Question #8ab70

I'm not really sure what you need here because the problem actually tells you what you must do in order to get the answer right--balance the number of electrons!

So know that aluminium cations, #"Al"^(3+)#, are being reduced to elemental aluminium, #"Al"#, so you can say that the reduction half-reaction looks like this

#"Al"^(3+) + 3"e"^(-) -> "Al"#

This takes place at the cathode.

Magnesium metal, on the other hand, is being oxidized to magnesium cations, #"Mg"^(2+)#, so you can say that the oxidation half-reaction looks like this

#"Mg" -> "Mg"^(2+) + 2"e"^(-)#

This takes place at the anode.

Now, in any redox reaction, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.

In other words, all the electrons that are used to reduce a species must come exclusively from the species that is being oxidized.

So every atom of magnesium provides #2# electrons when oxidized. Since every atom of aluminium requires #3# electrons to be reduced, you will need to have #3# atoms of magnesium for every #2# atoms of aluminium in order for the redox reaction to work.

#{("Al"^(3+) + 3"e"^(-) -> "Al" color(white)(aaaaaaaaaaa) | xx 2), (color(white)(aaaaaaa)"Mg" -> "Mg"^(2+) + 2"e"^(-) " " | xx 3) :}#

This means that you have

#{(2"Al"^(3+) + 6"e"^(-) -> 2"Al"), (color(white)(aaaaaaa)3"Mg" -> 3"Mg"^(2+) + 6"e"^(-)) :}#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#
#2"Al"^(3+) + color(red)(cancel(color(black)(6"e"^(-)))) + 3"Mg" -> 2"Al" + color(red)(cancel(color(black)(6"e"^(-)))) + 3"Mg"^(2+)#

which gets you

#2"Al"^(3+) + 3"Mg" -> 3"Mg"^(2+) + 2"Al"#

The galvanic cell would look something like this