InterviewSolution
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InterviewSolution
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What happens when: (i) Ferric chloride is added to potassium ferrocynide. (ii) Iron reacct with cold dilute nitric acid. (iii) Potassium ferricyanide is added to ferrous sulphate. (iv) Excess of potassium iodide is added to mercuric chloride. (v) Silver choloride is treated with aq. Sodium cyanide and the product thus formed is allowed to react with zinc in alkaline medium. |
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Answer» Solution :(i) Prussian blue is formed. `4FeCl_(3) + 3K_(4)Fe(CN)_(6) to Fe_(4)[Fe(CN)_(6)]_(3) + 12KCl` Prussian blue (Ferric Ferrocyanide) (ii) Ammonium nitrate is formed: `[Fe + 2HNO_(3) to Fe(NO_(3))_(2) + 2H] xx 4` `HNO_(3) + 8H to NH_(3) + 3H_(2)O` `NH_(3) + HNO_(3) to NH_(4)NO_(3)` ________________________ `4Fe + 10 HNO_(3) to 4Fe(NO_(3))_(4) + NH_(4)NO_(3) + 3H_(2)O` _____________________________ (iii) Ferrous ion is first oxidation to ferric ion while ferricyanide ion is reduced to ferrocyanide ions to form potassium ferric ferrocyanide (Tumbull.s blue) `Fe^(2+) + [Fe(CN)_(6)]^(2-) to Fe^(+3) + [Fe(CN)_(6)]^(4-)` `K^(+) + Fe^(3+) + [Fe(CN)_(6)]^(4-) to K^(+)Fe^(+3)[Fe(CN)_(6)]^(4-)` Potassium ferricferrocyanide (Tumbull.s blue) (IV) First scarlet precipitate is formed then dissolves in excess of potassium iodide forming a complex. `HCl_(2) + 2Kl to Hgl_(2) + 2KCl` `Hgl_(2) + 2Kl to K_(2)Hgl_(4)` Potassium tetraiodo mercurate (colorless) (v) AgCl dissolves in KCN forming a complex, potassium argento CYANIDE. The addition of ZINC precipitate silver. `AgCl + 2KCN to KAg(CN)_(2) + KCl` `2KAg(CN)_(2) + Zn to K_(2)Zn(CN)_(4) + 2Ag` (Potassium Zinocyanide). |
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