Chemistry Forum

Hi, I need an help to balance the following redox reaction in acid solution.
KClO₃ + H₂SO₄ --> K₂SO₄ + O₂ + ClO₂ + H₂O
I tried lots of times but I wasn't able to do that. Can you please give me an help? I will love also the explanation of the process, so I can understand it.
Thanks

Build Redox pair

ClO₃- / ClO₂ and O₂/ ClO₃-
In acidic condition we add H+ and develop water.

ClO₃- + 2 H+ => ClO₂ + H₂O

Equal the charge with Electron


ClO₃- + 2 H+ + e- => ClO₂ + H₂O Reduction finished

2 ClO₃- => 2 ClO₂ + O₂

Equal the charge with Electron

2 ClO₃- => 2 ClO₂ + O₂ + 2 e- Oxidation finished

Multiply the first equation by 2

2 ClO₃- + 4 H+ + 2e- => 2 ClO₂ +2 H₂O

Addition

4 ClO₃- + 4 H+ => 4 ClO₂ + O₂ + 2 H₂O

Finished

The spectator ion K+ and SO₄ 2- can added now if necessary.

4 KClO₃ + 2 H₂SO₄ => 2 K₂SO₄ + 4 ClO₂ + O₂ + 2 H₂O

finished

Can you explain me why does the second pair become:
2 ClO₃- => 2 ClO₂ + O₂ ?

Shouldn't it be like that?
ClO₃- => O₂

Above I wrote the second pair is

O₂/ ClO₃-

But you are right the pair should be O₂- and O₂.

The second equation should be

2 H₂O => O₂ + 4 H+ +4 e-

The first has be multiplied by 4

4 ClO₃- + 8 H+ + 4 e- => 4 ClO₂ + 4 H₂O

Addition gives

4 ClO₃- + 4 H+ => 4 ClO₂ + O₂ + 2 H₂O

The same result in the end.