How does the oxidation of NaI by sodium hypochlorite to give I+ work?
I think the reaction should be done in basic solution, not acidic solution.  "One equivalent each of sodium iodide (2.77 g) and sodium hydroxide (0.74 g) was added, and the solution was cooled to 0 C. Aqueous sodium hypochlorite (34.50 g, 4.0% NaOCl) was
added dropwise over 75 min at 0-3 C" J. Org. Chem. 1990,55, 5287-5291.

So perhaps:

$$\ce{ ClO- (aq) + I- (aq) + H2O (l)  -&gt; I+ (aq) + Cl- (aq) + 2OH- (aq)}$$



Edit: after looking at the cited article more closely, the hydroxide was only being added because of the acidity of the iodophenol product, so you could write

$$\ce{ ClO- (aq) + I- (aq) + 2H+(aq)  -&gt; I+ (aq) + Cl- (aq) + H2O (l)}$$

also.