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Newbie

Joined: 27 Sep 2018
Posts: 2

# 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

Distinguished Member

Joined: 27 Sep 2017
Posts: 269
Location: Berlin, Germany
 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

Newbie

Joined: 27 Sep 2018
Posts: 2
 Can you explain me why does the second pair become: 2 ClO3- => 2 ClO2 + O2 ? Shouldn't it be like that? ClO3- => O2

Distinguished Member

Joined: 27 Sep 2017
Posts: 269
Location: Berlin, Germany
 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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