Doc Brown's Chemistry - Advanced Level Inorganic Chemistry Periodic Table Revision Notes

Part 9. Group 7/17 The Halogens

9.4 The products of heating halide salts with conc. sulfuric acid and 9.5 Chemical tests for halogens and halide ions

The reaction of fluorides, chlorides, bromides and iodides with hot concentrated sulfuric acid is fully described and the observations fully explained. The chemical test methods for detecting and identifying halogens and the corresponding halide ions are described.

9.4 The effect of heating halide salts with concentrated sulfuric acid

• If salts such as sodium chloride, sodium bromide and sodium iodide are heated with conc. sulfuric acid, no simple pattern is observed, each reaction gives different products irrespective of the halogen element/compound formed. Similar results are obtained with potassium salts (just swap Na with a K) and most other metal salts of the halogens.

  • Conc. sulfuric acid is a viscous covalent liquid which can act as a mild oxidising agent with reducing agents.

  • In each case the metal ion is a spectator ion.

  • In writing the equations, it doesn't really matter whether you show the formation of sodium sulfate (Na₂SO₄) or sodium hydrogensulfate (NaHSO₄), I think the former is generally more convenient.

  • These are messy reactions and the equations are a 'theoretical summary' of what happens!

• (1) Sodium chloride

  • The less volatile sulfuric acid displaces the more 'volatile' colourless hydrogen chloride gas and leaving a white residue of sodium hydrogensulfate or sodium sulfate.

  • NaCl(s) + H₂SO₄(l) ==> NaHSO₄(s) + HCl(g)

  • or 2NaCl(s) + H₂SO₄(l) ==> Na₂SO₄(s) + 2HCl(g)

  • or Cl^–(s) + H₂SO₄(l) ==> HSO₄^–(s) + HCl(g)

  • Since it is not a redox reaction, non of the elements involved changes in oxidation state.

  • Sodium fluoride would behave in a similar way i.e. hydrogen fluoride gas, HF, would be released.

  • These are 'displacement' reaction, NOT a redox reaction.

  • However, in the case of bromides and iodides, the hydrogen halides are still formed, other more significant redox reactions take places in which the halide ions are oxidised to the halogen and the conc. sulfuric acid undergoes reduction. These redox reactions are now described below.

  • Note:

    • Other chloride salts e.g. potassium chloride etc. behave in the same way.

    • Sodium fluoride or potassium fluoride salts also behave in the same way e.g.

      • KF(s) + H₂SO₄(l) ==> KHSO₄(s) + HF(g)

      • The hydrogen fluoride gas released reacts with glass and will etch it.

• (2) Sodium bromide

  • With hot conc. sulfuric acid, the bromide ion in solid salts, can act as a reducing agent and the sulfuric acid is reduced to sulfur dioxide (sulfur(IV) oxide) and the bromide ion oxidised to free bromine. You observe the orange–brown vapour of the bromine as well as hydrogen bromide fumes and a white residue of sodium sulfate is formed.

  • This is a redox reaction.

  • 2NaBr(s) + 2H₂SO₄(l) ==> Na₂SO₄(s) + Br₂(g) + SO₂(g) + 2H₂O(l)

  • The half–cell reactions are:

    • (i) 2Br^– ==> Br₂ + 2e^– (oxidation, 2 electron loss)

    • (ii) H₂SO₄ + 2H⁺ + 2e^– ==> SO₂ + 2H₂O (reduction, 2 electron gain)

    • (i) balances (ii) in terms of electron/oxidation number change, so the ionic redox equation is

  • 2Br^– + H₂SO₄ + 2H⁺ ==> Br₂ + SO₂ + 2H₂O

  • The bromide ion is oxidised to bromine, half of the 'sulfate' essentially remains unchanged, but the other molecule of sulfuric acid is reduced to sulfur dioxide (sulfur dioxide, sulfur(IV) oxide).

  • Redox analysis

  • The oxidation state of bromine changes from –1 in Br^–, to 0 in Br₂ (oxidation, increase in oxidation state).

  • The oxidation state of sulfur changes from +6 in H₂SO₄, to +4 in SO₂ (reduction, decrease in oxidation state).

  • The sulfur undergoes 2 'units' of oxidation number decrease, hence for each of 2 bromide ions, an oxidation number increase of 1 per bromine atom.

    • NOTE: You can write the bromide oxidation reaction as if hydrogen bromide is first formed and then oxidised to bromine and the concentrated sulfuric acid reduced to sulfur(IV) oxide (sulfur dioxide).

    • NaBr(s) + H₂SO₄(l) ==> NaHSO₄(s) + HBr(g)

      • This explains why you get hydrogen bromide fumes too.

      • Its a simple non-redox 'acid' displacement reaction.

    • 2HBr(g) + H₂SO₄(l) ==> Br₂(g) + SO₂(g) + 2H₂O(l)

      • This is the effectively an alternative to the overall redox reaction.

    • I don't know how much of the HBr survives the oxidation?

• (3) Sodium iodide

  • With hot conc. sulfuric acid, the iodide ion in solid salts acts as an even stronger reducing agent than bromide, and the sulfuric acid is reduced to hydrogen sulfide gas and the iodide ion oxidised to free iodine. You observe the purple vapour of iodine, the strong smell of hydrogen sulfide (rotten eggs!) as well as hydrogen iodide fumes and a white residue of sodium sulfate is formed. The iodine will crystallise out on the upper cooler parts of the test tube.

  • This is a redox reaction.

  • 8NaI(s) + 5H₂SO₄(l) ==> 4Na₂SO₄(s) + 4I₂(g/s) + H₂S(g) + 4H₂O(l)

  • The half–cell reactions are:

    • (i) 2I^– ==> I₂ + 2e^– (oxidation, 2 electron loss)

    • (ii) H₂SO₄ + 8H⁺ + 8e^– ==> H₂S + 4H₂O (reduction, 8 electrons gained)

    • 4 x (i) balances (ii) in terms of electron/oxidation number change, so the ionic redox equation is

  • 8I^– + H₂SO₄ + 8H⁺ ==> 4I₂ + H₂S + 4H₂O

  • The iodide ion is oxidised to iodine, ⁴/₅ths of the 'sulfate' essentially remains unchanged, but the 5th molecule of sulfuric acid is reduced to hydrogen sulfide.

  • Apart from the smell, hydrogen sulfide gives a black precipitate with lead ethanoate paper.

  • Redox analysis

  • The oxidation state of iodine changes from –1 in I^–, to 0 in I₂ (oxidation, increase in oxidation state).

  • The oxidation state of sulfur changes from +6 in H₂SO₄, to –2 in H₂S (reduction, decrease in oxidation state).

  • The sulfur undergoes 8 'units' of oxidation number decrease, hence for each of 8 iodide ions, an oxidation number increase of 1 per iodine atom.

    • NOTE: You can write the iodide oxidation reaction as if hydrogen iodide is first formed and then oxidised to iodine and the concentrated sulfuric acid reduced to hydrogen sulfide (hydrogen sulfide).

    • NaI(s) + H₂SO₄(l) ==> NaHSO₄(s) + HI(g)

    • 8HI(g) + H₂SO₄(l) ==> 4I₂(g/s) + H₂S(g) + 4H₂O(l)

    • I don't know how much of the HI survives the oxidation?

    • I also wonder whether sulfur dioxide is also formed? as in the case of bromides.

• (4) Note the increasing reducing power as you descend the Group 7 Halogen halide salts

  • i.e. the halide ion is increasingly more easily oxidised.

  • This is reflected in the oxidation potentials for the half cell halogen/halide ion

  • E^θ_{F2(aq)/F–(aq)} = +2.87V > E^θ_{Cl2(aq)/Cl–(aq)} = +1.36V > E^θ_{Br2(aq)/Br–(aq)} = +1.09V > E^θ_{I2(aq)/I–(aq)} = +0.54V

  • The more positive the half–cell potential, the stronger the oxidising power (of the halogen)

    • and the weaker the reducing power of the corresponding halide ion.

    • Oxidising power: F₂ > Cl₂ > Br₂ > I₂

    • Reduction power: I^– > Br^– > Cl^– > F^–

9.5 Tests for halogens and halide Ions

Test for halogen	Test method	Test observations	Test chemistry and comments
Chlorine gas Cl2 A pungent green gas.	(i) Apply damp blue litmus. (Can use red litmus and just see bleaching effect.) (ii) A drop silver nitrate on the end of a glass rod into the gas.	(i) litmus turns red and then is bleached white. (ii) White precipitate.	(i) Non–metal, is acid in aqueous solution and a powerful oxidising agent (ii) It forms a small amount of chloride ion in water, so gives a positive result for the chloride test.
Bromine Br2 (l or aq) A dark red liquid – orange–brown fumes, yellow–orange aqueous solution. The other common orange–brown gas is nitrogen dioxide	(i) Shake with a liquid alkene. (ii) Mix with silver nitrate solution.	(ii) Decolourised. See alkene test. (ii) Cream ppt. of silver bromide as in the test for bromide.	(i) Forms a colourless organic dibromo–compound >C=C< + Br2 ==> >CBr–CBr< (ii) Ag+(aq) + Br–(aq) ==> AgBr(s)   Any soluble bromide gives a silver bromide precipitate.
Iodine (i) solid or (ii) solution A very dark solid	(i) Gently heat the dark coloured solid. (ii) Test aqueous solution or solid with starch solution.	(i) Gives brilliant purple vapour. (ii) A blue black colour.	(i) Iodine forms a distinctive coloured vapour. (ii) Forms a blue–black complex with starch and in biology the test is used to detect starch with iodine solution.

Tests for Halide Ions

In test (i) the silver nitrate is acidified with dilute nitric acid to prevent the precipitation of other non–halide silver salts.

The solubility tend for AgX in ammonia solution is (AgF water soluble) AgCl >> AgBr >> AgI

HCl(g) KF(s) + H2SO4(l) ==> KHSO4(s) + HF(g) 2NaBr(s) + 2H2SO4(l) ==> NaHSO4(s) + Br2(g) + SO2(g) + 2H2O(l) H2SO4 + 2H+ + 2e– ==> SO2 + 2H2O 2Br– + H2SO4 + 2H+ ==> Br2 + SO2 + 2H2O NaBr(s) + H2SO4(l) ==> NaHSO4(s) + HBr(g) 2HBr(g) + H2SO4(l) ==> Br2(g) + SO2(g) + 2H2O(l) 8NaI(s) + 5H2SO4(l) ==> 4Na2SO4(s) + 4I2(g/s) + H2S(g) + 4H2O(l) 8I– + H2SO4 + 8H+ ==> 4I2 + H2S + 4H2O NaI(s) + H2SO4(l) ==> NaHSO4(s) + HI(g) 8HI(g) + H2SO4(l) ==> 4I2(g/s) + H2S(g) + 4H2O(l)

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