Some reactions of the halide ions

AS Level Chemistry · Group 17. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

Test this topic freeAll AS Level Chemistry topics

The idea

A halide ion can lose an electron, so it can act as a reducing agent. Down the group the ion gets larger and its outer electrons are further from the nucleus and more shielded, so they are lost more easily. Reducing power increases: chloride is the weakest, iodide the strongest.

Concentrated sulfuric acid shows this clearly. With every solid halide it first makes the hydrogen halide. Chloride does nothing more. Bromide and iodide go on to reduce the sulfuric acid, and iodide reduces it furthest.

Halide ions are identified with aqueous silver nitrate. The colour of the silver halide precipitate, and whether it dissolves in aqueous ammonia, shows which halide is present.

Rules to remember

Common mistake

Students often mix up the two trends: the halogens become weaker oxidising agents down the group, but the halide ions become stronger reducing agents. Chloride ions cannot reduce concentrated sulfuric acid at all.

Worked example

Hydrogen iodide reduces concentrated sulfuric acid to hydrogen sulfide, H2S. Use oxidation numbers to find how many moles of HI react with one mole of H2SO4.

  1. Sulfur changes from +6 in H2SO4 to −2 in H2S, a gain of 8 electrons.
  2. Each iodide ion changes from −1 to 0 in I2, a loss of 1 electron.
  3. Electrons lost must equal electrons gained, so 8 HI are needed and 4 I2 are formed.
  4. Balance the rest: 8HI + H2SO4 → 4I2 + H2S + 4H2O (10 H and 4 O on each side).

Answer: 8 mol of HI react with 1 mol of H2SO4.

Practice questions

Try each one, then open the answer.

1. Which product of the reaction of sodium iodide with concentrated sulfuric acid shows that sulfur has been reduced from +6 to −2?

  1. A
    SO2
  2. B
    S
  3. C
    H2S
  4. D
    I2
Show answer

Answer: C. Iodide is a strong enough reducing agent to take sulfur all the way from +6 in H2SO4 to −2 in H2S (the bad-egg smell). SO2 is +4 and S is 0; I2 is the oxidation product.

2. Which list shows the halide ions in order of increasing strength as reducing agents?

  1. A
    I−, Br−, Cl−
  2. B
    Br−, Cl−, I−
  3. C
    Cl−, I−, Br−
  4. D
    Cl−, Br−, I−
Show answer

Answer: D. Down the group the halide ion is larger and its outer electrons are more shielded, so it loses electrons more easily. Iodide is the strongest reducing agent and chloride the weakest.

3. An aqueous halide gives a cream precipitate with silver nitrate which dissolves in concentrated but not in dilute aqueous ammonia. Which halide is present?

  1. A
    bromide
  2. B
    iodide
  3. C
    chloride
  4. D
    fluoride
Show answer

Answer: A. AgCl is white and dissolves in dilute ammonia; AgBr is cream and dissolves only in concentrated ammonia; AgI is yellow and insoluble in both. Silver fluoride is soluble, so fluoride gives no precipitate.

More questions on this topic

Keep going

← The chemical properties of the halogen elements and the hydrogen halidesThe reactions of chlorine →All rules on one pageStuck? Ask a question