We described the defining characteristics of oxidation–reduction, or redox, reactions. Most of the reactions we considered there were relatively simple, and balancing them was straightforward. When oxidation–reduction reactions occur in aqueous solution, however, the equations are more complex and can be more difficult to balance by inspection. Because a balanced chemical equation is the most important prerequisite for solving any stoichiometry problem, we need a method for balancing oxidation–reduction reactions in aqueous solution that is generally applicable. One such method usesoxidation states, and a second is referred to as the half-reaction method.
Balancing Redox Equations Using Oxidation States
To balance a redox equation using the oxidation state method, we conceptually separate the overall reaction into two parts: an oxidation—in which the atoms of one element lose electrons—and a reduction—in which the atoms of one element gain electrons. Consider, for example, the reaction of Cr2+(aq) with manganese dioxide (MnO2) in the presence of dilute acid. Equation \(\ref{20.2.1}\) is the net ionic equation for this reaction before balancing; the oxidation state of each element in each species has been assigned using the procedure described previously (in red above each element):
\[\overset{\color{red}{+2}}{Cr^{2+}} ( aq ) + \overset{\color{red}{+4}}{Mn} \overset{\color{red}{-2}}{O_2} ( aq ) + \overset{\color{red}{+1}} {H^{+}} ( aq ) \rightarrow \overset{\color{red}{+3}}{Cr^{3+}} ( aq ) + \overset{\color{red}{+2}}{Mn^{2+}}( aq ) + \overset{\color{red}{+1}} {H_2} \overset{\color{red}{-2}} {O} (l) \label{20.2.1} \]
Notice that chromium is oxidized from the +2 to the +3 oxidation state, while manganese is reduced from the +4 to the +2 oxidation state. We can write an equation for this reaction that shows only the atoms that are oxidized and reduced (ignoring the oxygen and hydrogen atoms):
\[\underbrace{Cr^{2+} \rightarrow Cr^{3+} + e^-}_{\text{oxidation with 1 electron lost}} \label{20.2.3}\]
and the reduction as
\[\underbrace{Mn^{4+} + 2e^- \rightarrow Mn^{2+}}_{\text{reduction with 2 electrons gained}} \label{20.2.4}\]
For the overall chemical equation to be balanced, the number of electrons lost by the reductant must equal the number gained by the oxidant. We must therefore multiply the oxidation and the reduction equations by appropriate coefficients to give us the same number of electrons in both. In this example, we must multiply the oxidation (Equation \ref{20.2.3}) by 2 to give
We then add the equations for the oxidation and the reduction and cancel the electrons on both sides of the equation, using the actual chemical forms of the reactants and products:
In a balanced redox reaction, the number of electrons lost by the reductant equals the number of electrons gained by the oxidant.
Although the electrons cancel and the metal atoms are balanced, the total charge on the left side of Equation \ref{20.2.7b} (+4) does not equal the charge on the right side (+8). Because the reaction is carried out in the presence of aqueous acid, we can add \(H^+\) as necessary to either side of the equation to balance the charge. By the same token, if the reaction were carried out in the presence of aqueous base, we could balance the charge by adding \(OH^−\) as necessary to either side of the equation to balance the charges.
In this case, adding four \(H^{+}\) ions to the left side of Equation \ref{20.2.7b} to give
Although the charges are now balanced in Equation \ref{20.2.8}, we have two oxygen atoms on the left side of the equation and none on the right. We can balance the oxygen atoms without affecting the overall charge balance by adding \(H_2O\) as necessary to either side of the equation. Here, we need to add two \(H_2O\) molecules to the right side of Equation \ref{20.2.8}:
Although we did not explicitly balance the hydrogen atoms, we can see by inspection that the overall chemical equation is now balanced with respect to all atoms and charge. All that remains is to check to make sure that we have not made a mistake. This procedure for balancing reactions is summarized below and illustrated in Example \(\PageIndex{1}\) below.
Reactions in basic solutions are balanced in exactly the same manner. To make sure you understand the procedure, consider Example \(\PageIndex{2}\).
As suggested in Examples \(\PageIndex{1}\) and \(\PageIndex{2}\), a wide variety of redox reactions are possible in aqueous solutions. The identity of the products obtained from a given set of reactants often depends on both the ratio of oxidant to reductant and whether the reaction is carried out in acidic or basic solution, which is one reason it can be difficult to predict the outcome of a reaction. Because oxidation–reduction reactions in solution are so common and so important, however, chemists have developed two general guidelines for predicting whether a redox reaction will occur and the identity of the products:
Compounds of elements in high oxidation states (such as \(\ce{ClO4^{−}}\), \(\ce{NO3^{−}}\), \(\ce{MnO4^{−}}\), \(\ce{Cr2O7^{2−}}\), and \(\ce{UF6}\)) tend to act as oxidants and become reduced in chemical reactions.
Compounds of elements in low oxidation states (such as \(\ce{CH4}\), \(\ce{NH3}\), \(\ce{H2S}\), and \(\ce{HI}\)) tend to act as reductants and become oxidized in chemical reactions.
When an aqueous solution of a compound that contains an element in a high oxidation state is mixed with an aqueous solution of a compound that contains an element in a low oxidation state, an oxidation–reduction reaction is likely to occur.
Species in high oxidation states act as oxidants, whereas species in low oxidation states act as reductants.
Summary
Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reduction equation. In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the oxidation state method, in which the overall reaction is separated into an oxidation equation and a reduction equation.