Sodium Peroxoborate reacting with Sulfuric Acid and Potassium Permanganate 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) -&gt;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

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

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

$$\ce{2KMnO4 +3H2O2 -&gt; 2MnO2  + 2KOH + 2H2O + 3O2}$$ 

Adding the two reactions together I get:

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