Unit 12: Chemical Properties of Halogens & Halides II
Investigating the reactions of halides with sulfuric acid and the preparation of chlorine gas.
12.7 Reactions of Conc. H₂SO₄ with Solid Halides
The reaction between a solid halide salt (like NaCl or KBr) and concentrated sulfuric acid provides a clear demonstration of the increasing reducing power of the halide ions down the group. Concentrated sulfuric acid is both a strong acid and a powerful oxidising agent.
With Solid Chloride (e.g., NaCl)
The chloride ion ($Cl^-$) is a weak reducing agent. It is not strong enough to reduce the
sulfur in H₂SO₄. Therefore, only a simple acid-base reaction occurs, producing steamy fumes
of hydrogen chloride gas.
$$ NaCl(s) + H_2SO_4(l) \rightarrow NaHSO_4(s) + HCl(g) $$
This is an acid-base reaction, not a redox reaction.
With Solid Bromide (e.g., KBr)
The bromide ion ($Br^-$) is a stronger reducing agent than chloride. It is strong enough to
reduce the sulfuric acid. First, an acid-base reaction produces hydrogen bromide (HBr).
Then, the HBr reduces the H₂SO₄ in a redox reaction.
1. Acid-Base: $KBr(s) + H_2SO_4(l) \rightarrow KHSO_4(s) + HBr(g)$
2. Redox: $2HBr(g) + H_2SO_4(l) \rightarrow Br_2(g) + SO_2(g) + 2H_2O(l)$
Observations include steamy fumes of HBr, brown fumes of bromine ($Br_2$), and a
colorless, choking gas, sulfur dioxide ($SO_2$).
With Solid Iodide (e.g., NaI)
The iodide ion ($I^-$) is the strongest reducing agent of the three. It reduces the sulfuric
acid much more extensively. The initial acid-base reaction produces hydrogen iodide (HI),
which then reduces the H₂SO₄ to multiple products, including sulfur dioxide ($SO_2$), solid
yellow sulfur (S), and even hydrogen sulfide ($H_2S$), a gas that smells of rotten eggs.
1. Acid-Base: $NaI(s) + H_2SO_4(l) \rightarrow NaHSO_4(s) + HI(g)$
2. Redox (multiple reactions): e.g., $8HI(g) + H_2SO_4(l) \rightarrow 4I_2(s/g) +
H_2S(g) + 4H_2O(l)$
Observations include steamy fumes of HI, purple vapor/grey solid of iodine ($I_2$), a
yellow solid (S), and the smell of rotten eggs ($H_2S$).
Solved Examples:
-
What is the only product observed when concentrated sulfuric acid reacts
with solid sodium chloride?
Solution: Steamy fumes of hydrogen chloride (HCl) gas. -
How do the reactions of KBr and KI with concentrated H₂SO₄ demonstrate that
iodide is a stronger reducing agent than bromide?
Solution: The bromide ion reduces sulfur from an oxidation state of +6 in H₂SO₄ to +4 in SO₂. The iodide ion is a stronger reducing agent and can reduce the sulfur even further, to 0 in solid sulfur (S) and -2 in hydrogen sulfide (H₂S). The greater range of reduction products shows iodide is the more powerful reducing agent. -
In the reaction between HBr and H₂SO₄, what is the change in oxidation state
for bromine and sulfur?
Solution: Bromine is oxidized from -1 in HBr to 0 in Br₂. Sulfur is reduced from +6 in H₂SO₄ to +4 in SO₂. -
Why is the reaction with sodium chloride not a redox reaction?
Solution: Because there are no changes in the oxidation states of any of the elements involved. It is a simple acid-base reaction where a proton is transferred.
12.8 Laboratory Preparation of Chlorine
Since the chloride ion is a weak reducing agent, a very strong oxidising agent is needed to oxidize it to chlorine gas ($Cl_2$). A common method in the laboratory is to react concentrated hydrochloric acid with a strong solid oxidising agent, such as manganese(IV) oxide ($MnO_2$) or potassium permanganate ($KMnO_4$).
Preparation using Manganese(IV) Oxide
Concentrated HCl is dripped onto solid $MnO_2$ and the mixture is gently heated.
$$ MnO_2(s) + 4HCl(aq) \rightarrow MnCl_2(aq) + Cl_2(g) + 2H_2O(l) $$
In this redox reaction:
- Manganese is reduced from an oxidation state of +4 in $MnO_2$ to +2 in $MnCl_2$.
- Some of the chlorine is oxidized from -1 in HCl to 0 in $Cl_2$.
Purification and Collection
The chlorine gas produced is impure. It is passed through:
- Water: To remove unreacted HCl gas (which is very soluble in water).
- Concentrated Sulfuric Acid: To dry the chlorine gas (H₂SO₄ is a dehydrating agent).
Solved Examples:
-
What is the role of manganese(IV) oxide in the preparation of
chlorine?
Solution: Manganese(IV) oxide acts as the oxidising agent. It takes electrons from the chloride ions, oxidizing them to chlorine gas. -
Why is the chlorine gas bubbled through water first, then concentrated
sulfuric acid, and not the other way around?
Solution: It is bubbled through water first to remove the HCl impurity. If it were bubbled through sulfuric acid first, it would be dried, but then bubbling it through water would make it wet again, defeating the purpose of the drying step. -
Why is chlorine collected by downward delivery?
Solution: The molar mass of chlorine ($Cl_2$) is approximately 71 g/mol, which is much greater than the average molar mass of air (~29 g/mol). Because it is denser than air, it will sink and displace the air upwards from a gas jar. -
What is the purpose of bubbling the product gas through concentrated
sulfuric acid?
Solution: To act as a dehydrating (drying) agent, removing any water vapor mixed with the chlorine gas. -
Write the half-equation for the oxidation of chloride ions in this
reaction.
Solution: $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$ -
What safety precaution is essential when preparing chlorine
gas?
Solution: Chlorine gas is toxic and corrosive. The entire preparation must be carried out in a well-ventilated fume hood.