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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?

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    $\begingroup$ I think $\ce{OH-}$ is too little $\endgroup$ – user6006786 Mar 7 '16 at 4:37
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    $\begingroup$ What is your reduction reaction? You didn't mention that anywhere. $\endgroup$ – ringo Mar 10 '16 at 5:28
  • $\begingroup$ I think that what @user6006786 suspects is right. Maybe you can measure the pH and use the Ksp to test the hypotesis. $\endgroup$ – user1420303 Mar 13 '16 at 1:52
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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.

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Your equations are right, but look at it this way: $\ce{2Al_{(s)} + 6H_2O_{(l)} -> 2 Al^{3+}_{(aq)} + 6 H^+_{(aq)} + 6 OH^{-}_{(aq)} + 6e^- <-> 2Al^{3+}_{(aq)} + H_2O_{(l)} + 6e^-}$.

The reaction produces equal amounts of hydrogen and hydroxide ions, and they reconvert to water. Thus, there is not as much hydroxide in the solution to precipitate.

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