Definition of Ion-Electron Method (balance)

The method It has a series of steps that must be carried out in order to establish a correct balance of the species. This procedure can be divided into the following stages:

1) Write the entire reaction we wish to balance. In turn, if possible, distinguish the species that make up the compounds and rewrite the reaction in its ionic form, with the charged species.

2) Write the half-reactions that make up the overall reaction. This involves putting the reactants and products into two different half-reactions and to identify which one is the oxidation and which of them the reduction. For this, we must understand that the species that loses electrons and remains positively charged, increases its oxidation state, therefore, it is the oxidation half-reaction. Meanwhile, the species that gains electrons decreases its oxidation state, so it is the reduction half-reaction.

3) Write the balanced half-reactions, this implies completing with the electrons in play and, If necessary, rewrite them so that the same amount of money is at stake in each one. electrons. For this, it may be necessary to find a minimum coefficient that allows equalization.

4) Write the global reaction as the sum of the previous half-reactions. If the above steps were done correctly, the electrons on either side of the reaction should cancel. Finally, the reaction is balanced.

Typical example

\(A{{l}_{\left( s \right)}}+CuS{{O}_{4}}_{\left( ac \right)}\to ~A{{l}_{2 }}{{\left( S{{O}_{4}} \right)}_{3}}_{\left( ac \right)}+~C{{u}_{\left( s \ right)}}~\)

1) We identify oxidation states:

• \(A{{l}_{\left( s \right)}}\) oxidizes when passing to \(A{{l}^{+3}}\) (First, Aluminum is in a state of oxidation 0 and goes to +3)

• \(C{{u}^{+2}}\) reduces to \(C{{u}_{\left( s \right)}}\) (First, Copper is in a state of oxidation +2 and goes to 0)

2) We ionize the compounds and identify oxidation and reduction reactions individually:

\(A{{l}_{\left( s \right)}}^{0}+~C{{u}^{+2}}_{\left( ac \right)}~\to ~A {{l}^{+3}}_{\left( ac \right)}+C{{u}_{\left( s \right)}}^{0}\)

Aluminum is the species that is being oxidized, while copper is the species that is being reduced.

3) This step consists of writing the balanced half-reactions:

• \(A{{l}_{\left( s \right)}}^{0}\to ~A{{l}^{+3}}_{\left( ac \right)}+3~ {{e}^{-}}~\) Oxidation

• \(C{{u}^{+2}}_{\left( ac \right)}+2~{{e}^{-}}\to ~C{{u}_{\left( s \right)}}^{0}~\) Reduction

4) If we observe, the half-reactions do not involve the same number of electrons in play, so we must balance them in such a way that the charges to be exchanged in both are equal:

• \(2~x~\left( A{{l}_{\left( s \right)}}^{0}\to ~A{{l}^{+3}}_{\left( ac \right)}+3~{{e}^{-}} \right)~\) Oxidation

• \(3~x~(C{{u}^{+2}}_{\left( ac \right)}+2~{{e}^{-}}\to ~C{{u}_ {\left( s \right)}}^{0})~\) Reduction

In abstract:

• \(2A{{l}_{\left( s \right)}}^{0}\to ~2A{{l}^{+3}}_{\left( ac \right)}+6~ {{e}^{-}}~\) Oxidation

• \(3C{{u}^{+2}}_{\left( ac \right)}+6~{{e}^{-}}\to ~3C{{u}_{\left( s \right)}}^{0}~\) Reduction

5) Finally, we will write the global balanced reaction, as the sum of the previous reactions:

\(2A{{l}_{\left( s \right)}}^{0}+~3C{{u}^{+2}}_{\left( ac \right)}\to ~2A{ {l}^{+3}}_{\left( ac \right)}+~3C{{u}_{\left( s \right)}}^{0}\)

We rewrite the equation above with the original compounds:

\(2A{{l}_{\left( s \right)}}+3CuS{{O}_{4}}_{\left( ac \right)}\to ~A{{l}_{2 }}{{\left( S{{O}_{4}} \right)}_{3}}_{\left( ac \right)}+~3C{{u}_{\left( s \ right)}}\)

There are two particular cases, where the reactions can occur in acidic or basic media. For those cases, the treatment it is somewhat different since it requires the addition of species that allow equalizing the reaction.

In the case of the acid medium, you must enter Water for the balance of oxygens and hydrogens and, therefore, we will see the presence of protons (H+) that will indicate the type of medium. While, in a basic medium, the addition of OH- (hydroxyl) may be required for the correct balancing.

Let's look at an example

\(Cu{{S}_{\left( ac \right)}}+HN{{O}_{3}}_{\left( ac \right)}\to ~Cu{{\left( N{ {O}_{3}} \right)}_{2}}_{\left( ac \right)}+~N{{O}_{2}}_{\left( g \right)}+S{{O}_{2}}_{\left( g \right)}+~ {{H}_{2}}{{O}_{\left( ac \right)}}\)

In the presence of nitric acid we are working in an acid medium.

1) First we will identify oxidation states:

• \(~{{S}^{-2}}\) is oxidized by passing to \({{S}^{+4}}\) (Firstly, the Sulfur is in oxidation state -2 and passes to +4)

• \({{N}^{+5}}\) is reduced when passing to \({{N}^{+4}}\) (First, Nitrogen is in oxidation state +5 and passes to + 4)

2) We ionize the compounds and identify oxidation and reduction reactions individually:

\({{S}^{-2}}_{\left( ac \right)}+~{{N}^{+5}}_{\left( ac \right)}~\to ~{{ S}^{+4}}_{\left( g \right)}+~{{N}^{+4}}_{\left( g \right)}\)

Sulfur is the species that is being oxidized, while Nitrogen is the species that is being reduced.

3) We write the balanced half-reactions:

• \(~\) \(2~{{H}_{2}}{{O}_{\left( ac \right)}}+~{{S}^{-2}}_{\left ( ac \right)}~\to ~S{{O}_{2}}_{\left( g \right)}+4{{H}^{+}}_{\left( ac \right) }+6~{{e}^{-}}\) Oxidation

• \(2{{H}^{+}}_{\left( ac \right)}+\) \(N{{O}_{3}}{{^{-}}_{\left( ac \right)}}+1~{{e}^{-}}~\to ~N{{O}_{2}}_{\left( g \right)}+~~{{H}_ {2}}{{O}_{\left( ac \right)}}~\) Reduction

As can be seen, the addition of water was necessary in the oxidation reaction for the correct balance of hydrogens and oxygens.

4) If we observe, the half-reactions do not involve the same number of electrons in play, so we must balance them in such a way that the charges to be exchanged in both are equal:

• \(~\) \(2~{{H}_{2}}{{O}_{\left( ac \right)}}+~{{S}^{-2}}_{\left ( ac \right)}~\to ~S{{O}_{2}}_{\left( g \right)}+4{{H}^{+}}_{\left( ac \right) }+6~{{e}^{-}}\) Oxidation

• \(12{{H}^{+}}_{\left( ac \right)}+\) \(6N{{O}_{3}}{{^{-}}_{\left( ac \right)}}+6~{{e}^{-}}~\to ~6N{{O}_{2}}_{\left( g \right)}+~~6{{H} 2}}{{O}_{\left( ac \right)}}~\) Reduction

5) Finally, we express the global balanced reaction, in response to the sum of the reactions addressed:

\(2~{{H}_{2}}{{O}_{\left( ac \right)}}+~{{S}^{-2}}_{\left( ac \right)} +~12{{H}^{+}}_{\left( ac \right)}+\) \(6N{{O}_{3}}{{^{-}}_{\left( ac \right)}}\to ~S{{O}_{2}}_{\left( g \right)}+4{{H}^{+}}_{\left( ac \right)}+ 6N{{O}_{2}}_{\left( g \right)}+~~6{{H}_{2}}{{O}_{\left( ac \right)}}\)

We rewrite the previous equation with the original compounds, taking into account that there are species, such as H+, which appear both in reactants and products and, therefore, part of them are cancel

$Cu\{\{S\}\_\{\backslash left(\; ac\; \backslash right)\}\}+8HN\{\{O\}\_\{3\}\}\_\{\backslash left(\; ac\; \backslash right)\}\backslash to\; ~Cu\{\{\backslash left(\; N\{\{O\; \}\_\{3\}\}\; \backslash right)\}\_\{2\}\}\_\{\backslash left(\; ac\; \backslash right)\}+~6N\{\{O\}\_\{2\}\}\_\{\backslash left(\; g\; \backslash right)\}+S\{\{O\}\_\{2\}\}\_\{\backslash left(\; g\; \backslash right)\}+~\; 4\{\{H\}\_\{2\}\}\{\{O\}\_\{\backslash left(\; ac\; \backslash right)\}\}\backslash )$