# Why "Mg"^(2+) +2"e"^(-) -> "MgO" ?

## From a web-tutorial on reduction potentials : Why should adding 2 electrons to ${\text{Mg}}^{2 +}$ result in $\text{MgO}$? AFAIK, in $\text{MgO}$ magnesium's oxidation state is also $\left(+ 2\right)$.

Feb 14, 2016

That's just a typo.

#### Explanation:

You are right, adding two electrons to a magnesium cation, ${\text{Mg}}^{2 +}$, would not result in the formation of magnesium oxide, $\text{MgO}$, it would result in the formation of magnesium metal, $\text{Mg}$.

You can oxidize magnesium metal to magnesium cations and reduce magnesium cations back to magnesium metal.

The correct reduction half-reaction would be

$\text{Mg"^(2+) + 2"e"^(-) -> "Mg}$ $\text{ "E^0 = -"2.38 V}$

So that is just a mistake they made when writing out the half-reaction.

For example, the synthesis of magnesium oxide, $\text{MgO}$, is a redox reaction in which oxygen gas oxidizes magnesium metal, while being reduced in the process.

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

Here you have the oxidation half-reaction

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

Each magnesium atom loses two electrons, so two magnesium atoms will lose a total of four electrons.

The reduction half-reaction is

${\text{O"_2 + 4"e"^(-) -> 2"O}}^{2 -}$

Each oxygen atom gains two electrons, so two oxygen atoms will gain a total of four electrons.