23.2: Oxidation – Reduction
Lesson Objectives
The student will:
 assign the correct oxidation number to any element in a compound or ion.
 identify the substance being oxidized, the substance being reduced, the oxidizing agent, and the reducing agent in an oxidationreduction equation.
Vocabulary
 oxidation
 oxidation number
 oxidizing agent
 redox reaction
 reducing agent
 reduction
Introduction
Many important chemical reactions involve the exchange of one or more electrons. The stoichiometric calculations that chemists make on chemical reactions require a balanced equation. The “inspection” method for balancing equations works well and quickly for many reactions, but when a more complex electron exchange is involved, a new method for equation balancing is needed.
Definition of Oxidation and Reduction
After Lavoisier, a substance was said to be oxidized when it reacted with oxygen. A reaction with oxygen was called oxidation. Today, the words “oxidized” and “oxidation” are still used for those situations, but these words have now acquired a much broader, second meaning. The broader sense of oxidation is defined as losing electrons. When a substance reacts with oxygen, it almost always loses electrons to the oxygen, so we are simply extending the term oxidation to mean losing electrons, whether it is to oxygen or to any other substance.
The other half of this process, the gaining of electrons, also needs a name. When an atom or an ion gains electrons, the positive charge on the particle is decreased. For example, if a neutral sulfur atom (charge of 0) gains two electrons, its charge becomes \begin{align*}2\end{align*}, and if an \begin{align*}Fe^{3+}\end{align*} ion gains an electron, its charge changes from \begin{align*}+3\end{align*} to \begin{align*}+2\end{align*}. In both cases, the charge on the particle is reduced by the gain of electrons. Reduction means the gain of electrons. In chemical systems, oxidation and reduction must occur simultaneously, so the number of electrons during oxidation must be the same as the number of electrons gained during reduction. In oxidationreduction reactions, electrons are transferred from one substance to another. Here’s an example of an oxidation–reduction reaction:


 \begin{align*}2 \ \mathrm{Ag}^+_{(aq)} + \mathrm{Cu}_{(s)} \rightarrow 2 \ \mathrm{Ag}_{(s)} + \mathrm{Cu}^{2+}_{(aq)} \end{align*}

In this reaction, the silver ions are gaining electrons to become silver atoms. Therefore, the silver ions are being reduced. The copper atoms are losing electrons to become copper ions and are being oxidized. Whenever a chemical reaction involves electrons being transferred from one substance to another, the reaction is an oxidation–reduction reaction. An oxidationreduction reaction can also be referred to as a redox reaction for short.
Oxidizing and Reducing Agents
When a substance is oxidized, it loses electrons. In chemical reactions, this requires that another substance take on those electrons and be reduced. Therefore, when a substance undergoes oxidation, it causes another substance to be reduced. The substance that caused another substance to be reduced is called a reducing agent. As you can see, the substance undergoing oxidation and the reducing agent are the same substance.
Similarly, when a substance gains electrons, it is reduced. By gaining electrons, it is causing some other substance to give up those electrons. Therefore, by undergoing reduction, the substance is causing another substance to be oxidized and is called an oxidizing agent. Again, the substance undergoing reduction and the oxidizing agent are the same substance.


 \begin{align*}2 \ \mathrm{Ag}^{+}_{(aq)} + \mathrm{Cu}_{(s)} \rightarrow 2 \ \mathrm{Ag}_{(s)} + \mathrm{Cu}^{2+}_{(aq)}\end{align*}

In the oxidationreduction reaction above, silver ions are being reduced and are the oxidizing agent. Similarly, copper atoms are being oxidized and are the reducing agent. These substances are always on the reactant side of the equation.
Oxidation Numbers
In order to balance oxidation and reduction reaction equations, it is necessary to have a bookkeeping system to keep track of the transferred electrons. The bookkeeping system chemists use to keep track of electrons in oxidationreduction reactions is called oxidation numbers. The assignment of oxidation numbers to all the atoms or ions in a reaction follows a set of rules. For the most part, these rules will have the oxidation number of a particle be the number of electrons the atom has gained or lost from its elemental state. For example, for a \begin{align*}\mathrm{Ca}^{2+}\end{align*} ion, the calcium ion has clearly lost two electrons from its elemental form, so its oxidation number is \begin{align*}+2\end{align*}. Similarly, it is clear that a fluoride ion, \begin{align*}\mathrm{F}^\end{align*}, has gained one electron from its elemental state and has an oxidation number of \begin{align*}1\end{align*}.
The first rule for assigning oxidation numbers is for substances in their elemental form. Substances in elemental form have oxidation numbers of zero. It is clear that substances in elemental form have neither gained nor lost any electrons from their elemental state (Table below).
Substance  Oxidation Number 

\begin{align*}\mathrm{Fe}\end{align*}  \begin{align*}0\end{align*} 
\begin{align*}\mathrm{Ag}\end{align*}  \begin{align*}0\end{align*} 
\begin{align*}\mathrm{H}_2 \end{align*}  \begin{align*}0\end{align*} 
\begin{align*}\mathrm{O}_2 \end{align*}  \begin{align*}0\end{align*} 
\begin{align*}\mathrm{Cl}_2 \end{align*}  \begin{align*}0\end{align*} 
\begin{align*}\mathrm{S}_8 \end{align*}  \begin{align*}0\end{align*} 
The second rule for assigning oxidation numbers is for monatomic ions. For monatomic ions, the oxidation number is the same as the charge on the ion. Again, it should be apparent that the charge on the ion is an indication of how many electrons have been gained or lost. Table below shows examples of oxidation numbers for monatomic ions.
Substance  Oxidation Number 

\begin{align*}\mathrm{Ca}^{2+}\end{align*}  \begin{align*}+2\end{align*} 
\begin{align*}\mathrm{Ag}^+\end{align*}  \begin{align*}+1\end{align*} 
\begin{align*}\mathrm{Fe}^{3+}\end{align*}  \begin{align*}+3\end{align*} 
\begin{align*}\mathrm{S}^{2}\end{align*}  \begin{align*}2\end{align*} 
\begin{align*}\mathrm{F}^\end{align*}  \begin{align*}1\end{align*} 
\begin{align*}\mathrm{Na}^+\end{align*}  \begin{align*}+1\end{align*} 
The third rule is for the atoms of family IA, the alkali metals, in compounds. Alkali metals always lose their single valence electron when they combine. Therefore, for IA metals, the oxidation number in compounds is \begin{align*}+1\end{align*} (see Table below for examples).
Substance  Oxidation Number for the Alkali Metal 

\begin{align*}\mathrm{NaCl}\end{align*}  \begin{align*}+1\end{align*} 
\begin{align*}\mathrm{Na}_2\mathrm{S}\end{align*}  \begin{align*}+1\end{align*} 
\begin{align*}\mathrm{Na}_2\mathrm{SO}_4 \end{align*}  \begin{align*}+1\end{align*} 
\begin{align*}\mathrm{NaC}_2\mathrm{H}_3\mathrm{O}_2\end{align*}  \begin{align*}+1\end{align*} 
\begin{align*}\mathrm{KClO}_4\end{align*}  \begin{align*}+1\end{align*} 
The fourth rule is for the atoms of family IIA, the alkali earth metals, in compounds. Alkali earth metals always lose both of their valence electrons when they combine chemically, so for IIA metals, the oxidation number in compounds is \begin{align*}+2\end{align*} (see Table below for examples).
Substance  Oxidation Number for the Alkali Earth Metal 

\begin{align*}\mathrm{MgF}_2 \end{align*}  \begin{align*}+2\end{align*} 
\begin{align*}\mathrm{Ca(NO}_3 )_2 \end{align*}  \begin{align*}+2\end{align*} 
\begin{align*}\mathrm{Mg(OH)}_2 \end{align*}  \begin{align*}+2\end{align*} 
\begin{align*}\mathrm{CaO}\end{align*}  \begin{align*}+2\end{align*} 
The fifth rule concerns hydrogen atoms when they are in compounds. In the great majority of compounds that hydrogen forms, it either completely or at least partially loses that electron. In compounds where hydrogen is the more electropositive atom, the oxidation number for hydrogen is \begin{align*}+1\end{align*} (see Table below for examples).
Substance  Oxidation Number for Hydrogen 

\begin{align*}\mathrm{HF}\end{align*}  \begin{align*}+1\end{align*} 
\begin{align*}\mathrm{H}_2\mathrm{O}\end{align*}  \begin{align*}+1\end{align*} 
\begin{align*}\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2\end{align*}  \begin{align*}+1\end{align*} 
\begin{align*}\mathrm{Mg(OH)}_2 \end{align*}  \begin{align*}+1\end{align*} 
\begin{align*}\mathrm{CH}_4 \end{align*}  \begin{align*}+1\end{align*} 
There is, however, an exception to this rule for hydrogen. It is possible for hydrogen to form compounds with some metals that are even more electropositive than hydrogen. In these cases, hydrogen becomes an electron acceptor instead of an electron donor. Active metals lose or partially lose their valence electrons to hydrogen. Since hydrogen is acting as the more electronegative element in these compounds, the compounds are named hydrides (see examples in Table below). In hydrides, the oxidation number of hydrogen is \begin{align*}1\end{align*}.
Substance  Oxidation Number for Hydrogen 

\begin{align*}\mathrm{LiH}\end{align*} (lithium hydride)  \begin{align*}1\end{align*} 
\begin{align*}\mathrm{NaH}\end{align*} (sodium hydride)  \begin{align*}1\end{align*} 
\begin{align*}\mathrm{MgH}_2 \end{align*} (magnesium hydride)  \begin{align*}1\end{align*} 
The sixth rule is about the oxidation number of oxygen in compounds. Oxygen is a very electronegative element that draws two electrons completely or partially from a bonding element in almost all of its compounds. Therefore, the oxidation number for oxygen in compounds is almost always \begin{align*}2\end{align*} (see examples in Table below).
Substance  Oxidation Number for Oxygen 

\begin{align*}\mathrm{H}_2\mathrm{O}\end{align*}  \begin{align*}2\end{align*} 
\begin{align*}\mathrm{MgO}\end{align*}  \begin{align*}2\end{align*} 
\begin{align*}\mathrm{H}_2\mathrm{SO}_4 \end{align*}  \begin{align*}2\end{align*} 
\begin{align*}\mathrm{Na}_2\mathrm{Cr}_2\mathrm{O}_7 \end{align*}  \begin{align*}2\end{align*} 
\begin{align*}\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 \end{align*}  \begin{align*}2\end{align*} 
Like hydrogen, there is an exception to the rule for oxygen. In a group of compounds named “peroxides” (hydrogen peroxide \begin{align*}\mathrm{H}_2\mathrm{O}_2\end{align*}, sodium peroxide \begin{align*}\mathrm{Na}_2\mathrm{O}_2\end{align*}, etc.), each of the oxygen atoms shares a bond with the other oxygen atom. Therefore, the oxygen atoms only accept one electron from the other element, so oxygen has an oxidation number of \begin{align*}1\end{align*} in peroxides (see examples in Table below).
Substance  Oxidation Number for Oxygen 

\begin{align*}\mathrm{H}_2\mathrm{O}_2 \end{align*}  \begin{align*}1\end{align*} 
\begin{align*}\mathrm{Na}_2\mathrm{O}_2 \end{align*}  \begin{align*}1\end{align*} 
Some other elements exhibit only one possible gain, loss, or sharing of electrons when they form compounds. Aluminum, for example, always loses or partially loses three electrons when it forms compounds, so its oxidation number in compounds is \begin{align*}+3\end{align*}. Zinc always loses or partially loses two electrons when it combines, so its oxidation number in compounds is \begin{align*}+2\end{align*}. The halogens (family VIIA) always gain one electron in binary compounds and would have an oxidation number of \begin{align*}1\end{align*} in these compounds. Some of the halogen atoms form compounds where there are three elements, and one of them is oxygen (\begin{align*}\mathrm{HClO}_3\end{align*}, for example). In these compounds, the halogen atom is almost never \begin{align*}1\end{align*}.
There will be quite a few atoms whose oxidation number may be different in different compounds. For these, you will have to calculate their oxidation numbers. To allow you make such calculations, there is a general rule that the sum of the oxidation numbers of all the atoms in a compound must be zero, and the sum of all the oxidation numbers of the atoms in a polyatomic ion must equal the charge on the ion.
Example:
What is the oxidation number of sulfur in \begin{align*}\mathrm{Na}_2\mathrm{SO}_4\end{align*} ?
Solution:
We have three pieces of information that will allow us to calculate the oxidation number of sulfur. We know that the sum of the oxidation numbers of all the atoms will equal zero.


 \begin{align*}2(\mathrm{Na}) + 1(\mathrm{S}) + 4(\mathrm{O}) = 0\end{align*}

We also know the oxidation number of sodium from rule \begin{align*}3\end{align*} is \begin{align*}+1\end{align*} and the oxidation number of oxygen from rule \begin{align*}6\end{align*} is \begin{align*}2\end{align*}. We simply plug these values into the equation and solve for the oxidation number of sulfur.


 \begin{align*}2(1) + 1(x) + 4(2) = 0\end{align*} so \begin{align*}x = +6\end{align*}

Example:
What is the oxidation number of chromium in \begin{align*}\mathrm{K}_2\mathrm{Cr}_2 \mathrm{O}_7 \end{align*}?
Solution:


 \begin{align*}2(\mathrm{K}) + 2(\mathrm{Cr}) + 7(\mathrm{O}) = 0\end{align*}



 \begin{align*}2(+1) + 2(x) + 7(2) = 0\end{align*}



 so \begin{align*}x = +6\end{align*}

Example:
What is the oxidation number of nitrogen in the nitrate ion, \begin{align*}\mathrm{NO}_3^\end{align*}?
Solution:


 \begin{align*}1(\mathrm{N}) + 3(\mathrm{O}) = 1\end{align*}



 \begin{align*}1(x) + 3(2) = 1\end{align*}



 so \begin{align*}x = +5\end{align*}

Example:
What is the oxidation number of phosphorus in the phosphate ion, \begin{align*}\mathrm{PO}_4^{3}\end{align*}?
Solution:


 \begin{align*}1(\mathrm{P}) + 4(\mathrm{O}) = 3\end{align*}



 \begin{align*}1(x) + 4(2) = 3\end{align*}



 so \begin{align*}x = +5\end{align*}

Example:
What is the oxidation number of iron in \begin{align*}\mathrm{Fe}_2\mathrm{O}_3\end{align*}?
Solution:


 \begin{align*}2(\mathrm{Fe}) + 3(\mathrm{O}) = 0\end{align*}



 \begin{align*}2(x) + 3(2) = 0\end{align*}



 so \begin{align*}x = +3\end{align*}

Example:
What is the oxidation number of iron in \begin{align*}\mathrm{Fe}_3\mathrm{O}_4 \end{align*}?
Solution:
This example was chosen specifically to make a point.


 \begin{align*}3(\mathrm{Fe}) + 4(\mathrm{O}) = 0\end{align*}



 \begin{align*}3(x) + 4(2) = 0\end{align*}



 so \begin{align*}x = \frac{8}{3}\end{align*}

In this case, we get an oxidation number that is not a whole number. Since this number supposedly represents the number of electrons gained or lost from the elemental state, we should feel uncomfortable about an oxidation number of \begin{align*}\frac{8} {3}\end{align*}. We know that an atom did not lose a fraction of an electron. The reason that the oxidation number appears to be fractional is because some of the iron atoms in \begin{align*}\mathrm{Fe}_3\mathrm{O}_4\end{align*} lost 3 electrons, and some lost 2. Another way to represent this molecule is as FeO•Fe_{2}O_{3}. In this case, it is easy to see that the Fe attached to one O is in a +2 oxidation state, whereas the \begin{align*}\mathrm{Fe}\end{align*} in the \begin{align*}\mathrm{Fe}_2\mathrm{O}_3\end{align*} is in the +3 oxidation state.
Lesson Summary
 Oxidation is the loss of electrons or a gain in oxidation number.
 Reduction is the gain of electrons or a decrease in oxidation number.
 The oxidizing agent is the substance causing oxidation and is itself reduced.
 The reducing agent is the substance causing reduction and is itself oxidized.
 Oxidation and reduction always occur simultaneously.
 A reaction involving the gain and loss of electrons is called an oxidationreduction or redox reaction.
 We use a bookkeeping system called oxidation numbers to keep track of the electrons gained and lost during a redox reaction.
Further Reading / Supplemental Links
This website offers a short animated video on electron transfer during oxidationreduction.
This web site offers more information on oxidationreduction and other aspects of electrochemistry.
Review Questions
 Indicate the oxidation numbers for each of the following atoms.
 \begin{align*}\mathrm{Mn}^{2+}\end{align*}
 \begin{align*}\mathrm{Al}\end{align*}
 \begin{align*}\mathrm{Al}\end{align*} in \begin{align*}\mathrm{Al}_2\mathrm{O}_3 \end{align*}
 \begin{align*}\mathrm{Br}\end{align*} in \begin{align*}\mathrm{NaBr}\end{align*}
 \begin{align*}\mathrm{Fe}\end{align*} in \begin{align*}\mathrm{Fe}_2\mathrm{O}_3 \end{align*}
 arsenic in \begin{align*}\mathrm{AsO}_4^{3}\end{align*}
 chlorine in \begin{align*}\mathrm{ClO}_4^\end{align*}
 sulfur in \begin{align*}\mathrm{H}_2\mathrm{SO}_3\end{align*}
 In the following reaction, identify the element that is being oxidized and the element that is being reduced: \begin{align*}\mathrm{MnO}_2 + 4 \ \mathrm{HCl} \rightarrow \mathrm{MnCl}_2 + \mathrm{Cl}_2 + 2 \ \mathrm{H}_2\mathrm{O}\end{align*}.
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