Q:

How can iodine(I) cations be produced in this reaction if no additional acid is added?

In laboratory, we had to synthesize an aromatic compound and identify it. The chemical reaction involved salicylamid, $\ce{NaI}$ and $\ce{NaClO}$. The solvent is ethanol.
First, we dissolved the salicylamid with the ethanol. Then we added the NaI. When the mixture was homogeneous, we let it cool down to 0 °C. Once the mixture reached that temperature, we quickly added the NaClO.
We figured out that for the salicylamid to react, we needed an $\ce{I+}$.
Here's how we think we can obtain the $\ce{I+}$:
$$\ce{NaI -> Na+ + I-}\tag{1}$$
$$\ce{NaClO -> Na+ + ClO-}\tag{2}$$
$$\ce{ClO- + I- + 2H+ -> I+ + Cl- + H2O}\tag{3}$$
It seems to be the easiest way to obtain the $\ce{I+}$. However, isn't it impossible for the ethanol to release an $\ce{H+}$? Isn't ethanol neutral? How can the ethanol reacts with the $\ce{NaI}$ and the $\ce{NaClO}$? 
PS: Sorry if the terms may not be the correct ones. English isn't my first language.
Edit: Since ethanol is only the solvent and isn't involved in the reaction, how can we obtain an $\ce{I+}$?
Is it like this?
$$\ce{2NaI + NaClO -> NaCl + Na2O + 2I+}\tag{1}$$

A:

There are multiple reasons why this works.

While ethanol may be a very weak acid, there should be a non-neglegible concentration of deprotonated ethanol molecules at any time which is enough to give a nonzero fraction of $\ce{I+}$.
As $\ce{I+}$ is consumed in the reaction, protons should again be released. Unfortunately, it is only one per $\ce{I+}$, so the process is not entirely catalytic. However:
Salicylamid is a phenol and phenols are acidic. The phenol will be deprotonated and supply even more excess protons.

You should also pretend you never wrote anything about $\ce{Na2O}$ in ethanolic solution. This will happen:
$$\ce{Na2O + 2 EtOH -> 2 NaOEt + H2O}$$

