An aqueous compound of an inorganic compound `(X)` shows the following reactions:
a. it decolourises and acidified `KMnO_(4)` solution accompanied by the evolution of oxygen.
b. it liberates `I_(2)` from an acidified `KI` solution.
c. It gives a brown precipitate with alkaline `KMnO_(4)` solution with evolution of oxygen.
d. It removes black stains from old oil paintings. Identify `(X)` and give chemical equations for the reaction at steps (a) to (d).

Explanation

(i) Compound (X) acts as a reducing agent both in acidic as well as in the basic medium because it decolourises an acidified solution of `KMnO_(4)` with evolution of `O_(2)` and produces a brown ppt. (probably of `MnO_(2)`) with evolution of `O_(2)` in the basic medium.
(ii) Compound (X) also acts as an oxidising agent since it liberates `I_(2)` from an acidified solution of KI. All these reactions of characteristic of `H_(2)O_(2)`. Therefore , compound (X) is `H_(2)O_(2)`
(iii) Further `H_(2)O_(2)` is known to remove black stains from old oil paintins due to oxidation of black PbS to with `PbSO_(4)`.
If compound (X) is `H_(2)O_(2)`, then all the reactions involved in this problem may be explained as follows :
`(i) 2KMnO_(4) + 3H_(2)SO_(4) + 5H_(2)O_(2) to 2K_(2)SO_(4) + 2MnSO_(4) = 3H_(2)O + 5O_(2)`
`(ii) 2KMnO_(4) + 3H_(2)O_(2) to underset("Brown ppt.")(2MnO_(2))+ 2KOH + 2H_(2)O+ 3O_(2)`
`(iii) 2KI + H_(2)SO_(4) + H_(2)O_(2)to K_(2)SO_(4) + I_(2) + 2H_(2)O`
`(iv) underset("Black")(PbS) + 4H_(2)O_(2) to underset("White")(PbSO_(4))+4H_(2)O`