9.1: Balancing Oxidation-Reduction Reactions

Electricity refers to a number of phenomena associated with the presence and flow of electric charge. Electricity includes such diverse things as lightning, static electricity, the current generated by a battery as it discharges, and many other influences on our daily lives. The flow or movement of charge is an electric current (Figure 9.1.1). Electrons or ions may carry the charge. The elementary unit of charge is the charge of a proton, which is equal in magnitude to the charge of an electron. The SI unit of charge is the coulomb (C) and the charge of a proton is 1.602 × 10⁻¹⁹ C. The presence of an electric charge generates an electric field. Electric current is the rate of flow of charge.

The SI unit for electrical current is the SI base unit called the ampere (A), which is a flow rate of 1 coulomb of charge per second (1 A = 1 C/s). An electric current flows in a path, called an electric circuit. In most chemical systems, it is necessary to maintain a closed path for current to flow. The flow of charge is generated by an electrical potential difference, or potential, between two points in the circuit. Electrical potential is the ability of the electric field to do work on the charge. The SI unit of electrical potential is the volt (V). When 1 coulomb of charge moves through a potential difference of 1 volt, it gains or loses 1 joule (J) of energy. Table 9.1.1 summarizes some of this information about electricity.

Electrochemistry studies oxidation-reduction reactions, which were first discussed in an earlier chapter, where we learned that oxidation was the loss of electrons and reduction was the gain of electrons. The reactions discussed tended to be rather simple, and conservation of mass (atom counting by type) and deriving a correctly balanced chemical equation were relatively simple. In this section, we will concentrate on the half-reaction method for balancing oxidation-reduction reactions. The use of half-reactions is important partly for balancing more complicated reactions and partly because many aspects of electrochemistry are easier to discuss in terms of half-reactions. There are alternate methods of balancing these reactions; however, there are no good alternatives to half-reactions for discussing what is occurring in many systems. The half-reaction method splits oxidation-reduction reactions into their oxidation “half” and reduction “half” to make finding the overall equation easier.

Electrochemical reactions frequently occur in solutions, which could be acidic, basic, or neutral. When balancing oxidation-reduction reactions, the nature of the solution may be important. It helps to see this in an actual problem. Consider the following unbalanced oxidation-reduction reaction in acidic solution:

[latex]\ce{MnO4-}(aq)+\ce{Fe^2+}(aq)⟶\ce{Mn^2+}(aq)+\ce{Fe^3+}(aq) \nonumber[/latex]

We can start by collecting the species we have so far into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction. Each of these half-reactions contain the same element in two different oxidation states. The Fe²⁺ has lost an electron to become Fe³⁺; therefore, the iron underwent oxidation. The reduction is not as obvious; however, the manganese gained five electrons to change from Mn⁷⁺ to Mn²⁺.

[latex]\begin{align*}
&\textrm{oxidation (unbalanced): }\ce{Fe^2+}(aq)⟶\ce{Fe^3+}(aq)\\
&\textrm{reduction (unbalanced): }\ce{MnO4-}(aq)⟶\ce{Mn^2+}(aq)
\end{align*} \nonumber[/latex]

In acidic solution, there are hydrogen ions present, which are often useful in balancing half-reactions. It may be necessary to use the hydrogen ions directly or as a reactant that may react with oxygen to generate water. Hydrogen ions are very important in acidic solutions where the reactants or products contain hydrogen and/or oxygen. In this example, the oxidation half-reaction involves neither hydrogen nor oxygen, so hydrogen ions are not necessary to the balancing. However, the reduction half-reaction does involve oxygen. It is necessary to use hydrogen ions to convert this oxygen to water.

[latex]\textrm{charge not balanced: }\ce{MnO4-}(aq)+\ce{8H+}(aq)⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l) \nonumber[/latex]

The situation is different in basic solution because the hydrogen ion concentration is lower and the hydroxide ion concentration is higher. After finishing this example, we will examine how basic solutions differ from acidic solutions. A neutral solution may be treated as acidic or basic, though treating it as acidic is usually easier.

The iron atoms in the oxidation half-reaction are balanced (mass balance); however, the charge is unbalanced, since the charges on the ions are not equal. It is necessary to use electrons to balance the charge. The way to balance the charge is by adding electrons to one side of the equation. Adding a single electron on the right side gives a balanced oxidation half-reaction:

[latex]\textrm{oxidation (balanced): }\ce{Fe^2+}(aq)⟶\ce{Fe^3+}(aq)+\ce{e-} \nonumber[/latex]

You should check the half-reaction for the number of each atom type and the total charge on each side of the equation. The charges include the actual charges of the ions times the number of ions and the charge on an electron times the number of electrons.

[latex]\begin{align*}
&\textrm{Fe: } \mathrm{Does\:(1×1)=(1×1)?\:Yes.}\\
&\textrm{Charge: } \mathrm{Does\:[1×(+2)]=[1×(+3)+1×(−1)]?\:Yes.}
\end{align*} \nonumber[/latex]

If the atoms and charges balance, the half-reaction is balanced. In oxidation half-reactions, electrons appear as products (on the right). As discussed in the earlier chapter, since iron underwent oxidation, iron is the reducing agent.

Now return to the reduction half-reaction equation:

[latex]\textrm{reduction (unbalanced): }\ce{MnO4-}(aq)+\ce{8H+}(aq)⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l) \nonumber[/latex]

The atoms are balanced (mass balance), so it is now necessary to check for charge balance. The total charge on the left of the reaction arrow is [(−1) × (1) + (8) × (+1)], or +7, while the total charge on the right side is [(1) × (+2) + (4) × (0)], or +2. The difference between +7 and +2 is five; therefore, it is necessary to add five electrons to the left side to achieve charge balance.

[latex]\textrm{Reduction (balanced): }\ce{MnO4-}(aq)+\ce{8H+}(aq)+\ce{5e-}⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l) \nonumber[/latex]

You should check this half-reaction for each atom type and for the charge, as well:

[latex]\begin{align*}
&\textrm{Mn: }\mathrm{Does\:(1×1)=(1×1)?\:Yes.}\\
&\textrm{H: }\mathrm{Does\:(8×1) =(4×2)?\:Yes.}\\
&\textrm{O: }\mathrm{Does\:(1×4) =(4×1)?\:Yes.}\\
&\textrm{Charge: }\mathrm{Does\:[1×(−1)+8×(+1)+5×(−1)]=[1×(+2)]?\:Yes.}
\end{align*} \nonumber[/latex]

Now that this half-reaction is balanced, it is easy to see it involves reduction because electrons were gained when [latex]\ce{MnO4-}[/latex] was reduced to Mn²⁺. In all reduction half-reactions, electrons appear as reactants (on the left side). As discussed in the earlier chapter, the species that was reduced, [latex]\ce{MnO4-}[/latex] in this case, is also called the oxidizing agent. We now have two balanced half-reactions.

[latex]\begin{align*}
&\textrm{oxidation: }\ce{Fe^2+}(aq)⟶\ce{Fe^3+}(aq)+\ce{e-}\\
&\textrm{reduction: }\ce{MnO4-}(aq)+\ce{8H+}(aq)+\ce{5e-}⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l)
\end{align*} \nonumber[/latex]

It is now necessary to combine the two halves to produce a whole reaction. The key to combining the half-reactions is the electrons. The electrons lost during oxidation must go somewhere. These electrons go to cause reduction. The number of electrons transferred from the oxidation half-reaction to the reduction half-reaction must be equal. There can be no missing or excess electrons. In this example, the oxidation half-reaction generates one electron, while the reduction half-reaction requires five. The lowest common multiple of one and five is five; therefore, it is necessary to multiply every term in the oxidation half-reaction by five and every term in the reduction half-reaction by one. (In this case, the multiplication of the reduction half-reaction generates no change; however, this will not always be the case.) The multiplication of the two half-reactions by the appropriate factor followed by addition of the two halves gives

[latex]\begin{align*}
&\textrm{oxidation: }5×(\ce{Fe^2+}(aq)⟶\ce{Fe^3+}(aq)+\ce{e-})\\
&\underline{\textrm{reduction: }\ce{MnO4-}(aq)+\ce{8H+}(aq)+\ce{5e-}⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l)}\\
&\textrm{overall: }\ce{5Fe^2+}(aq)+\ce{MnO4-}(aq)+\ce{8H+}(aq)⟶\ce{5Fe^3+}(aq)+\ce{Mn^2+}(aq)+\ce{4H2O}(l)
\end{align*} \nonumber[/latex]

The electrons do not appear in the final answer because the oxidation electrons are the same electrons as the reduction electrons and they “cancel.” Carefully check each side of the overall equation to verify everything was combined correctly:

[latex]\begin{align*}
&\textrm{Fe: }\mathrm{Does\:(5×1)=(5×1)?\:Yes.}\\
&\textrm{Mn: }\mathrm{Does\:(1×1)=(1×1)?\:Yes.}\\
&\textrm{H: }\mathrm{Does\:(8×1)=(4×2)?\:Yes.}\\
&\textrm{O: }\mathrm{Does\:(1×4)=(4×1)?\:Yes.}\\
&\textrm{Charge: }\mathrm{Does\:[5×(+2)+1×(−1)+8×(+1)]=[5×(+3)+1×(+2)]?\:Yes.}
\end{align*} \nonumber[/latex]

Everything checks, so this is the overall equation in acidic solution. If something does not check, the most common error occurs during the multiplication of the individual half-reactions.

Now suppose we wanted the solution to be basic. Recall that basic solutions have excess hydroxide ions. Some of these hydroxide ions will react with hydrogen ions to produce water. The simplest way to generate the balanced overall equation in basic solution is to start with the balanced equation in acidic solution, then “convert” it to the equation for basic solution. However, it is necessary to exercise caution when doing this, as many reactants behave differently under basic conditions and many metal ions will precipitate as the metal hydroxide. We just produced the following reaction, which we want to change to a basic reaction:

[latex]\ce{5Fe^2+}(aq)+\ce{MnO4-}(aq)+\ce{8H+}(aq)⟶\ce{5Fe^3+}(aq)+\ce{Mn^2+}(aq)+\ce{4H2O}(l) \nonumber[/latex]

However, under basic conditions, [latex]\ce{MnO4-}[/latex] normally reduces to MnO₂ and iron will be present as either Fe(OH)₂ or Fe(OH)₃. For these reasons, under basic conditions, this reaction will be

[latex]\ce{3Fe(OH)2}(s)+\ce{MnO4-}(aq)+\ce{2H2O}(l)⟶\ce{3Fe(OH)3}(s)+\ce{MnO2}(s)+\ce{OH-}(aq) \nonumber[/latex]

(Under very basic conditions [latex]\ce{MnO4-}[/latex] will reduce to [latex]\ce{MnO4^2-}[/latex], instead of MnO₂.)

It is still possible to balance any oxidation-reduction reaction as an acidic reaction and then, when necessary, convert the equation to a basic reaction. This will work if the acidic and basic reactants and products are the same or if the basic reactants and products are used before the conversion from acidic or basic. There are very few examples in which the acidic and basic reactions will involve the same reactants and products. However, balancing a basic reaction as acidic and then converting to basic will work. To convert to a basic reaction, it is necessary to add the same number of hydroxide ions to each side of the equation so that all the hydrogen ions (H⁺) are removed and mass balance is maintained. Hydrogen ion combines with hydroxide ion (OH⁻) to produce water.

Let us now try a basic equation. We will start with the following basic reaction:

[latex]\ce{Cl-}(aq)+\ce{MnO4-}(aq)⟶\ce{ClO3-}(aq)+\ce{MnO2}(s) \nonumber[/latex]

Balancing this as acid gives

[latex]\ce{Cl-}(aq)+\ce{2MnO4-}(aq)+\ce{2H+}(aq)⟶\ce{ClO3-}(aq)+\ce{2MnO2}(s)+\ce{H2O}(l) \nonumber[/latex]

In this case, it is necessary to add two hydroxide ions to each side of the equation to convert the two hydrogen ions on the left into water:

[latex]\ce{Cl-}(aq)+\ce{2MnO4-}(aq)+\ce{(2H+ + 2OH- )}(aq)⟶\ce{ClO3-}(aq)+\ce{2MnO2}(s)+\ce{H2O}(l)+\ce{2OH-}(aq) \nonumber[/latex]

[latex]\ce{Cl-}(aq)+\ce{2MnO4-}(aq)+\ce{(2H2O)}(l)⟶\ce{ClO3-}(aq)+\ce{2MnO2}(s)+\ce{H2O}(l)+\ce{2OH-}(aq) \nonumber[/latex]

Note that both sides of the equation show water. Simplifying should be done when necessary, and gives the desired equation. In this case, it is necessary to remove one H₂O from each side of the reaction arrows.

[latex]\ce{Cl-}(aq)+\ce{2MnO4-}(aq)+\ce{H2O}(l)⟶\ce{ClO3-}(aq)+\ce{2MnO2}(s)+\ce{2OH-}(aq) \nonumber[/latex]

Again, check each side of the overall equation to make sure there are no errors:

[latex]\begin{align*}
&\textrm{Cl: }\mathrm{Does\:(1×1)=(1×1)?\:Yes.}\\
&\textrm{Mn: }\mathrm{Does\:(2×1)=(2×1)?\:Yes.}\\
&\textrm{H: }\mathrm{Does\:(1×2)=(2×1)?\:Yes.}\\
&\textrm{O: }\mathrm{Does\:(2×4+1×1)=(3×1+2×2+2×1)?\:Yes.}\\
&\textrm{Charge: }\mathrm{Does\:[1×(−1)+2×(−1)]=[1×(−1)+2×(−1)]?\:Yes.}
\end{align*} \nonumber[/latex]

Everything checks, so this is the overall equation in basic solution.

Starting with the oxidation half-reaction, we can balance the chromium

[latex]\textrm{oxidation (unbalanced): }\ce{2Cr^3+}(aq)⟶\ce{Cr2O7^2-}(aq) \nonumber[/latex]

In acidic solution, we can use or generate hydrogen ions (H⁺). Adding seven water molecules to the left side provides the necessary oxygen; the “left over” hydrogen appears as 14 H⁺ on the right:

[latex]\textrm{oxidation (unbalanced): }\ce{2Cr^3+}(aq)+\ce{7H2O}(l)⟶\ce{Cr2O7^2-}(aq)+\ce{14H+}(aq) \nonumber[/latex]

The left side of the equation has a total charge of [2 × (+3) = +6], and the right side a total charge of [−2 + 14 × (+1) = +12]. The difference is six; adding six electrons to the right side produces a mass- and charge-balanced oxidation half-reaction (in acidic solution):

[latex]\textrm{oxidation (balanced): }\ce{2Cr^3+}(aq)+\ce{7H2O}(l)⟶\ce{Cr2O7^2-}(aq)+\ce{14H+}(aq)+\ce{6e-} \nonumber[/latex]

Checking the half-reaction:

[latex]\begin{align*}
&\textrm{Cr: }\mathrm{Does\:(2×1)=(1×2)?\:Yes.}\\
&\textrm{H: }\mathrm{Does\:(7×2)=(14×1)?\:Yes.}\\
&\textrm{O: }\mathrm{Does\:(7×1) =(1×7)?\:Yes.}\\
&\textrm{Charge: }\mathrm{Does\:[2×(+3)]=[1×(−2)+14×(+1)+6×(−1)]?\:Yes.}
\end{align*} \nonumber[/latex]

Now work on the reduction. It is necessary to convert the four oxygen atoms in the permanganate into four water molecules. To do this, add eight H⁺ to convert the oxygen into four water molecules:

[latex]\textrm{reduction (unbalanced): }\ce{MnO4-}(aq)+\ce{8H+}(aq)⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l) \nonumber[/latex]

Then add five electrons to the left side to balance the charge:

[latex]\textrm{reduction (balanced): }\ce{MnO4-}(aq)+\ce{8H+}(aq)+\ce{5e-}⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l) \nonumber[/latex]

Make sure to check the half-reaction:

[latex]\begin{align*}
&\textrm{Mn: }\mathrm{Does\:(1×1)=(1×1)?\:Yes.}\\
&\textrm{H: }\mathrm{Does\:(8×1)=(4×2)?\:Yes.}\\
&\textrm{O: }\mathrm{Does\:(1×4)=(4×1)?\:Yes.}\\
&\textrm{Charge: }\mathrm{Does\:[1×(−1)+8×(+1)+5×(−1)]=[1×(+2)]?\:Yes.}
\end{align*} \nonumber[/latex]

Collecting what we have so far:

[latex]\begin{align*}
&\textrm{oxidation: }\ce{2Cr^3+}(aq)+\ce{7H2O}(l)⟶\ce{Cr2O7^2-}(aq)+\ce{14H+}(aq)+\ce{6e-}\\
&\textrm{reduction: }\ce{MnO4-}(aq)+\ce{8H+}(aq)+\ce{5e-}⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l)
\end{align*} \nonumber[/latex]

The least common multiple for the electrons is 30, so multiply the oxidation half-reaction by five, the reduction half-reaction by six, combine, and simplify:

[latex]\ce{10Cr^3+}(aq)+\ce{35H2O}(l)+\ce{6MnO4-}(aq)+\ce{48H+}(aq)⟶\ce{5Cr2O7^2-}(aq)+\ce{70H+}(aq)+\ce{6Mn^2+}(aq)+\ce{24H2O}(l) \nonumber[/latex]
[latex]\ce{10Cr^3+}(aq)+\ce{11H2O}(l)+\ce{6MnO4-}(aq)⟶\ce{5Cr2O7^2-}(aq)+\ce{22H+}(aq)+\ce{6Mn^2+}(aq) \nonumber[/latex]

Checking each side of the equation:

[latex]\begin{align*}
&\textrm{Mn: }\mathrm{Does\:(6×1)=(6×1)?\:Yes.}\\
&\textrm{Cr: }\mathrm{Does\:(10×1)=(5×2)?\:Yes.}\\
&\textrm{H: }\mathrm{Does\:(11×2)=(22×1)?\:Yes.}\\
&\textrm{O: }\mathrm{Does\:(11×1+6×4)=(5×7)?\:Yes.}\\
&\textrm{Charge: }\mathrm{Does\:[10×(+3)+6×(−1)]=[5×(−2)+22×(+1)+6×(+2)]?\:Yes.}
\end{align*} \nonumber[/latex]

This is the balanced equation in acidic solution.