An inorganic compound liberates O₂ when heated, turns an acid solution of  KI brown and reduces acidified KMnO₄. The: substances is :

Hydrogen peroxide is H₂O₂ and acts as both a reducing and oxidizing agent.

The heating of hydrogen peroxide is:

$2{\mathrm{H}}_2{\mathrm{O}}_2\stackrel{∆}{\to }{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}$

The reaction of KI with hydrogen peroxide is shown below where hydrogen peroxide acts as a reducing agent and reduces iodine from -1 to 0. The reaction of KI with hydrogen peroxide turns the solution into an acidic medium.

${\mathrm{H}}_2{\mathrm{O}}_2+\stackrel{-1}{\mathrm{KI}}+2{\mathrm{H}}^{+}\to \stackrel{0}{{\mathrm{I}}_2}+{\mathrm{H}}_2\mathrm{O}+2{\mathrm{K}}^{+}$

The reaction of KMnO₄ with H₂O₂ is shown below where H₂O₂ acts as a reducing agent and reduces Mn from +7 to +2.

$3{\mathrm{H}}_2{\mathrm{SO}}_4+\underset{\mathrm{Purple} \mathrm{or} \mathrm{voilet} \mathrm{color}}{\stackrel{+7}{{\mathrm{KMnO}}_4}}+5{\mathrm{H}}_2{\mathrm{O}}_2\to 5{\mathrm{O}}_2+\underset{\mathrm{colorless}}{\stackrel{+2}{2{\mathrm{MnSO}}_4}}+8{\mathrm{H}}_2\mathrm{O}+{\mathrm{K}}_2{\mathrm{SO}}_4$