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?

  • 1
    $\begingroup$ I think $\ce{OH-}$ is too little $\endgroup$ Commented Mar 7, 2016 at 4:37
  • 1
    $\begingroup$ What is your reduction reaction? You didn't mention that anywhere. $\endgroup$
    – ringo
    Commented Mar 10, 2016 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. $\endgroup$ Commented Mar 13, 2016 at 1:52

2 Answers 2


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.


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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