Spontaneous reaction between chromium(III) and aluminum I'm trying to rationalize this problem. It's been a while since I've done redox problems. So here goes, and any input would be great. Basically - am I right? 

Here is what happens when you store chromium ion in an aluminum container: 

$\ce{Cr^3+ + Al -&gt; Cr + Al^3+ ~~~~~~E^{o}_{cell}=}$ ${+ 0.94}$$\ce{V}$

$\ce{Cr^3+}$: oxidizing agent 

$\ce{Al}$: reducing agent 

$\ce{Cr}$: conjugate reducing agent 

$\ce{Al^3+}$: conjugate oxidizing agent 

This reaction is spontaneous because we are going from strong to weak (just like in a spontaneous acid-base reaction). 

So if we are going from strong to weak, all these statements are true:

$\ce{Cr^3+}$ is a stronger oxidizing agent than $\ce{Al^3+}$. 

$\ce{Al}$ is a stronger reducing agent than $\ce{Cr}$.
Yes, your rationalization here is correct. Chromium ions will oxidize aluminum container and deplete chromium while contaminating the solution with aluminum ions. It's exactly the same process, as for instance, copper nitrate with an iron paper clip experiment that we have all seen in a chemistry class at some point. But here you have an aluminum container instead of a paper clip :)