$5 \mathrm{O}_{2}+2 \mathrm{Mn} \mathrm{SO}_{4}+8 \mathrm{H}_{2} \mathrm{O}+\mathrm{K}_{2} \mathrm{SO}_{4}$

In the above reaction, $K M n O_{4}$ oxidises $H_{2} O_{2}$ to $O_{2}$ and itself $\left[M n O_{4}^{-}\right]$ gets reduced to $M n^{2+}$ ion as $M n S O_{4}$ 

Hence, aqueous solution of $K M n O_{4} .$ with $H_{2} O_{2}$ yields $M n^{2+}$ and $O_{2}$ in acidic conditions.