How does #"AlCl"_3# behave as an acid in the presence of #"Cl"_2# in water? It has no protons... so is it a Lewis acid?

There is still an #H^(+)# transfer, but it is not very intuitive.

To do this in real life you typically just add some #HCl# in with aluminum solid to form #AlCl_3#, and #Cl^(-)# is then readily available from the #HCl# to catalyze the deprotonation process.

Let's go with your example of #AlCl_3#.

You're right, it is a Lewis acid. What aluminum does here is that it acts as an electrophile in a catalysis reaction early on, and then the chlorines on it become nucleophilic, grabbing protons from other water molecules.

This proton transfer looks something like the following.

If you have more #"Cl"^-#, then this repeats with #"AlCl"_2"OH"# and #"Cl"_2# to form #"Al"("OH")_3(s)# eventually.

Essentially, the protons come from water deprotonating itself, with a little help!

1. First, you're actually supposed to activate #AlCl_3# by reacting it with #Cl_2# to get #[AlCl_4]^(-)#.

2. Then, aluminum becomes negatively charged and the #Al-Cl# bonding electrons want to get off. They are donated to grab a proton from water.

3. At this point, #AlCl_3# is a lewis acid again, and now, there is #OH^(-)#, a nucleophile, and it can attack the aluminum center and displace a chlorine.

4. Then this lone chloride ion attacks another #AlCl_3# in solution (there is more than one), reforming #[AlCl_4]^(-)#, and the process repeats two more times.

In another representation, you have:

Acid activation

0. #color(highlight)(3AlCl_3) + 3Cl_2 => color(highlight)([3AlCl_4]^(-) + 3Cl^(+))#

Proton Transfer Process

1A. #[AlCl_4]^(-) + H-OH => color(highlight)(AlCl_3) + color(green)(OH^(-)) + color(blue)(HCl)#
2A. #color(green)(AlCl_3) + color(green)(OH^(-)) => color(green)(AlCl_2OH) + color(highlight)(Cl^(-))#
3A. # color(highlight)(AlCl_3 + Cl^(-)) => [AlCl_4]^(-)#

1B. #[AlCl_4]^(-) + H-OH => color(highlight)(AlCl_3) + color(green)(OH^(-)) + color(blue)(HCl)#
2B. #color(green)(AlCl_2OH) + color(green)(OH^(-)) => color(green)(AlCl(OH)_2) + color(highlight)(Cl^(-))#
3B. # color(highlight)(AlCl_3 + Cl^(-)) => [AlCl_4]^(-)#

1C. #[AlCl_4]^(-) + H-OH => color(highlight)(AlCl_3) + color(green)(OH^(-)) + color(blue)(HCl)#
2C. #color(green)(AlCl(OH)_2) + color(green)(OH^(-)) => color(green)(Al(OH)_3) + color(highlight)(Cl^(-))#
3C. # color(highlight)(AlCl_3 + Cl^(-)) => [AlCl_4]^(-)#

Green indicates the products you get and/or reuse.
Highlight indicates the reaction steps that re-activate #AlCl_3# for more.
Blue indicates the products that are why you still get protons in solution.

Once that's over, you actually have this finishing up the process:

#Al(OH)_3 + OH^(-) -> [Al(OH)_4]^(-)#
...deactivating the aluminum center.

And the protons you have in solution are from:

#3HCl -> 3H^(+) + 3Cl^(-)#

Finally:

#3Cl^(+) + 3Cl^(-) -> 3Cl_2#
...making all the weird #Cl^(+)# go away.

Overall, what you get is essentially the protons grabbed off of water, in solution with #[Al(OH)_4]^(-)#, #Cl_2#, and other pure water molecules leftover.

So, you do have #H^(+)# in the solution, in the end, but in an unconventional way.

(There shouldn't be #AlCl_3# anymore because otherwise the reaction would keep going since #Cl_2# is still available.)