If you came to this page before studying redox much, I'd ask you to consider coming here after some more study. Please trust me on this point.
Problem #1: P2H4 ---> PH3 + P4H2
Solution #1:
1) Break into half-reactions:
P2H4 ---> PH3P2H4 ---> P4H2
2) Balance in acidic solution:
2e¯ + 2H+ + P2H4 ---> 2PH32P2H4 ---> P4H2 + 6H+ + 6e¯
3) Make electrons equal:
6e¯ + 6H+ + 3P2H4 ---> 6PH32P2H4 ---> P4H2 + 6H+ + 6e¯
4) Add the two half-reactions together:
5P2H4 ---> 6PH3 + P4H2
Solution #2:
1) remove the hydrogen from each compound and make phosphorous "ions:"
P24¯ ---> P3¯ + P42¯
2) Break into half-reactions:
P24¯ ---> P3¯P24¯ ---> P42¯
3) Balance each half-reaction:
2e¯ + P24¯ ---> 2 P3¯2P24¯ ---> P42¯ + 6e¯
4) Make electrons equal:
6e¯ + 3P24¯ ---> 6P3¯2 P24¯ ---> P42¯ + 6 e¯
5) Add the two half-reactions together:
5P24¯ ---> 6P3¯ + P42¯
6) Add hydrogens back in:
5P2H4 ---> 6PH3 + P4H2
The first method is easy, the second is rather cool (at least I think so).
Problem #2:
ClO3¯(aq) + Cl¯(aq) ---> Cl2(g) + ClO2(aq) [acidic sol.]
Before balancing, allow me to point out the oxidation numbers for the Cl in each compound:
ClO3¯ (+5)
Cl¯ (-1)
Cl2 (0)
ClO2 (+4)
Solution #1:
1) Half-reactions:
Cl¯ ---> Cl2
ClO3¯ ---> ClO2
2) Balance them:
2Cl¯ ---> Cl2 + 2e¯
e¯ + 2H+ + ClO3¯ ---> ClO2 + H2O
3) Equalize electrons and combine the half-reactions:
4H+ + 2ClO3¯ + 2Cl¯ ---> Cl2 + 2ClO2 + 2H2O
4) If, perchance, you wanted a full molecular equation:
2HClO3 + 2HCl ---> Cl2 + 2ClO2 + 2H2O
Solution #2:
1) Half-reactions:
ClO3¯ ---> Cl2
Cl¯ ---> ClO2
2) Balance them:
10e¯ + 12H+ + 2ClO3¯ ---> Cl2 + 6H2O
2H2O + Cl¯ ---> ClO2 + 4H+ + 5e¯
3) Equalize electrons and combine the half-reactions:
I will leave you to satisfy yourself that the same net ionic equation results as that in Solution #1.
Problem #3:
ClO4¯(aq) + Cl¯(aq) ---> Cl2(g) + ClO¯(aq) [acidic sol.]
Solution #1:
1) Half-reactions:
Cl¯ ---> Cl2
ClO4¯ ---> ClO¯
2) Balance them:
2Cl¯ ---> Cl2 + 2e¯
6e¯ + 6H+ + ClO4¯ ---> ClO¯ + 3H2O
3) Equalize electrons:
6Cl¯ ---> 3Cl2 + 6e¯
6e¯ + 6H+ + ClO4¯ ---> ClO¯ + 3H2O
4) Add:
6HCl + ClO4¯ ---> 3Cl2 + ClO¯ + 3H2O
Solution #2:
1) Half-reactions:
Cl¯ ---> ClO¯
ClO4¯ ---> Cl2
2) Balance them:
H2O + Cl¯ ---> ClO¯ + 2H+ + 2e¯
14e¯ + 16H+ + 2ClO4¯ ---> Cl2 + 8H2O
3) Equalize electrons:
7H2O + 7Cl¯ ---> 7ClO¯ + 14H+ + 14e¯
14e¯ + 16H+ + 2ClO4¯ ---> Cl2 + 8H2O
4) Add:
7Cl¯ + 2ClO4¯ + 2H+ ---> 7ClO¯ + Cl2 + H2O
So, which one is the "correct" answer, the one that actually takes place in the real world? I believe the answer can be demonstrated to be Solution #1 based on Ecell values. However, the ChemTeam has never checked on this point (and probably never will).