Problem #1: OCN¯ + OCl¯ ---> CO3 2 ¯ + N2 + Cl¯
On initial inspection, this problem seems like it might require three half-reactions. However, it only needs two half-reactions. This is because the oxidation number on the C does not change. The C is +4 in OCN¯ and also +4 in the carbonate.
1) Here are the two half-reactions:
OCN¯ ---> CO3 2 ¯ + N2
OCl¯ ---> Cl¯
2) Balance them as if in acid solution:
4H2O + 2OCN¯ ---> 2CO3 2 ¯ + N2 + 8H + + 6e¯
2e¯ + 2H + + OCl¯ ---> Cl¯ + H2O
3) Equalize the electrons:
4H2O + 2OCN¯ ---> 2CO3 2 ¯ + N2 + 8H + + 6e¯
6e¯ + 6H + + 3OCl¯ ---> 3Cl¯ + 3H2O
H2O + 2OCN¯ + 3OCl¯ ---> 2CO3 2 ¯ + N2 + 3Cl¯ + 2H +
5) Convert to basic:
2OH¯ + 2OCN¯ + 3OCl¯ ---> 2CO3 2 ¯ + N2 + 3Cl¯ + H2O
Problem #2: Dentrification in soils and oceans occurs when the nitrate ion is reduced to nitrous oxide by anaerobic bacteria in the presence of water. Oxygen and the hydroxyl ion are also produced during this process. Write a balanced net-ionic equation for this reaction.
1) Reaction from the information in the question:
NO3¯ + H2O ---> N2O + O2 + OH¯
2) We balance by half-reactions (in basic solution):
NO3¯ ---> N2O
H2O ---> O2
8e¯ + 5H2O + 2NO3¯ ---> N2O + 10OH¯
4OH¯ ---> O2 + 2H2O + 4e¯
3) Equalize electrons:
8e¯ + 5H2O + 2NO3¯ ---> N2O + 10OH¯
8OH¯ ---> 2O2 + 4H2O + 8e¯
4) Add and eliminate electrons, water and hydroxide:
H2O + 2NO3¯ ---> N2O + 2O2 + 2OH¯
Note: I balanced the second half-reaction via the "balance in acid first" method:
Comment: ammonia is a base, consequently we balance in basic solution.