Balancing redox equations in acidic media: to use H+ or H3O+? Is it preferable to balance redox equations with  $\ce{H+}$ or $\ce{H3O+}$? I think the latter, because first, the bare hydrogen proton doesn't exist in water solution. So why write down something we know not to exist? 

I ask because in this problem, if we balance using both species, $\ce{H^+}$ and $\ce{H3O+}$, we achieve two different equations. Mass is conserved in both equations as well as charge; all the differs is one coefficient. 

So are the two equations equivalent in that they are still describing the same reaction? I think so, because sometimes I see textbook authors write these two equations for $K_\mathrm{w}$:

$$\ce{H2O &lt;=&gt; H+ + OH-}$$

$$\ce{2H2O &lt;=&gt; H3O+ + OH-}$$

However, if we were actually going to run an experiment and needed exact quantities, we would probably prefer the second reaction both above and below because the coefficients in these reactions better represent the chemistry of the system right? 

For example, which of the following is "better"?

$\ce{ClO_{3}^- + 6I^- + 6H^+ \leftrightharpoons 3I_2 + Cl^- + 3H_2O}$

$\ce{ClO_{3}^- + 6I^- + 6H_3^+O \leftrightharpoons 3I_2 + Cl^- + 9H_2O}$
This is a question of "style" so there is no correct answer.  In fact, just as $\ce{H^+}$ doesn't really exist in solution, neither does $\ce{H3O+}$ exist in solution.  In reality higher hydrates of $\ce{H^+}$ are what exist in solution. See the first paragraph in this Wikipedia article.  Again, both $\ce{H^+}$ and $\ce{H3O+}$ are just shorthand notations for reality, choose the style that you like best.