Question

Question #8ab70

Answer 1

See below.

Explanation:

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

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