H-atom and O-atom methods: from balancing redox reactions to determining the number of transferred electrons

H-atom method: procedures

When using the H-atom method, H, O, and H₂O are used as balancing devices. This method requires the use of a molecular chemical equation. If an ionic chemical equation is provided, it must be converted to a molecular chemical equation by adding proper spectator ions. H₂O can be omitted in the process optionally. When both half and overall chemical equations are balanced, H₂O will appear consequently through the working procedures to fulfill the conservational law of matter, which is one of the advantages of using the H-atom and O-atom methods. It is important to emphasize that these two methods are mathematically balancing methods. The H-atom and O-atom may not necessarily act as intermediates, and the H-atom and O-atom methods do not represent any reaction mechanisms.

1. Divide into two half reactions

2. Balance all atoms in the two half reactions

   1. Balance all other atoms except H and O

   2. Balance the oxygen atoms with O

   3. Balance the hydrogen atoms with H

   4. Add two H atoms for each O atom

   5. Convert two H atoms and one O atom to one H₂O molecule

3. Make the H-atom of the two half reactions equivalent

4. Combine the two half reactions

5. Simplify the overall chemical equation

In this process, each H-atom is equivalent to the gain/loss of one electron. In example 4, C₇H₈ + KMnO₄ + H₂SO₄ → C₇H₆O₂ + K₂SO₄ + MnSO₄, one half of the equation “C₇H₈ + 2H₂O → C₇H₆O₂ + 6H” releasing six H-atoms represents the loss of six electrons, and the other half equation “2KMnO₄ + 3H₂SO₄ + 10H → K₂SO₄ + 2MnSO₄ + 8H₂O” obtaining ten H-atoms represents the gain of 10 electrons. KMnO₄ and C₇H₈ function as oxidizing agent and reducing agent respectively.

In example 5, C₂H₆O + Cr₂O₇ ²⁻ + H⁺ → C₂H₄O₂ + Cr³⁺, the ionic equation is converted to a molecular chemical equation, C₂H₆O + K₂Cr₂O₇ + H₂SO₄ → C₂H₄O₂ + Cr₂(SO₄)₃ + K₂SO₄, by adding spectator ions K⁺ and SO₄ ²⁻. One half of the oxidation reaction “C₂H₆O + H₂O → C₂H₄O₂ + 4H” releasing four H-atoms represents the loss of four electrons, and the other half reduction reaction “K₂Cr₂O₇ + 4H₂SO₄ + 6H → Cr₂(SO₄)₃ + K₂SO₄ + 7H₂O” obtaining six H-atoms represents the gain of six electrons. K₂Cr₂O₇ and C₂H₆O act as oxidizing agent and reducing agent respectively.

In example 6, P₂I₄ + P₄ + H₂O → PH₄I + H₃PO₄, one half of the reduction reaction “2P₂I₄ + P₄ + 32H → 8PH₄I” gaining thirty-two H-atoms represents the gain of 32 electrons, and the other half oxidation reaction of “P₄ + 16H₂O → 4H₃PO₄ + 20H” releasing twenty H-atoms represents the loss of 20 electrons.

In example 7, Fe₃C + HNO₃ → Fe(NO₃)₃ + CO₂ + NO₂, one half of the oxidation reaction “Fe₃C + 9HNO₃ + 2H₂O → 3Fe(NO₃)₃ + CO₂ + 13H” releasing 13 H-atoms represents the loss of 13 electrons, and the other half reduction reaction of “HNO₃ + H → NO₂ + H₂O” gaining one H-atom represents the gain of one electron. Although example 7 contains uncertain oxidation numbers of a reactant Fe₃C, it can still be balanced by the H-atom method.

Stoichiometric number of transferred H-atoms in half redox reactions

For examples 4 through 7, the half redox reactions and their stoichiometric number of transferred H-atoms are presented and defined in Table 2.

According to the H-atom method, the gain/loss of one H is identical to the gain/loss of one electron (H → H⁺ + e⁻). The number of transferred H-atoms (TH) is equal to the number of transferred electrons (Te⁻) as shown in Table 3. TH can function as an electron-counting concept, Te⁻ = 1 × TH.

The O-atom method: procedures

The procedures of the O-atom method are similar to that of the H-atom method. The O-atom, rather than the H-atom, acts as the dominant balancing device. The differences in their procedures are illustrated in Step 2d, Step 2e, and Step 3.

1. Divide into two half reactions

2. Balance all atoms in the two half reactions

   1. Balance all other atoms except H and O

   2. Balance the oxygen atoms with O

   3. Balance the hydrogen atoms with H

   4. Add one O atom for two H atoms

   5. Convert one O atom and two H atoms to one H₂O molecule

3. Make the O-atom of the two half reactions equivalent

4. Combine the two half reactions

5. Simplify the overall chemical equation

The O-atom method: examples

In this process, each O-atom is equivalent to the gain/loss of two electrons. By using example 8, C₇H₈ + KMnO₄ + H₂SO₄ → C₇H₆O₂ + K₂SO₄ + MnSO₄, as a demonstration, in the half oxidation reaction of “C₇H₈ + 3O → C₇H₆O₂ + H₂O”, the gain of three O-atoms represents the loss of six electrons. In the other half reduction reaction of “2KMnO₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 8H₂O + 5O”, the loss of five O-atoms represents the gain of 10 electrons.

By using example 9, C₂H₆O + K₂Cr₂O₇ + H₂SO₄ → C₂H₄O₂ + Cr₂(SO₄)₃ + K₂SO₄, as a demonstration, in the half reaction of “C₂H₆O + 2O → C₂H₄O₂ + H₂O”, the gain of two O-atoms represents the loss of four electrons. In the other half reaction of “K₂Cr₂O₇ + 4H₂SO₄ → Cr₂(SO₄)₃ + K₂SO₄ + 4H₂O + 3O”, the release of three O-atoms represents the gain of six electrons.

By using example 10, P₂I₄ + P₄ + H₂O → PH₄I + H₃PO₄, as a demonstration, in the half reduction reaction of “2P₂I₄ + P₄ + 16H₂O → 8PH₄I + 16O”, the loss of sixteen O-atoms represents the gain of 32 electrons. In the other half reduction reaction of “8P₄ + 48H₂O + 80O → 32H₃PO₄”, the gain of 80 O-atoms represents the loss of 160 electrons.

By using example 11, Pb(N₃)₂ + Cr(MnO₄)₂ → Pb₃O₄ + NO + Cr₂O₃ + MnO₂, as a demonstration, in the half oxidation reaction of “3Pb(N₃)₂ + 22O → Pb₃O₄ + 18NO”, the gain of 22 O-atoms represents the loss of 44 electrons. In the other half reduction reaction of “2Cr(MnO₄)₂ → Cr₂O₃ + 4MnO₂ + 5O”, the loss of five O-atoms represents the gain of 10 electrons. Although example 11 contains multiple redox couples, it can still be balanced by the O-atom method.

Stoichiometric number of transferred O-atoms in half redox reactions

For examples 8 through 11, the half redox reactions and their stoichiometric number of transferred O-atoms are summarized and defined in Table 4.

According to the O-atom method, the gain/loss of one O is identical to the loss/gain of two electrons (O + 2e⁻ → O²⁻). The number of transferred O-atoms (TO) can be represented by Te⁻ as shown in Table 5. TO acts as an electron-counting concept, Te⁻ = −2 × TO.

Electron-counting concepts: ΔON, TH, and TO

The notion of electron transfer is the core of redox reactions. The mathematical relationship of Te⁻ = n × ΔON has been established and ΔON acts as an electron-counting concept in a half redox reaction (Yuen & Lau, 2022). TH and TO are demonstrated as electron-counting concepts by the H-atom and O-atom methods. The relationships among TH, TO, and ΔON are centered around Te⁻ as shown in Figure 1. The nature of the electron-counting concepts of TH, TO, and ΔON are summarized in Table 6.

Figure 1:

Mathematical relationships among Te⁻, ΔON, TH, and TO.

TH, TO, Te⁻, and ΔON in molecular and ionic chemical equations

The balanced half reduction reaction in molecular chemical equation (from example 4, H-atom method) is converted to ionic chemical equation as follows:

The balanced half reduction reaction in molecular chemical equation (from example 8, O-atom method) is converted to ionic chemical equation as follows:

The reduction reaction of “KMnO₄ + H₂SO₄ → K₂SO₄ + MnSO₄” can be balanced by having a gain of ten H-atoms, a loss of five O-atoms, or a gain of 10 electrons. In Table 7, the half reduction molecular chemical equations are demonstrated by using either TH or TO as an electron-counting concept, and then Te⁻ can be calculated; the converted ionic chemical equation are exhibited by using Te⁻, and then ΔON can be calculated.