halide ions

how does oxidising ability of halide ions change down the group?

• attraction between outer e^- and nucleus decreases down group as ions get bigger

• so oxidising ability decreases

give the eqns for halides reacting w/ concentrated sulfuric acid:

NaX + H₂SO₄ → NaHSO₄ + HX

give the further reaction for some halides reducing sulfuric acid:

2HX + H₂SO₄ → X₂ + SO₂ + 2H₂O

give the overall eqn for halides reacting w/ concentrated sulfuric acid and reducing it further:

2NaX + 3H₂SO₄ → 2NaHSO₄ + X₂ + SO₂ + 2H₂O

give the half eqns for some halides reducing sulfuric acid:

2X^- → X₂ + 2e^-

H₂SO₄ + 2H⁺ +2e^- → SO₂ + 2H₂O

name and explain the observations for the reactions of NaF and NaCl w/ H₂SO₄:

• HF and HCl are not strong enough reducing agents to reduce the H₂SO₄ further

• not a redox reaction - oxidation states of halogens and sulfur remain the same

• observation: misty white fumes of HF / HCl

name and explain the observations for the reaction of NaBr w/ H₂SO₄:

• HBr is a strong enough reducing agent to reduce H₂SO₄ further

• second eqn = redox: Br oxidised (from -1 to 0) and S reduced from (+6 to +4)

• observations: misty fumes of HBr, red-brown vapour Br₂, choking fumes of SO₂

what are the overall and ionic eqns for the overall reaction of sodium bromide w/ sulfuric acid, w/ the sulfuric acid being further reduced?

ionic: 2Br⁻ + 3SO₄^2- + 6H⁺ → 2HSO₄^- + Br₂ + SO₂ + 2H₂O

overall: 2NaBr + 3H₂SO₄ → 2NaHSO₄ + Br₂ + SO₂ + 2H₂O

give and explain the eqns for NaI reacting w/ sulfuric acid:

NaI_(s) + H₂SO_{4 (l)} → NaHSO_{4 (s)} + HI_(g)

HI can reduce H₂SO₄ further: 2HI_(aq) + H₂SO_{4 (l)} → I_{2 (s)} + SO_{2 (g)} + 2H₂O_(l)

HI is a very strong reducing agent and can reduce the SO₂ again to S and H₂S: 6HI_(g) + H₂SO_4(l) → 3I_2(s) + S_(s) + 4H₂O_(l)

6HI_(g) + SO_2(g)→ H₂S_(g) + 3I_2(s) + 2H₂O _(l)

give and explain the observations for NaI reacting w/ sulfuric acid:

• misty white fumes (HI)

• purple vapour (I₂)

• yellow solid formed (S)

• rotten egg smell (H₂S)

• black solid formed (I₂)

• choking fumes (SO₂)

give the overall and overall ionic eqns for NaI reacting w/ H₂SO₄ and the observations^_:

overall: 8NaI + 9H₂SO₄ → 4I₂ + 8NaHSO₄ + H₂S + 4H₂O
ionic: 8I⁻ + SO₄^2- + 10H⁺ → 4I₂ + H₂S + 4H₂O

observations: purple vapour released, black solid formed and a rotten egg smell

how can we test for halide ions in aqueous solution?

• add 10 drops of substance you are testing to a clean, dry test tube

• add approx 5 drops of dilute nitric acid and shake

• add 10 drops silver nitrate solution and record observations

• add dilute then concentrated ammonia and record further observations

give the general eqn for the reaction of halide ions w/ silver nitrate:

Ag⁺ (aq) + X^- (aq) → AgX (s)

give the results of the silver nitrate test:

AgCl - white ppt

AgBr - cream ppt

AgI - yellow ppt

give the results for the silver halide ppts dissolving in ammonia:

AgCl - dissolves in dilute ammonia to form a colourless solution

AgBr - dissolves in concentrated ammonia to form a colourless solution

AgI - does not dissolve