I made a galvanic cell with the anode electrode being aluminum and the electrolyte being aluminum sulfate. The salt bride was magnesium chloride. I am a little confused on what my oxidation equation would be for this set up.
So I did some research and I found out that aluminum sulfate in water will go to aluminum hydroxide and sulfuric acid, so I proposed this half cell equation:
$$\ce{2Al_{(s)} + Al2(SO4)3_{(aq)} + 6Cl- + 6H2O_{(l)} -> 3H2SO4_{(aq)} + 2Al(OH)3_{(s)} + 2AlCl3 + 6e-}$$
So this would then become:
$$\ce{2Al_{(s)} + 2Al^3+ + 3(SO4)^2- + 6Cl- + 6H+ + 6OH- -> 6H+ + 3(SO4)^2- + Al(OH)3_{(s)} + 2Al^3+ + 6Cl- + 6e-}$$
The spectator ions cancel off to give:
$$\ce{2Al_{(s)} + 6OH- -> 2Al(OH)3_{(s)} + 6e-}$$
This equation makes sense but aluminum hydroxide should form a precipitate. However, when I ran my cell, there was no visible precipitate, so it makes me question whether my anode half cell reaction is correct.
What am I doing wrong? Is the process I used to describe the half reaction right?