reaction of halide salts with conc. sulfuric acid

explanation of differing reducing power of halides

• reducing agents donate electrons

• reducing power of halides increases down group 7

• greater tendency to donate electrons

• ions get bigger therefore it is easier for outer electrons to be given away

• as pull from nucleus become smaller

fluoride and chloride

H₂SO₄ is not strong enough an oxidising agent to oxidise the chloride and fluoride ions. no redox reactions occur, only acid base reactions:

• NaF(s) + H2SO4 (l) →NaHSO4 (s) + HF(g)

• NaCl(s) + H2SO4(l) → NaHSO4(s) + HCl(g)

Observations: White steamy fumes of HCl and HF formed

bromide ions reaction

Br^- ions are stronger reducing agents

Acid-base step: NaBr(s) + H2SO4 (l) →NaHSO4 (s) + HBr(g)

Redox step: 2HBr + H2SO4 →Br2 (g) + SO2 (g) + 2H2O(l)

Observations: White steamy fumes of HBr formed, Brown fumes of Bromine

what does the oxidation state of sulfur change to during reaction with bromide ions

+6 → +4

iodide ions reaction

I^- ions are the strongest halide reducing agents

• NaI(s) + H₂SO₄ (l) → NaHSO₄ (s) + HI(g)

• 8HI + H₂SO₄ → 4I₂ (s) + H₂S(g) + 4H₂O(l)

Observations: White steamy fumes of HI formed, Black solid and purple fumes of Iodine formed, Yellow solid sulfur formed, H₂S (bad egg smell)

what does the oxidation state of sulfur change to during reaction with iodide ions

+6 → -2