This is “Oxidation-Reduction Reactions”, section 14.1 from the book Beginning Chemistry (v. 1.0). For details on it (including licensing), click here.
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Consider this chemical reaction:
Mg(s) + Cl2(g) → MgCl2The reactants are two electrically neutral elements; they have the same number of electrons as protons. The product, however, is ionic; it is composed of Mg2+ and Cl− ions. Somehow, the individual Mg atoms lose two electrons to make the Mg2+ ion, while the Cl atoms gain an electron to become Cl− ions. This reaction involves the transfer of electrons between atoms.
The process of losing and gaining electrons occurs simultaneously. However, mentally we can separate the two processes. OxidationThe loss of one or more electrons by an atom; an increase in oxidation number. is defined as the loss of one or more electrons by an atom. ReductionThe gain of one or more electrons by an atom; a decrease in oxidation number. is defined as the gain of one or more electrons by an atom. So oxidation and reduction always occur together; it is only mentally that we can separate them. Chemical reactions that involve the transfer of electrons are called oxidation-reduction (or redox) reactionsA chemical reaction that involves the transfer of electrons..
Redox reactions require that we keep track of the electrons assigned to each atom in a chemical reaction. How do we do that? We use oxidation numbersA number assigned to an atom that helps keep track of the number of electrons on the atom. to keep track of electrons in atoms. Oxidation numbers are assigned to atoms based on four rules. Oxidation numbers are not necessarily equal to the charge on the atom (although sometimes they can be); we must keep the concepts of charge and oxidation numbers separate.
The rules for assigning oxidation numbers to atoms are as follows:
Here are some examples for practice. In H2, both H atoms have an oxidation number of 0 by rule 1. In MgCl2, magnesium has an oxidation number of +2, while chlorine has an oxidation number of −1 by rule 2. In H2O, the H atoms each have an oxidation number of +1, while the O atom has an oxidation number of −2, even though hydrogen and oxygen do not exist as ions in this compound (rule 3). By contrast, by rule 3, each H atom in hydrogen peroxide (H2O2) has an oxidation number of +1, while each O atom has an oxidation number of −1. We can use rule 4 to determine oxidation numbers for the atoms in SO2. Each O atom has an oxidation number of −2; for the sum of the oxidation numbers to equal the charge on the species (which is zero), the S atom is assigned an oxidation number of +4. Does this mean that the sulfur atom has a 4+ charge on it? No, it means only that the S atom is assigned a +4 oxidation number by our rules of apportioning electrons among the atoms in a compound.
Assign oxidation numbers to the atoms in each substance.
Solution
Ca(NO3)2 can be separated into two parts: the Ca2+ ion and the NO3− ion. Considering these separately, the Ca2+ ion has an oxidation number of +2 by rule 2. Now consider the NO3− ion. Oxygen is assigned an oxidation number of −2, and there are three of them. According to rule 4, the sum of the oxidation numbers on all atoms must equal the charge on the species, so we have the simple algebraic equation
x + 3(−2) = −1where x is the oxidation number of the N atom and the −1 represents the charge on the species. Evaluating for x,
x + (−6) = −1 x = +5Thus the oxidation number on the N atom in the NO3− ion is +5.
Test Yourself
Assign oxidation numbers to the atoms in H3PO4.
Answer
H: +1; O: −2; P: +5
All redox reactions occur with a simultaneous change in the oxidation numbers of some atoms. At least two elements must change their oxidation numbers. When an oxidation number of an atom is increased in the course of a redox reaction, that atom is being oxidized. When an oxidation number of an atom is decreased in the course of a redox reaction, that atom is being reduced. Thus oxidation and reduction can also be defined in terms of increasing or decreasing oxidation numbers, respectively.
Identify what is being oxidized and reduced in this redox reaction.
2Na + Br2 → 2NaBrSolution
Both reactants are the elemental forms of their atoms, so the Na and Br atoms have oxidation numbers of 0. In the ionic product, the Na+ ions have an oxidation number of +1, while the Br− ions have an oxidation number of −1.
Sodium is increasing its oxidation number from 0 to +1, so it is being oxidized; bromine is decreasing its oxidation number from 0 to −1, so it is being reduced:
Because oxidation numbers are changing, this is a redox reaction. The total number of electrons being lost by sodium (two, one lost from each Na atom) is gained by bromine (two, one gained for each Br atom).
Test Yourself
Identify what is being oxidized and reduced in this redox reaction.
C + O2 → CO2Answer
C is being oxidized from 0 to +4; O is being reduced from 0 to −2.
Oxidation reactions can become quite complex, as attested by the following redox reaction:
To demonstrate that this is a redox reaction, the oxidation numbers of the species being oxidized and reduced are listed; can you determine what is being oxidized and what is being reduced? This is also an example of a net ionic reaction; spectator ions that do not change oxidation numbers are not displayed in the equation. Eventually, we will need to learn techniques for writing correct (i.e., balanced) redox reactions.
Iron is an essential mineral in our diet; iron-containing compounds like the heme protein in hemoglobin could not function without it. Most biological iron has the form of the Fe2+ ion; iron with other oxidation numbers is almost inconsequential in human biology (although the body does contain an enzyme to reduce Fe3+ to Fe2+, so Fe3+ must have some biological significance, albeit minor). To ensure that we ingest enough iron, many foods are enriched with iron. Although Fe2+ compounds are the most logical substances to use, some foods—bread and breakfast cereals are the most well-known examples—use “reduced iron” as an ingredient. Reduced iron is simply iron metal; iron is added as a fine metallic powder. The metallic iron is oxidized to Fe2+ in the digestive system and then absorbed by the body, but the question remains: Why are we ingesting metallic iron? Why not just use Fe2+ salts as an additive?
Many prepared foods list reduced iron in their ingredients list.
Although it is difficult to establish conclusive reasons, a search of scientific and medical literature suggests a few reasons. One reason is that fine iron filings do not affect the taste of the product. The size of the iron powder (several dozen micrometers) is not noticeable when chewing iron-supplemented foods, and the tongue does not detect any changes in flavor that can be detected when using Fe2+ salts. Fe2+ compounds can affect other properties of foodstuffs during preparation and cooking, like dough pliability, yeast growth, and color. Finally, of the common iron substances that might be used, metallic iron is the least expensive. These factors appear to be among the reasons why metallic iron is the supplement of choice in some foods.
Is this reaction a redox reaction? Explain your answer.
2K(s) + Br2(ℓ) → 2KBr(s)Is this reaction a redox reaction? Explain your answer.
2NaCl(aq) + Pb(NO3)2(aq) → 2NaNO3(aq) + PbCl2(s)Which substance loses electrons and which substance gains electrons in this reaction?
2Mg(s) + O2(g) → 2MgOWhich substance loses electrons and which substance gains electrons in this reaction?
16Fe(s) + 3S8(s) → 8Fe2S3(s)Which substance is oxidized and which substance is reduced in this reaction?
2Li(s) + O2(g) → Li2O2(s)Which substance is oxidized and which substance is reduced in this reaction?
2Fe(s) + 3I2(s) → 2FeI3(s)What are two different definitions of oxidation?
What are two different definitions of reduction?
Assign oxidation numbers to the atoms in each substance.
Assign oxidation numbers to the atoms in each substance.
Assign oxidation numbers to the atoms in each substance.
Assign oxidation numbers to the atoms in each substance.
Assign oxidation numbers to the atoms in each substance.
Assign oxidation numbers to the atoms in each substance.
Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.
2NO + Cl2 → 2NOClIdentify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.
Sr + SO3 → SrSO3Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.
2KrF2 + 2H2O → 2Kr + 4HF + O2Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.
SO3 + SCl2 → SOCl2 + SO2Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.
2Rb + MgCl2 → 2RbCl + MgIdentify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms.
2C8H18 + 25O2 → 16CO2 + 18H2Oyes because oxidation numbers are changing
lose: Mg; gain: O
oxidized: Li; reduced: O
increase in oxidation number; loss of electrons
oxidized: N; reduced: Cl
oxidized: O; reduced: Kr
oxidized: Rb; reduced: Mg