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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

    (B) Basic (alkaline) 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.

    (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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