Question

An aqueous solution of a compound $X$ shows the following reactions.
(i) It decolourises an acidified $KMn{O}_4$ solution accompanied by the evolution of oxygen.
(ii) It liberates iodine from an acidified potassium iodine solution.
(iii) It gives a brown precipitate with alkaline $KMn{O}_4$ solution with the evolution of oxygen.
(iv) It removes black stains from old oil paintings.
Identify $X$.

• A. $H_2{O}_2$
• B. $H_{2}O$
• C. $C{O}_2$
• D. None of these

Answer 1

The correct option is A $H_2{O}_2$

The compound $\left(X\right)$ is hydrogen peroxide $H_2{O}_2$
(i) Hydrogen peroxide reduces Mn(VII) to Mn(II). Hence potassium permangante is decolourized. Oxygen is evolved during the process.
$2Mn{O}_4^{-}+5H_2{O}_2+6H^{+}⟶2M{n}^{2+}+8H_{2}O+5O_2$
(ii) It reduces iodide ion to iodine in acidified potassium iodide solution.
$H_2{O}_2+2I+2H^{+}⟶I_2+2H_{2}O$
(iii) Hydrogen peroxide reduces $Mn\left(VII\right)$ to $Mn\left(II\right)$. The brown precipitate is due to formation of manganese dioxide. Oxygen is evolved during the process.
$2Mn{O}_4+3H_2{O}_2⟶2Mn{O}_2+3O_2+2H_{2}O+2O{H}^{-}$
(iv) In the oil painting the black stains are due to presence of lead sulphide.
Hydrogen peroxide converts lead sulphide to lead sulphate.
$PbS+4H_2{O}_2⟶PbS{O}_4+4H_{2}O$