balance redox reactions in acidic solution
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Erry215
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balance redox reactions in acidic solution


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
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ChenBeier
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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
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Erry215
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Can you explain me why does the second pair become:
2 ClO3- => 2 ClO2 + O2 ?

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