But you have also assumed in setting up the equations that there is no H+ or H2O
I didn't assume anything, I just checked it works.
If we had to involve these
What for? Reaction is already balanced.
Oh. So you just used the algebraic method to check that this is the only solution?
To originally find the reaction equation we have to keep an open mind that H
+ and H
2O could be involved (even if finally it turns out that neither is involved). To me this implies some kind of redox method of balancing. So, how did you initially find the reaction equation? As far as I can see we have 3 half-equations
S
2O
32- + 5 H
2O
2 SO
42- + 10 H
+ + 8e
-3 S
2O
32- + 3 H
2O
2 S
3O
62- + 6 H
+ + 8e
-O
2 + 4 H
+ + 4e
- 2 H
2O
If we consider the possibility (
pre-balancing) that H
+ and H
2O
might be involved in the final reaction equation, there is no way we could skip straight to that final reaction equation, since there will be infinite solutions. So what was the redox balancing method for this question?
When I originally did, I summed up the two oxidations, multiplied the reduction by 4 and then summed. Why? Just because it is the same method as on that wrong half-equations page linked above. It worked this time. But the logic behind this particular choice of coefficients is not apparent - why not multiply the first oxidation by 3, add on the second oxidation and then multiply the reduction by 8, for a random example?