Balancing Redox Reactions

1.0Introduction

In a redox (oxidation-reduction) reaction, electrons are transferred between chemical species. Balancing these reactions is crucial because both mass and charge must be conserved. The half-reaction method is an effective way to balance these equations, particularly for reactions in aqueous solutions. Here's an overview of the steps involved:

2.0Balancing Redox Equations

A balanced redox equation must satisfy two key criteria:

1. Atom Balance (Mass Balance):

• The number of each type of atom must be the same on both sides of the equation.

2. Charge Balance:

• The total charge on the reactant side must equal the total charge on the product side.

There are two main methods for balancing redox equations:

3.0Oxidation Number Change Method

This method, developed by Jonson, is based on the principle that the total increase in oxidation number (due to oxidation) must equal the total decrease in oxidation number (due to reduction).

Steps for Balancing Using the Oxidation Number Change Method:

1. Identify the Atoms Undergoing Oxidation and Reduction:

• Determine which atom in the oxidizing agent decreases in oxidation number (gains electrons).
• Identify the atom in the reducing agent that increases in oxidation number (loses electrons).

2. Write the Electron Transfer for Each Atom:

• Show the number of electrons gained in reduction and lost in oxidation.

3. Cross-Multiply Electron Changes:

• Multiply the oxidizing agent by the number of electrons lost by the reducing agent.
• Multiply the reducing agent by the number of electrons gained by the oxidizing agent.

4. Balance the Atoms That Undergo Oxidation and Reduction:

• Ensure that the number of atoms of elements involved in redox changes are equal on both sides.

5. Balance Oxygen Atoms:

• Add molecules to the side that lacks oxygen.

6. Balance Hydrogen Atoms:

• Add ceH+ ions to balance hydrogen in acidic solutions.

Balancing the Reaction Using the Oxidation Number Method

Solved Examples 1

Given Reaction:

Nitrate ions in an acidic medium oxidize magnesium to Mg²⁺ ions, while the nitrate ions are reduced to nitrous oxide (N₂O).

Step 1: Write the Skeleton Equation

$\text{Mg}+{\text{NO}}_3^{-}\to {\text{Mg}}^{2+}+{\text{N}}_{2}O+{\text{H}}_{2}O$

Step 2: Identify Changes in Oxidation Numbers

• Magnesium (Mg) is oxidized from 0 to +2.
• Nitrate ion (NO₃⁻) is reduced to N₂O, so the oxidation number of nitrogen decreases.

Step 3: Equalize the Number of Nitrogen Atoms

Multiply the nitrate ion by 2 to ensure that nitrogen atoms are balanced:

$\text{Mg}+2{\text{NO}}_3^{-}\to {\text{Mg}}^{2+}+{\text{N}}_{2}O+{\text{H}}_{2}O$

Step 4: Equalize the Oxidation Number Changes

To balance the overall oxidation and reduction, multiply Mg by 4 and NO₃⁻ by 1:

$4\text{Mg}+2{\text{NO}}_3^{-}\to 4{\text{Mg}}^{2+}+{\text{N}}_{2}O+{\text{H}}_{2}O$

Step 5: Balance Atoms Other Than Oxygen and Hydrogen

Ensure that the number of magnesium and nitrogen atoms are equal:

$4\text{Mg}+2{\text{NO}}_3^{-}\to 4{\text{Mg}}^{2+}+{\text{N}}_{2}O+{\text{H}}_{2}O$

Step 6: Balance Oxygen Atoms

Since there are 6 oxygen atoms in 2 NO₃⁻, add 4 H₂O molecules to balance oxygen:

$4\text{Mg}+2{\text{NO}}_3^{-}\to 4{\text{Mg}}^{2+}+{\text{N}}_{2}O+4{\text{H}}_{2}O$

Step 7: Balance Hydrogen Atoms in Acidic Medium

Add 10 H⁺ ions to balance the hydrogen atoms:

$4\text{Mg}+2{\text{NO}}_3^{-}+10{\text{H}}^{+}\to 4{\text{Mg}}^{2+}+{\text{N}}_{2}O+5{\text{H}}_{2}O$

Thus, the balanced redox reaction is:

$4\text{Mg}+2{\text{NO}}_3^{-}+10{\text{H}}^{+}\to 4{\text{Mg}}^{2+}+{\text{N}}_{2}O+5{\text{H}}_{2}O$

4.0Ion-Electron Method (Half-Cell Method)

The half-reaction method for balancing redox equations involves dividing the reaction into two half-reactions: one for oxidation and one for reduction. Each half-reaction is then balanced separately for both mass and charge. If needed, the equations are adjusted to ensure that the number of electrons lost in oxidation equals the number gained in reduction. Finally, the balanced half-reactions are combined to obtain the overall balanced redox equation.

The balancing process differs depending on the medium in which the reaction occurs:

1. Acidic Medium
2. Basic Medium

The Ion-Electron Method in Acidic Medium

Solved Example 2

Balance the following redox reaction:

${\text{FeSO}}_4+{\text{KMnO}}_4+{\text{H}}_2{\text{SO}}_4\to {\text{Fe}}_2({\text{SO}}_4{)}_3+{\text{MnSO}}_4+{\text{H}}_2\text{O}$

Step-by-Step Solution:

Step 1: Assign Oxidation Numbers

Identify the oxidation states of elements in the reaction to determine which species undergo oxidation and reduction.

Step 2: Convert to Ionic Form

Eliminate spectator ions (ions that do not participate in redox changes) to simplify the equation:

${\text{Fe}}^{2+}+{\text{MnO}}_4^{-}+{\text{H}}^{+}\to {\text{Fe}}^{3+}+{\text{Mn}}^{2+}+{\text{H}}_2\text{O}$

Step 3: Identify Oxidation and Reduction

• Oxidation: Fe²⁺ is oxidized to Fe³⁺.
• Reduction: MnO₄⁻ is reduced to Mn²⁺.

Step 4: Split the Reaction into Half-Reactions

Oxidation Half-Reaction:

${\text{Fe}}^{2+}\to {\text{Fe}}^{3+}$

Reduction Half-Reaction:

Step 5: Balance Atoms Other Than Oxygen and Hydrogen

Fe and Mn atoms are already balanced.

Step 6: Balance Oxygen and Hydrogen Atoms

• To balance oxygen, add H₂O molecules.
• To balance hydrogen, add H⁺ ions.

For the reduction reaction:

$8{\text{H}}^{+}+{\text{MnO}}_4^{-}\to {\text{Mn}}^{2+}+4{\text{H}}_2\text{O}$

Step 7: Balance Charge by Adding Electrons

For oxidation:

${\text{Fe}}^{2+}\to {\text{Fe}}^{3+}+e^{-}$

For reduction:

$5e^{-}+8{\text{H}}^{+}+{\text{MnO}}_4^{-}\to {\text{Mn}}^{2+}+4{\text{H}}_2\text{O}$

Step 8: Equalize Electrons in Both Half-Reactions

Multiply the oxidation half-reaction by 5 so that the number of electrons matches:

$5{\text{Fe}}^{2+}\to 5{\text{Fe}}^{3+}+5e^{-}$

Now, add both half-reactions:

$5{\text{Fe}}^{2+}+8{\text{H}}^{+}+{\text{MnO}}_4^{-}\to 5{\text{Fe}}^{3+}+{\text{Mn}}^{2+}+4{\text{H}}_2\text{O}$

At this stage, the equation is balanced in ionic form.

Step 9: Convert to Molecular Form

Reintroduce spectator ions (SO₄²⁻) that were removed in Step 2:

$5{\text{FeSO}}_4+{\text{KMnO}}_4+4{\text{H}}_2{\text{SO}}_4\to {\text{Fe}}_2({\text{SO}}_4{)}_3+{\text{MnSO}}_4+4{\text{H}}_2\text{O}$

Alternatively, doubling the equation:

$10{\text{FeSO}}_4+2{\text{KMnO}}_4+8{\text{H}}_2{\text{SO}}_4\to 5{\text{Fe}}_2({\text{SO}}_4{)}_3+2{\text{MnSO}}_4+8{\text{H}}_2\text{O}+{\text{K}}_2{\text{SO}}_4$

Final Balanced Equation:

$10{\text{FeSO}}_4+2{\text{KMnO}}_4+8{\text{H}}_2{\text{SO}}_4\to 5{\text{Fe}}_2({\text{SO}}_4{)}_3+2{\text{MnSO}}_4+8{\text{H}}_2\text{O}+{\text{K}}_2{\text{SO}}_4$

The Ion-Electron Method In Basic Medium

Solved Example 3

We are given the unbalanced ionic equation:

MnO₄⁻(aq) + C₂O₄²⁻(aq) ⟶ MnO₂(s) + CO₃²⁻(aq)

Step 1: Identify Oxidation and Reduction Half-Reactions

• Reduction Half-Reaction: Manganate ion (MnO₄⁻) is reduced to manganese dioxide (MnO₂).
• MnO₄⁻ ⟶ MnO₂
• Oxidation Half-Reaction: Oxalate ion (C₂O₄²⁻) is oxidized to carbonate ion (CO₃²⁻).
• C₂O₄²⁻ ⟶ CO₃²⁻

Step 2: Balance the Half-Reactions

Balancing the Oxidation Half-Reaction (C₂O₄²⁻ ⟶ CO₃²⁻)

1. Balance carbon atoms:
   Since oxalate has two carbon atoms, it forms two carbonate ions:
   C₂O₄²⁻ ⟶ 2CO₃²⁻
2. Balance oxygen atoms by adding H₂O:
   2H₂O + C₂O₄²⁻ ⟶ 2CO₃²⁻
3. Balance hydrogen atoms by adding H⁺:
   2H₂O + C₂O₄²⁻ ⟶ 2CO₃²⁻ + 4H⁺
4. Balance charge by adding electrons:
   2H₂O + C₂O₄²⁻ ⟶ 2CO₃²⁻ + 4H⁺ + 2e⁻

Balancing the Reduction Half-Reaction (MnO₄⁻ ⟶ MnO₂)

1. Balance manganese atoms:
   MnO₄⁻ ⟶ MnO₂
2. Balance oxygen atoms by adding H₂O:
   MnO₄⁻ ⟶ MnO₂ + 2H₂O
3. Balance hydrogen atoms by adding H⁺:
   4H⁺ + MnO₄⁻ ⟶ MnO₂ + 2H₂O
4. Balance charge by adding electrons:
   3e⁻ + 4H⁺ + MnO₄⁻ ⟶ MnO₂ + 2H₂O

Step 3: Equalize the Electron Transfer

To balance the electrons lost and gained, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:

2(3e⁻ + 4H⁺ + MnO₄⁻ ⟶ MnO₂ + 2H₂O)

3(2H₂O + C₂O₄²⁻ ⟶ 2CO₃²⁻ + 4H⁺ + 2e⁻)

This gives:

6e⁻ + 8H⁺ + 2MnO₄⁻ ⟶ 2MnO₂ + 4H₂O

6H₂O + 3C₂O₄²⁻ ⟶ 6CO₃²⁻ + 12H⁺ + 6e⁻

Step 4: Add the Two Half-Reactions

Adding both reactions together:

6e⁻ + 8H⁺ + 2MnO₄⁻ + 6H₂O + 3C₂O₄²⁻ ⟶ 2MnO₂ + 4H₂O + 6CO₃²⁻ + 12H⁺ + 6e⁻

Cancel electrons and common terms:

• The six electrons cancel.
• Eight H⁺ on the left cancel with 12H⁺ on the right, leaving 4H⁺ on the right.
• Four H₂O on the right cancel with six H₂O on the left, leaving 2H₂O on the left.

This simplifies to:

2MnO₄⁻ + 2H₂O + 3C₂O₄²⁻ ⟶ 2MnO₂ + 6CO₃²⁻ + 4H⁺

Step 5: Convert to Basic Solution

Since the reaction occurs in a basic medium, we neutralize H⁺ by adding OH⁻ to both sides:

• Add 4OH⁻ to both sides to neutralize 4H⁺:
  2MnO₄⁻ + 2H₂O + 3C₂O₄²⁻ + 4OH⁻ ⟶ 2MnO₂ + 6CO₃²⁻ + 4H⁺ + 4OH⁻
• Combine H⁺ and OH⁻ to form water:
  2MnO₄⁻ + 2H₂O + 3C₂O₄²⁻ + 4OH⁻ ⟶ 2MnO₂ + 6CO₃²⁻ + 4H₂O
• Cancel excess water molecules:
  2MnO₄⁻ + 3C₂O₄²⁻ + 2H₂O ⟶ 2MnO₂ + 6CO₃²⁻ + 2H₂O

Step 6: Final Balanced Equation with States

2MnO₄⁻(aq) + 3C₂O₄²⁻(aq) + 4OH⁻(aq) ⟶ 2MnO₂(s) + 6CO₃²⁻(aq) + 2H₂O(l)