- ==> CNO^- + 2e
3. Add H2O to balance.
2OH^- + CN^- ==> CNO^- + 2e + H2O
Have you caught on to how I do it? If not I can write up a quickie for you, you can print it out and save it until you have this down pat.
Balance equations by ion-electron method
Step 1. Determine oxidation states of each element and focus on those that change from left side to right side.
Step 2. Multiply, as appropriate so that the number of atoms you are considering are the same; e.g., Cr2O7^2- ==> Cr^3+.
Cr changes from +6 for EACH Cr on the left to +3 for EACH Cr on the right BUT we will NEVER balance the equation by talking about 2 atoms on the left and 1 on the right. Therefore, multiply Cr^3+ on the right by 2 so it looks this way.
Cr2O7^2- ==> 2Cr^3+.
Now we talk about total oxidation numbers. Cr on the left is +12 total and on the right is +6 total which means we add 6e on the left to balance the change in oxidation number.
Step 3. Count the charge on the left and right. Then
(a) if acidic solution, add H^+ to the appropriate side to balance the charge or
(b) if basic solution, add OH^- to the appropriate side to balance the charge.
Step 4. Then add H2O to the appropriate side to balance H and O.
Step 5. Do the other half equation the same way.
Step 6. Multiply each equation by a number to make electrons lost = electrons gained.
Step 7. Add the two multiplied equations.
Step 8. Cancel any ions common to both sides.
Step 9. If you wish to convert to a molecular equation, add cations to negative ions and anions to positive ions on each side of the equation.
Step 10. Check one last time to make sure th equation is balanced three ways.
a. atoms balance.
b. charges balance.
c. electrons lost = electrons gained.
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