What is the reduction half-reaction for #2Mg + O_2 -> 2MgO#?
1 Answer
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
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
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
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
#0 = (2+) + 2 xx (1-)#
Now, to get the balanced chemical equation, multiply the oxidation half-reaction by
#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))#