A few drops of water can initiate a reaction between iodine and aluminium.
https://www.youtube.com/watch?v=SKSU72-1ERc
How can this happen, since iodine is only slightly soluble in water?
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3 Answers
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A naked aluminum surface (freshly sanded) in air immediately grows about 4 nm of adherent aluminum oxide that passivates the surface. Iodine slightly dissolves in water and disproportionates
$$\ce{I2_{(s)} + H2O_{(l)} -> HOI_{(aq)} + I^{-}_{(aq)} + H+_{(aq)}}$$
Iodine is strongly solubilized by forming triiodide.
$$\ce{I2 + I- -> [I3]-}$$
The acids chew through the alumina surface barrier. When bare metal and triiodide meet, everything takes off, also igniting the bare metal in air.
http://www2.ucdsb.on.ca/tiss/stretton/database/inorganic_thermo.htm
$$\begin{alignat}{2}
&\ce{AlI3_{(s)}}\quad &&\Delta G_\text{f}^\circ = -300.8\ \mathrm{kJ/mol}\\
&\ce{Al2O3_{(s)}}\quad &&\Delta G_\text{f}^\circ = -1582.4\ \mathrm{kJ/mol}\\
\end{alignat}$$
Absent water the iodine must diffuse through the intact alumina barrier, react with naked aluminum, then swell and disrupt the surface passivation from underneath.
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It is very difficult to have a solid react with another solid due to the low contact area and the inability of the molecules to obtain the activation energy. Even though there is little iodine dissolved in the water, the contact area is greatly increased.
Once the reaction starts, the heat of the reaction provides the activation energy and the gaseous iodine (which makes the contact area large).
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On warming, the action of iodine on aluminum to a limited extent forms aluminum iodide:
$$\ce{Al + 3 I2 -> 2 AlI3}$$
But, in the presence of water, hydrolysis occurs:
$$\ce{2 AlI3 + 6 H2O -> 2 Al(OH)3 + 6 HI}$$
removing some water. Also, with the newly created $\ce{HI}$ (see the same Wikipedia link as above):
$$\ce{2 Al + 6 HI -> 2 AlI3 + 3 H2}$$
or, for all three reactions the net reaction is:
$$\ce{4Al + 6H2O + 3I2 -> 2Al(OH)3(s) + 3H2(g) + 2AlI3}$$
Upon adding 6 more $\ce{H2O}$'s to each side we get:
$$\ce{4Al + 12H2O + 3I2 -> 4Al(OH)3(s) + 3H2(g) + 6HI}$$
And adding 2 more $\ce{Al}$'s to each side gives:
$$\ce{6Al + 12H2O + 3I2 -> 4Al(OH)3(s) + 6 H2(g) + 2AlI3}$$
Note, the process can be repeated, but the quantities '$\ce{3I2}$' and '$\ce{2AlI3}$' will remain constant while the amount of $\ce{Al}$ and $\ce{H2O}$ increase. As such, the Iodine acts essentially as a catalyst (only a small amount is required to initiate the reaction regardless of the amount of water, and only that initial Iodine is consumed).
Note, energy arguments are generally uninformative on reaction mechanics and if one postulates that $\ce{HOI}$ (from $\ce{I2}$ plus water) attacks $\ce{Al2O3}$ layer, not likely as even $\ce{HOCl}$ but slowly reacts.
