Question

How do you balance `NaClO_3 -> NaCl + O_2`?

Answer 1

`2NaClO_3(s)->2NaCl(s)+3O_2(g)`
or
`NaClO_3(aq)+Na_2SO_3(aq)->NaCl(aq)+O_2(g)+Na_2SO_4(aq)`

Explanation:

To balance the reaction, if it is happening at the solid state (decomposition):

`2NaClO_3(s)->2NaCl(s)+3O_2(g)`

However, to balance the reaction, if it is happening in aqueous medium:

`NaClO_3(aq)->NaCl(aq)+O_2(g)`

We should recognize that this is a reduction half equation since the oxidation number of oxygen is increasing from `(-2)` in `NaClO_3` to `(0)` in `O_2`.

The oxidation numbers for `Na` and `Cl` are not changing and they are `(+1)` and `(-1)` respectively.

To balance this reaction then:
`NaClO_3(aq)+2H^(+)(aq)+2e^(-)->NaCl(aq)+O_2(g)+H_2O(l)`

Therefore, in order for this reaction to occur, you will need a reducing agent which will reduce the oxidizing agent `NaClO_3` such as for example `Na_2SO_3`.

The oxidation half equation will then be:

`SO_3^(2-)(aq)+H_2O(l)->SO_4^(2-)(aq)+2H^(+)(aq)+2e^(-)`

The redox reaction will thus be:

Oxidation: `SO_3^(2-)(aq)+cancel(color(green)(H_2O(l)))->SO_4^(2-)(aq) +cancel(color(blue)(2H^(+)(aq)))+cancel(color(red)(2e^(-)))`

Reduction: `NaClO_3(aq)+cancel(color(blue)(2H^(+)(aq)))+cancel(color(red)(2e^(-)))->NaCl(aq)+O_2(g)+cancel(color(green)(H_2O(l)))`

RedOx:
`NaClO_3(aq)+SO_3^(2-)(aq)->NaCl(aq)+O_2(g)+SO_4^(2-)(aq)`
or
`NaClO_3(aq)+Na_2SO_3(aq)->NaCl(aq)+O_2(g)+Na_2SO_4(aq)`

Here is a video that explains how to balance a redox reaction in acidic medium:
Balancing Redox Reactions | Acidic Medium.

And here is another one on how to balance a redox reaction in basic medium:
Balancing Redox Reactions | Basic Medium.