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.
What I would like you to do is balance the equation below without looking at the answers first. I'll give you a hint about two different ways to balance it.
Here is the reaction:
P2H4 ---> PH3 + P4H2
Balancing Method Number One: think about balancing it in acid solution. The answer.
Balancing Method Number Two: remove the hydrogen from each compound and make phosphorous "ions." Then balance it. The answer.
The first method is easy, the second is rather cool (at least I think so). The "ions" idea may confuse you a bit, that's OK. Think about it some and try it before looking at the answer. You might even discuss this equation with your teacher before looking at the answer.
Several years after writing the above, I had occasion to balance this redox reaction:
P4 ---> HPO32¯ + PH3
and decided to include it here. The solution follows just below.
1) Separate into half-reactions:
P4 ---> HPO32¯P4 ---> PH3
2) Solve both half-reactions by the "fake acid" method:
12e¯ + 12H+ + P4 ---> 4PH312H2O + P4 ---> 4HPO32¯ + 20H+ + 12e¯
3) Add the two half-reactions and eliminate duplicates:
12H2O + 2P4 ---> 4PH3 + 4HPO32¯ + 8H+
4) Allow the strong acid (H+) to combine with the base (HPO32¯) to form a weaker acid:
12H2O + 2P4 ---> 4PH3 + 4H3PO3
Here's another solution direction.
1) After balancing the two half-reactions, convert to basic:
12e¯ + 12H2O + P4 ---> 4PH3 + 12OH¯P4 + 20OH¯ ---> 4HPO32¯ + 8H2O + 12e¯
2) Add the two equations and eliminate eight duplicate water and 12 duplicate hydroxide:
4H2O + 2P4 + 8OH¯ ---> 4PH3 + 4HPO32¯
3) Add a bit of acid to this basic solution:
12H2O + 2P4 ---> 4PH3 + 4H3PO3
I added 8H+ to each side. On the left-hand side, it reacted with the hydroxide to form water. On the right-hand side, it reacted with the hydrogen phosphite ion to make phosphorous acid (NOT phosphoric acid).
A third interesting redox reaction:
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)
I'm going to balance it two ways.
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 balancing method #1.