19.1: Balancing Oxidation-Reduction Reactions in Acidic/Basic Solutions

Electrochemistry

Balancing Oxidation-Reduction Reactions in Acidic and Basic Solutions with Half Reaction Method:

If acidic, reaction takes place in solution where H₂O, H⁺, and e^– are available in solution.

Step 1: Split the equation up into a reduction half reaction and an oxidation half reaction. In a redox reaction, there is a reduction and an oxidation. Ignoring hydrogen and oxygen, be sure to match pairs with a common element, in other words, don't create elements out of thin air.

Step 2: Balance all elements that are not H or O

Step 3: Balance O atoms by adding H₂O's to one side of the equation.

Step 4: Balance H atoms by adding H⁺ ions to one side of the equation.

Step 5: Balance the electric charge by adding e^– to the more positive side. Note that you are not trying to get both sides equal to zero, just equal to each other. At this point, the reduction reaction has e^– as a reactant and the oxidation reaction has e^– as a product.

Step 6: Make sure the number of e^– that were added to the total reduction reaction is the same as the number of e^– that were added to the total oxidation reaction. If not, multiply either one or both total reactions by an integer that will make them equal.

Step 7: Combine the two half reactions and cancel species that appear on both sides and reduce the coefficients to the smallest whole numbers. No e^– should appear in the final equation.

Example:
Balance using the half reaction method. (acidic)

MnO₂ (aq) + HNO₂ (aq) → Mn²⁺ (aq) + NO₃^– (aq)

Step 1:
MnO₂ → Mn²⁺
HNO₂ → NO₃^–

Step 2:
Mn is already balanced.
N is already balanced.

Step 3:
MnO₂ → Mn²⁺ + 2H₂O
HNO₂ + H₂O → NO₃^–

Step 4:
MnO₂ + 4H⁺ → Mn²⁺ + 2H₂O
HNO₂ + H₂O → NO₃^– + 3H⁺

Step 5:
MnO₂ + 4H⁺ + 2e^– → Mn²⁺ + 2H₂O
HNO₂ + H₂O → NO₃^– + 3H⁺ + 2e^–

Step 6:
2e^– = 2e^–

Step 7:
MnO₂ + 4H⁺ + 2e^– + HNO₂ + H₂O → Mn²⁺ + 2H₂O + NO₃^– + 3H⁺ + 2e^–

=

H⁺ + MnO₂ + HNO₂ → Mn²⁺ + H₂O + NO₃^–


Example:
Balance using the half reaction method. (acidic)

Cr(OH)₄^– + H₂O₂ → CrO₄^2– + H₂O

Step1:
Cr(OH)₄^– → CrO₄^2–
H₂O₂ → H₂O

Step2:
Cr is already balanced.

Step3:
Cr(OH)₄^– → CrO₄^2–
H₂O₂ → H₂O + H₂O

Step4:
Cr(OH)₄^– → CrO₄^2– + 4H⁺
H₂O₂ + 2H⁺ → 2H₂O

Step5:
Cr(OH)₄^– → CrO₄^2– + 4H⁺ + 3e^–
H₂O₂ + 2H⁺ + 2e^– → 2H₂O

Step6:
(Cr(OH)₄^– → CrO₄^2– + 4H⁺ + 3e^–) x 2
(H₂O₂ + 2H⁺ + 2e^– → 2H₂O) x 3

=

2Cr(OH)₄^– → 2CrO₄^2– + 8H⁺ + 6e^–
3H₂O₂ + 6H⁺ + 6e^– → 6H₂O

Step7:
2Cr(OH)₄^– + 3H₂O₂ + 6H⁺ + 6e^– → 2CrO₄^2– + 82H⁺ + 6e^– + 6H₂O

=

2Cr(OH)₄^– + 3H₂O₂ → 2CrO₄^2– + 2H⁺ + 6H₂O


If basic, reaction takes place in solution where H₂O, OH^–, and e^– are available in solution. So when the reaction is in basic solution, the process of balancing by half reactions is exactly the same, except that two more steps are added on to the end.

Step 8: Take the number of H⁺ ions in the finished equation. Add that many OH^– ions to both sides of the equation.

Step 9: Every H⁺ will react with a OH^– to produce that many water molecules. Cancel repeating water molecules and reduce.

Example:
Balance using the half reaction method. (basic)

Bi(OH)₃ + Sn(OH)₃^– → Sn(OH)₆^2– + Bi

Step 1:
Bi(OH)₃ → Bi
Sn(OH)₃^– → Sn(OH)₆^2–

Step 2:
Bi and Sn are balanced.

Step 3:
Bi(OH)₃ → Bi + 3H₂O
Sn(OH)₃^– + 3H₂O → Sn(OH)₆^2–

Step 4:
Bi(OH)₃ + 3H⁺ → Bi + 3H₂O
Sn(OH)₃^– + 3H₂O → Sn(OH)₆^2– + 3 H⁺

Step 5:
Bi(OH)₃ + 3H⁺ + 3e^– → Bi + 3H₂O
Sn(OH)₃^– + 3H₂O → Sn(OH)₆^2– + 3 H⁺ + 2e^–

Step 6:
(Bi(OH)₃ + 3H⁺ + 3e^– → Bi + 3H₂O) x 2
(Sn(OH)₃^– + 3H₂O → Sn(OH)₆^2– + 3 H⁺ + 2e^–) x 3

=

2Bi(OH)₃ + 6H⁺ + 6e^– → 2Bi + 6H₂O
3Sn(OH)₃^– + 9H₂O → 3Sn(OH)₆^2– + 9H⁺ + 6e^–

Step 7:
2Bi(OH)₃ + 6H⁺ + 6e^– + 3Sn(OH)₃^– + 93H₂O → 2Bi + 6H₂O + 3Sn(OH)₆^2– + 93H⁺ + 6e^–

=

2Bi(OH)₃ + 3Sn(OH)₃^– + 3H₂O → 2Bi + 3Sn(OH)₆^2– + 3H⁺

Step 8:
2Bi(OH)₃ + 3Sn(OH)₃^– + 3H₂O + 3OH^– → 2Bi + 3Sn(OH)₆^2– + 3H⁺ + 3OH^–

Step 9:
2Bi(OH)₃ + 3Sn(OH)₃^– + 3H₂O + 3OH^– → 2Bi + 3Sn(OH)₆^2– + 3H₂O

=

2Bi(OH)₃ + 3Sn(OH)₃^– + 3OH^– → 2Bi + 3Sn(OH)₆^2–