# Writing Half Equations: Aqueous Solutions

## Key Concepts

To write a balanced oxidation or reduction reaction for a species in 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:

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
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.

4. Balance the charge by adding electrons (e) to the side deficient in negative charge.

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## Examples

1. Write a balanced half equation for the oxidation of Fe2+(aq) to Fe3+(aq)
1. Write the skeletal equation:
Fe2+(aq) → Fe3+(aq)
2. Balance atoms:
1. Balance all atoms other than H or O:
only 1 Fe2+ and 1 Fe3+ are required:
Fe2+(aq) → Fe3+(aq)
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 needed on the right hand side in order to balance the charge:
Fe2+(aq) → Fe3+(aq) + e

2. Write a balanced half equation for the reduction of Al3+(aq) to Al(s)
1. Write the skeletal equation:
Al3+(aq) → Al(s)
2. Balance atoms:
1. Balance all atoms other than H or O:
only 1 Al3+ and 1 Al are required:
Al3+(aq) → 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 needed on the left hand side in order to balance the charge:
Al3+(aq) + 3e → Al(s)

3. Write a balanced half equation for the reduction of Cr2O72-(aq) to Cr3+(aq) in acidic solution
1. Write the skeletal equation:
Cr2O72-(aq) → Cr3+(aq)
2. Balance atoms:
1. Balance all atoms other than H or O:
2 Cr atoms are on the left hand side.
1 Cr is present on the right hand side hand of the equation.
Balance the Cr atoms by multiplying Cr3+ by 2:
Cr2O72-(aq) → 2Cr3+(aq)
2. Balance O atoms by adding H2O to the side deficient in O
Cr2O72-(aq) → 2Cr3+(aq) + H2O

7 O atoms are present on the left hand side.
1 O atom on the right hand side of the equation.
Balance O atoms by multiplying H2O by 7
Cr2O72-(aq) → 2Cr3+(aq) + 7H2O
3. Balance H atoms as for an acidic solution by adding H+ to the side deficient in H
Cr2O72-(aq) + H+ → 2Cr3+(aq) + 7H2O

1 H atom on the left hand side.
14 H atoms on the right hand side of the equation:
Balance H atoms by multiplying H+ by 14
Cr2O72-(aq) + 14H+ → 2Cr3+(aq) + 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:
Cr2O72-(aq) + 14H+ + 6e → 2Cr3+(aq) + 7H2O

4. 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:
1. 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)
2. Balance O atoms by adding H2O to the side deficient in O
CrO42-(aq) → CrO2-(aq) + H2O

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
3. Balance H atoms as for a basic solution by adding H2O to the side deficient in H and OH- to the opposite side
CrO42-(aq) + H2O → CrO2-(aq) + 2H2O + OH-

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
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:
CrO42-(aq) + 2H2O → CrO2-(aq) + 4OH-
4. Balance charge:
charge on left hand side = 2-
charge on right hand side = 1- + 4- = 5-
difference in charge = 2- - 5- = 3+
3 electrons are required on the left hand side of the equation:
CrO42-(aq) + 2H2O + 3e → CrO2-(aq) + 4OH-

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