48 Balancing Redox Reactions in Acidic and Basic Solutions (M18Q2)

Learning Objectives

  • Balance oxidation-reduction reactions in neutral, acidic, or basic solutions using the half-reaction method.

| Key Concepts and Summary | Glossary | End of Section Exercises |

In this section, we will focus on the half-reaction method for balancing oxidation-reduction reactions. The use of half-reactions is important for balancing more complicated reactions and since many aspects of electrochemistry are easier to discuss in terms of half-reactions. The half-reaction method splits oxidation-reduction reactions into their oxidation “half” and reduction “half”  reactions and focuses on balancing each separately before recombining them to get one overall equation.

Electrochemical reactions frequently occur in solution, which could be acidic, basic, or neutral. When balancing oxidation-reduction reactions, the nature of the solution may be important. The following protocol can be used for balancing oxidation-reduction reactions under all conditions: however, note that step 5 only applies when the reaction is taking place in a basic environment.

Balancing Oxidation-Reduction Reactions Using the Half-Reaction Method

  1. Assign oxidation numbers and identify which species is oxidized and which is reduced.
  2. Write the oxidation and reduction half reactions, only including the compounds containing the element being oxidized or reduced.
  3. Balance each half-reaction.
    1. Balance elements other than O and H.
    2. Balance the O by adding H2O.
    3. Balance the H by adding H+.
    4. Balance the charges by adding electrons (e–).
    5. If the solution is basic, add enough OH– ions to BOTH sides to react with all of the H+. On the side of the reaction with both H+ and OH– ions, combine them and replace with H2O (H++ OH– → H2O).
  4. Multiply each half-reaction by an integer so that electrons gained are equal to electrons lost.
  5. Add half-reactions together.

Balancing an Oxidation-Reduction Reaction in an Acidic Environment

Consider the following unbalanced oxidation-reduction reaction in acidic solution and apply the above protocol in order to balance it:

MnO4–(aq)  +  Fe2+(aq)   →   Mn2+(aq)  +  Fe3+(aq)

Step 1: Assign oxidation numbers and identify which species is oxidized and which is reduced.

Fe2+ has lost an electron to become Fe3+; therefore, the iron is undergoing oxidation. The manganese in MnO4– gained five electrons to change from Mn7+ to Mn2+; therefore, the manganese is undergoing reduction.

Step 2: Write the oxidation and reduction half reactions, only including the compounds containing the element being oxidized or reduced.

Oxidation: Fe2+(aq)   →   Fe3+(aq)
Reduction: MnO4–(aq)   →   Mn2+(aq)

Step 3.1: Balance elements other than O and H.

Balanced in this regard so this step is not necessary in this situation: Fe2+(aq)   →   Fe3+(aq)
Balanced in this regard so this step is not necessary in this situation: MnO4–(aq)   →   Mn2+(aq)

Step 3.2: Balance the O by adding H2O.

No O in this reaction so leave it as is: Fe2+(aq)   →   Fe3+(aq)
Balance the 4 O on the reactant-side by adding 4 H2O to the product-side: MnO4–(aq)   →   Mn2+(aq)  +  4 H2O(l)

Step 3.3: Balance the H by adding H+.

No H in this reaction so leave it as is: Fe2+(aq)   →   Fe3+(aq)
Balance the 8 H on the product-side by adding 8 H+ to the reactant-side: MnO4–(aq)  +  8 H+(aq)   →   Mn2+(aq)  +  4 H2O(l)
Step 3.4: Balance the charges by adding electrons (e-)
Add 1 e– to the product side to give a +2 charge on both sides: Fe2+(aq)   →   Fe3+(aq)  +  e–
Add 5 e– to the reactant side to give a +2 charge on both sides: MnO4–(aq)  +  8 H+(aq)  +  5 e–   →   Mn2+(aq)  +  4 H2O(l)

Tip: At this point you should have electrons on opposite sides of each half-reaction, with the electrons on the product-side for the oxidation half-reaction and electrons on the reactant-side for the reduction reaction. If this isn’t the case, it is a good idea to go back and check your work!

Step 3.5: This reaction is taking place in an acidic environment so this step is not applicable.

Step 4: Multiply each half-reaction by an integer so that electrons gained are equal to electrons lost.

The lowest common number of electrons is 5 so multiply this reaction’s coefficients all by 5: 5 Fe2+(aq)   →   5 Fe3+(aq)  +  5 e–
This reaction does not need to be multiplied: MnO4–(aq)  +  8 H+(aq)  +  5 e–   →   Mn2+(aq)  +  4 H2O(l)

Step 5: Add half-reactions together

Add together to get final answer: 5 Fe2+(aq)  +  MnO4–(aq)  +  8 H+(aq)   →   5 Fe3+(aq)  +  Mn2+(aq)  +  4 H2O(l)

Balancing an Oxidation-Reduction Reaction in a Basic Environment

Now consider an oxidation-reduction reaction taking place in a basic solution:

NH3(aq)  +  ClO–(aq)   →   N2H4(aq)  +  Cl–(aq)

Step 1: Assign oxidation numbers and identify which species is oxidized and which is reduced.

Nitrogen’s oxidation number changes from -3 to -2 so it is undergoing oxidation. The chlorine’s oxidation number changes from +1 to -1 so it is undergoing reduction.

Step 2: Write the oxidation and reduction half reactions, only including the compounds containing the element being oxidized or reduced.

Oxidation: NH3(aq)   →   N2H4(aq))
Reduction: ClO–(aq)   →   Cl–(aq)

Step 3.1: Balance elements other than O and H.

Nitrogens need to be balanced so add a 2 coefficient to the reactant-side: 2 NH3(aq)   →   N2H4(aq)
Chlorines are balanced so this step is not necessary: ClO–(aq)   →   Cl–(aq)

Step 3.2: Balance the O by adding H2O.

No O in this reaction so leave it as is: 2 NH3(aq)   →   N2H4(aq)
Balance the O on the reactant-side by adding 1 H2O to the product-side: ClO–(aq)   →   Cl–(aq)  +  H2O(l)

Step 3.3: Balance the H by adding H+.

Balance the 6 H on the reactant-side by adding 2 additional H+ to the product-side: 2 NH3(aq)   →   N2H4(aq)  +  2 H+(aq)
Balance the 2 H on the product-side by adding 2 H+ to the reactant-side: ClO–(aq)  +  2 H+(aq)   →   Cl–(aq)  +  H2O(l)
Step 3.4: Balance the charges by adding electrons (e-)
Add 2 e– to the product side to give a 0 charge on both sides: 2 NH3(aq)   →   N2H4(aq)  +  2 H+(aq)  +  2 e–
Add 2 e– to the reactant side to give a 0 charge on both sides: ClO–(aq)  +  2 H+(aq)  +  2 e–   →   Cl–(aq)  +  H2O(l)

Step 3.5: If the solution is basic, add enough OH– ions to BOTH sides to react with all of the H+. On the side of the reaction with both H+ and OH– ions, combine them and replace with H2O (H++ OH– → H2O).

Add 2 OH– to the reactant and product side:

Combine H++ OH– when they are the same side to form H2O:

 2 NH3(aq)  +  2 OH–(aq)   →   N2H4(aq)  +  2 H+(aq) +  2 OH–(aq)  +  2 e–

2 NH3(aq)  +  2 OH–(aq)   →   N2H4(aq)  +  2 H2O(l)  +  2 e–
Add 2 OH– to the reactant and product side:

Combine H++ OH– when they are the same side to form H2O. Reduce H2O on both sides:

 ClO–(aq)  +  2 H+(aq)  +  2 OH–(aq)  +  2 e–   →   Cl–(aq)  +  H2O(l)  +  2 OH– (aq)

ClO–(aq)  +  H2O(l)  +  2 e–  →   Cl–(aq)  +  2 OH–(aq)

Step 4: Multiply each half-reaction by an integer so that electrons gained are equal to electrons lost.

There are an equal number of electrons so this step is not necessary: 2 NH3(aq)  +  2 OH–(aq)   →   N2H4(aq)  +  2 H2O(l)  +  2 e–
There are an equal number of electrons so this step is not necessary: ClO–(aq)  +  H2O(l)  +  2 e–   →   Cl–(aq)  +  2 OH–(aq)

Step 5: Add half-reactions together. Reduce OH– and H2O on both sides.

Add together:

Reduce to get final answer:

 2 NH3(aq)  +  2 OH–(aq)  +  ClO–(aq)  +  H2O(l)   →   N2H4(aq)  +  2 H2O(l)  +  Cl–(aq)  +  2 OH–(aq)

2 NH3(aq)  +  ClO–(aq)   →   N2H4(aq)  +  H2O(l)  +  Cl–(aq)

Key Concepts and Summary

When balancing reactions, we have often solely focused on achieving mass balance, where the number of each element on the reactant side must equal the number of that same element on the product side. In redox reactions, mass balance must be achieved along with charge balance. In order to balance redox reactions, we follow a series of steps, starting with assigning oxidation numbers and splitting the overall reaction into the oxidation and reduction half-reactions. After balancing all atoms and electrons, the final step is to ensure the reaction is balanced in acidic conditions or to follow one additional step to balance the reaction if it occurs in basic conditions.

Glossary

half-reaction method
method to balance redox reactions by splitting the reaction into the oxidation “half” and reduction “half” reactions, balancing the half-reactions separately, and then combining them for one overall balanced equation

Chemistry End of Section Exercises

  1. Balance the following reactions in acidic solution. Identify the species that undergoes oxidation, the species that undergoes reduction, the oxidizing agent, and the reducing agent in each of the reactions
    1. H2O2(aq) + Sn2+(aq) → H2O(l) + Sn4+(aq)
    2. PbO2(s) + Hg(l) → Hg22+(aq) + Pb2+(aq)
    3. Al(s) + Cr2O72-(aq) → Al3+(aq) + 2 Cr3+(aq)
  2. Balance the reactions from #1 in basic solution.

Answers to Chemistry End of Section Exercises

    1. 2 H+(aq) + H2O2(aq) + Sn2+(aq) → 2 H2O(l) + Sn4+(aq); Oxidized: Sn2+; Reduced: O; Oxidizing agent: H2O2; Reducing agent: Sn2+
    2. 4 H+(aq) + PbO2(s) + 2 Hg(l) → Pb2+(aq) + 2 H2O(l) + Hg22+(aq); Oxidized: Hg; Reduced: Pb; Oxidizing agent: PbO2; Reducing agent: Hg
    3. 14 H+(aq) + 2 Al(s) + Cr2O72-(aq) → 2 Al3+(aq) + 2 Cr3+(aq) + 7 H2O(l); Oxidized: Al; Reduced: Cr; Oxidizing agent: Cr2O72-; Reducing agent: Al
    1. H2O2(aq) + Sn2+(aq) → Sn4+(aq) + 2 OH–(aq)
    2. 2 H2O(l) + PbO2(s) + 2 Hg(l) → Pb2+(aq) + Hg22+(aq) + 4 OH–(aq)
    3. 2 Al(s) + Cr2O72-(aq) + 7 H2O(l) → 2 Al3+(aq) + 2 Cr3+(aq) + 14 OH–(aq)

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