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I recently did an experiment putting pieces of $\ce{Al}$ and $\ce{Fe}$ into copper(II) sulfate solution and I noticed when the piece of aluminium was reacting, tiny bubbles were coming out whereas with the piece of iron this didn't happen.
I think it was part of the bare aluminium that was reacting with water producing hydrogen gas but I'm not sure. Can anyone please explain this to me?
As Ivan Neretin noted in a comment, your supposition is absolutely correct. The aluminium metal is reactive enough with water that it forms hydrogen gas and aluminum hydroxide (or other aluminum oxyhydroxide species) where it's exposed to the solution:
$$ \begin{align} \ce{6H2O + 6e-} & \ce{->~ 6OH- + 3H2\uparrow} \\ \ce{2Al(s) + 6H2O } & \ce{->~2 Al(OH)3 + 6H+ + 6e-} \end{align} $$
Overall reaction:
$$ \ce{2Al(s) + 6H2O -> 2Al(OH)3 + 3H2\uparrow} $$
If the $\ce{CuSO4}$ solution is acidic enough, the $\ce{Al(OH)3}$ rapidly dissolves, exposing a fresh aluminium surface, ready for further reaction.
On the other hand, the redox reaction of $\ce{Fe(s)}$ and water is not favorable enough for hydrogen evolution to occur, which explains the lack of bubbles you observed. The single-displacement reaction between $\ce{Cu^2+}$ and $\ce{Fe(s)}$ is quite favorable, however, leading to paired reactions like the following:
$$ \begin{align} \ce{Fe(s)} & \ce{->~ Fe^2+ + 2e-} \\ \ce{Cu^2+ + 2e-} & \ce{->~ Cu(s)} \end{align} $$
Overall reaction:
$$ \ce{Fe(s) + Cu^2+ -> Cu(s) + Fe^2+} $$
This results in the rapid exchange of metallic Cu for metallic Fe on the iron surface, leading to a film of plated copper on the surface.
