Balancing Redox Reactions Using the Half Reaction Method

Many redox reactions occur in aqueous solutions or suspensions. In this medium most of the reactants and products exist as charged species (ions) and their reaction is often affected by the pH of the medium. The following provides examples of how these equations may be balanced systematically. The method that is used is called the ion-electron or "half-reaction" method.

Example 1 -- Balancing Redox Reactions Which Occur in Acidic Solution

Organic compounds, called alcohols, are readily oxidized by acidic solutions of dichromate ions. The following reaction, written in net ionic form, records this change. The oxidation states of each atom in each compound is listed in order to identify the species that are oxidized and reduced, respectively.

Cr2O72-  +  C2H6O     Cr3+       +       C2H4O
dichromate  ethanol              chromium(III)  acetaldehyde

An examination of the oxidation states, indicates that carbon is being oxidized, and chromium, is being reduced. To balance the equation, use the following steps:

Cr2O72-     Cr3+
    C2H6O     C2H4O
Cr2O72-     2 Cr3+
Cr2O72-    2 Cr3+  +  7 H2O
14 H+  +  Cr2O72-     2 Cr3+  +  7 H2O
14 H+  +  Cr2O72-     2 Cr3+  +  7 H2O
+12                                  +6
    C2H6O    C2H4O  +  2 H+
     0                    +2

    The electrons must always be added to that side which has the greater positive charge as shown below.

6 e-  +  14 H+  +  Cr2O72-     2 Cr3+  +  7 H2O
+6                                                +6
6 e-  +  14 H+  +  Cr2O72-     2 Cr3+  +  7 H2O
    3 x (C2H6O    C2H4O  +  2 H+  +  2 e-)

    The seventh and last step involves adding the two half reactions and reducing to the smallest whole number by cancelling species which on both sides of the arrow.

6 e+  14 H+  +  Cr2O72-  +  3 C2H6O    2 Cr3+  +  7 H2O  +  3 C2H4O  +  6 H+  +  6 e-
    Note that the above equation can be further simplified by subtracting out 6 e- and 6 H+ ions from both sides of the equation to give the final equation.
8 H+  +  Cr2O72-  +  3 C2H6O    2 Cr3+  +  7 H2O  +  3 C2H4O
    Note: the equation above is completely balanced in terms of having an equal number of atoms as well as charges.
Example 2 - Balancing Redox Reactions in Basic Solutions

The active ingredient in bleach is the hypochlorite (OCl-) ion. This ion is a powerful oxidizing agent which oxidizes many substances under basic conditions. A typical reaction is its behavior with iodide (I-) ions as shown below in net ionic form.

I-(aq)  +  OCl-(aq)    I2(s)  +  Cl-(aq) +  H2O(l)
Balancing redox equations in basic solutions is identical to that of acidic solutions except for the last few steps as shown below.
I-   I2
2 I-   I2
2 I-   I2
2 I-   I2
2 I-   I2  +  2 e-
2 e-  +  2 I-  +  2 H+  +  OCl-    I2  +  Cl-  +  H2O  +  2e-
    and subtract "like" terms from both sides of the equation. Subtracting "2e-" from both sides of the equation gives the net equation:
2 I-  +  2 H+  +  OCl-    I2  +  Cl-  +  H2O
2 OH- + 2 I- + 2 H+ + OCl-    I2 + Cl- + H2O + 2 OH-
2 H2O + 2 I- + OCl-    I2 + Cl- + H2O + 2 OH-
H2O + 2 I- + OCl-    I2 + Cl- + 2 OH-
    Note that both the atoms and charges are equal on both sides of the equation, and the presence of hydroxide ions (OH-) indicates that the reaction occurs in basic solution.