Reaction of iodide ions with concentrated sulfuric acid Flashcards
Reaction of iodide ions with concentrated sulfuric acid
The reaction of sodium iodide and concentrated sulfuric acid is:
H2SO4 (l) + NaI (s) → HI (g) + NaHSO4 (s)
H2SO4 (l) + NaI (s) → HI (g) + NaHSO4 (s)
2HI (g) + H2SO4 (l) → I2 (g) + SO2 (g) + 2H2O (l)
what happens during this reaction to the next
Hydrogen iodide decomposes readily
Sulfuric acid oxidises the hydrogen iodide to form several products:
The concentrated sulfuric acid oxidises HI and is itself reduced to sulfur dioxide gas
2HI (g) + H2SO4 (l) → I2 (g) + SO2 (g) + 2H2O (l)
6HI (g) + H2SO4 (l) → 3I2 (g) + S (s) + 4H2O (l)
what happens between these reactions
Iodine is seen as a violet/purple vapour in first reaction
The concentrated sulfuric acid oxidises HI and is itself reduced to sulfur:
6HI (g) + H2SO4 (l) → 3I2 (g) + S (s) + 4H2O (l)
8HI (g) + H2SO4 (l) → 4I2 (g) + H2S (s) + 4H2O (l)
what happens between these reactions
Sulfur is seen as a yellow solid in first reaction
The concentrated sulfuric acid oxidises HI and is itself reduced to hydrogen sulfide:
8HI (g) + H2SO4 (l) → 4I2 (g) + H2S (s) + 4H2O (l)
what happens this reaction
Hydrogen sulfide has a strong smell of bad eggs