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Half Reactions

Skills to develop

  • Explain what half reactions are.
  • Write a chemical equation to represent a (half) reduction reaction.
  • Write a chemical equation to represent an oxidation reaction.
  • Combine reduction and oxidation equations to explain chemical reactions.
  • Write two half reactions for an unbalanced overall chemical reaction.
  • Half Reactions

    A half reaction is a reduction or an oxidation reaction, such as

        2 H+ + 2 e- --> H2

    Balancing Redox Reaductions with Half Reactions

    Oxidation and reduction reactions are called redox reactions. The half reaction equations described earlier are useful in balancing redox reactions.

    Theoretically, an oxidation-reduction reaction takes place even when the reactants are well separated in space, as long as the flow of electrons and ions are facilitated by electrical connections (salt bridge and wire). A simple redox reaction of the type:

        Zn(s) + Cu2+(aq) --> Cu(s) + Zn2+(aq)

    may be carried out using a galvanic cell.

    This reaction may be written as two half-reactions and adding the two half reactions gives the overall equation representing a chemical process:

         Zn             --> Zn2+(aq) + 2 e-
    +) Cu2+(aq) + 2 e- --> Cu(s) ----------------------------------------- Zn(s) + Cu2+(aq) --> Cu(s) + Zn2+(aq)

    In this case, the electrons travel from the Zn electrode to the Cu electrode via the wire connecting the two electrodes. The ions travel in a solution or through a salt bridge to balance the charge in the electrolyte solutions.

    You have already seen the demonstration of this reaction in the past two weeks, and in the module called Battery.

    A Redox reaction may be balanced by first writing two half-reactions, and then canceling the electrons by adding them algebraically. You will learn to balance half-reaction equations in this tutorial.

    Guide for writing and balancing half-reaction equations

    1. identify the key element that undergoes an oxidation state change
    2. balance the number of atoms of the key element on both sides,
    3. add the appropriate number of electrons to compensate for the change of oxidation state.
    4. add H+ (in acid medium), or OH- (in basic medium), to balance the charge on both sides of the half-reactions; and H2O, if necessary, to balance the equations.


    Some examples are given to illustrate how we use half reactions to describe and balance some reduction and oxidation (redox) reactions.

    As the first example, consider the reaction in an acid solution: H2O2 + I- -> I2 + H2O

    1. I- is oxidized (oxidation state increases from -1 to 0).
      O (oxygen) is reduced (oxidation state decreases from -1 to -2).
    2. The two half-reactions with balanced number of key atoms are:
      2 I- --> I2; <--- (oxidation)
      H2O2 --> 2 H2O (reduction)
    3. Add electrons to compensate for the changes of oxidation state 2 I- --> I2 + 2 e- (oxidation)
      H2O2 + 2 e- --> 2 H2O (reduction)
    4. Obviously, H+ should be added to the reduction half-reaction, and the balanced equations are: 2 I- --> I2; <--- (oxidation)
      H2O2 + 2H+ --> 2 H2O (reduction)

    As the second example, consider the reaction in a basic solution: ClO2 + OH- --> ClO2- + ClO3-
    1. In this reaction, Cl from ClO2 is both oxidized and reduced.
    2. The two half-reactions are: ClO2 --> ClO2-; (reduction)
      ClO2 --> ClO3-; (oxidation)
    3. Add electrons to compensate for the oxidation changes: ClO2 + e- --> ClO2-; (reduced, 4 -> 3 for Cl)
      ClO2 --> ClO3- + e-; (oxidized, 4 -> 5)
    4. Add H+, OH-, or H2O to balance both equations results in ClO2 + e- --> ClO2-
      ClO2 + 2 OH- --> ClO3- + e- + H2O
      Now add the two half reactions together to give the overall reaction: 2 ClO2 + 2 OH- --> ClO2- + ClO3- + H2O

    As a third example, consider the reaction in an acidic solution: HS2O3- --> S + HSO4-

    You may think that the two sulfur atoms in the formula are identical, but they are different. You have to understand the chemistry of these ions and then start to investigate their chemical reaction. The structure of HS2O3- may be compared to that of HSO4-:

           S     O-       O     O-
            \\ /           \\ /
              S              S
            // \           // \
           O    OH        O    OH
    Thus, one of the two S atoms has an oxidation state of -2, and we represent this S atom by (=S) to indicate that it is attached to another S atom by a double bond (=).
    1. In this reaction, one S atom goes from -2 to 0, whereas the oxidation state of the other S atom does not change. You have to assume that the S atom is oxidized by a reducing agent, H2O.
    2. Only the key elements are given on the left in the half-reactions: HS(=S)O3- --> S + ...(HSO4-)
      H2O --> H2(g) + ...(HSO4-)
    3. Add electrons to compensate for the oxidation changes: HS(=S)O3- --> S + 2 e- + HSO4-
      H2O + 2 e- --> H2(g) + HSO4-
    4. Combining the two half-reactions gives the following balanced chemical equation. HS(=S)O3- + H2O --> S + H2(g) + HSO4-

    Consider the following reaction in acidic solution as a 4th example: S(=S)O32- + I2 --> I- + S4O62-

    The chemistry of the above reaction is complicated, but you don't have to worry about that at this time. You may use the above method even if you do not know the structure of these species.

    1. The S atoms are oxidized. For convenience, the best way is to assume the average oxidation state for both S atoms. S oxidized (oxidation state (+2) -> (+10/4 or +2.5)
      I reduced ( 0 -> -1)
    2. Write the half-reactions and balance the key elements: 2 S2O32- --> S4O62-
      I2 --> 2 I-
    3. Add electron to compensate the oxidation state changes: 2 S2O32- --> S4O62- + 2 e-
      I2 + 2 e- --> 2 I-
    4. Since the half-reactions are balanced with respect to charges and number of atoms, no further work is required. Just add the two equations and get a balanced equation. 2 S2O32- + I2 --> 2 I- + S4O62-

    You have read four (4) examples now, and to make the skill yours, it is your turn to practice. Have fun!! There are some questions in the dialogue to help you, but pracitice will make the skills yours.

    Confidence Building Questions