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

Jun 12, 2017

${\text{O"_ 2 + 4"e"^(-) -> 2"O}}^{2 -}$

#### 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

${\stackrel{\textcolor{b l u e}{0}}{\text{Mg")_ ((s)) + stackrel(color(blue)(0))("O") _ (2(g)) -> stackrel(color(blue)(+2))("Mg")stackrel(color(blue)(-2))("O}}}_{\left(s\right)}$

Now, notice that the oxidation state of oxygen goes from $\textcolor{b l u e}{0}$ on the reactants' side to $\textcolor{b l u e}{- 2}$ 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

$2 \times 0 + 4 \times \left(1 -\right) = 2 \times \left(2 -\right)$

On the other hand, the oxidation state of magnesium is going from $\textcolor{b l u e}{0}$ on the reactants' side to $\textcolor{b l u e}{+ 2}$ 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

$0 = \left(2 +\right) + 2 \times \left(1 -\right)$

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"^(-)
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

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

$2 {\text{Mg"_ ((s)) + "O"_ (2(g)) -> 2"MgO}}_{\left(s\right)}$