Writing Half Equations for Aqueous Solutions Under Acidic Conditions Chemistry Tutorial

Key Concepts

To write a balanced oxidation or reduction reaction for a species in acidic 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 by adding the appropriate number of H⁺ to the side deficient in H

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

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Writing Balanced Half-Equations for Reduction Reactions Under Acidic Conditions

Recall that a reduction reaction involves a species gaining electrons, that is, electrons will be a reactant in the half-equation.

reactants + electrons → products

Recall also that an aqueous acidic solution contains water molecules, H₂O, and protons (H⁺).⁽¹⁾, so these are the species we can use to balance the number of atoms of each element in the half-equation for a reduction reaction.

After all the atoms have been balanced we will balance the charge by adding electrons to the side with the greatest positive charge (deficient in electrons).
Remember that for a reduction reaction electrons should be a reactant in the equation.

Oxidising agents, or oxidants, are the species that will be reduced.
Oxidising agents (oxidants) you may be familiar with are the dichromate ion (Cr₂O₇^2-_(aq)) and the permanganate ion (MnO₄^-_(aq)).

• Purple permanganate, MnO₄^-_(aq), is an oxidising agent, it is reduced to colourless manganese(2+), Mn²⁺_(aq).
• Orange dichromate, Cr₂O₇^2-_(aq), is an oxidising agent, it is reduced to green chromium(3+), Cr³⁺_(aq).

Worked Example: Write a balanced half equation for the reduction of orange Cr₂O₇^2-_(aq) to green Cr³⁺_(aq) in acidic solution

1. Write the skeletal equation:

   Cr₂O₇^2-_(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:

   Cr₂O₇^2-_(aq) → 2Cr³⁺_(aq)

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

   There are no O atoms on the right hand side of the equation, so we add H₂O to the right hand side:

   Cr₂O₇^2-_(aq) → 2Cr³⁺_(aq) + H₂O

   Check to see if the O atoms are balanced:

   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

   Cr₂O₇^2-_(aq) → 2Cr³⁺_(aq) + 7H₂O

   (iii) Balance H atoms as for an acidic solution by adding H⁺ to the side deficient in H

   There are no H atoms on the left hand side of the equation, so we add H⁺ to the left hand side:

   Cr₂O₇^2-_(aq) + H⁺ → 2Cr³⁺_(aq) + 7H₂O

   Check to see of the H atoms are balanced:

   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

   Cr₂O₇^2-_(aq) + 14H⁺ → 2Cr³⁺_(aq) + 7H₂O

3. Balance charge:

   charge on left hand side = 2- (14 × 1+) = 12+

   charge on right hand side = 2 × 3+ = 6+

   difference in charge = 12+ - 6+ = 6+

   So 6 electrons (6e^-)are required on the left hand side of the equation:

   Cr₂O₇^2-_(aq) + 14H⁺ + 6e^- → 2Cr³⁺_(aq) + 7H₂O

The balanced half equation for the reduction of orange Cr₂O₇^2-_(aq) to green Cr³⁺_(aq) in acidic solution :

Cr₂O₇^2-_(aq) + 14H⁺ + 6e^- → 2Cr³⁺_(aq) + 7H₂O

Worked Example: Write a balanced half equation for the reduction of purple MnO₄^-_(aq) to colourless Mn²⁺_(aq) in acidic solution

1. Write the skeletal equation:

   MnO₄^-_(aq) → Mn²⁺_(aq)

2. Balance atoms:

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

   1 Mn atom is on the left hand side.

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

   The Mn atoms are balanced. The skeletal equation remains unchanged.

   MnO₄^-_(aq) → Mn²⁺_(aq)

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

   There are no O atoms on the right hand side of the equation, but 4 O atoms on the left hand side, so we add H₂O to the right hand side:

   MnO₄^-_(aq) → Mn²⁺_(aq) + H₂O

   Check to see if the O atoms are balanced:

   4 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 4

   MnO₄^-_(aq) → Mn²⁺_(aq) + 4H₂O

   (iii) Balance H atoms as for an acidic solution by adding H⁺ to the side deficient in H

   There are no H atoms on the left hand side of the equation, so we add H⁺ to the left hand side:

   MnO₄^-_(aq) + H⁺ → Mn²⁺_(aq) + 4H₂O

   Check to see of the H atoms are balanced:

   1 H atom on the left hand side.

   8 H atoms on the right hand side of the equation:

   Balance H atoms by multiplying H⁺ by 8

   MnO₄^-_(aq) + 8H⁺ → Mn²⁺_(aq) + 4H₂O

3. Balance charge:

   charge on left hand side = 1- (8 × 1+) = 7+

   charge on right hand side = 2+

   difference in charge = 7+ - 2+ = 5+

   So 5 electrons (5e^-) are required on the left hand side of the equation:

   MnO₄^-_(aq) + 8H⁺ + 5e^- → Mn²⁺_(aq) + 4H₂O

The balanced half equation for the reduction of purple MnO₄^-_(aq) to colourless Mn²⁺_(aq) in acidic solution :

MnO₄^-_(aq) + 8H⁺ + 5e^- → Mn²⁺_(aq) + 4H₂O

Writing Balanced Half-Equations for Oxidation Reactions Under Acidic Conditions

Recall that an oxidation reaction involves a species losing electrons, that is, electrons will be a product in the half-equation.

reactants → products + electrons

Recall also that an aqueous acidic solution contains water molecules, H₂O, and protons (H⁺).⁽¹⁾, so these are the species we can use to balance the number of atoms of each element in the half-equation for an oxidation reaction.

After all the atoms have been balanced we will balance the charge by adding electrons to the side with the greatest positive charge (deficient in electrons).
Remember that for an oxidation reaction electrons should be a product in the equation.

Reducing agents, or reductants, are the species that will be oxidised.
In the laboratory, simple ions such as metal ions such as Fe²⁺ and Sn²⁺ and non-metal ions such as I^- can be used as reducing agents. You should already be able to balance half-equations for these oxidation reactions.
Sulfur dioxide (SO₂) can also be used as a reducing agent.
SO₂ will be oxidised to SO₄^2-_(aq).

Worked Example: Write a balanced half equation for the oxidation of SO₂ to SO₄^2-_(aq) in acidic solution

1. Write the skeletal equation:

   SO_2(aq) → SO₄^-_(aq)

2. Balance atoms:

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

   1 S atom is on the left hand side.

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

   The S atoms are balanced. The skeletal equation remains unchanged.

   SO_2(aq) → SO₄^-_(aq)

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

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

   SO_2(aq) + H₂O → SO₄^-_(aq)

   Check to see if the O atoms are balanced:

   3 O atoms are present on the left hand side.

   4 O atom on the right hand side of the equation.

   Balance O atoms by multiplying H₂O by 2

   SO_2(aq) + 2H₂O → SO₄^-_(aq)

   (iii) Balance H atoms as for an acidic solution by adding H⁺ to the side deficient in H

   There are no H atoms on the right hand side of the equation, so we add H⁺ to the right hand side:

   SO_2(aq) + 2H₂O → SO₄^-_(aq) + H⁺

   Check to see of the H atoms are balanced:

   4 H atom on the left hand side.

   1 H atoms on the right hand side of the equation:

   Balance H atoms by multiplying H⁺ by 4

   SO_2(aq) + 2H₂O → SO₄^-_(aq) + 4H⁺

3. Balance charge:

   charge on left hand side = 0

   charge on right hand side = 1- (4 × 1+) = 3+

   difference in charge = 3+ - 0 = 3+

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

   SO_2(aq) + 2H₂O → SO₄^-_(aq) + 4H⁺ + 3e^-

The balanced half equation for the oxidation of SO₂ to SO₄^2-_(aq) in acidic solution:

SO_2(aq) + 2H₂O → SO₄^-_(aq) + 4H⁺ + 3e^-

Writing Half-Equations for the Oxidation of Organic Compounds Under Acidic Conditions

Recall that an oxidation reaction involves a species losing electrons, that is, electrons will be a product in the half-equation.

reactants → products + electrons

Recall also that an aqueous acidic solution contains water molecules, H₂O, and protons (H⁺).⁽¹⁾, so these are the species we can use to balance the number of atoms of each element in the half-equation for an oxidation reaction.

After all the atoms have been balanced we will balance the charge by adding electrons to the side with the greatest positive charge (deficient in electrons).
Remember that for an oxidation reaction electrons should be a product in the equation.

You may already be familiar with oxidation of alkanols:

• oxidation of primary alkanols produces alkanals then alkanoic acids.
• oxidation of secondary alkanols produces alkanones
• tertiary alkanols cannot be readily oxidised

We can write balanced half-equations for the oxidation of an aqueous solution of alkanol, for example ethanol, to an alkanoic acid, for example acetic acid (ethanoic acid), under acidic conditions.

Worked Example: Write a balanced half-equation for the oxidation of ethanol, C₂H₅OH_(aq), to acetic acid (ethanoic acid), CH₃COOH_(aq), in acidic solution

1. Write the skeletal equation:

   C₂H₅OH_(aq) → CH₃COOH_(aq)

2. Balance atoms:

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

   2 C atoms are on the left hand side.

   2 C atoms are present on the right hand side hand of the equation.

   The C atoms are balanced. The skeletal equation remains unchanged.

   C₂H₅OH_(aq) → CH₃COOH_(aq)

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

   There are 2 O atoms on the right hand side of the equation, but only 1 O atom on the left hand side, so we add H₂O to the left hand side:

   C₂H₅OH_(aq) + H₂O → CH₃COOH_(aq)

   Check to see if the O atoms are balanced:

   2 O atoms are present on the left hand side.

   2 O atom on the right hand side of the equation.

   The O atoms are balanced so the skeletal equation remains unchanged.

   C₂H₅OH_(aq) + H₂O → CH₃COOH_(aq)

   (iii) Balance H atoms as for an acidic solution by adding H⁺ to the side deficient in H

   There are 8 H atoms on the left hand side of the equation but only 4 H atoms on the right hand side of the equation, so we add H⁺ to the right hand side:

   C₂H₅OH_(aq) + H₂O → CH₃COOH_(aq) + H⁺

   Check to see of the H atoms are balanced:

   8 H atom on the left hand side.

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

   Balance H atoms by multiplying H⁺ by 4

   C₂H₅OH_(aq) + H₂O → CH₃COOH_(aq) + 4H⁺

3. Balance charge:

   charge on left hand side = 0

   charge on right hand side = 4 × 1+ = 4+

   difference in charge = 4+ - 0 = 4+

   So 4 electrons (4e^-) are required on the right hand side of the equation:

   C₂H₅OH_(aq) + H₂O → CH₃COOH_(aq) + 4H⁺ + 4e^-

The balanced half-equation for the oxidation of ethanol, C₂H₅OH_(aq), to acetic acid (ethanoic acid), CH₃COOH_(aq), in acidic solution:

C₂H₅OH_(aq) + H₂O → CH₃COOH_(aq) + 4H⁺ + 4e^-