# Is it necessary to break the equation into half reactions in the oxidation number method?

Apr 28, 2015

No, you don't use half-reactions in the oxidation number method.

Here's how you use the oxidation number to balance the following equation.

$\text{MnO"_4^(-) + "HSO"_3^(-) → "Mn"^(2+) + "SO"_4^(2-)+ "H"_2"O}$

Step 1. Identify the atoms that change oxidation number

Left hand side: $\text{Mn} = + 7$; $\text{O} = - 2$; $\text{H} = + 1$; $\text{S} = + 4$
Right hand side: $\text{Mn} = + 2$; $\text{S} = + 6$; $\text{O} = - 2$; $\text{H} = + 1$

The changes in oxidation number are:
$\text{Mn}$: +7 → +2; Change = -5
$\text{S}$: +4 → +6; Change = +2

Step 2. Equalize the changes in oxidation number

You need 5 atoms of $\text{S}$ for every 2 atoms of $\text{Mn}$. This gives us total changes of +10 and -10.

Step 3. Insert coefficients to get these numbers

$\textcolor{red}{2} {\text{MnO"_4^(-) + color(red)(5)"HSO"_3^(-) → color(red)(2)"Mn"^(2+) + color(red)(5)"SO}}_{4}^{2 -}$

Step 4. Balance $\text{O}$ by adding $\text{H"_2"O}$ molecules to the appropriate side

$\textcolor{red}{2} \text{MnO"_4^(-) + color(red)(5)"HSO"_3^(-) → color(red)(2)"Mn"^(2+) + color(red)(5)"SO"_4^(2-)+ color(blue)(3)"H"_2"O}$

Step 5. Balance H by adding H⁺ ions to the appropriate side

$\textcolor{red}{2} \text{MnO"_4^(-) + color(red)(5)"HSO"_3^(-) + color(green)(1)"H"^+ → color(red)(2)"Mn"^(2+) + color(red)(5)"SO"_4^(2-)+ color(blue)(3)"H"_2"O}$

And the equation is balanced.