Redox reaction is a core concept in teaching and learning chemistry (Goodstein, 1970). The oxidation number method has been widely used for balancing redox reactions, and the determination of the oxidation number is a required step in the process (Herndon, 1997; Chang & Goldsby, 2013; Tro, 2014; Generalic & Vladislavic, 2018). However, the act of balancing organic redox reactions often poses problems for students (Lockwood, 1961; Jensen, 2009). In this article, we discuss a new method, the proton method, that can be applied to both molecular and ionic chemical equations. The method can be used for not only balancing redox equations but also determining the number of transferred electrons and the change of oxidation numbers.

Proton method: Procedures for balancing half reactions

This section shows how the proton method is used to balance both atoms and charges. H⁺, O, H₂O, and e^– are used as vehicles in two half reactions.

Step 1. Balance atoms.

Step 1.1. Balance all other atoms except H and O.

Step 1.2. Balance hydrogen atoms with H⁺.

Step 1.3. Balance oxygen atoms with O.

Step 1.4. Add two H⁺ atoms for each O atom.

Step 1.5. Convert two H⁺ and one O to one H₂O molecule.

Step 2. Balance charges.

Step 2.1. Count the number of charges on both sides.

Step 2.2. Add electrons to make charges on both sides

equivalent.

Step 3. Convert H⁺ to OH^– (optional step when working in a basic medium).

Step 3.1. Add one OH^– for each H⁺.

Step 3.2. Convert one OH^– and one H⁺ to one H₂O

molecule.

Step 3.3. Simplify an overall equation.

Examples for balancing molecular half reactions

Example 1

Convert the equation CH₃CH₂OH → CH₃COOH to C₂H₆O → C₂H₄O₂.

Step 1. Balance atoms.

C₂H₆O → C₂H₄O₂

C₂H₆O + O → C₂H₄O₂ + 2H⁺

C₂H₆O + O + 2H⁺ → C₂H₄O₂ + 2H⁺ + 2H⁺

C₂H₆O + H₂O → C₂H₄O₂ + 4H⁺

Step 2. Balance charges.

C₂H₆O + H₂O → C₂H₄O₂ + 4H⁺

(0) (+4)

C₂H₆O + H₂O → C₂H₄O₂ + 4H⁺ + 4e^–

Example 2

Convert the equation CH₃CH₂CH₃ → CO₂ to C₃H₈ → CO₂.

Step 1. Balance atoms.

C₃H₈ → CO₂

C₃H₈ → 3CO₂

C₃H₈ + 6O → 3CO₂ + 8H⁺

C₃H₈ + 6O + 12H⁺ → 3CO₂ + 8H⁺ + 12H⁺

C₃H₈ + 6H₂O → 3CO₂ + 20H⁺

Step 2. Balance charges.

C₃H₈ + 6H₂O → 3CO₂ + 20H⁺

(0) (+20)

C₃H₈ + 6H₂O → 3CO₂ + 20H⁺ + 20e^–

Examples for balancing ionic half reactions

Example 3

Given an ionic equation Cr₂O₇^2- → Cr³⁺ (in acidic medium):

Step 1. Balance atoms.

Cr₂O₇^2- → Cr³⁺

Cr₂O₇^2- → 2Cr³⁺

Cr₂O₇^2- → 2Cr³⁺ + 7O

Cr₂O₇^2- + 14H⁺ → 2Cr³⁺ + 7O + 14H⁺

Cr₂O₇^2- + 14H⁺ → 2Cr³⁺ + 7H₂O

Step 2. Balance charges.

Cr₂O₇^2- + 14H⁺ → 2Cr³⁺ + 7H₂O

(+12) (+6)

Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O

Example 4

Given an ionic equation MnO₄^– → MnO₂ (in basic medium):

Step 1. Balance atoms.

MnO₄^– → MnO₂

MnO₄^– → MnO₂ + 2O

MnO₄^– + 4H⁺ → MnO₂ + 2O + 4H⁺

MnO₄^– + 4H⁺ → MnO₂ + 2H₂O

Step 2. Balance charges.

MnO₄^– + 4H⁺ → MnO₂ + 2H₂O

(+3) (0)

MnO₄^– + 4H⁺ + 3e^– → MnO₂ + 2H₂O

Step 3. Convert H⁺ to OH^– (in basic medium).

MnO₄^– + 4H⁺ + 3e^– → MnO₂ + 2H₂O

MnO₄^– + 4H⁺ + 3e^– + 4OH^– → MnO₂ + 2H₂O + 4OH^-

MnO₄^– + 4H₂O +3e^– → MnO₂ + 2H₂O + 4OH^–

MnO₄^– + 2H₂O + 3e^– → MnO₂ + 4OH^–

Determination of Te

First, one needs to balance atoms and charges, then the number of transferred electrons can be determined accordingly. In Example 1, the reaction of C₂H₆O + H₂O → C₂H₄O₂ + 4H⁺ + 4e^–, a loss of four electrons identifies an oxidation reaction. In Example 3, the reaction of Cr₂O₇^2- + 14H⁺ + 6e^– → 2Cr³⁺ + 7H₂O, a gain of six electrons identifies a reduction reaction. An oxidation reaction can be represented by Te^– > 0 (positive value), and a reduction reaction can be represented by Te^– < 0 (negative value). These half-reaction examples are quantified and defined in Table 1.

Relationship among Te

Based on Examples 1 through 4, the values of Te^–, n, and ΔON are shown in Table 2. In Table 2, Te^–, n, and ΔON denote the number of transferred electrons, the number of atoms with oxidation number change, and the change of mean oxidation numbers, respectively. ΔON can be counted by the difference between the oxidation number of an atom on the product side (ON_f) and the oxidation number of an atom on the reactant side (ON_i), and it is demonstrated by ΔON = ON_f – ON_i. An oxidation reaction can be represented by ΔON > 0 (positive value), and a reduction reaction can be represented by ΔON < 0 (negative value).

In a balanced half reaction, the relationship among Te^–, n, and ΔON can be deducted, and the mathematical equation is shown as follows:

Te^– = n ΔON

Proton method: Procedures for balancing overall redox reactions

An overall number of transferred electrons is counted by the least common multiple (LCM) of two half redox reactions. The procedures for balancing an overall reaction are as follows:

Step 1. Divide into two half reactions.

Step 2. Balance all atoms in two half reactions.

Step 3. Balance charges of two half reactions by adding electrons.

Step 4. Make electrons of two half reactions equivalent.

Step 5. Combine and simplify two half reactions.

Step 6. Convert H⁺ to OH^– (optional step when working in basic medium).

Examples for balancing overall redox reactions

Example 5

Convert the ionic chemical equation CH₃CH₂OH + Cr₂O₇^2- → CH₃COOH + Cr³⁺ (in acidic medium) to C₂H₆O + Cr₂O₇^2- → C₂H₄O₂ + Cr³⁺.

Step 1. Divide into two half reactions:

C₂H₆O → C₂H₄O₂

Cr₂O₇^2- → Cr³⁺

Step 2. Balance all atoms in two half reactions.

C₂H₆O → C₂H₄O₂

C₂H₆O + O → C₂H₄O₂ + 2H⁺

C₂H₆O + O + 2H⁺ → C₂H₄O₂ + 2H⁺ + 2H⁺

C₂H₆O + H₂O → C₂H₄O₂ + 4H⁺

Cr₂O₇^2-→ 2Cr³⁺

Cr₂O₇^2- → 2Cr³⁺ + 7O

Cr₂O₇^2- + 14H⁺ → 2Cr³⁺ + 7O + 14H⁺

Cr₂O₇^2- + 14H⁺ → 2Cr³⁺ + 7H₂O

Step 3. Balance charges of two half reactions by adding electrons.

C₂H₆O + H₂O → C₂H₄O₂ + 4H⁺ + 4e^–

Cr₂O₇^2- + 14H⁺ + 6e^– → 2Cr³⁺ + 7H₂O

Step 4. Make electrons of two half reactions equivalent (LCM = 12).

(C₂H₆O + H₂O → C₂H₄O₂ + 4H⁺ + 4e^–) × 3

(Cr₂O₇^2- + 14H⁺ + 6e^– → 2Cr³⁺ + 7H₂O) × 2

Step 5. Combine and simplify two half reactions.

3C₂H₆O + 3H₂O + 2Cr₂O₇^2- + 28H⁺ + 12e^– → 3C₂H₄O₂ + 12H⁺ + 4Cr³⁺ + 14H₂O + 12e^–

3C₂H₆O + 2Cr₂O₇^2- + 16H⁺ → 3C₂H₄O₂ + 4Cr³⁺ + 11H₂O

Example 6

Convert molecular chemical equation CH₃CH₂CH₃ + O₂ → CO₂ + H₂O to C₃H₈ + O₂ → CO₂ + H₂O.

Step 1. Divide into two half reactions.

C₃H₈ → CO₂

O₂ → H₂O

Step 2. Balance all atoms in two half reactions.

C₃H₈ → CO₂

C₃H₈ → 3CO₂

C₃H₈ + 6O + 12H⁺ → 3CO₂ + 8H⁺ + 12H⁺

C₃H₈ + 6H₂O → 3CO₂ + 20H⁺

O₂ → H₂O

O₂ → 2H₂O

O₂ + 4H⁺ → 2H₂O

Step 3. Balance charges of two half reactions by adding electrons.

C₃H₈ + 6H₂O → 3CO₂ + 20H⁺ + 20e^–

O₂ + 4H⁺ + 4e^– → 2H₂O

Step 4. Make electrons of the two half reactions equivalent (LCM = 20).

(C₃H₈ + 6H₂O → 3CO₂ + 20H⁺ + 20e^–) × 1

(O₂ + 4H⁺ + 4e^– → 2H₂O) × 5

Step 5. Combine and simplify two half reactions.

C₃H₈ + 6H₂O + 5O₂ + 20H⁺ + 20e^– → 3CO₂ + 20H⁺ +10H₂O + 20e^–

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Example 7

Convert ionic chemical equation CH₃CHO + Cu²⁺ → CH₃COO^– + Cu₂O (in basic medium) to C₂H₄O + Cu²⁺ → C₂H₃O₂^– + Cu₂O.

Step 1. Divide into two half reactions.

C₂H₄O → C₂H₃O₂^–

Cu²⁺ → Cu₂O

Step 2. Balance all atoms in two half reactions.

C₂H₄O → C₂H₃O₂^–

C₂H₄O + O → C₂H₃O₂^– + H⁺

C₂H₄O + O + 2H⁺ → C₂H₃O₂^– + H⁺ + 2H⁺

C₂H₄O + H₂O → C₂H₃O₂^– + 3H⁺

Cu²⁺ → Cu₂O

2Cu²⁺ + O → Cu₂O

2Cu²⁺ + O + 2H⁺ → Cu₂O + 2H⁺

2Cu²⁺ + H₂O → Cu₂O + 2H⁺

Step 3. Add electrons to balance charges of half reactions.

C₂H₄O + H₂O → C₂H₃O₂^– + 3H⁺ + 2e^–

2Cu²⁺ + H₂O + 2e^– → Cu₂O + 2H⁺

Step 4. Make electrons of two half reactions equivalent (LCM = 2).

(C₂H₄O + H₂O → C₂H₃O₂^– + 3H⁺ + 2e^–) ×١

(2Cu²⁺ + H₂O + 2e^– → Cu₂O + 2H⁺) ×1

Step 5. Combine and simplify into two half reactions.

C₂H₄O + H₂O + 2Cu²⁺ + H₂O + 2e^– → C₂H₃O₂^– + 3H⁺ + Cu₂O + 2H⁺ + 2e^–

C₂H₄O + 2Cu²⁺ + 2H₂O → C₂H₃O₂^– + Cu₂O + 5H⁺

Step 6. Convert H⁺ to OH^–.

C₂H₄O + 2Cu²⁺ + 2H₂O → C₂H₃O₂^– + Cu₂O + 5H⁺

C₂H₄O + 2Cu²⁺ + 2H₂O + 5OH^– → C₂H₃O₂^– + Cu₂O + 5H⁺ + 5OH^-

C₂H₄O + 2Cu²⁺ + 2H₂O + 5OH^– → C₂H₃O₂^– + Cu₂O + 5H₂O

C₂H₄O + 2Cu²⁺ + 5OH^– → C₂H₃O₂^– + Cu₂O + 3H₂O

Calculation of

By using the oxidation method, the oxidation numbers of atoms are first assigned, then the change of oxidation numbers and the number of transferred electrons are counted, and finally the given equation is balanced. The proton method works in a reverse direction. By balancing an equation, the number of transferred electrons can be determined first, then the change of oxidation numbers can be calculated. Here are two half reactions from Example 7 that demonstrate calculations:

Example 7a

Given the balanced half redox reaction: Reduction 2Cu²⁺ + H₂O + 2e^– → Cu₂O + 2H⁺

In 2Cu²⁺ + H₂O + 2e^– → Cu₂O + 2H⁺, there are two copper atoms gaining two electrons in the half reduction reaction. The change of mean copper oxidation numbers (ΔON_Cu) from Cu²⁺ (ON_i) to Cu₂O (ON_f) equals –1.

Example 7b

Given the balanced half redox reaction: Oxidation

C₂H₄O + H₂O → C₂H₃O₂^– + 3H⁺ + 2e^–

In the balanced half equation of C₂H₄O + H₂O → C₂H₃O₂^– + 3H⁺ + 2e^–, two carbon atoms carry the oxidation numbers change and lose two electrons. The calculated change of mean carbon oxidation numbers (ΔONc) equals +1.

Example 8

Determine the change of mean carbon oxidation numbers (ΔONc) for a redox couple of C₆H₅CH₃ and C₆H₅COOH.

Step 1. Balance atoms.

C₆H₅CH₃ → C₆H₅COOH

C₇H₈ → C₇H₆O₂

C₇H₈ + 2O → C₇H₆O₂ + 2H⁺

C₇H₈ + 2O + 4H⁺ → C₇H₆O₂ + 2H⁺ + 4H⁺

C₇H₈ + 2H₂O → C₇H₆O₂ + 6H⁺

Step 2. Balance charges.

C₇H₈ + 2H₂O → C₇H₆O₂ + 6H⁺

C₇H₈ + 2H₂O → C₇H₆O₂ + 6H⁺ + 6e^–

Step 3. Determine change of mean carbon oxidation numbers.

Example 9

Given a half redox reaction C₆H₁₂O₆ → CO₂, determine the change of mean carbon oxidation numbers (ΔONc).

Step 1. Balance atoms.

C₆H₁₂O₆ → CO₂

C₆H₁₂O₆ → 6CO₂

C₆H₁₂O₆ + 6O → 6CO₂ + 12H⁺

C₆H₁₂O₆ + 6O + 12H⁺ → 6CO₂ + 12H⁺ + 12H⁺

C₆H₁₂O₆ + 6H₂O → 6CO₂ + 24H⁺

Step 2. Balance charges.

C₆H₁₂O₆ + 6H₂O → 6CO₂ + 24H⁺

C₆H₁₂O₆ + 6H₂O → 6CO₂ + 24H⁺ + 24e^–

Step 3. Determine change of mean carbon oxidation numbers.

C₆H₁₂O₆ + 6H₂O → 6CO₂ + 24H⁺ + 24e^– n_C = 6; Te^– = +24

Conclusion

By using a non-oxidation number approach, the proton method can be used to balance atoms and charges in redox reactions for both molecular and ionic chemical equations. Regarding a balanced half redox equation, the mathematic equation of Te^– = n ΔON is established. This offers a new approach to count the change of mean oxidation numbers for an assigned atom in a balanced half reaction.

Pong Kau Yuen (pongkauyuen@yahoo.com) is an adjunct professor in the Department of Chemistry at Texas Southern University in Houston, Texas, and the Chairperson of the Macau Chemical Society, Macao. Cheng Man Diana Lau is a member of the Macau Chemical Society, Macao.