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Balancing Half Equations

Key Concepts

Write a skeletal equation for the oxidation or reduction equation based on the information provided.
Balance the half-reaction equation according to the following sequence:
1. Balance all atoms other than H and O by inspection
2. Balance O atoms by adding H₂O to the appropriate side
3. Balance the H atoms. The way this is done depends on whether the solution is acidic or basic:
  - Acidic Solution:
    add the appropriate number of H⁺ to the side deficient in H
  - Basic Solution:
    add one H₂O molecule to the side deficient in H AND one OH^- ion to the opposite side, for each H atom needed.
    You may need to cancel out H₂O molecules duplicated on each side at this point.
4. Balance the charge by adding electrons (e) to the side deficient in negative charge.

Two half-reaction equations (one oxidation and one reduction) can be added together (as for calculating cell EMF).
In this case, multiply each balanced half equation by an appropriate number in order to balance the electrons lost in the oxidation reaction with the electrons gained in the reduction reaction.
Subtract, or cancel out, anything that appears on both sides. The electrons should cancel out in this step.

Examples

Write a balanced half equation for the oxidation of Fe²⁺ to Fe³⁺
1. Write the skeletal equation: Fe²⁺ -----> Fe³⁺
2. Balance atoms:
  1. Balance all atoms other than H or O: only 1 Fe²⁺ and 1 Fe³⁺ are required:
    Fe²⁺ -----> Fe³⁺
  2. Balance O atoms: none present
  3. Balance H atoms: none present
  4. Balance charge: left hand side charge = +2, right hand side charge = +3, 1 electron is required on the right hand side in order to balance the charge:
    Fe²⁺ -----> Fe³⁺ + e
Write a balanced half equation for the reduction of Al³⁺ to Al(s)
1. Write the skeletal equation: Al³⁺ -----> Al(s)
2. Balance atoms:
  1. Balance all atoms other than H or O: only 1 Al³⁺ and 1 Al are required:
    Al³⁺ -----> Al(s)
  2. Balance O atoms: none present
  3. Balance H atoms: none present
  4. Balance charge: left hand side charge = +3, right hand side charge = 0, 3 electrons are required on the left hand side in order to balance the charge:
    Al³⁺ + 3e -----> Al(s)
Write a balanced half equation for the reduction of Cr₂O₇^2- to Cr³⁺ in acidic solution
1. Write the skeletal equation: Cr₂O₇^2- -----> Cr³⁺
2. Balance atoms:
  1. Balance all atoms other than H or O:
    2 Cr atoms are on the left hand side, only 1 Cr is present on the right hand side hand of the equation.
    Cr₂O₇^2- -----> 2Cr³⁺
  2. Balance O atoms by adding H₂O to the side deficient in O
    Cr₂O₇^2- -----> 2Cr³⁺ + H₂O
    7 O atoms are present on the left hand side, 1 on the right hand side of the equation:
    Cr₂O₇^2- -----> 2Cr³⁺ + 7H₂O
  3. Balance H atoms as for an acidic solution by adding H⁺ to the side deficient in H
    Cr₂O₇^2- + H⁺ -----> 2Cr³⁺ + 7H₂O
    There is 1 H atom on the left hand side, and 14 H atoms on the right hand side of the equation:
    Cr₂O₇^2- + 14H⁺ -----> 2Cr³⁺ + 7H₂O
  4. Balance charge:
    charge on left hand side = -2 + 14 = +12
    charge on right hand side = 2 x +3 = +6
    difference in charge = +12 - +6 = +6 So 6 electrons are required on the left hand side of the equation:
    Cr₂O₇^2- + 14H⁺ + 6e -----> 2Cr³⁺ + 7H₂O
Write a balanced half equation for the reduction of CrO₄^2- to CrO₂^- in basic solution
1. Write the skeletal equation: CrO₄^2- -----> CrO₂^-
2. Balance atoms:
  1. Balance all atoms other than H or O:
    1 Cr atom is on the left hand side, 1 Cr is present on the right hand side hand of the equation.
    CrO₄^2- -----> CrO₂^-
  2. Balance O atoms by adding H₂O to the side deficient in O
    CrO₄^2- -----> CrO₂^- + H₂O
    4 O atoms are present on the left hand side, 3 on the right hand side of the equation:
    CrO₄^2- -----> CrO₂^- + 2H₂O
  3. Balance H atoms as for a basic solution by adding H₂O to the side deficient in H and OH^- to the opposite side
    CrO₄^2- + H₂O-----> CrO₂^- + 2H₂O + OH^-
    There are 2 H atoms on the left hand side, and 5 H atoms on the right hand side of the equation:
    CrO₄^2- + 4H₂O-----> CrO₂^- + 2H₂O + 4OH^-
    Cancel out H₂O molecules appearing on both sides of the equation:
    CrO₄^2- + 2H₂O-----> CrO₂^- + 4OH^-
  4. Balance charge:
    charge on left hand side = -2
    charge on right hand side = -1 + -4 = -5
    difference in charge = -2 - -5 = +3 So 3 electrons are required on the left hand side of the equation:
    CrO₄^2- + 2H₂O + 3e -----> CrO₂^- + 4OH^-