 # Writing Half Equations For Acidic and Basic Aqueous Solutions Tutorial

## 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:
(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.
The way this is done depends on whether the solution is acidic or basic (alkaline):

(A) Acidic Solution:

Add the appropriate number of H+ to the side deficient in H

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.

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

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## Worked Example: Half-Equations for Simple Ions in Neutral Solution

Question: Write a balanced half equation for the oxidation of Fe2+(aq) to Fe3+(aq)

Solution:

1. Write the skeletal equation:

Fe2+(aq) → Fe3+(aq)

2. Balance atoms:
(i) Balance all atoms other than H or O:
only 1 Fe2+ and 1 Fe3+ are required:

Fe2+(aq) → Fe3+(aq)

(ii) Balance O atoms: none present

(iii) Balance H atoms: none present

(iv) 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-

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## Worked Example: Half-Equations for Acidic Solution

Question: Write a balanced half equation for the reduction of Cr2O72-(aq) to Cr3+(aq) in acidic solution

Solution:

1. Write the skeletal equation:

Cr2O72-(aq) → Cr3+(aq)

2. Balance atoms:
(i) 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)

(ii) 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

(iii) 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

(iv) 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

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## Worked Example: Half-Equations for Basic Solution

Question: Write a balanced half equation for the reduction of CrO42-(aq) to CrO2-(aq) in basic solution

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

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

(iii) 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
(4 on the left hand side, 2 on the right hand side, so 4-2 = 2 on the left hand side and none on the right hand side):

CrO42-(aq) + 2H2O → CrO2-(aq) + 4OH-

(iv) 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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