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

From a web-tutorial on reduction potentials :

enter image source here

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

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

1 Answer
Stefan V.
Feb 14, 2016

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.

Related questions
See all questions in Oxidation and Reduction Reactions
Impact of this question
21711 views around the world
You can reuse this answer
Creative Commons License