Reaction of potassium iodide in presence of silver ion
Answers
The net ionic reaction is useful because it clearly shows which ions participate in the reaction and which ones remain in solution as "spectator ions." The ions that participate in the reaction form a precipitate, an insoluble solid. The animation below shows the reaction between potassium iodide and silver nitrate to form silver iodide as a precipitate. The potassium and nitrate ions stay in solution. The same product (silver iodide) could be formed using many other spectator ions, since they are not part of the net ionic equation.
Molecular Equation: KI (aq) + AgNO3 (aq)  AgI (s) + KNO3 (aq)
Net Ionic Equation: I- (aq) + Ag+ (aq)  AgI (s)
The phase diagram of the KI+AgI system has been redetermined. Two intermediate compounds have been identified. K2AgI3 is stable up to 130°C. At higher temperatures it undergoes a solid state disproportionation reaction to KAg4I5 and KI. KAg4I5 is stable between its incongruent melting point (253°C) and 38°C. At lower temperatures it disproportionates to β-AgI and K2AgI3. A eutectic between KAg4I5 and KI occurs at a nominal composition of 29.5 mole % KI and 238°C. The compound KAg4I5 has an exceptionally high ionic conductivity for a solid, reaching 0.31 ohm–1 cm–1 at the incongruent melting point. The current is carried by the Ag+ ions. Thermodynamic data derived from e.m.f. measurements indicate that the formation of KAg4I5 from K2AgI3 and AgI is accompanied by an entropy gain, implying an unusually high degree of disorder in KAg4I5.
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