# 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 CrO42-(aq) to CrO2-(aq) in basic solution

1. Write the skeletal equation:

CrO42-(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:

CrO42-(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:

CrO42-(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:

CrO42-(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:

CrO42-(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

CrO42-(aq) + 4H2O → CrO2-(aq) + 2H2O + OH-

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

CrO42-(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:

CrO42-(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:

CrO42-(aq) + 2H2O + 3e- → CrO2-(aq) + 4OH-

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## 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-

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