# How do you balance disproportionation reactions?

Feb 28, 2014

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:

$\text{Cl"_2 + "NaOH" → "NaCl" + "NaClO"_3 + "H"_2"O}$

The oxidation number of $\text{Cl}$ in ${\text{Cl}}_{2}$ is 0.

It decreases to -1 in $\text{NaCl}$ (reduction) and increases to +5 in ${\text{NaClO}}_{3}$ (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: $\text{Cl = 0"; "Na = +1"; "O = -2"; "H = +1}$.
Right hand side: $\text{Na = +1"; "Cl in NaCl = -1"; "Cl in NaClO"_3 = "+5"; "H = +1"; "O = -2}$.

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

$\text{Cl}$ changes from 0 to -1 in $\text{NaCl}$; change = -1

$\text{Cl}$ changes from 0 to +5 in ${\text{NaClO}}_{3}$; change = +5

c. Adjust coefficients of $\text{NaCl}$ and ${\text{NaClO}}_{3}$ to balance the changes in oxidation number.

$\text{Cl"_2 + "NaOH" → color(red)(5)"NaCl" + color(red)(1)"NaClO"_3 + "H"_2"O}$

d. Balance atoms other than $\text{O}$ and $\text{H}$.

$\textcolor{b l u e}{3} \text{Cl"_2 + color(blue)(6)"NaOH" → color(red)(5)"NaCl" + color(red)(1)"NaClO"_3 + "H"_2"O}$

e: Balance $\text{O}$ and $\text{H}$

$\textcolor{b l u e}{3} \text{Cl"_2 + color(blue)(6)"NaOH" → color(red)(5)"NaCl" + color(red)(1)"NaClO"_3 + color(green)(3)"H"_2"O}$

g. The balanced equation is

$3 \text{Cl"_2 + 6"NaOH" → 5"NaCl" + "NaClO"_3 + 3"H"_2"O}$