Halogens: Disproportionation and Halide Ions

Spec Mapping — OCR H432 Module 3.1.3 — Group 7 (the halogens), covering the concept of disproportionation (the simultaneous oxidation and reduction of the same element), the disproportionation of chlorine with water (Cl₂ + H₂O → HClO + HCl) and its application to water treatment, the disproportionation of chlorine with cold dilute aqueous sodium hydroxide (Cl₂ + 2NaOH → NaCl + NaClO + H₂O) and its application to household bleach production, and the trend in reducing ability of halide ions (I⁻ > Br⁻ > Cl⁻ > F⁻) (refer to the official OCR H432 specification document for exact wording).

Disproportionation is OCR's canonical example of a non-obvious redox — a reaction in which the same element is simultaneously oxidised and reduced. The chemistry looks puzzling at first sight (how can something be both oxidising and reducing agent?), but the oxidation-number book-keeping makes it transparent: one chlorine atom in Cl₂ goes up to +1, the other goes down to −1; the average is still 0, but the two atoms have ended in different oxidation states. The same concept appears in industrial water treatment (chlorination of drinking water and swimming pools — the active sterilising species is HClO, not Cl₂ directly) and in household bleach production (Cl₂ + cold dilute NaOH → NaClO + NaCl). This lesson develops fluency in disproportionation and pivots to the mirror-image trend of the previous lesson: while halogen oxidising power decreases down Group 7, halide reducing power increases. The next lesson (halide ions with concentrated sulfuric acid) is the experimental demonstration of that reducing-power trend.

Key Definition: Disproportionation is a redox reaction in which atoms of the same element are simultaneously oxidised and reduced. One species (the starting material) ends up as two products with different oxidation numbers — one higher and one lower than the starting oxidation number.

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Disproportionation — definition and recognition

The defining feature of disproportionation is that the same element changes oxidation state in both directions during one reaction. For an atom labelled X with starting oxidation number $n$n:

$\text{some X (ox state } n) \rightarrow \text{ X (ox state } > n) + \text{X (ox state } < n)$some X (ox state n)→ X (ox state >n)+X (ox state <n)

If only one of those changes occurs, it is not disproportionation — it is just a "normal" redox reaction in which X is an oxidising or reducing agent. Both must occur, on the same element, for the disproportionation label to apply.

Quick example: hydrogen peroxide

$2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2 \quad \text{(disproportionation of O)}$2H2​O2​→2H2​O+O2​(disproportionation of O)

Oxidation-number assignment:

• O in H₂O₂ (peroxide): −1
• O in H₂O (water): −2 ← reduced from −1 (gained an electron)
• O in O₂ (elemental dioxygen): 0 ← oxidised from −1 (lost an electron)

Same element (oxygen), same reaction, both oxidised and reduced — this is the formal disproportionation.

Why it happens

Disproportionation is favoured when an element has an intermediate oxidation state that is thermodynamically less stable than the combination of higher and lower states. For chlorine: the oxidation state 0 (in Cl₂) is intermediate between +1 (HClO, ClO⁻) and −1 (HCl, Cl⁻). With water or hydroxide present to capture the products, the chemistry shifts towards the more stable +1 / −1 combination. Hydrogen peroxide is similar: the −1 oxidation state of oxygen is unstable to splitting into 0 (O₂) and −2 (H₂O).

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Chlorine + water (water treatment chemistry)

Chlorine dissolves slightly in water and undergoes disproportionation to give chloric(I) acid and hydrochloric acid:

$\text{Cl}_2\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{HClO(aq)} + \text{HCl(aq)}$Cl2​(aq)+H2​O(l)⇌HClO(aq)+HCl(aq)

Or, with both acids fully ionised, the equivalent ionic equation:

$\text{Cl}_2\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{HClO(aq)} + \text{H}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$Cl2​(aq)+H2​O(l)⇌HClO(aq)+H+(aq)+Cl−(aq)

Oxidation numbers of chlorine:

Species	Cl oxidation number	Role
Cl₂	0	Starting material
HClO (chloric(I) acid)	+1	Oxidised half
HCl / Cl⁻	−1	Reduced half

Half of the chlorine atoms in Cl₂ are oxidised (0 → +1), and the other half are reduced (0 → −1) — the textbook disproportionation pattern.

Why it matters — pathogen kill in water treatment

HClO (chloric(I) acid, also written HOCl) is the active sterilising species in chlorinated water. It is a small, neutral, lipid-soluble molecule that penetrates bacterial cell membranes far more readily than the bulky ClO⁻ ion. Once inside, HClO acts as a powerful oxidising agent on essential biomolecules — denaturing proteins (especially enzymes), oxidising nucleic acid bases, and disrupting cellular machinery. Bacteria and viruses are killed within seconds at HClO concentrations of ~1 ppm (1 mg/L).

The Cl₂ + H₂O ⇌ HClO + HCl equilibrium acts as a slow-release HClO reservoir in the pipe network — as HClO is consumed by oxidising microbes, more Cl₂ disproportionates to top up the supply. This is why chlorination delivers continuous sterilisation along the whole water distribution network, not just at the treatment plant.

Chlorination of drinking water has saved more lives than any single chemical intervention in public health history. Pre-chlorination (introduced widely from ~1900) typhoid fever and cholera were common in industrial cities; after chlorination, both became rare in chlorinated supplies. The global health benefit of chlorination is estimated at hundreds of millions of lives over the last century.

Risks and benefits — a balanced view

Benefit	Risk
Kills waterborne bacteria and viruses (cholera, typhoid, hepatitis A, polio)	Cl₂ is toxic at high concentration (was used as a war gas in WWI)
Easy to produce and store (Cl₂ gas, NaOCl bleach, Ca(ClO)₂ HTH tablets)	Can react with organic matter in water to form trihalomethanes (THMs) such as CHCl₃ (carcinogenic at high doses)
Long-lasting residual effect in pipes (vs ozone, which decays quickly)	Taste and smell can be unpleasant
Millions of lives saved worldwide	Some communities prefer alternative disinfection (ozone, UV) where THM formation is a concern

The international consensus is that the public-health benefits of chlorination massively outweigh the risks for routine drinking water supply, especially in low- and middle-income countries where alternative methods are too expensive. THM levels are tightly regulated (typically < 100 µg/L total THM in drinking water).

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Chlorine + cold dilute NaOH (bleach production)

When chlorine is bubbled through cold dilute aqueous sodium hydroxide at room temperature:

$\text{Cl}_2\text{(g)} + 2\text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{NaClO(aq)} + \text{H}_2\text{O(l)}$Cl2​(g)+2NaOH(aq)→NaCl(aq)+NaClO(aq)+H2​O(l)

Oxidation numbers of chlorine:

Species	Cl oxidation number	Role
Cl₂	0	Starting material
NaCl	−1	Reduced (gained electron)
NaClO (sodium chlorate(I))	+1	Oxidised (lost electron)

Again this is disproportionation — Cl 0 → ±1. The product mixture is household bleach (sometimes called "sodium hypochlorite solution"), an aqueous mix of NaClO (the active ingredient that does the bleaching/disinfecting) and the spectator NaCl. Concentrations vary: industrial bleach is ~12 % NaClO; household bleach is typically 4–6 %.

NaClO is a powerful oxidising agent used as:

• A bleach for textiles and paper (oxidises chromophores in dyes and lignin).
• A disinfectant for surfaces, swimming pools (in concentrated form), and washing machine cleaning.
• A deodorising agent by oxidising organic odour molecules.

Conditions matter for this reaction. With hot concentrated NaOH, chlorine disproportionates further to chlorate(V):

$3\text{Cl}_2 + 6\text{NaOH (hot, conc)} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}$3Cl2​+6NaOH (hot, conc)→5NaCl+NaClO3​+3H2​O

NaClO₃ (sodium chlorate(V)) is used as a weedkiller. However, this hot-concentrated reaction is not required for OCR AS-level — focus on the cold dilute reaction giving NaClO.

flowchart TD
  A[Cl2 gas] --> B[+ H2O cold]
  A --> C[+ cold dilute NaOH]
  A --> D[+ hot concentrated NaOH not OCR AS]
  B --> B1[HClO + HCl - disproportionation - water treatment]
  C --> C1[NaCl + NaClO + H2O - disproportionation - household bleach]
  D --> D1[NaCl + NaClO3 - further disproportionation - weedkiller]

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Reducing ability of halide ions — opposite trend to oxidising power of halogens

The previous lesson established: oxidising ability of halogens decreases down Group 7 (F₂ > Cl₂ > Br₂ > I₂). The mirror-image statement for halide ions is:

Reducing ability of halide ions increases down the group: I⁻ > Br⁻ > Cl⁻ > F⁻

A reducing agent loses electrons; halide ions lose one electron each to become halogen atoms (which combine into X₂):

$2\text{X}^-\text{(aq)} \rightarrow \text{X}_2 + 2\text{e}^-$2X−(aq)→X2​+2e−

Down the group:

• Ionic radius increases (F⁻ ~133 pm, Cl⁻ ~181 pm, Br⁻ ~196 pm, I⁻ ~220 pm) — the outer electron is progressively further from the nucleus.
• Shielding increases — more inner shells screen the outer electron from the nuclear charge.
• The outer electron is therefore less tightly held and progressively easier to lose.
• Nuclear charge also rises, but distance and shielding dominate (same four-factor argument as for IE).

Hence iodide ions (I⁻) are the strongest halide reducing agents at A-Level; fluoride ions (F⁻) are the weakest (F is the most electronegative element; F⁻ holds onto its outer electrons very tightly).

Trend summary table

Halide	Ionic radius / pm	Reducing power	Outer electron sub-shell
F⁻	133	Weakest	2p
Cl⁻	181	Weak	3p
Br⁻	196	Moderate	4p
I⁻	220	Strongest	5p

Halides X⁻ — reducing agents F⁻ weakest reducing agent Cl⁻ Br⁻ I⁻ strongest reducing agent Increases down group

This is the trend that the next lesson (halide ions with concentrated sulfuric acid) will demonstrate experimentally — Br⁻ partially reduces H₂SO₄ (S goes +6 → +4 in SO₂), while I⁻ reduces it all the way to H₂S (S goes +6 → −2), confirming I⁻'s dramatically higher reducing power.

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Worked examples

Example 1 — disproportionation in cold dilute NaOH

Q: Write a balanced equation (with state symbols) for the reaction of chlorine with cold dilute aqueous sodium hydroxide. State the change in oxidation number of each chlorine atom in the products. Explain why this is a disproportionation reaction.

Answer: $\text{Cl}_2\text{(g)} + 2\text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{NaClO(aq)} + \text{H}_2\text{O(l)}$Cl2​(g)+2NaOH(aq)→NaCl(aq)+NaClO(aq)+H2​O(l)

Oxidation-number changes for chlorine:

• One Cl: 0 → −1 (in NaCl) — reduced (gained an electron)
• Other Cl: 0 → +1 (in NaClO) — oxidised (lost an electron)

Why disproportionation: both the oxidation and reduction involve the same element (chlorine). One Cl atom in Cl₂ is reduced to Cl⁻ and the other is oxidised to ClO⁻, so chlorine is simultaneously the oxidising agent and the reducing agent in this reaction. The starting Cl₂ (oxidation number 0) is split into two products with oxidation numbers higher and lower than the starting value.

Example 2 — disproportionation in water

Q: Write an ionic equation for the reaction of chlorine with water and identify (i) the species oxidised, (ii) the species reduced, (iii) the active sterilising agent in chlorinated drinking water.

Answer: $\text{Cl}_2\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{HClO(aq)} + \text{H}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$Cl2​(aq)+H2​O(l)⇌HClO(aq)+H+(aq)+Cl−(aq)

(i) Species oxidised: half of the chlorine atoms in Cl₂ (oxidation state 0 → +1 in HClO). (ii) Species reduced: the other half of the chlorine atoms in Cl₂ (0 → −1 in Cl⁻). (iii) The active sterilising agent is HClO (chloric(I) acid) — a small, neutral, lipid-soluble oxidising molecule that penetrates bacterial cells and damages essential biomolecules.

Example 3 — proving disproportionation with oxidation numbers

Q: Sodium thiosulfate Na₂S₂O₃ disproportionates in acidic solution: $\text{S}_2\text{O}_3^{2-} + 2\text{H}^+ \rightarrow \text{S} + \text{SO}_2 + \text{H}_2\text{O}$S2​O32−​+2H+→S+SO2​+H2​O. Verify that this is disproportionation by assigning oxidation states to each sulfur.

Answer: Average oxidation state of S in S₂O₃²⁻: sum of all = −2, so 2(S) + 3(−2) = −2, giving S average = +2. In the product S(s), oxidation state of S is 0 (reduced from +2). In SO₂, oxidation state of S is +4 (oxidised from +2). Same element (S), same reaction, both oxidised and reduced → disproportionation confirmed. (Note: this is not on the OCR AS specification but is an analogous worked example to consolidate the technique.)

Example 4 — halide reducing-power argument

Q: Explain, in terms of atomic structure, why iodide ions are better reducing agents than chloride ions.