Balancing redox equations

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Intros
Lessons
  1. Balancing redox equations
  2. Two methods to balance redox.
  3. Balancing redox using half-equations.
  4. Balancing redox using oxidation number.
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Examples
Lessons
  1. Balance and complete the following redox equations using either oxidation numbers or half-equations. All are in acidic conditions.
    1. Sn2+ + IO3-\enspace \enspace Sn4+ + I-
    2. Cr3+ + BiO3-\enspace \enspace Bi2+ + Cr2O72-
    3. MnO4- + SO2 \enspace \enspace Mn2+ + HSO4-
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Topic Notes
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In this lesson, we will learn:

  • To recall the two methods for completing full redox equations.
  • How to complete redox equations using oxidation numbers.
  • How to complete redox equations using half-equations.

Notes:

  • We have seen two ways to work with redox reactions:
    • We can split the redox reaction in half to show the reduction and oxidation processes separately. This gives us two half-equations.
      These half-equations show the electrons lost or gained in the separate processes, and the number of electrons must match in the overall equation.
    • We can assign oxidation numbers to the atoms involved in the reaction. Some oxidation numbers will change from reactants to products as atoms are either reduced or oxidized. The overall change in oxidation numbers must be zero in the overall equation to balance out.

  • This will depend on your curriculum, but generally to start a redox question, you will only be given compounds containing major atoms.
    In practice, this means an unbalanced equation with no H+ or H2O present. You must complete the redox equation from this point. In this lesson we will practice completing and balancing them.

  • Worked example using half-equations:

    Cr(OH)3 + Cl2 \enspace \enspace CrO42- + Cl-

    This equation needs to be split into its half equations. This is done by separating the species with major atoms into separate equations:

    \quad Equation 1: \quad \qquad \qquad \quad Cr(OH)3 \enspace \enspace CrO42- \qquad \qquad \qquad \qquad \qquad \qquad
    Equation 2:   \quad \qquad \qquad \qquad \quad \; Cl2 \enspace \enspace Cl- \qquad \qquad \qquad \qquad \qquad \qquad

    First for equation 1, balance the major atoms. The Cr is already balanced so no change:

    Cr(OH)3\enspace \enspace CrO42-

    Then equation 2 where the Cl atoms need to be balanced:

    Cl2\enspace \enspace 2CL-

    Next, balance the oxygen atoms:

    Equation 1: \quad \qquad \qquad \, Cr(OH)3 + H2O\enspace \enspace CrO42-\qquad \qquad \qquad \qquad \qquad \qquad
    Equation 2:   \quad \qquad \qquad \qquad \quad \; Cl2 \enspace \enspace 2Cl- \qquad \qquad \qquad \qquad \qquad \qquad \qquad

    Next, balance the hydrogen atoms:

    \enspace \: Equation 1: \quad \qquad \enspace Cr(OH)3 + H2O \enspace \enspace CrO42-+ 5 H+ \qquad \qquad \qquad \qquad \qquad \qquad
    Equation 2:   \quad \qquad \qquad \qquad \quad \; Cl2 \enspace \enspace 2Cl- \qquad \qquad \qquad \qquad \qquad \qquad \qquad

    Finally, balance the charge:

    Equation 1: \quad \qquad \qquad Cr(OH)3 + H2O \enspace \enspace CrO42- + 5 H+ + 3e- \qquad \qquad \qquad
    \qquad Equation 2: \quad \qquad \qquad \qquad \quad \enspace \, 2e- + Cl2 \enspace \enspace 2Cl- \qquad \qquad \qquad \qquad \qquad \qquad

    Now with two complete half-equations, we need to re-combine them with an equal number of electrons on both sides. Three are transferred in equation 1, only two are transferred in equation 2, so if we multiply both equations to have six on both sides we will balance:

    Equation 1: \quad \qquad \qquad 2 x [ Cr(OH)3 + H2O\enspace \enspace CrO42- + 5H+ + 3e- ] \qquad \qquad \qquad
    \qquad Equation 2: \quad \qquad \qquad \qquad \quad \enspace \, 3 x [ 2e- + Cl2\enspace \enspace 2Cl- ] \qquad \qquad \qquad \qquad \qquad \qquad

    Multiplying and combining these equations with the electrons removed gives us a final balanced redox equation. This is the final equation; atoms and charge must balance now!

    Full equation: \quad \qquad \qquad 2Cr(OH)3 + 2H2O + 3Cl2\enspace \enspace 2CrO42- + 10H+ + 6Cl- \qquad \qquad \qquad

    Worked example using oxidation numbers:

    SO32- + Cr2O72-\enspace \enspace SO42- + Cr3+

    The total change in oxidation number must be zero. First, calculate oxidation numbers of the major atoms in the reactants and products.

    Reactants:
    Sulfur in SO32-, oxidation state +4 (-6 due to oxygen, +4 in sulfur gives 2- overall charge).
    Chromium in Cr2O72-, Oxidation state +6 (-14 due to oxygen, +6 for two Cr atoms gives 2- overall charge).

    Products:
    Sulfur in SO42-, oxidation state +6 (-8 due to oxygen, +6 in sulfur gives 2- overall charge).
    Chromium in Cr3+ ion, oxidation state +3.
    Δ\Delta OS for sulfur = +2 (from +4 to +6)
    Δ\Delta OS for chromium = -3 x 2 Cr atoms (from +6 to +3) = -6

    Δ\Delta OS must balance to zero, therefore:
    3 x SO32-\enspace \enspace SO42- gives total Δ\Delta OS of +6
    1 x Cr2O72-\enspace \enspace Cr3+ gives total Δ\Delta OS of -6
    This balances the oxidation state.
    Balance major atoms:

    3SO32- + Cr2O72-\enspace \enspace 3SO42- + 2Cr3+

    Now, balance oxygen with H2O:

    3SO32- + Cr2O72-\enspace \enspace 3SO42- + 2Cr3+ + 4H2O

    Next, balance hydrogen with H+:

    3SO32- + Cr2O72- + 8H+\enspace \enspace 3SO42- + 2Cr3+ + 4H2O

    Check that the charge is balanced:

    3SO32- + Cr2O72- + 8H+\enspace \enspace 3SO42- + 2Cr3+ + 4H2O