I know how to solve redox reactions, but this one is really confusing me since I can't really write half reactions for it.
$\ce{ClO3- -> Cl2 + O2}$
The answer is apparently:
$\ce{2H+ + 2ClO3- -> 5/2 O2 + Cl2 + H2O}$
This makes sense. I understand that 12 electrons would be lost from oxygen if it all went to $\ce{O2}$ but only 10 electrons are gained by the two chlorine(V)'s so you have to use a water to make the charge balance work out. I just don't see how you'd come to this conclusion on your own.
