Although oxidation and reduction proceed simultaneously and an oxidation-reduction reaction can be shown in a single equation, the processes of oxidation and reduction are often shown as separate equations known as half-reactions. We encountered several examples of half-reactions in Section 8.3 during the introduction to oxidation, but we did not use the term itself. The half-reactions for the oxidation of sodium and magnesium are:

Na Na⁺ + e^-

Mg Mg²⁺ + 2 e^-

In these oxidation half-reactions, electrons are found as products. Similarly, we have already encountered reduction half-reactions for chlorine and oxygen:

Cl₂ + 2 e^- 2 Cl^-

O₂ + 4 e^- 2 O^2-

In a reduction half-reaction, the electrons are reactants.

An oxidation-reduction reaction results from the combination of an oxidation half-reaction with a reduction half-reaction. The reaction of iron with copper(II) ion (discussed in Section 14.1) combines the two half-reactions:

oxidation: Fe(s) Fe²⁺(aq) + 2 e^-

reduction: Cu²⁺(aq) + 2 e^- Cu(s)

The equations have been labeled oxidation and reduction. Note that electrons are a product (have been lost) in the oxidation half-reaction and are a reactant (have been gained) in the reduction half-reaction.

By adding together an oxidation half-reaction and a reduction half-reaction, the net ionic equation for an oxidation-reduction is obtained. This equation will be balanced if the numbers of electrons in the two half-reactions are equal. Both half-reactions shown above involve two electrons. Adding them together gives the balanced net ionic equation for the overall reaction.

If the number of electrons in the two half-reactions is not the same, as, for example, in the reaction of aluminum with hydrogen ion, each equation must be multiplied by an appropriate factor. The unbalanced equation for this reaction is:

Al(s) + H⁺ Al³⁺ + H₂(g)

The half-reactions are:

oxidation: Al(s) Al³⁺(aq) + 3 e^-

reduction: 2 H⁺(aq) + 2 e^- H₂(g)

Three electrons are lost in the oxidation half-reaction, but only two are gained in the reduction half-reaction. Before we can add them to obtain the net ionic equation for the overall reaction, we must multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3. We then obtain

oxidation: 2 Al(s) 2 Al³⁺(aq) + 6 e^-

reduction: 6 H⁺(aq) + 6 e^- 3 H₂(g)

which can be added to give the balanced equation:

Example: Write the equation for the reaction of zinc with hydrochloric acid. Isolate the half-reactions involved. Show that the balanced net ionic equation can e obtained from these half-reactions. Solution The equation is: Zn + 2 HCl H2 + ZnCl2 Zinc's oxidation number changes from 0 to +2; it is oxidized. Zn Zn2+ + 2 e- Hydrogen's oxidation number changes from +1 to 0; it is reduced. Because the product hydrogen is a diatomic gas, we must use two hydrogen ions as reactants. 2 H+ + 2 e- H2 Both half-reactions use two electrons; therefore, they can be added as is to give: