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# 23.3: Balancing Redox Equations Using the Oxidation Number Method

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

The student will:

• determine which substances in a redox equation are changing their oxidation state.
• balance redox equations using the oxidation number method.

## Introduction

The method you have used previously to balance equations was by inspection, that is, it was a trial and error method. You would keep trying numbers, using intelligent guesses, until you get the simplest whole number ratio that balances. That method works very well and very quickly for simpler reactions, but with the more complex redox reactions, that method will not work. As an example, here is the unbalanced net ionic equation for the reaction of silver ions with copper atoms.

Ag+(aq)+Cu(s)Ag(s)+Cu2+(aq)\begin{align*}\mathrm{Ag}^+_{(aq)} + \mathrm{Cu}_{(s)} \rightarrow \mathrm{Ag}_{(s)} + \mathrm{Cu}^{2+}_{(aq)}\end{align*}

By the inspection method, this equation would be balanced as is. This is not the correct answer, however, for this reaction. The properly balanced equation for this reaction is actually:

2 Ag+(aq)+Cu(s)2 Ag(s)+Cu2+(aq)\begin{align*}2 \ \mathrm{Ag}^+_{(aq)} + \mathrm{Cu}_{(s)} \rightarrow 2 \ \mathrm{Ag}_{(s)} + \mathrm{Cu}^{2+}_{(aq)}\end{align*}

In non-redox reactions, all that is required is to balance the number of atoms or molecules of each species. In redox reactions, you must balance not only the number of atoms, ions, or molecules of each species, but you must also balance the gain and loss of electrons. Balancing redox reactions requires a process that may have up to ten steps.

## Change of Oxidation State

A redox reaction occurs when an atom in the products has a different oxidation number than it had in the reactants. Since oxidation is always accompanied by reduction, there will usually be two species that change their oxidation state between the reactant side and the product side of the equation. In the following equation, oxidation numbers have been assigned to each atom in the equation.

When we examine the oxidation number of each atom on the two sides of the equation, we discover that the oxidation state of nitrogen has changed from +5 to +2. Nitrogen has been reduced by gaining three electrons. The oxidation state of sulfur has also changed, going from +2 to 0. Sulfur, therefore, has been oxidized by losing two electrons. This is a redox reaction.

## Balancing Redox Equations

The following example illustrates the steps involved in balancing redox equations.

Example:

Balance the following redox equation:

Zn+HNO3Zn(NO3)2+NO2+H2O\begin{align*}\mathrm{Zn} + \mathrm{HNO}_3 \rightarrow \mathrm{Zn(NO}_3)_2 + \mathrm{NO}_2 + \mathrm{H}_2\mathrm{O}\end{align*}

Solution:

Step 1: Assign oxidation numbers to all the atoms in the reaction.

Step 2: Determine which atom is being oxidized and write a half-reaction for the oxidation process, showing the species containing the atom being oxidized and the product containing that atom.

ZnZn(NO3)2\begin{align*} \mathrm{Zn} \rightarrow \mathrm{Zn(NO}_3)_2\end{align*}

Step 3: Determine which atom is being reduced and write a half-reaction for the reduction process, showing the species containing the atom being reduced and the product containing that atom.

HNO3NO2\begin{align*} \mathrm{HNO}_3 \rightarrow \mathrm{NO}_2\end{align*}

Step 4: If the atoms being oxidized and reduced are not already balanced in the half-reactions, balance them.

In this case, the atoms being oxidized and reduced are already balanced.

Step 5: Add the appropriate number of electrons to each half-reaction needed to bring about the reduction and oxidation.

Since the zinc with oxidation number of 0 is losing two electrons to become the zinc with oxidation number of +2,

ZnZn(NO3)2+2 e\begin{align*}\mathrm{Zn} \rightarrow \mathrm{Zn(NO}_3)_2 + 2 \ e^-\end{align*}

Since the nitrogen with oxidation number +5 is gaining one electron to become the nitrogen with oxidation number +4,

HNO3+eNO2\begin{align*}\mathrm{HNO}_3 + e^- \rightarrow \mathrm{NO}_2\end{align*}

Step 6: Determine the lowest common multiple (LCM) of the number of electrons in each half-reaction and multiply each half-reaction by whatever multiplier is necessary to make the number of electrons in the half-reaction equal to the LCM.

In this case the LCM is 2, so the first half-reaction needs no multiplier, and the second half-reaction is multiplied by 2.

ZnZn(NO3)2+2 e\begin{align*}\mathrm{Zn} \rightarrow \mathrm{Zn(NO}_3)_2 + 2 \ e^-\end{align*}
2 HNO3+2 e2 NO2\begin{align*}2 \ \mathrm{HNO}_3 + 2 \ \mathrm{e}^- \rightarrow 2 \ \mathrm{NO}_2\end{align*}

Step 7: Add the two half-reactions and cancel equal numbers of any species that appears on both sides.

Zn+2 HNO3+2 eZn(NO3)2+2 e+2 NO2\begin{align*}Zn + 2 \ \mathrm{HNO}_3 + 2 \ e^- \rightarrow \mathrm{Zn(NO}_3)_2 + 2 \ e^- + 2 \ \mathrm{NO}_2\end{align*}
Zn+2 HNO3Zn(NO3)2+2 NO2\begin{align*}\mathrm{Zn} + 2 \ \mathrm{HNO}_3 \rightarrow \mathrm{Zn(NO}_3)_2 + 2 \ \mathrm{NO}_2\end{align*}

Step 8: Check the final equation to make sure it is balanced in terms of both atoms and charge.

Example:

Balance the following redox equation:

Au3++MnAu+Mn2+\begin{align*}\mathrm{Au}^{3+} + \mathrm{Mn} \rightarrow \mathrm{Au} + \mathrm{Mn}^{2+}\end{align*}

Solution:

Step 1: Assign oxidation numbers to all the atoms in the reaction.

Step 2: Determine which atom is being oxidized and write a half-reaction for the oxidation process, showing the species containing the atom being oxidized and the product containing that atom.

MnMn2+\begin{align*}\mathrm{Mn} \rightarrow \mathrm{Mn}^{2+}\end{align*}

Step 3: Determine which atom is being reduced and write a half-reaction for the reduction process, showing the species containing the atom being reduced and the product containing that atom.

Au3+Au\begin{align*}\mathrm{Au}^{3+} \rightarrow \mathrm{Au}\end{align*}

Step 4: If the atoms being oxidized and reduced are not already balanced in the half-reactions, balance them.

In this case, the atoms being oxidized and reduced are already balanced.

Step 5: Add the appropriate number of electrons to each half-reaction needed to bring about the reduction and oxidation.

Since manganese with oxidation number of 0 is losing two electrons to become manganese with oxidation number of +2,

MnMn2++2 e\begin{align*}\mathrm{Mn} \rightarrow \mathrm{Mn}^{2+} + 2 \ e^-\end{align*}

Since gold with oxidation number +3 is gaining three electron to become gold with oxidation number 0,

Au3++3 eAu\begin{align*}\mathrm{Au}^{3+} + 3 \ e^- \rightarrow \mathrm{Au}\end{align*}

Step 6: Determine the lowest common multiple (LCM) of the number of electrons in each half-reaction and multiply each half-reaction by whatever multiplier is necessary to make the number of electrons in the half-reaction equal to the LCM.

In this case the LCM is 6, so the first half-reaction needs to be multiplied by 3 and the second half-reaction needs to be multiplied by 2.

3 Mn3 Mn2++6 e\begin{align*}3 \ \mathrm{Mn} \rightarrow 3 \ \mathrm{Mn}^{2+} + 6 \ e^-\end{align*}
2 Au3++6 e2 Au\begin{align*}2 \ \mathrm{Au}^{3+} + 6 \ e^- \rightarrow 2 \ \mathrm{Au}\end{align*}

Step 7: Add the two half-reactions and cancel equal numbers of any species that appears on both sides.

3 Mn+2 Au3++6 e3 Mn2++6 e+2 Au\begin{align*}3 \ \mathrm{Mn} + 2 \ \mathrm{Au}^{3+} + 6 \ e^- \rightarrow 3 \ \mathrm{Mn}^{2+} + 6 \ e^- + 2 \ \mathrm{Au}\end{align*}
3 Mn+2 Au3+3 Mn2++2 Au\begin{align*}3 \ \mathrm{Mn} + 2 \ \mathrm{Au}^{3+} \rightarrow 3 \ \mathrm{Mn}^{2+} + 2 \ \mathrm{Au}\end{align*}

Step 8: Check the final equation to make sure it is balanced in terms of both atoms and charge.

Redox reactions often occur in solutions that are acidic or basic. When the reaction occurs in acidic or basic solution, there are a few more steps in the balancing process. In an acidic solution, hydrogen ions are available in the solution. In a basic solution, hydroxide ions are available in the solution. When a redox reaction occurs in acid solution, add hydrogen ions and water to the equation wherever they are needed to balance. When the redox reaction occurs in base solution, add hydroxide ions and water wherever needed to balance.

Example:

Balance the following redox equation. It is in an acid solution.

MnO4+H++ClMn2++Cl2\begin{align*}\mathrm{MnO}_4^- + \mathrm{H}^+ + \mathrm{Cl}^- \rightarrow \mathrm{Mn}^{2+} + \mathrm{Cl}_2\end{align*}

Step 1: Assign oxidation numbers.

Step 2: Determine which atom is being oxidized and write a half-reaction for the oxidation process, showing the species containing the atom being oxidized and the product containing that atom.

The chlorine whose oxidation number is -1 is being oxidized to chlorine with an oxidation number of 0.

ClCl2\begin{align*}\mathrm{Cl}^- \rightarrow \mathrm{Cl}_2\end{align*}

Step 3: Determine which atom is being reduced and write a half-reaction for the reduction process, showing the species containing the atom being reduced and the product containing that atom.

The manganese with +7 oxidation number is being reduced to manganese with oxidation number +2.

MnO4Mn2+\begin{align*}\mathrm{MnO}_4^- \rightarrow \mathrm{Mn}^{2+}\end{align*}

Step 4: If the atoms being oxidized and reduced are not already balanced in the half-reactions, balance them.

In this case, the manganese is balanced in the reduction half-reaction so nothing is needed, but the chlorine is not balanced in the oxidation half-reaction so a coefficient of 2 must be entered in the reactant side of that half-reaction.

2 ClCl2\begin{align*}2 \ \mathrm{Cl}^- \rightarrow \mathrm{Cl}_2\end{align*}

Step 5: Add the appropriate number of electrons to each half-reaction needed to bring about the reduction and oxidation.

2 ClCl2+2 e\begin{align*}2 \ \mathrm{Cl}^- \rightarrow \mathrm{Cl}_2 + 2 \ e^-\end{align*}
MnO4+5 eMn2+\begin{align*}\mathrm{MnO}_4^- + 5 \ e^- \rightarrow \mathrm{Mn}^{2+}\end{align*}

Step 6: Balance all other atoms in each half-reaction except hydrogen and oxygen.

In this case, all other atoms except hydrogen and oxygen are already balanced.

Step 7: In a redox reaction in acid solution, balance the hydrogen and oxygen in each half-reaction by adding water where needed to balance the oxygen and then add hydrogen ions where needed to balance the hydrogen.

2 ClCl2+2 e\begin{align*}2 \ \mathrm{Cl}^- \rightarrow \mathrm{Cl}_2 + 2 \ e^-\end{align*}
MnO4+5 e+8 H+Mn2++4 H2O\begin{align*}\mathrm{MnO}_4^- + 5 \ e^- + 8 \ \mathrm{H}^+ \rightarrow \mathrm{Mn}^{2+} + 4 \ \mathrm{H}_2\mathrm{O}\end{align*}

In this half-reaction, 4 water are needed to balance the oxygen and 8 H+ are needed to balance the hydrogen. After this step, the charge in the half-reactions should be balanced. In this case, the oxidation half-reaction has a net -2 charge on each side of the equation and the reduction half-reaction has a net +2 charge on each side of the equation. Therefore, the charge is balanced.

Step 8: Determine the lowest common multiple (LCM) of the number of electrons in each half-reaction and multiply each half-reaction by whatever multiplier is necessary to make the number of electrons in the half-reaction equal to the LCM.

In this case the LCM is 10, so the first half-reaction needs to be multiplied by 5, and the second half-reaction needs to be multiplied by 2.

10 Cl5 Cl2+10 e\begin{align*}10 \ \mathrm{Cl}^- \rightarrow 5 \ \mathrm{Cl}_2 + 10 \ e^-\end{align*}
2 MnO4+10 e+16 H+2 Mn2++8 H2O\begin{align*}2 \ \mathrm{MnO}_4^- + 10 \ e^- + 16 \ \mathrm{H}^+ \rightarrow 2 \ \mathrm{Mn}^{2+} + 8 \ \mathrm{H}_2\mathrm{O}\end{align*}

Step 9: Add the two half-reactions and cancel equal numbers of any species that appears on both sides.

10 Cl+2 MnO4+10 e+16 H+5 Cl2+10 e+2 Mn2++8 H2O\begin{align*}10 \ \mathrm{Cl}^- + 2 \ \mathrm{MnO}_4^- + 10 \ e^- + 16 \ \mathrm{H}^+ \rightarrow 5 \ \mathrm{Cl}_2 + 10 \ e^- + 2 \ \mathrm{Mn}^{2+} + 8 \ \mathrm{H}_2\mathrm{O}\end{align*}
10 Cl+2 MnO4+16 H+5 Cl2+2 Mn2++8 H2O\begin{align*}10 \ \mathrm{Cl}^- + 2 \ \mathrm{MnO}_4^- + 16 \ \mathrm{H}^+ \rightarrow 5 \ \mathrm{Cl}_2 + 2 \ \mathrm{Mn}^{2+} + 8 \ \mathrm{H}_2\mathrm{O}\end{align*}

Step 10: Check the final equation to make sure it is balanced in terms of both atoms and charge.

Example:

Balance the following equation. It is in a basic solution.

\begin{align*}\mathrm{MnO}_4^{2-} + \mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow \mathrm{MnO}_2 + \mathrm{CO}_3^{2-}\end{align*}

Step 1: Assign oxidation numbers.

Step 2: Determine which atom is being oxidized and write a half-reaction for the oxidation process, showing the species containing the atom being oxidized and the product containing that atom.

The carbon with oxidation number +3 is losing one electron and becoming carbon with an oxidation number of +4.

\begin{align*}\mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow \mathrm{CO}_3^{2-}\end{align*}

Step 3: Determine which atom is being reduced and write a half-reaction for the reduction process, showing the species containing the atom being reduced and the product containing that atom.

The manganese atoms with oxidation numbers of +6 and accepting two electrons and becoming manganese atoms with an oxidation number of +4.

\begin{align*}\mathrm{MnO}_4^{2-} \rightarrow \mathrm{MnO}_2\end{align*}

Step 4: If the atoms being oxidized and reduced are not already balanced in the half-reactions, balance them.

In this case, the oxidation half-equation requires a coefficient of 2 in front of the carbonate ion. In the reduction half-equation, the manganese is already balanced.

\begin{align*}\mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow 2 \ \mathrm{CO}_3^{2-}\end{align*}
\begin{align*}\mathrm{MnO}_4^{2-} \rightarrow \mathrm{MnO}_2\end{align*}

Step 5: Add the appropriate number of electrons to each half-reaction needed to bring about the reduction and oxidation.

\begin{align*}\mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow 2 \ \mathrm{CO}_3^{2-} + 2 \ e^-\end{align*}

Each carbon is giving up one electron, and since there are two carbon atoms in the half-equation, two electrons are required.

\begin{align*}\mathrm{MnO}_4^{2-} + 2 \ e^- \rightarrow \mathrm{MnO}_2\end{align*}

Step 6: Balance all other atoms in each half-reaction except hydrogen and oxygen.

With the exception of hydrogen and oxygen, all atoms in the equations are balanced.

Step 7: In a redox reaction in basic solution, balance the charge first by adding OH- and then balance the oxygen by adding H2O. The hydrogen should balance automatically.

\begin{align*}\mathrm{C}_2\mathrm{O}_4^{2-} + 4 \ \mathrm{OH}^- \rightarrow 2 \ \mathrm{CO}_3^{2-} + 2 \ \mathrm{H}_2\mathrm{O} + 2 \ e^-\end{align*}
\begin{align*}\mathrm{MnO}_4^{2-} + 2 \ \mathrm{H}_2\mathrm{O} + 2 \ e^- \rightarrow \mathrm{MnO}_2 + 4 \ \mathrm{OH}^-\end{align*}

Step 8: Determine the lowest common multiple (LCM) of the number of electrons in each half-reaction and multiply each half-reaction by whatever multiplier is necessary to make the number of electrons in the half-reaction equal to the LCM.

In this case the LCM is 2, so neither half-equation needs to be multiplied.

Step 9: Add the two half-reactions and cancel equal numbers of any species that appears on both sides.

\begin{align*}\mathrm{C}_2\mathrm{O}_4^{2-} + \mathrm{OH}^- + \mathrm{MnO}_4^{2-} + 2 \ \mathrm{H}_2\mathrm{O} + 2 \ e^- \rightarrow 2 \ \mathrm{CO}_3^{2-} + 2 \ \mathrm{H}_2\mathrm{O} + 2 \ e^- + \mathrm{MnO}_2 + 4 \ \mathrm{OH}^-\end{align*}
\begin{align*}\mathrm{C}_2\mathrm{O}_4^{2-} + \mathrm{MnO}_4^{2-} \rightarrow 2 \ \mathrm{CO}_3^{2-} + \mathrm{MnO}_2\end{align*}

Step 10: Check the final equation to make sure it is balanced in terms of both atoms and charge.

This video shows the process in balancing a redox equation (3g): http://www.youtube.com/watch?v=TBmwhTzc41o (8:19).

## Lesson Summary

• The following are the steps to balance a redox reaction.
1. Assign oxidation numbers for all atoms in the reaction.
2. Determine which atom is being oxidized and write a half-reaction for the oxidation process, showing the species containing the atom being oxidized and the product containing that atom.
3. Determine which atom is being reduced and write a half-reaction for the reduction process, showing the species containing the atom being reduced and the product containing that atom.
4. If the atoms being oxidized and reduced are not already balanced in the half-reactions, balance them.
5. Add the appropriate number of electrons to each half-reaction needed to bring about the reduction and oxidation.
6. Balance all other atoms in each half-reaction except H and O.
7. Balance the H and O according to either (a) or (b) depending on whether the reaction is acidic or basic.
(a) If the reaction is acidic, add \begin{align*}\mathrm{H}_2\mathrm{O}\end{align*} and \begin{align*}\mathrm{H}^+\end{align*}. Balance O first by adding \begin{align*}\mathrm{H}_2\mathrm{O}\end{align*}, then balance H by adding \begin{align*}\mathrm{H}^+\end{align*}. Charge should now be balanced.
(b) If the reaction is basic, add \begin{align*}\mathrm{OH}^-\end{align*} and \begin{align*}\mathrm{H}_2\mathrm{O}\end{align*}. Balance charge first by adding \begin{align*}\mathrm{OH}^-\end{align*}, then balance O by adding \begin{align*}\mathrm{H}_2\mathrm{O}\end{align*}. The H should now be balanced.
1. Determine the lowest common multiple (LCM) of the number of electrons in each half-reaction and multiply each half-reaction by whatever multiplier is necessary to make the number of electrons in the half-reaction equal to the LCM.
2. Add the two half-reactions and cancel those species that are common to both sides.
3. Check the equation to be sure that it is balanced in terms of both atoms and charge.

## Review Questions

1. Balance the following equation using the oxidation number method.

\begin{align*}\mathrm{HNO}_3 + \mathrm{Br}_2 \rightarrow \mathrm{HBrO}_3 + \mathrm{NO}_2 + \mathrm{H}_2\mathrm{O}\end{align*}

1. Balance the following equation using the oxidation number method.

\begin{align*}\mathrm{Zn} + \mathrm{HNO}_3 \rightarrow \mathrm{Zn(NO}_3 )_2 + \mathrm{NO}_2 + \mathrm{H}_2\mathrm{O}\end{align*}

1. In terms of electron gain and loss, explain why chlorine undergoes both oxidation and reduction in the following unbalanced reaction.

\begin{align*}\mathrm{Cl}_2 + \mathrm{KOH} \rightarrow \mathrm{KCl} + \mathrm{KClO}_3 + \mathrm{H}_2\mathrm{O}\end{align*}

1. Balance the equation in the previous problem.

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