Galvanic cell chemistry 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:

$$\ce{2Al_{(s)} + Al2(SO4)3_{(aq)} + 6Cl- + 6H2O_{(l)} -&gt; 3H2SO4_{(aq)} + 2Al(OH)3_{(s)} + 2AlCl3 + 6e-}$$

So this would then become:

$$\ce{2Al_{(s)} + 2Al^3+ + 3(SO4)^2- + 6Cl- + 6H+ + 6OH- -&gt; 
6H+ + 3(SO4)^2- + Al(OH)3_{(s)} + 2Al^3+ + 6Cl- + 6e-}$$

The spectator ions cancel off to give:

$$\ce{2Al_{(s)} + 6OH- -&gt; 2Al(OH)3_{(s)} + 6e-}$$

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?
The aluminum may not hydrolyze all the way to a neutral hydroxide.  It could form hydroxo complexes that remain in solution, especially if the solution is made acidic by the same hydrolysis.

Consider a half-reaction in which the Al is combined with only one or two hydroxide ions.

Some other tips: 

1)  Since you have an acidic solution don't put hydroxide ions on the left.  Put water and balance with hydrogen ions on the right.

2)  Sulfate ion is a weak base.  It will combine with hydrogen ions to make bisulfate.  So add sulfate on the left and incorporate the hydrogen ions noted above into bisulfate ions.