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Chemistry LibreTexts

18.02: Balancing Redox Reactions By Inspection

  • Page ID
    178244
  • Learning Objectives

    • Learn to balanced simple redox reactions by inspection.
    • Learn to balance complex redox reactions by the half reaction method.
    • Use the solvent, or parts of it, as a reactant or a product in balancing a redox reaction.

    Balancing simple redox reactions can be a straightforward matter of going back and forth between products and reactants. For example, in the redox reaction of Na and Cl2:

    \[\ce{Na + Cl2 → NaCl}\]

    it should be immediately clear that the Cl atoms are not balanced. We can fix this by putting the coefficient 2 in front of the product:

    \[\ce{Na + Cl2 → 2NaCl}\]

    However, now the sodium is unbalanced. This can be fixed by including the coefficient 2 in front of the Na reactant:

    \[\ce{2Na + Cl2 → 2NaCl}\]

    Let's check the charge on both sides of the equation:

    On the left hand side we pure sodium, so the oxidation number is zero. We also have elemental chlorine, with an oxidation number of zero. On the right hand side we have 2 sodium ions, Na+, so a +2 charge, but we have 2 chloride ions as well, Cl-, so a -2 charge. Since the overall charge on both the left hand side and the right hand side is zero this reaction is now balanced. That was fairly straightforward; we say that we are able to balance the reaction by inspection. Many simple redox reactions can be balanced by inspection.

    Example \(\PageIndex{1}\):

    Balance this redox reaction by inspection.

    SO2 + O2 → SO3

    Solution

    There is one S atom on both sides of the equation, so the sulfur is balanced. However, the reactant side has four O atoms while the product side has three. Clearly we need more O atoms on the product side, so let us start by including the coefficient 2 on the SO3:

    SO2 + O2 → 2SO3

    This now gives us six O atoms on the product side, and it also imbalances the S atoms. We can balance both the elements by adding coefficient 2 on the SO2 on the reactant side:

    2SO2 + O2 → 2SO3

    This gives us two S atoms on both sides and a total of six O atoms on both sides of the chemical equation. As the total charge on both sides of the chemical equation is zero, and therefore equal, this redox reaction is now balanced.

    Exercise \(\PageIndex{1}\)

    Balance this redox reaction by inspection.

    Al + O2 → Al2O3

    Answer

    4Al + 3O2 → 2Al2O3

    The first thing you should do when encountering an unbalanced redox reaction is to try to balance it by inspection.