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Which fo the following will release a gaseous product from Pb_3O_4 ? |
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Answer» `HCl` `Pb_3O_4` is actually a stoichiometric mixture fo 2 mol fo `PBO` and ` 1 ` mol of ` PbO_2`. In `PbO_2`, lead is present in ` +4` oxidation STATE (an unstable state) while in ` PbO`, lead is present in state (an unstable state) while in `PbO`, lead is present in `+2` oxidation state (a stable state). Since lead hds a strong tendency to change from ` +4` state to` +2` state,`PbO_(2)` can act as an oxidizing agent (oxidant). THEREFORE, it oxidizes `Cl^(-)` ion fo ` HCl` into chlorine. On the other hand, `PbO` is a basic oxide. Therefore, it can react with hydrochloric acid `(HCl)` to FORM salt `(PbCl_2)` and water. Hence, the above reaction can be split into two reactions namely : (a) Redox reaction `overset(+4)(Pb) O_(2)+4Hoverset(-1)(Cl) rarr overset(+2)(Pb)Cl_(2)+overset(0)(C)l_(2)+2H_(2)O` (b) Acid-base reaction `2PbO + 4HCl rarr 2PbCl_2 +2H_2O` Since ` HNO_3` ( N is present in the highest oxidation state, +5) itself is an oxidizing agent, there is no reaction between ` PbO_2` and ` HNO_3`, However, an acid-base reaction takes place between ` PbO` and ` HNO_3` : ` 2PbO+4HNO_3 rarr 2Pb (NO_3)_2 +2 H_2O` Thus, it is the passive nature of `PbO_2` against ` HNO_3` that makes the reaction different from the one that follows with ` HCl`. |
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