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Half Reactions

Skills to develop

Half Reactions

A half reaction is a reduction or an oxidation reaction. For example, the following are half reactions.

2 H+ + 2 e- ® H2
MnO4- + 5 e- + 8 H+ ® Mn2+ + 4H2O
Zn ® Zn2+ + 2 e-
Cu ® Cu2+ + 2 e-
A half reaction does not occur by itself, at least two such reactions must be coupled so that the electron released by one reactant is accepted by another in order to complete the reaction. Thus, oxidation and reduction reactions must take place simultaneously in a system, and this type of reactions is called oxidation reduction reaction or simply redox reaction. For example, Zn + Cu2+ ® Zn2+ + Cu is such a redox reaction, Zn being oxidized, and Cu2+ being reduced. Redox reactions take place in battery operations.

Half-reaction equations are useful for balancing redox reaction equations.

Balancing Redox Reactions with Half Reactions

Oxidation and reduction reactions are called 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 diagram of a galvanic cell.

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.
Example 1. Balance the two half reactions for 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 of oxygen)
  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)

Example 2. Balance two half reactions for 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

Example 3. Balance two half reactions for 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-

Example 4. Balance the two half reaction equations for the following reaction in acidic solution: 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-

The four (4) examples above illustrate how to break down a task in steps. You will need the practice in order to master the skills. Take any redox chemical reaction equation and try to balance the two half reactions. The Confidence Building Questions below have redox reaction equations for you to practice.

The following questions requires one steps at a time, but you may take any question and follow the four steps as illustrated in the above examples.

Confidence Building Questions

The skill to balance redox equations can be broken down in several steps. Each question above involvs a small step. You should review the steps and acquire all the skill needed. These questions also shows that you can be tested for skills to balance equations by one of these questions.