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1.8: Balancing Redox Reactions

  • Page ID
    201534
  • In any reaction, it is useful to quantify things. How much product will there be? How much of each reactant do we need to add? What ratios do we need? There are an awful lot of reactions for which this process is straightforward, but sometimes it can be tricky. Redox reactions are sometimes on the tricky side (although certainly not always). For that reason, it's good to have a reliable method for balancing redox reactions: determining the ratios of reactants needed to give the products in the proper amounts.

    Suppose, for example, we have a reaction in which silver oxide (Ag2O) reacts with manganese ion (Mn3+) to produce manganese dioxide and silver. That's:

    \(\ce{Mn^{3+} + Ag2O -> MnO2 + Ag}\)

    What would the balanced reaction look like?

    The first thing to do is make sure you are working with one half-reaction at a time. So that's:

    \(\ce{Mn^{3+} -> MnO2}\)

    and

    \(\ce{Ag2O -> Ag}\)

    We start out by balancing the atoms involved, one at a time. First we look at the atoms other than hydrogen or oxygen, and we just balance those by adding the right coefficient.

    \(\ce{Mn^{3+} -> MnO2}\)

    and

    \(\ce{Ag2O -> Ag}\)

    Second, we balance the oxygen atoms by adding water to one side or the other.

    \(\ce{Mn^{3+} + 2H2O -> MnO2}\)

    and

    \(\ce{Ag2O -> 2Ag + H2O}\)

    Third, we balance any hydrogens by adding protons.

    \(\ce{Mn^{3+} + 2H2O -> 4H^{+} + MnO2}\)

    and

    \(\ce{Ag2O + 2H^{+} -> 2Ag + H2O}\)

    Fourth, we balance the charge by adding electrons.

    \(\ce{Mn^{3+} + 2H2O -> 4H^{+} + MnO2 + e^{-}}\)

    and

    \(\ce{Ag2O + 2H^{+} + 2e^{-} -> 2Ag+ H2O}\)

    Fifth, we multiply so that the number of electrons is the same in both reactions.

    \(2 \times (\ce{Mn^{3+} + 2H2O -> 4H^{+} + MnO2 + e^{-}})\)

    or

    \(\ce{2Mn^{3+} + 4H2O -> 8H^{+} + 2MnO2 + 2e^{-}}\)

    and

    \(\ce{Ag2O + 2H^{+} + 2e^{-} -> 2Ag + H2O}\)

    Sixth, we simply add these two reactions together. The reaction arrow functions like an equals sign. The left side adds to the left side, and the right side adds to the right.

    \(\ce{2Mn^{3+} + 4H2O + Ag2O + 2H^{+} + 2e^{-} -> 8H^{+} + 2MnO2 + 2e^{-} + 2Ag + H2O}\)

    At that point, gratifyingly, the equation simplifies. Notice that we have added the same number of electrons to each side; they cancel out. That's perfect, because it means we have supplied just the right number of electrons from one half reaction to satisfy the other half reaction.

    \(\ce{2Mn^{3+} + 4H2O + Ag2O + 2H^{+} -> 8H^{+} + 2MnO2 + 2Ag + H2O}\)

    Also if we subtract one water from each side, things get slightly simpler.

    \(\ce{2Mn^{3+} + 3H2O + Ag2O + 2H^{+} -> 8H^{+} + 2MnO2 + 2Ag}\)

    Subtracting two protons from each side makes it simpler still.

    \(\ce{2Mn^{3+} + 3H2O + Ag2O -> 6H^{+} + 2MnO2 + 2Ag}\)

    This method works for any redox reaction, no matter how complicated.

    Exercise \(\PageIndex{1}\)

    Balance the following reactions.

    1. \(\ce{Cu + MoO2 -> Cu2O + Mo}\)
    2. \(\ce{NH2OH + Ag2O -> N2 + Ag}\)
    3. \(\ce{Fe3O4 + CO -> Fe + CO2}\)
    4. \(\ce{I2 + MnO4^{-} -> IO3^{-} + MnO2}\)
    5. \(\ce{H3Mo7O24 + S2O3^{2-} -> Mo + So3^{2-}}\)
    Answer a)

    a) Cu --> Cu2O MoO2 --> Mo

    2 Cu --> Cu2O MoO2 --> Mo

    2 Cu + H2O --> Cu2O MoO2 --> Mo + 2 H2O

    2 Cu + H2O --> Cu2O + 2 H+ 4H+ + MoO2 --> Mo + 2 H2O

    2 Cu + H2O --> Cu2O + 2 H+ + 2e- 4e- + 4H+ + MoO2 --> Mo + 2 H2O

    2x (2 Cu + H2O --> Cu2O + 2 H+ + 2e- ) 4e- + 4H+ + MoO2 --> Mo + 2 H2O

    adding:

    4 Cu + 2 H2O --> 2 Cu2O + 4 H+ + 4e-

    4e- + 4H+ + MoO2 --> Mo + 2 H2O

    equals

    4 Cu + MoO2 --> 2 Cu2O + Mo

    Answer b)

    b) NH2OH --> N2 Ag2O --> Ag

    2 NH2OH --> N2 Ag2O --> 2 Ag

    2 NH2OH --> N2 + 2 H2O Ag2O --> 2 Ag + H2O

    2 NH2OH --> N2 + 2 H2O + 2 H+ 2 H+ + Ag2O --> 2 Ag + H2O

    2 NH2OH --> N2 + 2 H2O + 2 H+ + 2 e- 2 H+ + Ag2O + 2 e- --> 2 Ag + H2O

    adding:

    2 NH2OH --> N2 + 2 H2O + 2 H+ + 2 e-

    2 H+ + Ag2O + 2 e- --> 2 Ag + H2O

    equals

    2 NH2OH + Ag2O --> N2 + 2 Ag + 3 H2O

    Answer c)

    c) Fe3O4 --> Fe CO --> CO2

    Fe3O4 --> 3 Fe CO --> CO2

    Fe3O4 --> 3 Fe + 4 H2O H2O + CO --> CO2

    8 H+ + Fe3O4 --> 3 Fe + 4 H2O H2O + CO --> CO2 + 2 H+

    8 e- + 8 H+ + Fe3O4 --> 3 Fe + 4 H2O H2O + CO --> CO2 + 2 H+ + 2 e-

    8 e- + 8 H+ + Fe3O4 --> 3 Fe + 4 H2O 4x (H2O + CO --> CO2 + 2 H+ + 2 e- )

    adding:

    8 e- + 8 H+ + Fe3O4 --> 3 Fe + 4 H2O

    4 H2O + 4 CO --> 4 CO2 + 8 H+ + 8 e-

    equals

    Fe3O4 + 4 CO --> 3 Fe + 4 CO2

    Answer d)

    d) I2 --> IO3- MnO4- --> MnO2

    I2 --> 2 IO3- MnO4- --> MnO2

    6 H2O + I2 --> 2 IO3- MnO4- --> MnO2 + 2 H2O

    6 H2O + I2 --> 2 IO3- + 12 H+ 4H+ + MnO4- --> MnO2 + 2 H2O

    6 H2O + I2 --> 2 IO3- + 12 H+ + 10 e- 3 e- + 4H+ + MnO4- --> MnO2 + 2 H2O

    3x (6 H2O + I2 --> 2 IO3- + 12 H+ + 10 e-) 10x ( 3 e- + 4H+ + MnO4- --> MnO2 + 2 H2O)

    adding

    18 H2O + 3 I2 --> 6 IO3- + 36 H+ + 30 e-

    30 e- + 40 H+ + 10 MnO4- --> 10 MnO2 + 20 H2O

    equals

    3 I2 + 4 H+ + 10 MnO4- --> 6 IO3- + 10 MnO2 + 2 H2O

    Answer e)

    e) H3Mo7O24 --> Mo S2O32- --> SO32-

    H3Mo7O24 --> 7 Mo S2O32- --> 2 SO32-

    H3Mo7O24 --> 7 Mo + 24 H2O 3H2O + S2O32- --> 2 SO32-

    45 H+ + H3Mo7O24 --> 7 Mo + 24 H2O 3H2O + S2O32- --> 2 SO32- + 6 H+

    45 e- + 45 H+ + H3Mo7O24 --> 7 Mo + 24 H2O 3H2O + S2O32- --> 2 SO32- + 6 H+ + 4 e-

    4 x (45 e- + 21 H+ + H3Mo7O24 --> 7 Mo + 24 H2O) 45 x (3H2O + S2O32- --> 2 SO32- + 6 H+ + 4 e- )

    adding:

    180 e- + 180 H+ + 4 H3Mo7O24 --> 28 Mo + 96 H2O

    135 H2O + 45 S2O32- --> 90 SO32- + 270 H+ + 180 e-

    equals

    4 H3Mo7O24 + 45 S2O32- + 39 H2O --> 28 Mo + 90 SO32- + 90 H+

    checking:

    28 Mo --> 28 Mo; 90 S --> 90 S; 90 H --> 90 H; 270 O --> 270 O

    In the event that the reaction is described as occuring under basic conditions, we can simply "neutralize" our protons at the end, by adding hydroxide to both sides.

    \(\ce{2Mn^{3+} + 3H2O + Ag2O + 6 OH^{-} -> 6H^{+} + 6OH^{-} + 2MnO2 + 2Ag}\)

    Which of course means

    \(\ce{2Mn^{3+} + 3H2O + Ag2O + 6OH^{-} -> 6H2O + 2MnO2 + 2Ag}\)

    Simplifying to

    \(\ce{2Mn^{3+} + Ag2O + 6OH^{-} -> 3H2O + 2MnO2 + 2Ag}\)

    Exercise \(\PageIndex{2}\)

    Balance the following reactions under basic conditions.

    1. \(\ce{Fe(OH)2 + N2H4 -> Fe2O3 + NH4^{+}}\)
    2. \(\ce{MnO4^{-} + V^{3+} -> HMnO4^{-} + VO2^{+}}\)
    Answer a)

    a) Fe(OH)2 --> Fe2O3 N2H4 --> NH4+

    2 Fe(OH)2 --> Fe2O3 N2H4 --> 2 NH4+

    2 Fe(OH)2 --> Fe2O3 + H2O N2H4 --> 2 NH4+

    2 Fe(OH)2 --> Fe2O3 + H2O + 2 H+ 4 H+ + N2H4 --> 2 NH4+

    2 Fe(OH)2 --> Fe2O3 + H2O + 2 H+ + 2 e- 2 e- + 4 H+ + N2H4 --> 2 NH4+

    adding:

    2 Fe(OH)2 --> Fe2O3 + H2O + 2 H+ + 2 e-

    2 e- + 4 H+ + N2H4 --> 2 NH4+

    equals:

    2 Fe(OH)2 + 2 H+ + N2H4 --> Fe2O3 + H2O + 2 NH4+

    in basic conditions:

    2 Fe(OH)2 + 2 H+ + 2 -OH + N2H4 --> Fe2O3 + H2O + 2 NH4+ + 2 -OH

    2 Fe(OH)2 + 2 H2O + N2H4 --> Fe2O3 + H2O + 2 NH4+ + 2 -OH

    2 Fe(OH)2 + H2O + N2H4 --> Fe2O3 + 2 NH4+ + 2 -OH

    Answer b)

    b) MnO4- --> HMnO4- V3+ --> VO2+

    MnO4- --> HMnO4- V3+ + H2O --> VO2+

    H+ + MnO4- --> HMnO4- V3+ + H2O --> VO2+ + 2 H+

    e- + H+ + MnO4- --> HMnO4- V3+ + H2O --> VO2+ + 2 H+ + e-

    adding:

    e- + H+ + MnO4- --> HMnO4-

    V3+ + H2O --> VO2+ + 2 H+ + e-

    equals:

    MnO4- + V3+ + H2O --> HMnO4- + VO2+ + H+

    under basic conditions:

    MnO4- + V3+ + H2O + -OH --> HMnO4- + VO2+ + H+ + -OH

    MnO4- + V3+ + H2O + -OH --> HMnO4- + VO2+ + H2O

    MnO4- + V3+ + -OH --> HMnO4- + VO2+