[_cheers]

If I am balancing using the half reaction method, how would I balance something with 2 reactants and 3 products or 1 reactant and 2 products?

say: P₄ ---> PH₃ + HPO₃ ^-2 (BASIC)

or S₂O₃-² + H₂O₂ ---> S₃O₆-² + SO₄ ^-2 + H₂0 (BASIC SOL'N)

[_cheers]

OK, Never mind....I got it, I was doing it all wrong and lucking out on the other answers...but just to confirm, when I'm balancing an ion say 2H₂SO₄ ^-2, I'm using -4 (2 x -2) as the oxidation #, not -8 (4 x -2), right?

[Borek]

2H₂SO₄^-2

No such ion, so it is hard to tell whether your thinking is correct or not...

[_cheers]

sorry, I was just making one up, not thinking if it was an actual ion or not. Using the an ion in the second equation (first post), 2S₂0₃-², the ox. # would be -4 not -6 in balancing the half reaction?

[chiralic]

Suggestions for: P₄  ---> PH₃ + HPO₃⁼

Use this:
P₄  ---> PH₃
P₄  --->  HPO₃⁼

Suggestion for: S₂O₃⁼ + H₂O₂ ---> S₃O₆⁼  + SO₄⁼ + H₂0

Use this:

S₂O₃⁼ ----> S₃O₆⁼  + SO₄⁼
H₂O₂ ----> H₂0

Read this to get more information about HOW TO Balance Redox equation (for basic media and another post by Borek in the same thread you'll get information HOW TO for acid and basic media)

[Borek]

If I am balancing using the half reaction method, how would I balance something with 2 reactants and 3 products or 1 reactant and 2 products?

just to confirm, when I'm balancing an ion say 2H₂SO₄ ^-2, I'm using -4 (2 x -2) as the oxidation #, not -8 (4 x -2), right?

On the second read - why do you need ON if you are balancing by half reactions?

[_cheers]

Thanks Chiralic

Sorry Borek, I dont think you got what I was saying. I know I dont explain it well. Thanks anyway