# 4.19: Classifying Chemical Reactions: Oxidation-Reduction Reactions


Learning Objectives
• Define oxidation reaction.
• Identify the unique characteristic of an oxidation reaction.
• Define reduction reaction.
• Identify the unique characteristic of a reduction reaction.
• Define half-reaction.

Recall that, during a chemical reaction, a chemical change occurs by breaking the bonds within a substance, rearranging the atoms that had previously been connected, and then generating new bonds, in order to produce a new chemical.  Scientists predict that over 1018, or one quintillion, unique chemical interactions have the potential to initiate a viable reaction.  As the amount of time that would be required to physically study this number of reactions is prohibitively large, chemists use currently-available data and observations to find reactivity themes that can be applied to make predictions about chemical combinations that have not yet been investigated.  The previous sections of this chapter have presented five reaction classification patterns that vary either the relative locations or types of chemicals that are present within a given reaction.  Therefore, these classifications are assigned by considering how the atoms that are involved in a chemical reaction have rearranged.

Chemists have identified a final classification pattern that corresponds to the bonding changes that occur during a chemical reaction.  Recall that a bond is defined as a lasting association between atoms or ions that results in the formation of a compound.  More specifically, an ionic bond is produced when a cation, which is a positively-charged particle, exists in close physical proximity to a negatively-charged anion, creating an electrostatic attractive force.  During a chemical reaction, these bonds must be broken and then regenerated in different configurations, in order to produce a new chemical.  More specifically, when ionic bonds are altered during a chemical reaction, the particles that are associated with those bonds are forced to gain or lose valence electrons and, therefore, transition between their atomic and ionic configurations, which are defined to be net-neutral and charged, respectively.

For example, consider the following chemical equation.

$$\ce{Ca} \left( s \right) + \ce{S} \left( s \right) \rightarrow \ce{CaS} \left( s \right)$$

Based on the information that is presented in Section 4.14, this reaction is classified as a combination because only a single compound, calcium sulfide, CaS, is represented on the right side of the reaction arrow.  Note that both of the reactants, calcium, Ca, and sulfur, S, participate in this reaction in their elemental forms.  The product, calcium sulfide, CaS, is an ionic compound, which can only be formed when two ions interact with one another.  Therefore, in order to successfully combine to form this product, calcium and sulfur must ionize to form Ca+2 and S–2, respectively, in accordance with the rules that were established in Chapter 3.

Recall that the charge of an ion is determined based on the number of valence electrons that its corresponding atom must gain or lose, in order to achieve a stable electron configuration.  A calcium atom contains two valence electrons.  Therefore, in order achieve an octet configuration, calcium would need to gain six electrons.  However, gaining more than three electrons is energetically-unfavorable and will not occur.  Alternatively, an atom can lose its valence electrons, in order to achieve an octet configuration.  Because calcium only has two valence electrons, losing both of them is possible, as doing so would not exceed the maximum loss-limit of three electrons.  Therefore, the initially-neutral calcium atom, Ca, loses both of its valence electrons, resulting in the formation of a calcium cation, Ca+2, as shown below.

$$\ce{Ca} \rightarrow \ce{Ca^{+2}}$$

However, as stated in Section 4.12, the Law of Conservation of Matter mandates that particles cannot be created or destroyed in the course of a chemical reaction.  Therefore, the electrons that are lost during the ionization process must also be represented in this chemical equation.  Because two electrons, which are symbolized as e, are released from the original calcium atom, these particles are generated as products and, therefore, should be written on the right side of the reaction arrow, as shown below.

$$\ce{Ca} \rightarrow \ce{Ca^{+2}} + \ce{2 e^{–}}$$

After incorporating these electrons into this equation, the combined products of this reaction are net-neutral, overall.  Because the reactant is a neutral atom, the total charges on each side of the reaction arrow are equal, and this reaction has achieved charge-balance.  Therefore, the equation that is shown above is the chemically-correct representation of an oxidation reaction half-reaction.  Oxidation is defined as the process in which a particle loses valence electrons.  This equation is a half-reaction, as it represents transition between the atomic and ionic configuration of one-half of the elements that participate in the overall chemical reaction.

The process described in the previous paragraphs can also be applied to derive a half-reaction for sulfur, which is the second element that participates in the overall chemical reaction.  A sulfur atom contains six valence electrons.  Therefore, in order achieve an octet configuration, sulfur would need to gain two electrons.  Because gaining this number of electrons would not exceed the maximum gain-limit of three electrons, the initially-neutral sulfur atom gains two additional valence electrons, resulting in the formation of a sulfide anion, S–2, as shown below.

$$\ce{S} \rightarrow \ce{S^{–2}}$$

However, in order to uphold the Law of Conservation of Matter, the electrons that are gained during the ionization process must also be represented in this chemical equation.  Because two electrons, which are symbolized as e, are inserted into the original sulfur atom, the resultant particles are incorporated as reactants and, therefore, should be written on the left side of the reaction arrow, as shown below.

$$\ce{S} + \ce{2 e^{–}} \rightarrow \ce{S^{–2}}$$

After incorporating these electrons into the equation, the combined reactants of this reaction have an overall –2 charge.  Because the product is an anion that bears a –2 charge, the total charges on each side of the reaction arrow are equal, and this reaction has achieved charge-balance.  Therefore, the equation that is shown above is the chemically-correct representation of a reduction reaction half-reaction.  Reduction is defined as the process in which a particle gains valence electrons.

In order to bring about the transformation that is represented in the original chemical equation, both an oxidation and a reduction must occur.  As a result, the overall chemical reaction is classified as an oxidation-reduction, or "redox," reaction.  Recall that this reaction was originally classified as a combination, because only a single compound is represented on the right side of the reaction arrow.  Therefore, if both the atomic and electronic characteristics of the reactants are altered during the course of a reaction, a single chemical change can be assigned more than one reaction classification.

4.19: Classifying Chemical Reactions: Oxidation-Reduction Reactions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.