Writing Half Equations for Basic Aqueous Solutions Chemistry Tutorial

Key Concepts

To write a balanced oxidation or reduction reaction for a species in basic or alkaline aqueous solution:

1. Write a skeletal equation for the oxidation or reduction equation based on the information provided.
2. Balance the half-reaction equation according to the following sequence:

   (i) Balance all atoms other than H and O by inspection

   (ii) Balance O atoms by adding H₂O to the appropriate side

   (iii) Balance the H atoms:

   ⚛ add H₂O molecule to the side deficient in H

   AND

   ⚛ add one OH^- ion to the opposite side, for each H atom needed.

   You may need to cancel out H₂O molecules duplicated on both sides of the equation at this point.

3. Balance the charge by adding electrons (e^-) to the side deficient in negative charge.

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Worked Example: Writing a balanced half-equation for a reduction reaction in basic solution

Write a balanced half equation for the reduction of CrO₄^2-_(aq) to CrO₂^-_(aq) in basic solution

1. Write the skeletal equation:

   CrO₄^2-_(aq) → CrO₂^-_(aq)

2. Balance atoms:

   (i) Balance all atoms other than H or O:

   1 Cr atom is on the left hand side

   1 Cr atom is present on the right hand side hand of the equation.

   Cr atoms are balanced:

   CrO₄^2-_(aq) → CrO₂^-_(aq)

   (ii) Balance O atoms by adding H₂O to the side deficient in O

   There are 4 O atoms on the left hand side but only 2 on the right hand side so we add H₂O to the right hand side of the equation:

   CrO₄^2-_(aq) → CrO₂^-_(aq) + H₂O

   Check to see if the O atoms are balanced:

   4 O atoms are present on the left hand side

   3 O atoms on the right hand side of the equation:

   Balance the O atoms by multiplying H₂O by 2:

   CrO₄^2-_(aq) → CrO₂^-_(aq) + 2H₂O

   (iii) Balance H atoms as for a basic solution by adding H₂O to the side deficient in H and OH^- to the opposite side

   Left hand side of the equation has no H atoms so add H₂O to the left hand side, AND, add OH^- to the right hand side:

   CrO₄^2-_(aq) + H₂O → CrO₂^-_(aq) + 2H₂O + OH^-

   Balance the number of H atoms:

   2 H atoms on the left hand side.

   5 H atoms on the right hand side of the equation.

   Balance the H atoms by multiplying H₂O on the left hand side by 4

   CrO₄^2-_(aq) + 4H₂O → CrO₂^-_(aq) + 2H₂O + OH^-

   and multiply OH^- on the right hand side by 4 to balance the equation:

   CrO₄^2-_(aq) + 4H₂O → CrO₂^-_(aq) + 2H₂O + 4OH^-

   Cancel out H₂O molecules appearing on both sides of the equation:

   Remove 2 water molecules from both sides of the equation:

   CrO₄^2-_(aq) + 2H₂O → CrO₂^-_(aq) + 4OH^-

3. Balance charge:

   charge on left hand side = 2-

   charge on right hand side = 1- + 4- = 5-

   difference in charge = 2- - 5- = 3+

   3 electrons (3e^-) are required on the left hand side of the equation:

   CrO₄^2-_(aq) + 2H₂O + 3e^- → CrO₂^-_(aq) + 4OH^-

Worked Example: Writing a balanced half-equation for an oxidation reaction in basic solution

Write a balanced half equation for the oxidation of Fe(OH)₂ to Fe(OH)₃ in basic solution

1. Write the skeletal equation:

   Fe(OH)_2(aq) → Fe(OH)_3(aq)

2. Balance atoms:

   (i) Balance all atoms other than H or O:

   1 Fe atom is on the left hand side

   1 Fe atom is present on the right hand side hand of the equation.

   Fe atoms are balanced, so the skeletal equation remains unchanged:

   Fe(OH)_2(aq) → Fe(OH)_3(aq)

   (ii) Balance O atoms by adding H₂O to the side deficient in O

   There are 2 O atoms on the left hand side but there are 3 on the right hand side so we add H₂O to the left hand side of the equation:

   Fe(OH)_2(aq) + H₂O → Fe(OH)_3(aq)

   Check to see if the O atoms are balanced:

   3 O atoms are present on the left hand side

   3 O atoms on the right hand side of the equation:

   O atoms are balanced so the skeletal equation remains the same:

   Fe(OH)_2(aq) + H₂O → Fe(OH)_3(aq)

   (iii) Balance H atoms as for a basic solution by adding H₂O to the side deficient in H and OH^- to the opposite side

   Left hand side of the equation has 4 H atoms but the right hand side has only 3 H atoms so we add H₂O to the right hand side, AND, we add OH^- to the left hand side:

   Fe(OH)_2(aq) + H₂O + OH^- → Fe(OH)_3(aq) + H₂O

   Balance the number of H atoms:

   5 H atoms on the left hand side.

   5 H atoms on the right hand side of the equation.

   H atoms are balanced so the skeletal equation remains unchanged:

   Fe(OH)_2(aq) + H₂O + OH^- → Fe(OH)_3(aq) + H₂O

   Cancel out H₂O molecules appearing on both sides of the equation:

   Remove 1 water molecule from both sides of the equation:

   Fe(OH)_2(aq) + OH^- → Fe(OH)_3(aq)

3. Balance charge:

   charge on left hand side = 1-

   charge on right hand side = 0

   difference in charge = 1- 0 = 1-

   1 electron (e^-) is required on the right hand side of the equation:

   Fe(OH)_2(aq) + OH^- → Fe(OH)_3(aq) + e^-