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4.2: Balancing Redox Reactions

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  • In studying redox chemistry, it is important to begin by learning to balance electrochemical reactions. Simple redox reactions (for example, H2 + I2 → 2 HI) can be balanced by inspection, but for more complex reactions it is helpful to have a foolproof, systematic method. The ion-electron method allows one to balance redox reactions regardless of their complexity. We illustrate this method with two examples.


    Example 1:
    I- is oxidized to IO3- by MnO4-, which is reduced to Mn2+.

    How can this reaction be balanced? In the ion-electron method we follow a series of four steps:

    Step 1A: Write out the (unbalanced) reaction and identify the elements that are undergoing redox.

    MnO4- + I- → IO3- + Mn2+ (The elements undergoing redox are Mn and I)

    Step 1B: Separate the reaction into two half reactions, balancing the element undergoing redox in each.

    MnO4- → Mn2+
    I- → IO3-

    Step 2A: Balance the oxygen atoms by adding water to one side of each half reaction.

    MnO4-→ Mn2+ + 4H2O
    3H2O + I- → IO3-

    Step 2B: Balance the hydrogen atoms by adding H+ ions.

    8H+ + MnO4-→ Mn2+ + 4H2O

    The left side has a net charge of +7 and the right side has a net charge of +2

    3H2O + I- → IO3- + 6H+

    The left side has a net charge of -1 and the right side has a net charge of +5

    Step 2C: Balance the overall charge by adding electrons

    8H+ + 5e- + MnO4-→ Mn2+ + 4H2O

    The left side has a charge of +2 while the right side has a charge of +2. They are balanced.

    3H2O + I- → IO3- + 6H+ + 6e-

    The left side has a charge of -1 while the right side has a charge of -1. They are balanced.
    Note: We did not need to explicitly determine the oxidation states of Mn or I to arrive at the correct number of electrons in each half reaction.

    Step 3: Combine the half reactions so that there are equal numbers of electrons on the left and right sides

    6 (8H+ + 5e- + MnO4-→ Mn2+ + 4H2O)
    5 (3H2O + I- → IO3- + 6H+ + 6e-)

    48H+ + 30e- + 15H2O + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 24H2O + 30e- + 30H+
    Cancel the H+electrons, and water:
    48H+ + 30e- + 15H2O + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 24H2O + 30e- + 30H+


    The overall balanced reaction is therefore:
    18H+ + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 9H2O

    Check your work by making sure that all elements and charges are balanced.

    Step 4: If the reaction occurs under basic conditions, we add OH- to each side to cancel H+
    18H+ + 18OH- + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 9H2O + 18OH-

    The 18H+ + 18OH- will become 18H2O so the overall balanced reaction is:

    9H2O + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 18OH-

    Again, it is a good idea to check and make sure that all of the elements are balanced, and that the charge is the same on both sides. If this is not the case, you need to find the error in one of the earlier steps.


    Example 2:
    Redox reaction of S2O32- and H2O2
    S2O32- + H2O2 → S4O62- + H2O

    Which elements are undergoing redox? S and O

    Step 1: Write out half reactions, balancing the element undergoing redox

    2S2O32- → S4O62-
    H2O2 → 2H2O

    Step 2A: Balance oxygen (already balanced)

    Step 2B: Balance hydrogen:

    2S2O32- → S4O62-
    H2O2 + 2H+ → 2H2O

    Step 2C: Balance charge by adding electrons:

    2S2O32- → S4O62- + 2e-
    H2O2 + 2H+ + 2e- → 2H2O

    Step 3: Combine the half reactions so that there are equal numbers of electrons on the left and right sides (already equal)
    Overall balanced reaction:

    2S2O32- + H2O2 + 2H+ → S4O62- + 2H2O

    Note that again, we did not need to know the formal oxidation states of S or O in the reactants and products in order to balance the reaction. In this case, assigning the oxidation states would be rather complex, because S2O32- and S4O62- both contain sulfur in more than one oxidation state.