Chlorine and NaOH Revision Notes | A Level AQA Chemistry

1. Why Chlorine and NaOH Matter

Chlorine gas reacting with sodium hydroxide solution is one of the most industrially significant reactions in Group 7 chemistry. The product — sodium chlorate(I) — is the active ingredient in household bleach and is central to water treatment worldwide.

There is also a fascinating redox story hidden inside this reaction: chlorine simultaneously increases and decreases its oxidation state. That makes it a textbook example of disproportionation — a concept AQA examiners test repeatedly. By the end of this lesson you will be able to write the equation, assign oxidation states, justify the disproportionation label, and link the product to its real-world uses.

2. The Reaction of Chlorine with Cold, Dilute Aqueous NaOH

When chlorine gas is bubbled through cold, dilute sodium hydroxide solution, the following reaction occurs:

Notice the conditions carefully: cold and dilute. If the NaOH is hot and concentrated, chlorine disproportionates further to give sodium chlorate(V), $\text{NaClO}_3$. AQA only requires you to know the cold, dilute version, but being aware of the contrast shows examiner-level understanding.

The equation is already balanced — check it: 2 Cl atoms on the left (in Cl2) become 2 Cl atoms in products (one in NaCl, one in NaOCl). Sodium: 2 on each side. Oxygen: 1 on each side. Hydrogen: 2 on each side. State symbols are essential in AQA mark schemes, so always include them.

3. Disproportionation: The Redox Story Inside the Equation

Here is where the chemistry becomes genuinely interesting. Look at the oxidation state of chlorine in each species across the equation.

Worked example: Assigning Oxidation States to Confirm Disproportionation

Show, using oxidation states, that the reaction of Cl2 with NaOH is a disproportionation reaction.

1. $\text{Cl}_2$: element in standard state, so oxidation state = $0$ — Any element in its standard state has an oxidation state of zero. Both Cl atoms in Cl2 are at 0.
2. $\text{NaCl}$: Na is $+1$, overall charge is $0$, so Cl = $-1$ — Chlorine has been reduced from 0 to −1. It has gained electrons.
3. $\text{NaOCl}$: Na = $+1$, O = $-2$, overall charge = $0$ So: $+1 + \text{Cl} + (-2) = 0 \Rightarrow \text{Cl} = +1$ — Chlorine has been oxidised from 0 to +1. It has lost electrons.
4. Cl: $0 \rightarrow -1$ (reduction) AND $0 \rightarrow +1$ (oxidation) — The same element (Cl) in the same reactant (Cl2) is simultaneously oxidised and reduced. This is the definition of disproportionation.

Answer: Chlorine is oxidised from 0 to +1 (in NaOCl) and reduced from 0 to −1 (in NaCl). Because the same element undergoes both processes simultaneously, this is a disproportionation reaction.

A common exam trap here is to state that 'chlorine is oxidised' or 'chlorine is reduced' without specifying which chlorine product corresponds to which process. Always name the product alongside the oxidation state change to secure the mark.

4. Uses of the Solution Formed

The solution produced — containing sodium chlorate(I), NaOCl — has two major uses that AQA expects you to know. Both stem from the oxidising power of the $\text{OCl}^-$ ion.

• Household bleach: NaOCl solution bleaches coloured materials by oxidising the dye molecules, destroying their chromophore (the part responsible for colour absorption).
• Water treatment and disinfection: NaOCl kills bacteria and other microorganisms in drinking water and swimming pools by oxidising biological molecules essential to microbial life.
• Industrial disinfectant: used to sterilise surfaces and equipment in food processing and healthcare settings.

In an exam 'describe' or 'suggest' question about uses, naming the product as sodium chlorate(I) or NaOCl and linking it to its oxidising action will score marks. Vague answers such as 'it cleans things' will not.

5. Identifying Cations — Group 2 and Ammonium

The second specification requirement for this lesson links to qualitative analysis: identifying Group 2 cations ($\text{Mg}^{2+}$, $\text{Ca}^{2+}$, $\text{Sr}^{2+}$, $\text{Ba}^{2+}$) and the ammonium ion ($\text{NH}_4^+$) using sodium hydroxide solution. This connects directly to the NaOH chemistry you have already studied — NaOH is the key reagent in both contexts. Note that this lesson focuses specifically on Group 2 and ammonium ion tests as required by this part of the AQA specification; flame tests for Group 1 and Group 2 cations, and the identification of transition metal cations using NaOH, are covered in separate parts of the course.

When NaOH(aq) is added to a solution containing a Group 2 cation, a metal hydroxide precipitate forms. The colour and solubility of that precipitate identify the cation. For ammonium ions, warming with NaOH releases ammonia gas, which is detected by its smell and by damp red litmus paper turning blue.

A key distinction to note: all Group 2 hydroxide precipitates are white, so colour alone cannot differentiate between $\text{Mg}^{2+}$, $\text{Ca}^{2+}$, $\text{Sr}^{2+}$, and $\text{Ba}^{2+}$. In practice, flame tests or additional reagents (such as dilute sulphuric acid to test for a white $\text{BaSO}_4$ precipitate) are used alongside NaOH to confirm identity. AQA questions may ask you to describe the observation and write the ionic equation — practise both.

For ammonium identification, the ionic equation is straightforward: $\text{NH}_4^+(aq) + \text{OH}^-(aq) \rightarrow \text{NH}_3(g) + \text{H}_2\text{O}(l)$. The critical observation is the alkaline gas turning damp red litmus paper blue — state this explicitly in exam answers rather than writing 'gas is produced'.

6. Exam Technique and Common Mistakes

AQA questions on this topic tend to fall into three categories: writing and balancing the equation for Cl2 + NaOH, explaining disproportionation using oxidation states, and describing observations for cation identification. Each has its own mark-winning requirements.

7. Bringing It Together: A Retrieval Check

You have now covered the full reaction of Cl2 with cold, dilute NaOH, the disproportionation argument, the uses of NaOCl, and the identification of Group 2 and ammonium cations using NaOH. This final check tests whether you can apply all of these ideas together — the kind of multi-step reasoning AQA rewards in longer questions.

The core thread running through this entire lesson is the dual nature of chlorine's reactivity: it can act as both an oxidising and a reducing agent within the same reaction, and the product of that reaction — NaOCl — is itself a powerful oxidising agent with wide practical application. Mastering the oxidation state arithmetic and the precise exam language for disproportionation will serve you across multiple AQA topics, including the broader Group 7 section and redox chemistry.