Question

What is the reduction half-reaction for \[2Mg+{{O}_{2}}\to 2MgO\] ?

Answer 1

Hint: We know that the oxidation – reduction half reaction is also known as redox reaction. In this reaction the oxidation reaction and reduction reaction takes place at the same time. The oxidation is the loss of electrons (addition of oxygen) and reduction is the gain of electrons (loss of oxygen).

Complete answer:
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 \[{{\overset{0}{\mathop{Mg}}\,}_{\left( s \right)}}+{{\overset{0}{\mathop{O}}\,}_{2\left( g \right)}}\to \overset{+2}{\mathop{Mg}}\,{{\overset{-2}{\mathop{O}}\,}_{\left( s \right)}}\]. Now, notice that the oxidation state of oxygen goes from 0 on the reactants' side to \[-2~\] on the products' side, which means that oxygen is being reduced. The reduction half-reaction will look like this \[{{\overset{0}{\mathop{O}}\,}_{2}}+4{{e}^{-}}\to 2{{\overset{-2}{\mathop{O}}\,}^{2-}}\]
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 \[2\times 0+4\times \left( 1- \right)=2\times \left( 2- \right)\]
On the other hand, the oxidation state of magnesium is going from \[~0~\] on the reactants' side to \[+2~\] on the products' side, which means that magnesium is being oxidized. The oxidation half-reaction will look like this \[\overset{0}{\mathop{Mg}}\,\to \overset{+2}{\mathop{M}}\,{{g}^{2+}}+2{{e}^{-}}\]
Here every atom of magnesium loses 2 electrons. The charge is balanced because you have \[0=\left( 2+ \right)+2\times \left( 1- \right)\]
Now, to get the balanced chemical equation, multiply the oxidation half-reaction by two to get equal numbers of electrons lost in oxidation half-reaction and gained in the reduction half-reaction. \[\overset{0}{\mathop{Mg}}\,\to \overset{+2}{\mathop{M}}\,{{g}^{2+}}+2{{e}^{-}}\] Multiplying with two we get; \[\overset{0}{\mathop{2Mg}}\,\to \overset{+2}{\mathop{2M}}\,{{g}^{2+}}+4{{e}^{-}}\]
Now add the two half-reactions to get \[\left[ \overset{0}{\mathop{{{O}_{2}}}}\,+4{{e}^{-}}\to 2\overset{-2}{\mathop{O}}\,{{~}^{2-}} \right]+\left[ \overset{0}{\mathop{2Mg}}\,\to \overset{+2}{\mathop{2M}}\,{{g}^{2+}}+4{{e}^{-}} \right]\]
\[{{\overset{0}{\mathop{O}}\,}_{2}}+4{{e}^{-}}+2\overset{0}{\mathop{Mg}}\,\to 2\overset{-2}{\mathop{O}}\,{{~}^{2-}}+2\overset{+2}{\mathop{Mg}}\,{{~}^{2+}}+4{{e}^{-}}\]which is equivalent to \[2M{{g}_{\left( s \right)}}+{{O}_{2\left( g \right)}}\to 2Mg{{O}_{\left( s \right)}}.\]

Note:
Remember that in the oxidation reaction, the oxidation state of the substance increases from the reactant side to the product side and in the reduction reaction, the oxidation state of the substance decreases from the reactant side to the product side.