How do you KNOW when to add H2O and H+ to redox rxns to balance them?

Question

Ex:
Br2(l) + SO2(g)--->Br^-(aq)+SO4^2-(aq)
You end up adding H2O to the Br2 side
and H+ to the So4^2 side

BUT....
I2(aq)+S2O3^2^-(aq)---->I^-(aq)+S3O6^2-(aq)
You dont.
Why dont you add it to this one too???
Im so confused!!!
Please help. thanks:)

Answer 1

What the guy above me said is correct. Just be careful and always check your answer. I've noticed when im trying to bal eq that when you start balancing the O2 and it seems like it doesnt work... then you should add H2O. :)

Answer 2

Lancenigo di Villorba (TV), Italy

ABOUT YOUR QUESTIONs
It is clear, these chemical equations need Mass and Electrical Charge Balancement!!
Since they belong to REDOX REACTIONs, I distinguish ANODIC HALF-REACTION (e.g. it interests the Reducing Reactive) from CATHODIC HALF-REACTION (e.g. it interests the OXIDIZER One).
I must verify that the Chemical Element's Atoms present in all the Compounds are Mass's Balanced, elsewhere I play the Stoichiometric Coefficients in order to save this Condition.
I calculate the "Oxidation Numbers" so I find the Atoms involved in the REDOX PHENOMENA : hence, I can add EXCHANGED ELECTRONs in order to Balance the Oxidation Numbers.
Now, I can reason separately on the Half-Reactions instead the Whole Reaction one.
Since they run in aqueous solutions, I decide if they stay in ACIDIC MEDIA or ALKALINE Ones.
In the former case, I must add HYDROGEN IONs in order to HYDROGEN ATOM's BALANCEMENT while I will add WATER MOLECULEs in order to OXYGEN ATOM's BALANCEMENT.
I balanced all two the Half-Reactions, so I can add ANODIC HALF-REACTION to the CATHODIC ONE, so in a fashion to erase the EXCHANGED ELECTRONs.
In the second case, I must add WATER MOLECULEs in order to HYDROGEN ATOM's BALANCEMENT while I will add OXYDRIL IONs in order to OXYGEN ATOM's BALANCEMENT.

EXAMPLE 1)
Br2(l) + SO2(g) ---> Br-(aq) + SO4--(aq)
The two Half-Reactions are the following ones
i) Br2(l) + 2e ---> 2 Br-(aq)
VERY EASY ONE!
ii) SO2(g) ---> SO4--(aq) + 2e
IT IS STILL UNBALANCED ONE!!
I suspect the second one run in an ACIDIC MEDIUM, so I add in a former time WATER MOLECULEs (e.g. OXYGEN ATOM's BALACEMENT)
ii) SO2(g) + 2 H2O(aq) ---> SO4--(aq) + 2e
hence HYDROGEN IONs (e.g. HYDROGEN ATOM's BALANCEMENT)
ii) SO2(g) + 2 H2O(aq) ---> SO4--(aq) + 4 H+(aq) + 2e
Now, I sum i) Half-Reaction to ii) One in a fashion erasing the EXCHANGED ELECTRONs
i+ii) Br2(l) + SO2(g) + 2 H2O(aq) --->
---> 2 Br-(aq) + SO4--(aq) + 4 H+(aq)
§§§§§§§§ Br2(l) + SO2(g) + 2 H2O(aq) --->
---> 2 HBr(aq) + H2SO4(aq) §§§§§§§§

EXAMPLE 1)
I2(l) + S2O3--(g) ---> I-(aq) + S4O6--(aq)
The two Half-Reactions are the following ones
iii) I2(l) + 2e ---> 2 I-(aq)
VERY EASY ONE!
iv) S2O3--(aq) ---> S4O6--(aq) + 2e
IT IS STILL UNBALANCED ONE!!
I must verify the SULPHUR ATOM's BALANCEMENT
iv) 2 S2O3--(aq) ---> S4O6--(aq) + 2e
Now, I sum iii) Half-Reaction to iv) One in a fashion erasing the EXCHANGED ELECTRONs
iii+iv) I2(l) + 2 S2O3--(aq) ---> 2 I-(aq) + S4O6--(aq)
§§§§§ I2(l) + 2 S2O3--(aq) ---> 2 I-(aq) + S4O6--(aq) §§§§§§

I hope this helps you.