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

From a web-tutorial on reduction potentials :

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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.

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