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This is one of the reactions I had to undertake in a lab. I think the reason for the test in the first place is to test for the peroxoborate anion in the solution.

From my understanding what's happening is that sodium peroxoborate will liberate $\ce{H2O2}$ when reacted with sulfuric acid. Adding the $\ce{KMnO4}$, the hydrogen will react to reduce $\ce{Mn(VII) ->Mn(II)} $ and form $\ce{O}$

I'm having difficulty interpreting this into balanced chemical equations. This is what I've tried but not sure if it's correct

$$\ce{Na2[B2(O2)2(OH)4] + H2SO4 ->Na2SO4 +B2O3 +2H2O2 +H2O}$$

This equation shows the liberated $\ce{H2O2}$ but not sure how to write a reaction now where $\ce{KMnO4}$ is added.

Any help would be much appreciated. I've been strugging with this for a while and there isn't much information online.

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  • $\begingroup$ typo I edited it there $\endgroup$
    – Patrick Moloney
    Sep 20, 2017 at 19:00
  • $\begingroup$ I'd rather write that product as H3BO3. $\endgroup$
    – Ivan Neretin
    Sep 20, 2017 at 19:06
  • $\begingroup$ What about this : $$\ce{Na2[B2(O2)2(OH)4] + H2SO4 +2H2O ->Na2SO4 +H3BO3 +2H2O2 }$$ $\endgroup$
    – Patrick Moloney
    Sep 20, 2017 at 19:30
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    $\begingroup$ Good. Now write the reaction between H2O2 and KMnO4, then add both reactions together. $\endgroup$
    – Ivan Neretin
    Sep 20, 2017 at 19:32

1 Answer 1

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Well $\ce{KMnO4} $ and $\ce{H2O2}$ react as follows:

$$\ce{2KMnO4 +3H2O2 -> 2MnO2 + 2KOH + 2H2O + 3O2}$$

Adding the two reactions together I get:

$$\ce{Na2[B2(O2)2(OH)4] + H2SO4 + 2KMnO4 + H2O2 ->Na2SO4 + 2H3BO3 + 3O2 + 2MnO2 + 2KOH}$$

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  • $\begingroup$ You can't have acid and alkali together. $\endgroup$
    – Ivan Neretin
    Sep 20, 2017 at 20:02
  • $\begingroup$ do I need to have a salt and water in the reactants? $\endgroup$
    – Patrick Moloney
    Sep 20, 2017 at 20:19
  • $\begingroup$ It is H3BO3 and K2SO4 in the products, I'd say. Also, it is probably MnSO4 instead of MnO2. Other than that, everything's mostly right. $\endgroup$
    – Ivan Neretin
    Sep 20, 2017 at 20:57
  • $\begingroup$ I appreciate the patients, this is what I have so far and tbh I've spent over 3 hours trying to find resources for this, $\endgroup$
    – Patrick Moloney
    Sep 20, 2017 at 21:04
  • $\begingroup$ $$\ce{Na2[B2(O2)2(OH)4] + H2SO4 + 2KMnO4 + H2O2 -> H3BO3 + MnSO4 + 2KO + H2O + 2NaOH + 3O2}$$ $\endgroup$
    – Patrick Moloney
    Sep 20, 2017 at 21:08

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