Oxidation-reduction balancing. What is redox balancing?

O balancing an oxidation-reduction equation it is based on the equality of the number of electrons given away with the number of electrons received. A simple method of performing this balancing is given by the following steps:

Let's see in practice how to apply these steps, through the following example:

Reaction between an aqueous solution of potassium permanganate and hydrochloric acid:

kmnO₄ + HCl → KCl + MnCl₂ + Cl₂ + H₂O

*1st step:Determine oxidation numbers:

This step is important because we usually cannot quickly visualize which species undergo oxidation and reduction.

_{+1 +7 -2 +1 -1 +1 -1 +2 -1 0 +1 -2}
kmnO₄ + HCl → KCl + MnCl₂ + Cl₂ + H₂O

*2nd step:Determination of oxidation and reduction variation:

Note that manganese (Mn) is reduced and chlorine (Cl) is oxidized.

MnCl₂ = ∆Nox = 5

Cl₂ = ∆Nox = 2

In the case of chlorine, we can note that HCl gave rise to 3 compounds (KCl, MnCl₂, and Cl₂), but what interests us is the Cl_2, because it is your Nox that has suffered variation. Each chlorine that forms Cl

₂ lose 1 electron; how it takes 2 chlorines to form each Cl₂, then two electrons are lost.

3rd step:Inversion of ∆ values:

In this step, the values ​​of ∆ are exchanged between the mentioned species, becoming their coefficients:

MnCl₂ = ∆Nox = 5 → 5 will be the coefficient of Cl₂

Cl₂ = ∆Nox = 2→ 2 will be the coefficient of MnCl₂

kmnO₄ + HCl → KCl + 2 MnCl₂ + 5 Cl₂ + H₂O

At this point it is already possible to know two coefficients of the equation.

Observation: normally, in most reactions, this reversal of values ​​is performed on the 1st member. But, as a general rule, this should be done in the member that has the greatest number of atoms that undergo redox. If this criterion cannot be met, we invert the values ​​for the member with the highest number of chemical species. This is what was done here, as the 2nd member has more substances.

4th step: Trial balancing:

kmnO₄ + HCl → KCl + 2 MnCl₂ + 5 Cl₂ + H₂O

• Since in the second member there are two manganese atoms, as shown by the coefficient, in the first there must also be. So we have:

2 kmnO₄ + HCl → KCl + 2 MnCl₂ + 5 Cl₂ + H₂O

• Thus, the amount of potassium (K) in the 1st member was 2, which will be the same coefficient for this atom in the second member:

2 kmnO₄ + HCl → 2 KCl + 2 MnCl₂ + 5 Cl₂ + H₂O

• The amount of chlorine (Cl) in the 2nd member is 16 in total, so the HCl coefficient of the 1st member will be:

2 kmnO₄ + 16 HCl → 2 KCl + 2 MnCl₂ + 5 Cl₂ + H₂O

• The number of hydrogens in the 1st member is 16, hence the coefficient of water (H₂O) of the 2nd member will be equal to 8, since the multiplication of the hydrogen index (2) by 8 is equal to 16:

2 kmnO₄ + 16 HCl → 2 KCl + 2 MnCl₂ + 5 Cl₂ + 8 H₂O

• To check if the equation is correctly balanced we can see two criteria:

1st) Check if the amount of each atom in the two members is equal:

2 kmnO₄ + 16 HCl →2 KCl + 2 MnCl₂ + 5 Cl₂ + 8 H₂O

K =2K =2

Mn = 2 Mn = 2

Cl = 16  Cl = 16

H = 16 H = 16

O = 8 O = 8

2nd) See if the total number of lost electrons is equal to the total number of electrons received:

By Jennifer Fogaça
Graduated in Chemistry

Source: Brazil School - https://brasilescola.uol.com.br/quimica/balanceamento-por-oxirreducao.htm