Halide Ions with Concentrated Sulfuric Acid

Spec Mapping — OCR H432 Module 3.1.3 — Group 7 (the halogens), covering the reactions of solid sodium halides (NaF, NaCl, NaBr, NaI) with concentrated sulfuric acid, the classification of each as acid-base or acid-base-plus-redox, the identification of the sulfur-containing reduction products (SO₂, S, H₂S), and the use of these reactions as experimental evidence for the trend in reducing ability of halide ions down Group 17 (refer to the official OCR H432 specification document for exact wording). This is OCR-specific content — AQA and Edexcel approach Group 7 differently.

The reactions of solid sodium halides with concentrated sulfuric acid are the OCR-distinguishing experiment of Module 3.1.3 — a sequence of test-tube reactions that simultaneously demonstrate (i) the trend in halide reducing power introduced in the previous lesson and (ii) the dual nature of H₂SO₄ as both a Brønsted acid and a moderate oxidising agent. The reactions form a clear progression: NaF and NaCl give only the acid-base reaction (no redox); NaBr gives acid-base plus one step of redox (S +6 → +4 in SO₂); NaI gives acid-base plus three steps of redox (S +6 → +4 in SO₂, +6 → 0 in S, +6 → −2 in H₂S). The progression of products tracks the strength of each halide as a reducing agent — F⁻ and Cl⁻ are too weak to reduce H₂SO₄, Br⁻ can push S from +6 to +4, and I⁻ can push S all the way to −2. The dramatic experimental signatures (rotten-egg smell of H₂S, yellow sulfur precipitate, violet I₂ vapour) make this one of the most memorable OCR demonstrations.

Key Insight: concentrated sulfuric acid is a strong acid (protonates X⁻ to give HX) and a moderate oxidising agent (S in H₂SO₄ is in oxidation state +6 and can be reduced). Whether the redox step happens depends on whether the halide is a strong enough reducing agent to attack the S(+6).

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Why this reaction matters

Concentrated H₂SO₄ does two things when it meets a solid sodium halide NaX:

1. Acid-base (Brønsted–Lowry): H₂SO₄ protonates the halide ion (X⁻) to form the gaseous hydrogen halide HX. This step happens with all four halides (NaF, NaCl, NaBr, NaI) because all halides are basic enough to accept a proton from H₂SO₄.
2. Redox: if X⁻ is a strong enough reducing agent, it then reduces H₂SO₄ to a lower oxidation state of sulfur (e.g. SO₂ at S = +4, elemental S at 0, H₂S at S = −2). The redox step only happens with the more reducing halides (Br⁻ partly; I⁻ fully).

The reducing-ability trend established in the previous lesson — I⁻ > Br⁻ > Cl⁻ > F⁻ — therefore manifests as a progressively richer mixture of redox products in the test tube as you move down the group.

flowchart TD
  A[Conc H2SO4 added to solid NaX] --> B[Brønsted acid-base always]
  B --> C[NaHSO4 + HX gas]
  C --> D{Is X- a strong enough reducing agent?}
  D -->|F- or Cl-| E[NO — stop here; only misty fumes of HX]
  D -->|Br-| F[YES — partial redox]
  D -->|I-| G[YES — extensive redox]
  F --> F1[Br2 vapour and SO2]
  G --> G1[I2 vapour, S yellow solid, H2S rotten-egg smell, SO2]

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Summary table — observations and oxidation-state book-keeping

Halide	Acid-base?	Redox?	S product	Cl-/Br-/I- result	Observations
F⁻ (NaF)	Yes	No	H₂SO₄ unchanged	F stays −1	Misty white HF fumes only
Cl⁻ (NaCl)	Yes	No	H₂SO₄ unchanged	Cl stays −1	Misty white HCl fumes only
Br⁻ (NaBr)	Yes	Partial	SO₂ (S +4)	Br: −1 → 0 (Br₂)	HBr fumes, orange/brown Br₂ vapour, choking SO₂
I⁻ (NaI)	Yes	Extensive	SO₂, S, H₂S (S +4, 0, −2)	I: −1 → 0 (I₂)	Brief HI fumes, black/purple I₂, yellow S, rotten-egg H₂S, possibly SO₂

Reducing ability of the halide ion increases F⁻ < Cl⁻ < Br⁻ < I⁻, so the reaction with H₂SO₄ becomes progressively more redox-dominated as you descend the group.

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1. NaF with concentrated H₂SO₄ — acid-base only

F⁻ is a very weak reducing agent because F is the most electronegative element in the entire periodic table; F⁻ holds onto its outer (lone-pair) electrons with extremely high effective nuclear charge. The reaction with H₂SO₄ is therefore acid-base only:

$\text{NaF(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HF(g)}$NaF(s)+H2​SO4​(l)→NaHSO4​(s)+HF(g)

Observations:

• Steamy/misty white fumes of HF gas escaping from the test tube.
• No colour change in the solid.
• No SO₂ smell, no sulfur precipitate, no Br₂ or I₂ colour.

Oxidation-number tracking: F stays at −1; S stays at +6 (in NaHSO₄). No oxidation states change — this is not a redox reaction, just a simple Brønsted–Lowry proton transfer.

HF is a covalent gas (b.p. 19.5 °C) that fumes in moist air because it dissolves rapidly in atmospheric water vapour to form hydrofluoric acid mist. HF is highly toxic and corrosive to glass, so this reaction is rarely performed in school labs — it is demonstrated chiefly via film or written description.

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2. NaCl with concentrated H₂SO₄ — acid-base only

Cl⁻ is a slightly stronger reducing agent than F⁻ but still too weak to reduce concentrated H₂SO₄. The reaction is again acid-base only:

$\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HCl(g)}$NaCl(s)+H2​SO4​(l)→NaHSO4​(s)+HCl(g)

Observations:

• Steamy/misty white fumes of HCl gas (turns damp blue litmus red).
• No colour change in the solid.
• No SO₂, no sulfur, no H₂S, no Cl₂.

Industrially, this reaction was once a major source of HCl gas for the chemical industry (the "Leblanc process" of the early 1800s used it on a vast scale). Oxidation numbers unchanged: Cl stays −1, S stays +6 — no redox.

Pedagogically, the take-home is: Cl⁻ cannot reduce H₂SO₄ at all. This is direct experimental evidence that Cl⁻ sits on the "weak reducing agent" side of the halide series.

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3. NaBr with concentrated H₂SO₄ — acid-base THEN partial redox

Br⁻ is a strong enough reducing agent to partially reduce concentrated H₂SO₄, taking S from oxidation state +6 to +4 (in SO₂). Two steps occur consecutively in the same test tube:

Step 1 (acid-base): $\text{NaBr(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HBr(g)}$NaBr(s)+H2​SO4​(l)→NaHSO4​(s)+HBr(g)

Step 2 (redox): the HBr (or unprotonated Br⁻) then reduces H₂SO₄: $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)}$2HBr(g)+H2​SO4​(l)→Br2​(g)+SO2​(g)+2H2​O(l)

Or equivalently, in net ionic form (combining both steps): $2\text{Br}^-\text{(s in NaBr)} + 3\text{H}_2\text{SO}_4\text{(l)} \rightarrow 2\text{HSO}_4^-\text{(s)} + \text{Br}_2\text{(g)} + \text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$2Br−(s in NaBr)+3H2​SO4​(l)→2HSO4−​(s)+Br2​(g)+SO2​(g)+2H2​O(l)

Oxidation-number changes:

• Br: −1 → 0 (oxidised; appears as Br₂)
• S: +6 → +4 (reduced; appears as SO₂)

Observations:

• Initial steamy fumes of HBr.
• Orange/brown vapour of Br₂ (the redox product) — the most diagnostic observation.
• Choking, acidic gas (SO₂) — turns damp blue litmus red and has a sharp, throat-irritating smell.
• Dark red-brown liquid (Br₂(l)) may collect on the test-tube wall.

The redox step does not progress further because Br⁻ is not a strong enough reducing agent to push S below +4. The system halts at SO₂.

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4. NaI with concentrated H₂SO₄ — vigorous, multi-step redox

I⁻ is the strongest halide reducing agent. With concentrated H₂SO₄, the redox cascade proceeds through three successive reductions of sulfur:

Step 1 (acid-base, brief): $\text{NaI(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HI(g)}$NaI(s)+H2​SO4​(l)→NaHSO4​(s)+HI(g)

Step 2 (redox, S +6 → +4 in SO₂): $2\text{HI(g)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{I}_2\text{(s)} + \text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$2HI(g)+H2​SO4​(l)→I2​(s)+SO2​(g)+2H2​O(l)

Step 3 (further redox, S +6 → 0 in S): $6\text{HI(g)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow 3\text{I}_2\text{(s)} + \text{S(s)} + 4\text{H}_2\text{O(l)}$6HI(g)+H2​SO4​(l)→3I2​(s)+S(s)+4H2​O(l)

Step 4 (deepest redox, S +6 → −2 in H₂S): $8\text{HI(g)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow 4\text{I}_2\text{(s)} + \text{H}_2\text{S(g)} + 4\text{H}_2\text{O(l)}$8HI(g)+H2​SO4​(l)→4I2​(s)+H2​S(g)+4H2​O(l)

Oxidation-number changes:

• I: −1 → 0 (in I₂)
• S: +6 → +4 (in SO₂), +6 → 0 (in S), +6 → −2 (in H₂S)

The progression +4, 0, −2 demonstrates increasingly deep reduction of sulfur as more electrons are supplied by the highly reducing I⁻ ions. The total electron transfer in the S +6 → −2 step is 8 electrons per S atom — a striking measure of how strong a reducing agent iodide is.

Observations:

• Brief HI(g) fumes initially.
• Black solid / purple vapour of I₂ (most prominent visible feature).
• Yellow solid deposit of sulfur (S) on the test-tube wall.
• Rotten-egg smell of H₂S gas — pungent, characteristic, and quite unpleasant.
• Possibly some SO₂ fumes (choking acidic gas) early in the reaction.

The combination of all four observations (I₂ + S + H₂S + SO₂) is the unmistakable signature of NaI + concentrated H₂SO₄ and is one of the most striking demonstrations in OCR practical work.

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Linking back to reducing ability

The pattern is a direct consequence of the reducing-power trend (F⁻ < Cl⁻ < Br⁻ < I⁻):

• F⁻ and Cl⁻ are too weak to reduce S(+6) at all — they only do the acid-base step.
• Br⁻ is strong enough to push S one step down (+6 → +4 in SO₂) but no further.
• I⁻ can push S all the way down to −2 (in H₂S), an 8-electron reduction per S atom.

The experimental observation that you smell rotten egg (H₂S) only with iodide, choking SO₂ with bromide and iodide, and neither with chloride or fluoride is the most direct evidence at A-Level for the down-group trend in halide reducing ability. This is why OCR makes this reaction central — it is the experimental anchor of the whole "Group 7 reducing-power" concept.

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Worked examples

Example 1 — NaI + conc H₂SO₄ in full

Q: Write equations (with state symbols) and describe the observations when concentrated sulfuric acid is added to solid potassium iodide. Identify the reduced sulfur products and explain what they tell you about the reducing ability of iodide ions.

Answer:

Step 1 (acid-base): $\text{KI(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{KHSO}_4\text{(s)} + \text{HI(g)}$KI(s)+H2​SO4​(l)→KHSO4​(s)+HI(g).

Step 2 (redox, S +6 → +4): $2\text{HI(g)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{I}_2\text{(s)} + \text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$2HI(g)+H2​SO4​(l)→I2​(s)+SO2​(g)+2H2​O(l).

Step 3 (redox, S +6 → 0): $6\text{HI(g)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow 3\text{I}_2\text{(s)} + \text{S(s)} + 4\text{H}_2\text{O(l)}$6HI(g)+H2​SO4​(l)→3I2​(s)+S(s)+4H2​O(l).

Step 4 (redox, S +6 → −2): $8\text{HI(g)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow 4\text{I}_2\text{(s)} + \text{H}_2\text{S(g)} + 4\text{H}_2\text{O(l)}$8HI(g)+H2​SO4​(l)→4I2​(s)+H2​S(g)+4H2​O(l).

Observations: brief HI fumes, black solid and purple vapour of I₂, yellow deposit of sulfur, choking acidic gas (SO₂), pungent rotten-egg smell of H₂S.

Reducing ability: the sulfur in H₂SO₄ is reduced all the way from +6 to −2 — a change of 8 units in oxidation state, requiring 8 electrons per S atom to be supplied by I⁻ ions. I⁻ is therefore a very strong reducing agent — much stronger than Br⁻ (which only reaches S +4 / SO₂) and immeasurably stronger than Cl⁻ and F⁻ (which do no redox at all). The sequence experimentally confirms I⁻ > Br⁻ > Cl⁻ > F⁻ as the order of halide reducing power.

Example 2 — NaBr + conc H₂SO₄

Q: A student adds a few drops of concentrated H₂SO₄ to solid NaBr. Describe and explain the observations, and write the two key balanced equations.

Answer: Two reactions occur. Step 1: $\text{NaBr(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HBr(g)}$NaBr(s)+H2​SO4​(l)→NaHSO4​(s)+HBr(g) — this is the acid-base step, giving misty white HBr fumes. Step 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)}$2HBr(g)+H2​SO4​(l)→Br2​(g)+SO2​(g)+2H2​O(l) — this is the redox step; Br⁻ (in HBr) is oxidised from −1 to 0 (in Br₂), and S in H₂SO₄ is reduced from +6 to +4 (in SO₂). Observations: initial misty HBr fumes, orange-brown Br₂ vapour, choking SO₂ gas (turns damp blue litmus red). The reaction halts at SO₂ because Br⁻ is not a strong enough reducing agent to push S below +4.

Example 3 — identifying an unknown halide