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An aqueous solution of an inorganic compound (X) shows the following reactions : (i) It decolourises an acidifed KMnO_(4) solution accomopained by evolution of oxygen . (ii) It libirates iodine from an acidified potassium iodide solution . (iii) It gives brown precipitate with an alkaline KMnO_(4) solution with the evolution of oxygen. (iv) It removes black stains from oil paintings. Identify (X) and write equations for the reactions involved. |
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Answer» Solution :(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` |
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