Question

How do you balance disproportionation reactions?

Answer 1

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"`