7.1: Balancing Redox Reactions

Balancing Redox Reactions

In studying redox chemistry, it is important to remember balancing electrochemical reactions. Simple redox reactions (for example, H₂ + I₂ → 2 HI) can be balanced by inspection, but for more complex reactions it is helpful to have a systematic method. There are many different methods for doing this. The focus in this section will be on the ion-electron method allows one to balance redox reactions regardless of their complexity and allows for identification of the species changing valence states in the reaction. We illustrate this method with two examples. For a review of how to determine oxidation states see Oxidation States (Oxidation Numbers).

Steps to balance a redox reaction

1. Identify what is being oxidized and reduced and divide the reaction into two half reactions
2. For each half reaction, balance all the atoms EXCEPT oxygen and hydrogen.
3. For each half reaction, balance oxygen by adding H₂O
4. For each half reaction, balance hydrogen by adding H⁺
5. For each half reaction, balance charge by adding electrons
6. Add the two half reaction together. If needed, multiply half reactions by coefficients so that there are an equal number of electrons in both half reactions
7. If the reaction is in BASIC solution, cancel any H⁺ by adding OH^- to both sides or the reaction

Example \(\PageIndex{1}\)

Determine the balanced redox reaction that occurs when I^- reacts with MnO₄^- to form IO₃^- and Mn²⁺ in basic solution.

Solution

Following the steps above:

Step 1. The I^- is oxidized and the MnO₄^- is reduced so the two half reactions are:

MnO₄^- → Mn²⁺

I^- → IO₃^-

Step 2. The Mn and I are already balanced in the half reactions.

Step 3. Add H₂O to balance the O in each half reaction:

MnO₄^- → Mn²⁺ + 4 H₂O

3 H₂O + I^- → IO₃^-

Step 4. Add H⁺ to balance H in each half reaction:

8 H⁺ + MnO₄^- → Mn²⁺ + 4 H₂O

3 H₂O + I^- → IO₃^- + 6 H⁺

Step 5. Add e^- to balance the charge in each half reaction:

8 H⁺ + MnO₄^- + 5 e^- → Mn²⁺ + 4 H₂O

3 H₂O + I^- → IO₃^- + 6 H⁺ + 6 e^-

Notice the electrons are on opposite sides in the two half reactions

Step 6. Multiply the half reactions by coefficients and add together. All the electrons should cancel in the final reaction:

Multiplying the top by 6 and bottom by 5 will give 30 electrons in each reaction

6(8 H⁺ + MnO₄^- + 5 e^- → Mn²⁺ + 4 H₂O)

5(3 H₂O + I^- → IO₃^- + 6 H⁺ + 6 e^-)

Add and cancel common species

48 H⁺ + 6 MnO₄^- + 15 H₂O + 5 I^- + 30 e^- → 6 Mn²⁺ + 5 IO₃^- + 30 H⁺ + 30 e^- + 24 H₂O

18 H⁺ + 6 MnO₄^- + 5 I^- → 6 Mn²⁺ + 5 IO₃^- + 9 H₂O

Step 7. The reaction is in basic solution so add OH^- to each side:

18 OH^- + 18 H⁺ + 6 MnO₄^- + 5 I^- → 6 Mn²⁺ + 5 IO₃^- + 9 H₂O + 18 OH^-

OH^- + H⁺ → H₂O

9 H₂O + 6 MnO₄^- + 5 I^- → 6 Mn²⁺ + 5 IO₃^- + 18 OH^-

Example \(\PageIndex{2}\)

Balance the redox reaction that occurs when S₂O₃^2- reacts with H₂O₂ to form S₄O₆^2- and water in acidic solution.

Solution

Following the steps above:

Step 1. The S₂O₃^2- is oxidized and the H₂O₂ is reduced so the two half reactions are:

S₂O₃^2- → S₄O₆^2-

H₂O₂ → H₂O

Step 2. Balance the S.

2 S₂O₃^2- → S₄O₆^2-

H₂O₂ → H₂O

Step 3. Add H₂O to balance the O in each half reaction:

2 S₂O₃^2- → S₄O₆^2-

H₂O₂ → 2 H₂O

Step 4. Add H⁺ to balance H in each half reaction:

2 S₂O₃^2- → S₄O₆^2-

2H⁺ + H₂O₂ → 2 H₂O

Step 5. Add e^- to balance the charge in each half reaction:

2 S₂O₃^2- → S₄O₆^2- + 2 e^-

2H⁺ + 2 e^- + H₂O₂ → 2 H₂O

Notice the electrons are on opposite sides in the two half reactions

Step 6. The number of electrons is already the same on both sides. Add and cancel common species:

2 S₂O₃^2- + 2H⁺ + 2 e^- + H₂O₂ → S₄O₆^2- + 2 e^- + 2 H₂O

2 S₂O₃^2- + 2H⁺ + H₂O₂ → S₄O₆^2- + 2 H₂O

Note that we did not need to know the oxidation states of S or O in the reactants and products in order to balance the reaction. In this case, assigning the valence states is a bit complex, because S₄O₆^2- contains sulfur in more than one oxidation state.