Self-Reduction Reactions

THE auto-oxi-reduction or disproportionate reaction is a type of redox reaction in which the same chemical element undergoes oxidation and reduction. Let's look at two examples of this type of reaction and how to balance them using the redox method:

1st Example:

AT THE₂^- + H⁺ → NO₃^- + NO + H₂O

• By calculating the oxidation numbers (NOX) of all atoms and ions involved in the reaction, it is possible to verify who oxidized and who reacted:

_{+3 -2 +1 +5 -2 +2 -2 +1 -2}
AT THE₂^- + H⁺ → NO₃^- + NO + H₂O

• Note that nitrogen was the species that both reacted and oxidized:

• To carry out the redox balancing of this reaction, we have to relate the NOX to the products, not the reagents:

AT THE₃^- =∆Nox = 5 - 3 = 2

NO=∆Nox = 3 - 2 = 1

• Inverting the ∆NOX by the coefficients, we have:

AT THE₃^- =∆NOX= 2 → 2 will be the coefficient of NO

NO=∆NOX= 1→ 1 will be the coefficient of NO₃^-

AT THE₂^- + H⁺ → 1 AT THE₃^- + 2 NO+H₂O

• With this, we already know that there are 3 N in the product, so the coefficient of NO₂^- will be 3:

3 NO₂^- + H⁺ → 1 NO₃^- + 2 NO + H₂O

• To determine the coefficients of H⁺ and from H₂O, remember that the number of electrons received equals the same amount of electrons donated; thus, the reagent charge will be equal to the product charge. In this way, we can make the following scheme:

Based on this information, we have that the total charge of the reagents is equal to x – 3 and the product is equal to -1. As stated, the charges of the two have to be equal. As we already have the total load of products, we can perform a simple calculation to know what the value of x will be:

x -3 = -1

x = -1 +3

x = 2

Thus, the coefficient of H⁺ is 2 and, consequently, that of H₂The will be 1:

3 NO₂^- + 2 H⁺ → 1 NO₃^- + 2 NO + 1 H₂O

2nd Example:

In this case, it was the S that underwent reduction and oxidation at the same time. Thus, as previously done, we can relate the NOX to the products and invert their values, assigning the coefficients to them:

At₂ S=∆Nox = 4 – (-2) = 6 → 6 will be the coefficient of Na₂ ONLY₄

At₂ ONLY₄=∆Nox = 6 - 4 = 2 → 2 will be the coefficient of Na₂ s

At₂ ONLY₃→ 2 At₂ Y+ 6 At₂ ONLY₄

Since there are 8 sulfurs in the 2nd limb, the Na coefficient₂ ONLY₃ will be 8:

8 In₂ ONLY₃→ 2 In₂ S + 6 In₂ ONLY₄

By Jennifer Fogaça
Graduated in Chemistry

Source: Brazil School - https://brasilescola.uol.com.br/quimica/reacoes-auto-oxirreducao.htm