Balancing Redox Reactions Using the 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.

Cr₂O₇^2-  +  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:

• First, divide the equation into two halves; one will be an oxidation half-reaction and the other a reduction half- reaction, by grouping appropriate species.  The nature of each will become evident in subsequent steps.

Cr₂O₇^2-     Cr³⁺

C₂H₆O     C₂H₄O

• Second, if necessary, balance all elements except oxygen and hydrogen in both equations by inspection.   In other words, balance the non-hydrogen and non-oxygen atoms only.  By following this guideline in the example above, only the chromium reaction needs to be balanced by placing the coefficient, 2 , in front of Cr⁺³ as shown below.

Cr₂O₇^2-     2 Cr³⁺

C₂H₆O    C₂H₄O
 
• The third step involves balancing oxygen atoms.  To do this, add water (H₂O) molecules. Use 1 molecule of water for each oxygen atom that needs to be balanced.  Add the appropriate number of water molecules to that side of the equation required to balance the oxygen atoms as shown below.

Cr₂O₇^2-    2 Cr³⁺  +  7 H₂O

C₂H₆O    C₂H₄O
 
• The fourth step involves balancing the hydrogen atoms. To do this one must use hydrogen ions (H⁺). Use one (1) H⁺ ion for every hydrogen atom that needs to balanced. Add the appropriate number of hydrogen ions to that side of the equation required to balance the hydrogen atoms as shown below

14 H⁺  +  Cr₂O₇^2-     2 Cr³⁺  +  7 H₂O

C₂H₆O    C₂H₄O  +  2 H⁺
 
• The fifth step involves the balancing charges. This is done by adding electrons (e-). Each electron has a charge equal to (-1). To determine the number of electrons required, find the net charge of each side the equation.

14 H⁺  +  Cr₂O₇^2-     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⁺  +  Cr₂O₇^2-     2 Cr³⁺  +  7 H₂O
+6                                                +6

C₂H₆O    C₂H₄O  +  2 H⁺  +  2 e^-
 0                    0

note: the net charge on each side of the equation does not have to equal zero.

The chromium reaction can now be identified as the reduction half-reaction and the ethanol/acetaldehyde as the oxidation half-reaction. The reduction half-reaction requires 6 e-, while the oxidation half-reaction produces 2 e-.
 

• The sixth step involves multiplying each half-reaction by the smallest whole number that is required to equalize the number of electrons gained by reduction with the number of electrons produced by oxidation. Using this guideline, the oxidation half reaction must be multiplied by "3" to give the 6 electrons required by the reduction half-reaction.

6 e^-  +  14 H⁺  +  Cr₂O₇^2-     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⁺  +  Cr₂O₇^2-  +  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⁺  +  Cr₂O₇^2-  +  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.

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)

• First, divide the equation into two halves by grouping appropriate species.

I^-   I₂

OCl^-   Cl^-
 
• Second, if needed, balance both equations, by inspection ignoring any oxygen and hydrogen atoms.

2 I^-   I₂

OCl^-   Cl^-
 
• Third, balance the oxygen atoms using water molecules .

2 I^-   I₂

OCl^-   Cl^-  +  H₂O
 
• Fourth, balance any hydrogen atoms by using an (H+) for each hydrogen atom

2 I^-   I₂

2H⁺  +  OCl^-   Cl^-  +  H₂O
 
• Fifth, use electrons (e-) to equalize the net charge on both sides of the equation. Note; each electron (e-) represents a charge of (-1).

2 I^-   I₂  +  2 e^-

2e^-  +  2H⁺  +  OCl^-   Cl^-  +  H₂O
 
• Sixth, equalize the number of electrons lost with the number of electrons gained by multiplying by an appropriate small whole number.  Since the number of electrons lost equals the number of electrons gained we do not have to do anything.

 
• Add the two equations, as shown below.

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

• To indicate the fact that the reaction takes place in a basic solution, one must now add one (OH-) unit for every (H+) present in the equation. The OH- ions must be added to both sides of the equation as shown below.

2 OH^- + 2 I^- + 2 H⁺ + OCl^-    I₂ + Cl^- + H₂O + 2 OH^-

• Then, on that side of the equation which contains both (OH^-) and (H⁺) ions, combine them to form H₂O. Note, combining the 2 OH^- with the 2 H⁺ ions above gives 2 HOH or 2 H₂O molecules as written below.

2 H₂O + 2 I^- + OCl^-    I₂ + Cl^- + H₂O + 2 OH^-

• Simplify the equation by subtracting out water molecules, to obtain the final, balanced equation.

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.