Question

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

From a web-tutorial on reduction potentials :

Why should adding 2 electrons to `"Mg"^(2+)` result in `"MgO"`?

AFAIK, in `"MgO"` magnesium's oxidation state is also `(+2)`.

Answer 1

That's just a typo.

Explanation:

You are right, adding two electrons to a magnesium cation, `"Mg"^(2+)`, would not result in the formation of magnesium oxide, `"MgO"`, it would result in the formation of magnesium metal, `"Mg"`.

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

The correct reduction half-reaction would be

`"Mg"^(2+) + 2"e"^(-) -> "Mg"` `" "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, `"MgO"`, is a redox reaction in which oxygen gas oxidizes magnesium metal, while being reduced in the process.

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

Here you have the oxidation half-reaction

`2"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

`"O"_2 + 4"e"^(-) -> 2"O"^(2-)`

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