This lesson explores the chemical reactions of halogens and halide ions, which demonstrate the trends in their oxidising and reducing abilities. We will focus on two major areas: halogen displacement reactions in aqueous solution, and the reactions of solid halides with concentrated sulfuric acid.
A reaction in which a more reactive element displaces a less reactive element from its compound in solution.
1. Halogen Displacement Reactions
When an aqueous halogen is mixed with a solution containing halide ions, a displacement reaction occurs if the added halogen is a stronger oxidising agent than the halogen in the compound. A more reactive halogen displaces a less reactive halide ion.
For example, chlorine (\(\text{Cl}_2\)) is a stronger oxidising agent than bromine (\(\text{Br}_2\)) and iodine (\(\text{I}_2\)). Therefore, chlorine will displace bromide and iodide ions from their solutions:
Reaction with Bromide:
\[ \text{Cl}_2\text{(aq)} + 2\text{KBr(aq)} \rightarrow 2\text{KCl(aq)} + \text{Br}_2\text{(aq)} \]
\[ \text{Cl}_2\text{(aq)} + 2\text{Br}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{Br}_2\text{(aq)} \quad \text{[Ionic Equation]} \]
Observations: The colourless solution turns orange due to the formation of aqueous bromine (\(\text{Br}_2\)).
Reaction with Iodide:
\[ \text{Cl}_2\text{(aq)} + 2\text{KI(aq)} \rightarrow 2\text{KCl(aq)} + \text{I}_2\text{(aq)} \]
\[ \text{Cl}_2\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{I}_2\text{(aq)} \quad \text{[Ionic Equation]} \]
Observations: The colourless solution turns brown due to the formation of aqueous iodine (\(\text{I}_2\)).
Bromine (\(\text{Br}_2\)) is a stronger oxidising agent than iodine (\(\text{I}_2\)), so it can displace iodide ions:
\[ \text{Br}_2\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Br}^-\text{(aq)} + \text{I}_2\text{(aq)} \]
Observations: The orange bromine solution turns brown as iodine is formed.
Iodine (\(\text{I}_2\)) is the weakest oxidising agent of the three, so it cannot displace fluoride, chloride, or bromide ions. There is no reaction in these cases.
2. Reactions of Halide Ions with Concentrated Sulfuric Acid
Solid sodium halides react with concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)). These reactions demonstrate the trend in the reducing power of the halide ions. Concentrated sulfuric acid is a strong acid and a moderate oxidising agent.
🔑 Key Principle: Sulfuric Acid Reactions
All halide salts initially undergo an acid-base reaction with concentrated sulfuric acid to produce toxic hydrogen halide gas. However, if the halide ion is a strong enough reducing agent, it will then reduce the sulfuric acid. The reducing power increases in the order: \(\text{Cl}^- < \text{Br}^- < \text{I}^-\).
A reducing agent is an electron donor that reduces another species while undergoing oxidation itself.
A) Sodium Chloride (\(\text{NaCl}\))
Chloride ions (\(\text{Cl}^-\)) are weak reducing agents. When solid sodium chloride reacts with concentrated sulfuric acid, only an acid-base (proton transfer) reaction occurs. No redox reaction takes place because chloride ions cannot reduce sulfuric acid.
\[ \text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HCl(g)} \]
Observations: White steamy/misty fumes of hydrogen chloride (\(\text{HCl}\)) gas are observed. The sulfur in \(\text{H}_2\text{SO}_4\) remains in the \(+6\) oxidation state.
B) Sodium Bromide (\(\text{NaBr}\))
Bromide ions (\(\text{Br}^-\)) are stronger reducing agents than chloride ions. The reaction occurs in two stages:
Stage 1 (Acid-Base):
\[ \text{NaBr(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HBr(g)} \]
White steamy fumes of hydrogen bromide (\(\text{HBr}\)) are produced initially.
Stage 2 (Redox):
The \(\text{HBr}\) reacts further. Bromide ions reduce the sulfur in sulfuric acid from oxidation state \(+6\) to \(+4\) (in sulfur dioxide, \(\text{SO}_2\)). The bromide ions are oxidised to bromine gas (\(\text{Br}_2\)):
\[ 2\text{HBr(g)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{Br}_2\text{(g)} + \text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)} \]
Observations: Orange/brown fumes of bromine (\(\text{Br}_2\)) vapour and a colourless, pungent-smelling gas (sulfur dioxide, \(\text{SO}_2\)) are produced.
C) Sodium Iodide (\(\text{NaI}\))
Iodide ions (\(\text{I}^-\)) are very powerful reducing agents. They reduce the sulfur in sulfuric acid even further: from \(+6\) down to \(+4\) (\(\text{SO}_2\)), \(0\) (solid sulfur, \(\text{S}\)), and \(-2\) (hydrogen sulfide, \(\text{H}_2\text{S}\)).
Stage 1 (Acid-Base):
\[ \text{NaI(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HI(g)} \]
White steamy fumes of hydrogen iodide (\(\text{HI}\)) are produced initially.
Stage 2 (Redox Reactions):
The \(\text{HI}\) reduces sulfuric acid in several concurrent redox reactions:
- Reduction to Sulfur Dioxide (\(\text{SO}_2\)), sulfur state \(+4\):
\[ 2\text{HI(g)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{I}_2\text{(g)} + \text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)} \]
- Reduction to Sulfur (\(\text{S}\)), sulfur state \(0\):
\[ 6\text{HI(g)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow 3\text{I}_2\text{(g)} + \text{S(s)} + 4\text{H}_2\text{O(l)} \]
- Reduction to Hydrogen Sulfide (\(\text{H}_2\text{S}\)), sulfur state \(-2\):
\[ 8\text{HI(g)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow 4\text{I}_2\text{(g)} + \text{H}_2\text{S(g)} + 4\text{H}_2\text{O(l)} \]
Observations: Purple/violet vapour of iodine (\(\text{I}_2\)) is seen (which can also condense as dark grey solids), a yellow solid (sulfur, \(\text{S}\)) forms, and a gas with a strong smell of rotten eggs (hydrogen sulfide, \(\text{H}_2\text{S}\)) is produced.
The observations and products of these sulfuric acid reactions are extremely popular in exams. Remember: \(\text{NaCl}\) only produces white steamy fumes (\(\text{HCl}\)). It does NOT undergo redox. \(\text{NaBr}\) produces orange fumes (\(\text{Br}_2\)) and a choking gas (\(\text{SO}_2\)). \(\text{NaI}\) produces purple fumes/dark solid (\(\text{I}_2\)), a yellow solid (sulfur, \(\text{S}\)), and a gas with a rotten egg smell (\(\text{H}_2\text{S}\)). You must associate each observation with its correct chemical species.
Answer:
Equation 1 (Acid-Base):
\[ \text{NaBr(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HBr(g)} \]
Equation 2 (Redox):
\[ 2\text{HBr(g)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{Br}_2\text{(g)} + \text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)} \]
Reducing Agent:
The \(\text{HBr}\) (or the \(\text{Br}^-\) ion) acts as the reducing agent. It reduces the sulfur in \(\text{H}_2\text{SO}_4\) from an oxidation state of \(+6\) to \(+4\) in \(\text{SO}_2\), while being oxidised from \(-1\) (in \(\text{HBr}\)) to \(0\) (in \(\text{Br}_2\)).
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