How do you balance disproportionation reactions?

1 Answer
Feb 28, 2014

Answer:

You can use the oxidation number method to balance an equation for a disproportionation reaction.

Explanation:

In a disproportionation reaction, the same substance forms two different redox products.

For example, chlorine gas reacts with dilute sodium hydroxide to form sodium chloride, sodium chlorate, and water:

#"Cl"_2 + "NaOH" → "NaCl" + "NaClO"_3 + "H"_2"O"#

The oxidation number of #"Cl"# in #"Cl"_2# is 0.

It decreases to -1 in #"NaCl"# (reduction) and increases to +5 in #"NaClO"_3# (oxidation).

The steps to balance the equation are:

a. Determine the oxidation numbers of each atom on both sides of the equation.

Left hand side: #"Cl = 0"; "Na = +1"; "O = -2"; "H = +1"#.
Right hand side: #"Na = +1"; "Cl in NaCl = -1"; "Cl in NaClO"_3 = "+5"; "H = +1"; "O = -2"#.

b. Identify the atoms for which the oxidation number changes.

#"Cl"# changes from 0 to -1 in #"NaCl"#; change = -1

#"Cl"# changes from 0 to +5 in #"NaClO"_3#; change = +5

c. Adjust coefficients of #"NaCl"# and #"NaClO"_3# to balance the changes in oxidation number.

#"Cl"_2 + "NaOH" → color(red)(5)"NaCl" + color(red)(1)"NaClO"_3 + "H"_2"O"#

d. Balance atoms other than #"O"# and #"H"#.

#color(blue)(3)"Cl"_2 + color(blue)(6)"NaOH" → color(red)(5)"NaCl" + color(red)(1)"NaClO"_3 + "H"_2"O"#

e: Balance #"O"# and #"H""#

#color(blue)(3)"Cl"_2 + color(blue)(6)"NaOH" → color(red)(5)"NaCl" + color(red)(1)"NaClO"_3 + color(green)(3)"H"_2"O"#

g. The balanced equation is

#3"Cl"_2 + 6"NaOH" → 5"NaCl" + "NaClO"_3 + 3"H"_2"O"#