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## From Balancing Redox Reactions to Determining Change of Oxidation Numbers

By Pong Kau Yuen and Cheng Man Diana Lau

##### Redox reaction is a core concept in teaching and learning chemistry. This article explores a new method for balancing organic redox reactions that requires the balancing of both atoms and charges. The H+, O, H2O, and e– are used as balanced vehicles in two half reactions. A non-oxidation number approach can be applied to both molecular and ionic equations. The article also provides standard operating procedures and examples. The number of transferred electrons is first determined by balancing a half redox reaction; consequently; the change of oxidation numbers can be calculated. The mathematical equation of Te– = n Te– = n ΔON is established, and the change of oxidation numbers (ΔON) can be counted by the number of transferred electrons (Te–) and the number of atoms with oxidation numbers change (n). By using this mathematical equation as a new approach, students can conveniently calculate the change of mean oxidation numbers for an assigned atom in a half redox reaction.

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 Cr2O72- → Cr3+ (in acidic medium):

Step 1. Balance atoms.

Cr2O72- → Cr3+

Cr2O72- → 2Cr3+

Cr2O72- → 2Cr3+ + 7O

Cr2O72- + 14H+ → 2Cr3+ + 7O + 14H+

Cr2O72- + 14H+ → 2Cr3+ + 7H2O

Step 2. Balance charges.

Cr2O72- + 14H+ → 2Cr3+ + 7H2O

(+12) (+6)

Cr2O72- + 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 Cr2O72- + 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 (ONf) and the oxidation number of an atom on the reactant side (ONi), and it is demonstrated by ΔON = ONf – ONi. 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 + Cr2O72- → CH3COOH + Cr3+ (in acidic medium) to C2H6O + Cr2O72- → C2H4O2 + Cr3+.

Step 1. Divide into two half reactions:

C2H6O → C2H4O2

Cr2O72- → 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+

Cr2O72-→ 2Cr3+

Cr2O72- → 2Cr3+ + 7O

Cr2O72- + 14H+ → 2Cr3+ + 7O + 14H+

Cr2O72- + 14H+ → 2Cr3+ + 7H2O

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

C2H6O + H2O → C2H4O2 + 4H+ + 4e

Cr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O

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

(C2H6O + H2O → C2H4O2 + 4H+ + 4e) × 3

(Cr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O) × 2

Step 5. Combine and simplify two half reactions.

3C2H6O + 3H2O + 2Cr2O72- + 28H+ + 12e → 3C2H4O2 + 12H+ + 4Cr3+ + 14H2O + 12e

3C2H6O + 2Cr2O72- + 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 (ΔONCu) from Cu2+ (ONi) to Cu2O (ONf) 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+ + 24enC = 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.

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