Chlorine Reactions with Water Revision Notes | A Level AQA Chemistry

1. Why Chlorine and Water Chemistry Matters

Chlorine is added to drinking water supplies across the UK to kill harmful microorganisms. The chemistry that makes this work — and the chemistry that eventually destroys the chlorine in sunlight — are both on your AQA specification, and both hinge on the same element doing two very different things at once.

This lesson covers two distinct reactions of chlorine with water, the anion identification tests you need for the exam, and the conceptual depth — oxidation states, disproportionation, equilibrium — that separates a grade 7 answer from a grade 9 one.

2. Chlorine Dissolving in Water: The Disproportionation Reaction

When chlorine gas is bubbled into water, it does not dissolve passively. It reacts. The product is a pale yellow-green solution called chlorine water, and the key equation is:

Notice the equilibrium sign: this reaction does not go to completion. At room temperature, only a small proportion of the dissolved chlorine reacts, so chlorine water contains a mixture of $Cl_2(aq)$, $H^+(aq)$, $Cl^-(aq)$, and $HClO(aq)$.

The AQA specification also requires you to know the overall ionic equation for this equilibrium, written in terms of chlorate(I) ions rather than chloric(I) acid. This form shows the full ionisation of $HClO$ and represents the same equilibrium in water:

To confirm this is a redox reaction, track the oxidation states of chlorine. In $Cl_2$, each chlorine atom has oxidation state 0. In $Cl^-$, the oxidation state is -1 (reduction). In $ClO^-$, the oxidation state is +1 (oxidation). Both changes occur in the same equation — that is the hallmark of disproportionation.

Both $HClO$ (chloric(I) acid / hypochlorous acid) and $ClO^-$ (chlorate(I) ion) are active disinfecting and bleaching species in chlorine water. $HClO$ is actually the more potent disinfectant of the two; the relative proportions of each depend on the pH of the solution. Together, they oxidise the pigments in dyes and kill bacteria by oxidising their cell membranes. This is why chlorine is used in water treatment and why bleach (sodium chlorate(I) solution) is such an effective cleaning agent.

3. The Second Reaction: Chlorine, Water, and Sunlight

The disproportionation reaction above is not the only thing that happens when chlorine meets water. In the presence of ultraviolet light — sunlight — a completely different reaction takes place, and the products are entirely different.

This reaction is why chlorine water stored in a clear bottle gradually loses its pale yellow-green colour and its bleaching power when left on a sunny windowsill. The UV light drives the decomposition of $HClO$ (or $Cl_2$ dissolved in water), producing $O_2$ gas and $Cl^-$ ions. Once the chlorine has been converted to chloride ions, the disinfecting ability is lost.

Notice that in this reaction, chlorine is only reduced — from oxidation state 0 in $Cl_2$ to -1 in $Cl^-$. The oxygen in water is oxidised from -2 in $H_2O$ to 0 in $O_2$. This is a straightforward redox reaction, not a disproportionation, because two different elements are changing oxidation state.

4. Comparing the Two Reactions: A Side-by-Side Analysis

Both reactions involve $Cl_2$ and $H_2O$ as reactants, but the conditions and products are completely different. Keeping these two reactions distinct is one of the most common points of confusion in this topic, so a direct comparison is worth building into your revision.

The key distinction to anchor in your memory: the dark reaction is a disproportionation because chlorine goes to two different oxidation states simultaneously. The sunlight reaction is not a disproportionation — chlorine only goes in one direction (reduction to $Cl^-$), while oxygen in water is oxidised to $O_2$.

5. Identifying Anions: Group 7 Halide Ions, $OH^-$, $CO_3^{2-}$, and $SO_4^{2-}$

The AQA specification requires you to identify a range of anions using chemical tests. These tests appear regularly in both written papers and practical assessments, so precision in the reagents, observations, and conclusions is essential.

The tests for halide ions connect directly to the chlorine chemistry above: when you add silver nitrate solution to a solution containing $Cl^-$ ions (such as chlorine water), you get a white precipitate of silver chloride. This is the same $Cl^-$ ion produced in both reactions of chlorine with water.

5a. Testing for Halide Ions ($Cl^-$, $Br^-$, $I^-$)

The test uses dilute nitric acid followed by silver nitrate solution ($AgNO_3$). Nitric acid is added first to remove any interfering anions such as $CO_3^{2-}$ or $SO_3^{2-}$ that would also precipitate with silver ions and give a false result.

The ionic equations for these precipitations follow the same pattern. For chloride ions as an example:

The ammonia solubility test is the key to distinguishing between the three halide precipitates when the colours are ambiguous. White $AgCl$ dissolves in both dilute and concentrated ammonia. Cream $AgBr$ dissolves only in concentrated ammonia. Yellow $AgI$ is insoluble in both. This trend reflects the decreasing solubility of the silver halides as you go down Group 7.

5b. Testing for $OH^-$, $CO_3^{2-}$, and $SO_4^{2-}$

Beyond halide ions, the specification requires tests for three further anions. Each uses a specific reagent and produces a characteristic observation.

• $OH^-$ (hydroxide ion): Add universal indicator or litmus — solution turns blue/purple, confirming alkaline conditions. Alternatively, add a few drops of the solution to damp red litmus paper — it turns blue. For a more rigorous test, add the solution to an ammonium salt and warm — ammonia gas is evolved, turning damp red litmus blue.
• $CO_3^{2-}$ (carbonate ion): Add dilute hydrochloric acid — effervescence occurs and the gas produced turns limewater ($Ca(OH)_2$ solution) milky. The gas is $CO_2$. Ionic equation: $CO_3^{2-}(aq) + 2H^+(aq) \rightarrow H_2O(l) + CO_2(g)$.
• $SO_4^{2-}$ (sulphate ion): Add dilute hydrochloric acid (to remove interfering ions such as $CO_3^{2-}$ and $SO_3^{2-}$), then add barium chloride solution ($BaCl_2$). A white precipitate of barium sulphate ($BaSO_4$) forms. Ionic equation: $Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)$.

A critical point for the sulphate test: dilute hydrochloric acid must be added before the barium chloride, not after. If you add $BaCl_2$ first to a solution containing $CO_3^{2-}$, you will get a white precipitate of $BaCO_3$, which is insoluble and will give a false positive for sulphate. Adding acid first dissolves any carbonate, leaving only sulphate to react with barium ions.

6. Worked Example: Identifying an Unknown Anion

Exam questions on anion identification often present you with a series of test results and ask you to identify the ion present, or ask you to design a test sequence. The step-by-step below models the reasoning process an examiner expects.

Worked example: Identifying an Unknown Anion from Test Results

A student has a solution of an unknown sodium salt. They carry out the following tests: (1) Adding dilute nitric acid produces no effervescence. (2) Adding silver nitrate solution after the acid produces a cream precipitate. (3) Adding dilute ammonia solution to the precipitate — it does not dissolve. (4) Adding concentrated ammonia solution — the precipitate dissolves. Identify the anion present and write the ionic equation for the precipitate forming.

1. No $CO_2$ gas produced → $CO_3^{2-}$ is absent. — Carbonate ions react with dilute acid to produce $CO_2$ gas (effervescence). No gas means no carbonate present.
2. Cream precipitate → likely $AgBr$, suggesting $Br^-$ is present. (White = $AgCl$; yellow = $AgI$; cream = $AgBr$.) — The colour of the silver halide precipitate is the first indicator of which halide ion is present. Cream narrows it to bromide.
3. $AgCl$ dissolves in dilute $NH_3$; $AgBr$ does not. Insolubility in dilute $NH_3$ is consistent with $AgBr$. — This rules out chloride ($AgCl$ would dissolve here) and is consistent with bromide.
4. $AgBr$ dissolves in concentrated $NH_3$; $AgI$ does not. Dissolving in concentrated $NH_3$ confirms $AgBr$. — This rules out iodide ($AgI$ would remain insoluble) and confirms the precipitate is $AgBr$, so the anion is $Br^-$.
5. $$Ag^+(aq) + Br^-(aq) \rightarrow AgBr(s)$$ — State symbols are required. The precipitate is silver bromide, a cream solid. The spectator ions ($Na^+$ and $NO_3^-$) are omitted from the ionic equation.

Answer: The anion is bromide, $Br^-$. Ionic equation: $Ag^+(aq) + Br^-(aq) \rightarrow AgBr(s)$ (cream precipitate, insoluble in dilute $NH_3$, soluble in concentrated $NH_3$).

7. Exam Technique and High-Mark Responses

The chlorine-water reactions and anion tests appear in a range of question styles on AQA papers. The marks are often lost not through lack of knowledge but through imprecise language, missing state symbols, or confusing the two chlorine reactions. The guidance below targets the most common mark-losing patterns.

For 'explain' questions about the disproportionation reaction, examiners want to see oxidation state tracking. A response that says 'chlorine is both oxidised and reduced' without specifying the oxidation states and the species involved will typically earn only one mark out of two or three. Always state: the starting oxidation state (0 in $Cl_2$), the oxidation state in the reduced product (-1 in $Cl^-$), and the oxidation state in the oxidised product (+1 in $ClO^-$).

For 'describe' questions about anion tests, the AQA mark scheme typically awards one mark for the correct reagent and one mark for the correct observation. Saying 'add silver nitrate' without specifying that dilute nitric acid is added first, or without stating the colour of the precipitate, will cost marks. Practise writing test descriptions in the format: reagent added → observation → conclusion.

For longer 'evaluate' or 'explain' questions about water treatment with chlorine, structure your answer around the chemistry: chlorine disproportionates in water to form $ClO^-$, which is the active disinfecting species. The equilibrium nature of the reaction means some $Cl_2$ remains dissolved, which also has disinfecting properties. The photochemical decomposition in sunlight is a limitation — it destroys the active species over time, which is why water treatment plants must maintain chlorine levels continuously.

The two reactions of chlorine with water — disproportionation in the dark to give $Cl^-$ and $ClO^-$, and photochemical decomposition in sunlight to give $Cl^-$ and $O_2$ — together with the systematic anion identification tests, form a coherent and examinable set of ideas. Master the oxidation state tracking, the correct reagents and observations for each anion test, and the distinction between equilibrium and complete reactions, and you have the tools to answer any question this topic can generate.