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

1 Answer
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.

"MnO"_4^(-) + "HSO"_3^(-) → "Mn"^(2+) + "SO"_4^(2-)+ "H"_2"O"

Step 1. Identify the atoms that change oxidation number

Left hand side: "Mn" =+7; "O" = -2; "H" = +1; "S" = +4
Right hand side: "Mn" = +2; "S" = +6; "O" = -2; "H" = +1

The changes in oxidation number are:
"Mn": +7 → +2; Change = -5
"S": +4 → +6; Change = +2

Step 2. Equalize the changes in oxidation number

You need 5 atoms of "S" for every 2 atoms of "Mn". This gives us total changes of +10 and -10.

Step 3. Insert coefficients to get these numbers

color(red)(2)"MnO"_4^(-) + color(red)(5)"HSO"_3^(-) → color(red)(2)"Mn"^(2+) + color(red)(5)"SO"_4^(2-)

Step 4. Balance "O" by adding "H"_2"O" molecules to the appropriate side

color(red)(2)"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

color(red)(2)"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.