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

1 Answer
Jun 12, 2017

Answer:

#"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(color(blue)(0))("Mg")_ ((s)) + stackrel(color(blue)(0))("O") _ (2(g)) -> stackrel(color(blue)(+2))("Mg")stackrel(color(blue)(-2))("O")_ ((s))#

Now, notice that the oxidation state of oxygen goes from #color(blue)(0)# on the reactants' side to #color(blue)(-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 xx 0 + 4 xx (1-) = 2 xx (2-)#

On the other hand, the oxidation state of magnesium is going from #color(blue)(0)# on the reactants' side to #color(blue)(+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 = (2+) + 2 xx (1-)#

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"^(-)#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaa)/color(white)(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"Mg"_ ((s)) + "O"_ (2(g)) -> 2"MgO"_ ((s))#