# Question #8ab70

Jul 19, 2017

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, ${\text{Al}}^{3 +}$, are being reduced to elemental aluminium, $\text{Al}$, so you can say that the reduction half-reaction looks like this

$\text{Al"^(3+) + 3"e"^(-) -> "Al}$

This takes place at the cathode.

Magnesium metal, on the other hand, is being oxidized to magnesium cations, ${\text{Mg}}^{2 +}$, so you can say that the oxidation half-reaction looks like this

${\text{Mg" -> "Mg"^(2+) + 2"e}}^{-}$

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.

$\left\{\begin{matrix}\text{Al"^(3+) + 3"e"^(-) -> "Al" color(white)(aaaaaaaaaaa) | xx 2 \\ color(white)(aaaaaaa)"Mg" -> "Mg"^(2+) + 2"e"^(-) " } | \times 3\end{matrix}\right.$

This means that you have

$\left\{\begin{matrix}2 {\text{Al"^(3+) + 6"e"^(-) -> 2"Al" \\ color(white)(aaaaaaa)3"Mg" -> 3"Mg"^(2+) + 6"e}}^{-}\end{matrix}\right.$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$
$2 {\text{Al"^(3+) + color(red)(cancel(color(black)(6"e"^(-)))) + 3"Mg" -> 2"Al" + color(red)(cancel(color(black)(6"e"^(-)))) + 3"Mg}}^{2 +}$

which gets you

$2 \text{Al"^(3+) + 3"Mg" -> 3"Mg"^(2+) + 2"Al}$

The galvanic cell would look something like this 