Potassium Iodide is used as a test for oxidising agent. When mixed with the oxidising agent it will turn brown.Example: Fluorine
Iodide is yellow in colour.
They are not soluble, therefore they do not precipitate or form a color....a.k.a....no reaction...
The color of aqueous iron(II) sulfate is typically pale green. When potassium iodide is added, it reacts with iron(II) ions to form a dark brown precipitate of iron(II) iodide.
The dark red-brown color change occurs because the chlorine oxidizes iodide ions to form iodine, which reacts with excess iodide ions to form triiodide ions. These triiodide ions absorb light in the visible range, resulting in the dark red-brown color observed in the solution.
The flame of potassium iodide is typically violet or purple in color.
Sodium peroxoborate is tested with sulfuric acid and potassium iodide to determine the presence of peroxides. When sodium peroxoborate reacts with sulfuric acid and potassium iodide, oxygen gas is released, causing iodine to be liberated from the potassium iodide. The presence of iodine can be observed by a color change from clear to brown or blue-black due to the formation of iodine.
They are not soluble, therefore they do not precipitate or form a color....a.k.a....no reaction...
The color of aqueous iron(II) sulfate is typically pale green. When potassium iodide is added, it reacts with iron(II) ions to form a dark brown precipitate of iron(II) iodide.
When excess potassium iodide is reacted with chlorine, the solution changes from colorless to dark brown due to the formation of Iodine (I2). Excess potassium iodide reacts with chlorine to form iodine, which imparts the dark brown color to the solution.
The dark red-brown color change occurs because the chlorine oxidizes iodide ions to form iodine, which reacts with excess iodide ions to form triiodide ions. These triiodide ions absorb light in the visible range, resulting in the dark red-brown color observed in the solution.
The flame of potassium iodide is typically violet or purple in color.
Sodium peroxoborate is tested with sulfuric acid and potassium iodide to determine the presence of peroxides. When sodium peroxoborate reacts with sulfuric acid and potassium iodide, oxygen gas is released, causing iodine to be liberated from the potassium iodide. The presence of iodine can be observed by a color change from clear to brown or blue-black due to the formation of iodine.
Chlorine gas reacts with the potassium iodide in the moist starch iodide paper to produce potassium chloride and iodine. The iodine then reacts with the starch in the paper to form a blue complex. This color change is used as a test for the presence of chlorine gas.
When ferric chloride (FeCl3) is added to a solution of potassium iodide (KI), it reacts to form iron(III) iodide (FeI3) and potassium chloride (KCl). The iron(III) iodide produced is a brownish-red color, indicating the presence of the Fe3+ ion. This reaction can be represented by the following chemical equation: 2FeCl3 + 6KI -> 2FeI3 + 6KCl
When potassium iodide is oxidized, it forms iodine. Potassium iodide loses electrons hence it is oxidized. This reaction can be observed by the change in color of the solution from colorless to brown/yellow due to the formation of iodine.
When iron sulfate reacts with potassium iodide, a double displacement reaction occurs where potassium sulfate and iron(II) iodide are formed. The balanced chemical equation for this reaction is FeSO4 + 2KI → FeI2 + K2SO4. This reaction is characterized by a color change from yellow (iron sulfate) to brown (iron(II) iodide).
When bleach is mixed with potassium iodide, it undergoes a redox reaction. The bleach oxidizes the iodide ions to produce iodine, which can be observed as a brown color. This reaction can be used to test for the presence of bleach in a solution.
The endpoint color of sodium hypochlocrite and potassium iodide is not listed.