Reactions of the halide ions
Reactions of the halide ions
- A halide ion can give away an electron — so it is a reducing agent.
- This power increases down the group.
- Two key tests identify which halide is present.
Practice
The reducing power of the halide ions increases down the group because:
Reducing means losing an electron; bigger ions lose theirs more easily, so reducing power rises down the group.
Reducing power
- A halide ion ($\text{Cl}^{-}$, $\text{Br}^{-}$, $\text{I}^{-}$) acts as a reducing agent (it loses an electron).
- This power increases down the group, because a larger ion holds its outer electron less tightly.
Practice
Match each halide to its silver halide precipitate colour.
AgCl is white, AgBr cream, AgI yellow; ammonia solubility confirms (Cl dissolves in dilute, I insoluble).
Practice
Silver iodide (AgI) is:
AgI is yellow and does not dissolve in ammonia, distinguishing it from AgCl (white) and AgBr (cream).
Silver nitrate test
Add silver nitrate, then ammonia, and read the precipitate:
| Halide | Precipitate | In ammonia |
|---|---|---|
| $\text{Cl}^{-}$ | white | dissolves in dilute ammonia |
| $\text{Br}^{-}$ | cream | dissolves in concentrated ammonia |
| $\text{I}^{-}$ | yellow | insoluble |
Practice
With concentrated sulfuric acid, iodide gives I₂ plus H₂S because:
I⁻ is the strongest reducer, so it reduces the sulfuric acid all the way to H₂S (and forms I₂); Cl⁻ only gives HCl.
With concentrated sulfuric acid
- All give the hydrogen halide first; the lower halides are then oxidised (they are stronger reducing agents):
- chloride: only $\text{HCl}$ (no redox).
- bromide: also brown $\text{Br}_2$ and $\text{SO}_2$.
- iodide: $\text{I}_2$ plus smelly $\text{H}_2\text{S}$ and $\text{SO}_2$ ($\text{I}^{-}$ is the strongest reducing agent).
You've got it
Key idea
- halide ions are reducing agents; power increases down the group (larger ion, electron held loosely)
- silver nitrate test: AgCl white, AgBr cream, AgI yellow (confirm with ammonia solubility)
- conc. H₂SO₄: Cl⁻ → HCl only; Br⁻ → Br₂ + SO₂; I⁻ → I₂ + H₂S + SO₂ (strongest reducer)