Understanding Aqueous Solutions

[LowWattBulb]

Hey all! Hope you'll bear with me. I'm an adult who finally has an opportunity to re-invest and further his science educaiton. My last formal educaiton was University level General Chemistry about 15 years ago. I'm hoping to be able to go back and do some organic chemistry courses in the near future. So in the mean time I'm doing some self-study.

So I was experimenting with some copper sufate in an aquaeous solution. I then added some aluminum and precipitate out the copper, and then I outdid myself and decanted the aluminum sulfate and added magnesium, precipitating out what I assume is aluminum powder.

My focus is on understanding the exact nature of the reactions going on. Now, I understand, copper sulfate when added to the solution forms an equilibrium reaction, ionizing into Cu+ and SO4-. Now here's what I'm trying to understand. When I add the aluminum, I'm guessing a redox reaction happens, with the copper being reduced from Cu+ to Cu, which then comes out of solution, and the aluminum is oxidized from Al to Al2-. If someone could help me visualize how exactly the electrons are transferred, and how this relates to the electronegativity of copper, aluminum, and magnesium, I would really appreciate it.

[elementcollector1]

Unfortunately, that second precipitation is likely aluminum hydroxide, as the finely powdered stuff can react (when fresh) with water.

The electronegativity series only applies here in that Mg is more electropositive (less electronegative) than aluminum, and aluminum more so than copper.

You're correct on the nature of the reaction, and as for the electron transfer, you'd need something like a balanced equation.

3CuSO4 + 2Al -> 3 Cu + Al2(SO4)3

Something like that?

Now given the oxidation states you mentioned earlier, work out how many electrons got transferred.