Question

What is the reduction half-reaction for `2Mg + O_2 -> 2MgO`?

Answer 1

`"O"_ 2 + 4"e"^(-) -> 2"O"^(2-)`

Explanation:

Start by assigning oxidation numbers to all the atoms that take part in the reaction--it's actually a good idea to start with the unbalanced chemical equation

`stackrel(color(blue)(0))("Mg")_ ((s)) + stackrel(color(blue)(0))("O") _ (2(g)) -> stackrel(color(blue)(+2))("Mg")stackrel(color(blue)(-2))("O")_ ((s))`

Now, notice that the oxidation state of oxygen goes from `color(blue)(0)` on the reactants' side to `color(blue)(-2)` on the products' side, which means that oxygen is being reduced.

The reduction half-reaction will look like this

`stackrel(color(blue)(0))("O") _ 2 + 4"e"^(-) -> 2stackrel(color(blue)(-2))("O") ""^(2-)`

Here every atom of oxygen takes in `2` electrons, which means that a molecule of oxygen will take in `4` electrons.

Notice that the charge is balanced because you have

`2 xx 0 + 4 xx (1-) = 2 xx (2-)`

On the other hand, the oxidation state of magnesium is going from `color(blue)(0)` on the reactants' side to `color(blue)(+2)` on the products' side, which means that magnesium is being oxidized.

The oxidation half-reaction will look like this

`stackrel(color(blue)(0))("Mg") -> stackrel(color(blue)(+2))("Mg") ""^(2+) + 2"e"^(-)`

Here every atom of magnesium loses `2` electrons. The charge is balanced because you have

`0 = (2+) + 2 xx (1-)`

Now, to get the balanced chemical equation, multiply the oxidation half-reaction by `2` to get equal numbers of electrons lost in oxidation half-reaction and gained in the reduction half-reaction.

`color(white)(a)stackrel(color(blue)(0))("Mg") -> stackrel(color(blue)(+2))("Mg") ""^(2+) + 2"e"^(-)" " |xx 2`

`2stackrel(color(blue)(0))("Mg") -> 2stackrel(color(blue)(+2))("Mg") ""^(2+) + 4"e"^(-)`

Now add the two half-reactions to get

`stackrel(color(blue)(0))("O") _ 2 + 4"e"^(-) -> 2stackrel(color(blue)(-2))("O") ""^(2-)`
`color(white)(aaaa)2stackrel(color(blue)(0))("Mg") -> 2stackrel(color(blue)(+2))("Mg") ""^(2+) + 4"e"^(-)`
`color(white)(aaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)`

`stackrel(color(blue)(0))("O") _ 2 + color(red)(cancel(color(black)(4"e"^(-)))) + 2stackrel(color(blue)(0))("Mg") -> 2stackrel(color(blue)(-2))("O") ""^(2-) + 2stackrel(color(blue)(+2))("Mg") ""^(2+) + color(red)(cancel(color(black)(4"e"^(-))))`

which is equivalent to

`2"Mg"_ ((s)) + "O"_ (2(g)) -> 2"MgO"_ ((s))`