Topic 8 · page 27 of 29

Halides with concentrated sulfuric acid

Every solid sodium halide gives a steamy hydrogen halide with concentrated sulfuric acid, but only HBr and HI go on to reduce the acid. Iodide reduces sulfur furthest, from +6 to −2, so the reducing ability is I⁻ > Br⁻ > Cl⁻.

14 min read8.278.27(i)

Read first: Building half-equations

Two reactions in one 8.278.27(i)

Add concentrated sulfuric acid to a solid sodium halide. Every halide first shows an acid–base reaction, in which the sulfuric acid gives an H⁺ to the halide ion to make a hydrogen halide gas. This is not redox.

NaX(s)+HX2SOX4(l)→NaHSOX4(s)+HX(g) \ce{NaX(s) + H2SO4(l) -> NaHSO4(s) + HX(g)}

The hydrogen halide gas meets moist air and forms steamy white fumes. For bromide and iodide a second, redox reaction follows. The hydrogen halide acts as a reducing agent, and sulfuric acid is reduced.

A ladder of sulfur oxidation numbers with the sulfur species in each: plus 6 in sulfuric acid, plus 4 in sulfur dioxide, 0 in sulfur, minus 2 in hydrogen sulfide. Next to the ladder, three columns for chloride, bromide and iodide show how far down it each reaches: chloride stays at plus 6, bromide reaches plus 4 only, iodide reaches plus 4, 0 and minus 2.
Chloride does not reduce sulfur. Bromide reduces it from +6 to +4. Iodide reduces it all the way to −2.

What you see 8.278.27(i)

Solid halide Observations Sulfur product Sulfur: change in oxidation number
NaCl steamy white fumes of HCl; no other change none none (+6 stays +6)
NaBr steamy fumes of HBr; orange-brown fumes of Br₂; colourless, choking gas, SO₂ SO₂ +6 → +4
NaI steamy fumes of HI; purple fumes and black solid, I₂; yellow solid, S; rotten-egg smell, H₂S SO₂, S and H₂S +6 → +4, 0 and −2

The equations:

Chloride (acid–base only):

NaCl(s)+HX2SOX4(l)→NaHSOX4(s)+HCl(g) \ce{NaCl(s) + H2SO4(l) -> NaHSO4(s) + HCl(g)}

Bromide: the first step, then the redox step.

NaBr(s)+HX2SOX4(l)→NaHSOX4(s)+HBr(g) \ce{NaBr(s) + H2SO4(l) -> NaHSO4(s) + HBr(g)}
2 HBr+HX2SOX4→BrX2+SOX2+2 HX2O \ce{2HBr + H2SO4 -> Br2 + SO2 + 2H2O}

Iodide: the first step, then three redox steps.

NaI(s)+HX2SOX4(l)→NaHSOX4(s)+HI(g) \ce{NaI(s) + H2SO4(l) -> NaHSO4(s) + HI(g)}
2 HI+HX2SOX4→IX2+SOX2+2 HX2O \ce{2HI + H2SO4 -> I2 + SO2 + 2H2O}
6 HI+HX2SOX4→3 IX2+S+4 HX2O \ce{6HI + H2SO4 -> 3I2 + S + 4H2O}
8 HI+HX2SOX4→4 IX2+HX2S+4 HX2O \ce{8HI + H2SO4 -> 4I2 + H2S + 4H2O}

These come from half-equations. The halide is oxidised, and the sulfuric acid is reduced, as page 6 showed:

Oxidation (halide) Reduction (sulfuric acid)
2 BrX−→BrX2+2 eX−\ce{2Br- -> Br2 + 2e-} HX2SOX4+2 HX++2 eX−→SOX2+2 HX2O\ce{H2SO4 + 2H+ + 2e- -> SO2 + 2H2O}
2 IX−→IX2+2 eX−\ce{2I- -> I2 + 2e-} HX2SOX4+2 HX++2 eX−→SOX2+2 HX2O\ce{H2SO4 + 2H+ + 2e- -> SO2 + 2H2O}
HX2SOX4+6 HX++6 eX−→S+4 HX2O\ce{H2SO4 + 6H+ + 6e- -> S + 4H2O}
HX2SOX4+8 HX++8 eX−→HX2S+4 HX2O\ce{H2SO4 + 8H+ + 8e- -> H2S + 4H2O}

Worked exampleBuilding the iodide to H₂S equation

Combine the half-equations that give H₂S.

  1. Reduction: HX2SOX4+8 HX++8 eX−→HX2S+4 HX2O\ce{H2SO4 + 8H+ + 8e- -> H2S + 4H2O} (S +6 → −2, 8 electrons)
  2. Oxidation: 2 IX−→IX2+2 eX−\ce{2I- -> I2 + 2e-}. Multiply by 4: 8 IX−→4 IX2+8 eX−\ce{8I- -> 4I2 + 8e-}
  3. Add: HX2SOX4+8 HX++8 IX−→HX2S+4 IX2+4 HX2O\ce{H2SO4 + 8H+ + 8I- -> H2S + 4I2 + 4H2O}
  4. Charge: left +8 − 8 = 0; right 0. ✓ This is the ionic version of 8 HI+HX2SOX4→4 IX2+HX2S+4 HX2O\ce{8HI + H2SO4 -> 4I2 + H2S + 4H2O}.

Explaining the trend 8.278.27(i)

Down the group the ion has more shells, so the outer electron is further from the nucleus and more shielded. It is held less strongly, so it is lost more easily. A stronger reducing agent can reduce concentrated sulfuric acid further.

Check yourself

  1. State what you see when concentrated sulfuric acid is added to solid sodium chloride, and write the equation.
  2. State two observations that show a redox reaction when concentrated sulfuric acid is added to solid sodium bromide, and write the equation for the redox step.
  3. Name the three sulfur-containing products formed with sodium iodide and give the oxidation number of sulfur in each.
  4. Explain why the reaction with sodium chloride is not a redox reaction.
  5. Explain the trend in reducing ability of the hydrogen halides.
  6. Write the half-equation for the reduction of sulfuric acid to sulfur, and use it to construct the equation for HI forming sulfur.

Answers cover these until you have tried the questions

  1. Steamy white fumes of hydrogen chloride; no other change. NaCl(s)+HX2SOX4(l)→NaHSOX4(s)+HCl(g)\ce{NaCl(s) + H2SO4(l) -> NaHSO4(s) + HCl(g)}
  2. Orange-brown fumes of bromine and a colourless choking gas (sulfur dioxide). 2 HBr+HX2SOX4→BrX2+SOX2+2 HX2O\ce{2HBr + H2SO4 -> Br2 + SO2 + 2H2O}
  3. SO₂ (S +4), S (0), H₂S (−2).
  4. No element changes oxidation number: Na +1, Cl −1, H +1, S +6 and O −2 are the same before and after. It is an acid–base reaction (H₂SO₄ gives H⁺ to Cl⁻).
  5. Down the group the halide ion is larger, with more shells and more shielding, so the outer electron is further from the nucleus and held less strongly. It is lost more easily, so reducing ability increases: HI > HBr > HCl.
  6. HX2SOX4+6 HX++6 eX−→S+4 HX2O\ce{H2SO4 + 6H+ + 6e- -> S + 4H2O}. Combine with 6 IX−→3 IX2+6 eX−\ce{6I- -> 3I2 + 6e-}: HX2SOX4+6 HX++6 IX−→S+3 IX2+4 HX2O\ce{H2SO4 + 6H+ + 6I- -> S + 3I2 + 4H2O}, which is the ionic form of 6 HI+HX2SOX4→3 IX2+S+4 HX2O\ce{6HI + H2SO4 -> 3I2 + S + 4H2O}.