Question

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

Answer 1

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.