# 22.3: Balancing Redox Reactions

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Balance a redox equation using the oxidation-number-change method.
• Balance a redox equation by breaking it down into oxidation and reduction half-reactions and using the half-reaction method.

## Check Your Understanding

### Recalling Prior Knowledge

• How are oxidation numbers determined?
• What is a half-reaction and how can two half-reactions be written from a single redox reaction?

You learned how to balance equations by inspection in an earlier chapter. Redox reactions can be considerably more complex, and balancing solely by inspection can sometimes be nearly impossible. Here you will learn two systematic processes for balancing redox equations.

## Oxidation-Number-Change Method

One way to balance redox reactions is by keeping track of the extent to which the oxidation numbers change for each of the atoms. For the oxidation-number-change method, start with the unbalanced skeleton equation. The example below is for the reaction of iron(III) oxide with carbon monoxide. This reaction takes place in blast furnaces during the processing of iron ore into metallic iron (Figure below).

Fe2O3(s)+CO(g)Fe(s)+CO2(g)\begin{align*}\text{Fe}_2\text{O}_3(s) + \text{CO}(g) \rightarrow \text{Fe}(s) + \text{CO}_2(g)\end{align*}

Step 1: Assign oxidation numbers to each of the atoms in the equation.

Fe2+3O32(s)+C+2O2(g)Fe0(s)+C+4O22(g)\begin{align*}\overset{+3}{\text{Fe}_2}\overset{-2}{\text{O}_3}(s) + \overset{+2}{\text{C}}\overset{-2}{\text{O}}(g) \rightarrow \overset{0}{\text{Fe}}(s) + \overset{+4}{\text{C}}\overset{-2}{\text{O}_2}(g)\end{align*}

Step 2: Identify the atoms that are oxidized and those that are reduced. In the above equation, the carbon atom is being oxidized, since its oxidation number increases from +2 to +4. The iron atom is being reduced, since its oxidation number decreases from +3 to 0.

Step 3: Use a line to connect the atoms that are undergoing a change in oxidation number, and indicate by how many units each oxidation number is changing.

The carbon atom’s oxidation number increases by 2, while the iron atom’s oxidation number decreases by 3. As written, the number of electrons lost does not equal the number of electrons gained. In a balanced redox equation, these values must be equal, so coefficients must be used to make the total increase in oxidation number for all atoms equal to the total decrease in oxidation number for all atoms.

Step 4: Use coefficients to make the total increase in oxidation number equal to the total decrease in oxidation number. In this case, the least common multiple of 2 and 3 is 6. So the oxidation-number increase should be multiplied by 3, while the oxidation-number decrease should be multiplied by 2. These coefficients are applied to the corresponding formulas in the equation. A 3 is placed in front of CO and CO2, while a 2 is placed in front of Fe on the right side of the equation. The Fe2O3 does not require a coefficient because the subscript of 2 indicates that there are already two iron atoms.

Step 5: Check to see if the equation is balanced in terms of both atoms and charge. Occasionally, a coefficient may need to be placed in front of a formula that was not involved in the redox process. In the current example, the equation is now balanced.

Fe2O3(s)+3CO(g)2Fe(s)+3CO2(g)\begin{align*}\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g)\end{align*}

A blast furnace is where iron ore is processed and turned into iron metal. First, air is blown through a mixture of iron ore and coke (carbon). The carbon monoxide produced reduces the Fe3+ ions in the iron ore to metallic iron.

## Half-Reaction Method

Another method for balancing redox reactions uses half-reactions. Recall that half-reactions treat oxidation and reduction as two separate processes that are occurring simultaneously. The half-reaction method tends to work better than the oxidation-number method when the substances in the reaction are dissolved in water. The aqueous solution is typically either acidic or basic, so hydrogen ions or hydroxide ions can be added if necessary to balance the atoms and charge.

In general, the half-reactions are first balanced separately by looking just at the number of each atom. Then, electrons are added to each half-reaction in order to balance the charges. Then, all of the species in each half-reaction are multiplied by some factor so that the number of electrons lost in the oxidation is equal to the number of electrons gained in the reduction. Finally, the two half-reactions are added back together. The following example shows the oxidation of Fe2+ ions to Fe3+ ions by dichromate (Cr2O72−) in acidic solution. The dichromate ions (Figure below) are reduced to Cr3+ ions.

Potassium dichromate (K2Cr2O7) is a striking orange color. The dichromate ion is a good oxidizing agent because of the +6 oxidation state of the chromium.

Step 1: Write the unbalanced ionic equation.

Fe2+(aq)+Cr2O27(aq)Fe3+(aq)+Cr3+(aq)\begin{align*}\text{Fe}^{2+}(aq) + \text{Cr}_2\text{O}_7^{2-}(aq) \rightarrow \text{Fe}^{3+}(aq) + \text{Cr}^{3+}(aq)\end{align*}

Notice that the equation is far from balanced, as there are no oxygen atoms on the right side. This will be resolved by the balancing method.

Step 2: Write separate half-reactions for the oxidation and the reduction processes. Determine the oxidation numbers first, if necessary.

Oxidation: Fe2+(aq)Fe3+(aq)\begin{align*}\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq)\end{align*}
Reduction: Cr2+6O27(aq)Cr3+(aq)\begin{align*}\overset{+6}{\text{Cr}_2}\text{O}_7^{2-}(aq) \rightarrow \text{Cr}^{3+}(aq)\end{align*}

Step 3: Balance the atoms in each half-reaction, but ignore hydrogen and oxygen. In the oxidation half-reaction above, the iron atoms are already balanced. The reduction half-reaction needs to be balanced as follows:

Cr2O27(aq)2Cr3+(aq)\begin{align*}\text{Cr}_2\text{O}_7^{2-}(aq) \rightarrow 2\text{Cr}^{3+}(aq)\end{align*}

Step 4: Balance oxygen atoms by adding water molecules to the appropriate side of the equation. For the reduction half-reaction above, seven H2O molecules will be added to the product side.

Cr2O27(aq)2Cr3+(aq)+7H2O(l)\begin{align*}\text{Cr}_2\text{O}_7^{2-}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l)\end{align*}

Now, the hydrogen atoms need to be balanced. In an acidic medium, add hydrogen ions to balance. In this example, fourteen H+ ions will be added to the reactant side.

14H+(aq)+Cr2O27(aq)2Cr3+(aq)+7H2O(l)\begin{align*}14\text{H}^+(aq) + \text{Cr}_2\text{O}_7^{2-}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l)\end{align*}

Step 5: Balance the charges by adding electrons to each half-reaction. For the oxidation half-reaction, the electrons will need to be added to the product side. For the reduction half-reaction, the electrons will be added to the reactant side. By adding one electron to the product side of the oxidation half-reaction, there is a 2+ total charge on both sides.

Fe2+(aq)Fe3+(aq)+e\begin{align*}\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + \text{e}^-\end{align*}

There is a total charge of 12+ on the reactant side of the reduction half-reaction (14 – 2). The product side has a total charge of 6+ due to the two chromium ions (2 × 3). To balance the charge, six electrons need to be added to the reactant side.

6e+14H+(aq)+Cr2O27(aq)2Cr3+(aq)+7H2O(l)\begin{align*}6\text{e}^- + 14\text{H}^+(aq) + \text{Cr}_2\text{O}_7^{2-}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l)\end{align*}

Now equalize the electrons by multiplying everything in one or both equations by a coefficient. In this example, the oxidation half-reaction will be multiplied by six.

6Fe2+(aq)6Fe3+(aq)+6e\begin{align*}6\text{Fe}^{2+}(aq) \rightarrow 6\text{Fe}^{3+}(aq) + 6\text{e}^-\end{align*}

Step 6: Add the two half-reactions together. The electrons must cancel. Balance any remaining substances by inspection. Cancel out some or all of the H2O molecules or H+ ions if they appear on both sides of the equation.

6Fe2+(aq)6e+14H+(aq)+Cr2O27(aq)14H+(aq)+6Fe2+(aq)+Cr2O27(aq)6Fe3+(aq)+6e2Cr3+(aq)+7H2O(l)6Fe3+(aq)+2Cr3+(aq)+7H2O(l)\begin{align*} 6\text{Fe}^{2+}(aq) & \rightarrow 6\text{Fe}^{3+}(aq) + \cancel{6\text{e}^-} \\ \cancel{6\text{e}^-} + 14\text{H}^+(aq) + \text{Cr}_2\text{O}_7^{2-}(aq) & \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) \\ \hline 14\text{H}^+(aq) + 6\text{Fe}^{2+}(aq) + \text{Cr}_2\text{O}_7^{2-}(aq) & \rightarrow 6\text{Fe}^{3+}(aq) + 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) \end{align*}

Step 7: Check the balancing. In the above equation, there are 14 H, 6 Fe, 2 Cr, and 7 O on both sides. The net charge is 24+ on both sides. The equation is balanced.

### Reactions in Basic Solution

For reactions that occur in basic solution rather than acidic solution, the steps are primarily the same. However, after finishing step 6, add an equal number of OH ions to both sides of the equation. Combine the H+ and OH to make H2O and cancel out any water molecules that appear on both sides. Continuing with the example above, we would get the following three steps:

1. Adding the hydroxide ions: 14OH(aq)+14H+(aq)+6Fe2+(aq)+Cr2O27(aq)6Fe3+(aq)+2Cr3+(aq)+7H2O(l)+14OH(aq)\begin{align*}14\text{OH}^-(aq) + 14\text{H}^+(aq) + 6\text{Fe}^{2+}(aq) + \text{Cr}_2\text{O}_7^{2-}(aq) \rightarrow 6\text{Fe}^{3+}(aq) + 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) + 14\text{OH}^-(aq)\end{align*}
2. Combining the hydrogen ions and hydroxide ions to make water: 14H2O(l)+6Fe2+(aq)+Cr2O27(aq)6Fe3+(aq)+2Cr3+(aq)+7H2O(l)+14OH(aq)\begin{align*}14\text{H}_2\text{O}(l) + 6\text{Fe}^{2+}(aq) + \text{Cr}_2\text{O}_7^{2-}(aq) \rightarrow 6\text{Fe}^{3+}(aq) + 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) + 14\text{OH}^-(aq)\end{align*}
3. Canceling out seven water molecules from both sides to get the final equation: 7H2O(l)+6Fe2+(aq)+Cr2O27(aq)6Fe3+(aq)+2Cr3+(aq)+14OH(aq)\begin{align*}7\text{H}_2\text{O}(l) + 6\text{Fe}^{2+}(aq) + \text{Cr}_2\text{O}_7^{2-}(aq) \rightarrow 6\text{Fe}^{3+}(aq) + 2\text{Cr}^{3+}(aq) + 14\text{OH}^-(aq)\end{align*}

The equation is still balanced by atoms and by charge, but the presence of hydroxide ions rather than hydrogen ions means that the reaction takes place in basic solution. Typically, most redox reactions will actually only proceed in one type of solution or the other. The oxidation of Fe2+ by Cr2O72− does not readily occur in basic solution, but it was balanced that way to demonstrate the method.

In summary, the choice of which balancing method to use depends on the kind of reaction. The oxidation-number method works best if the oxidized and reduced species appear only once on each side of the equation and if no acids or bases are present. The half-reaction method is more versatile and works well for reactions involving ions in aqueous solution.

## Lesson Summary

• In the oxidation-number-change method for balancing a redox reaction, the increase in oxidation number by one atom is compared to the decrease in oxidation number of another atom. The total number of electrons lost in the oxidation is then made equal to the total number of electrons gained in the reduction.
• To balance a redox reaction by the half-reaction method, write separate half-reactions for the oxidation and reduction processes. Balance by atoms, adding in water and hydrogen ions if necessary. Put in electrons to balance by charge, equalize the electrons in the two reactions, and add them back together. For basic solutions, add hydroxide ions.

## Lesson Review Questions

### Reviewing Concepts

1. When balancing an equation by the oxidation-number-change method, what must be true about the total change in oxidation number of the oxidized and reduced species?
2. When balancing an equation by the half-reaction method, what must be true about the number of electrons lost in the oxidation compared to the number of electrons gained in the reduction?
3. Which method works well for molecular substances?
4. Which method works well for reactions between ions in aqueous solution?

### Problems

1. Balance the reactions below using the oxidation-number-change method.
1. WO3(s) + H2(g) → W(s) + H2O(g)
2. SbCl5(aq) + KI(aq) → SbCl3(aq) + KCl(aq) +I2(s)
3. Bi2S3(s) + HNO3(aq) → Bi(NO3)3(aq) + NO(g) + S(s) + H2O(l)
4. Cl2(g) + KOH(aq) → KClO3(aq) + KCl(aq) + H2O(l)
2. Balance the reactions below using the half-reaction method. All are in acidic solution.
1. MnO4-(aq) + Cl-(aq) → Mn2+(aq) + Cl2(g)
2. H2O2(l) + Fe2+(aq) → Fe3+(aq) + H2O(l)
3. Cr2O72-(aq) + C2O42-(aq) → Cr3+(aq) + CO2(g)
3. Balance the reactions below using the half-reaction method. All are in basic solution.
1. CN-(aq) + MnO4-(aq) → CNO-(aq) + MnO2(s)
2. Bi(OH)3(s) + SnO22-(aq) → SnO32-(aq) + Bi(s)
3. Br2(l) → BrO3-(aq) + Br-(aq)
4. Oxalic acid (H2C2O4) is present in many plants and vegetables. In acidic solution, oxalic acid can be oxidized by the permanganate ion (MnO4) according to the following unbalanced equation.
1. Balance the equation.
2. In a redox titration, 15.0 mL of 0.100 M MnO4 fully reacts with 23.6 mL of a C2O42− solution. What is the concentration of the oxalate ion?

## Points to Consider

A battery is a device that uses a redox reaction to turn chemical energy into electrical energy.

• Why can't a redox reaction directly generate electricity?
• What is an electrochemical cell?

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Date Created:
Mar 29, 2013