The reactions of chlorine

1. Overview

In Cl₂, chlorine has oxidation number 0, partway between its lowest value (−1) and its highest (+7). So in one reaction some chlorine atoms can be oxidised while others are reduced. This section covers two such reactions with aqueous sodium hydroxide, cold and hot, which give different products, and the reaction with water that makes drinking water safe. It builds on oxidation numbers and disproportionation from 6.1.

Key Definitions

• Disproportionation: a redox reaction in which atoms of one element, all starting in the same oxidation state, are both oxidised and reduced, so some of them end in a higher oxidation state and some in a lower one.
• Sodium chlorate(I), NaClO: the salt formed, with sodium chloride and water, when chlorine reacts with cold aqueous sodium hydroxide; chlorine is +1 in it, and its solution is bleach.
• Sodium chlorate(V), NaClO₃: the salt formed, with sodium chloride and water, when chlorine reacts with hot aqueous sodium hydroxide; chlorine is +5 in it.
• Chloric(I) acid, HOCl: the weak acid formed, with hydrochloric acid, when chlorine reacts with water; chlorine is +1 in it, and it is one of the two active species that kill bacteria.
• Chlorate(I) ion, ClO⁻: the ion formed when chloric(I) acid partly dissociates, also present in bleach; chlorine is +1 in it, and it is the other active species that kills bacteria.
• Active species (in water purification): the species that actually kill bacteria in chlorinated water, HOCl and ClO⁻; the chloride ions formed at the same time do not.

Content

One element, two directions

When chlorine reacts with aqueous sodium hydroxide, chlorine is the only element whose oxidation number changes (sodium stays +1, oxygen −2, hydrogen +1).

• Some chlorine atoms are reduced to chloride ions, Cl⁻, at −1.
• The rest are oxidised to an ion in which chlorine is bonded to oxygen. Oxygen is more electronegative (3.5, against 3.0 for chlorine, on the data booklet's Pauling scale), so this chlorine has a positive oxidation number.

The temperature decides which oxidised ion forms. The cold reaction usually uses dilute alkali and the hot one concentrated alkali, but the condition you must always state is "cold" or "hot".

Chlorine with cold aqueous sodium hydroxide

Bubble chlorine into cold, dilute NaOH(aq), at room temperature or below. The pale green chlorine is absorbed, giving a colourless solution:

Cl₂(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l)

Ionic equation: Cl₂(g) + 2OH⁻(aq) → Cl⁻(aq) + ClO⁻(aq) + H₂O(l)

The products are sodium chloride and sodium chlorate(I), NaClO (also called sodium hypochlorite). Writing $x$ for chlorine's oxidation number, in ClO⁻ $x + (-2) = -1$, so chlorine is +1.

Chlorine in	Oxidation number	Change per atom
Cl₂	0	start
Cl⁻ (NaCl)	−1	reduced, gains 1 e⁻
ClO⁻ (NaClO)	+1	oxidised, loses 1 e⁻

For every Cl₂ that reacts, one chlorine atom goes down by 1 and the other goes up by 1. The changes balance one to one, so NaCl and NaClO form in equal amounts. Chlorine is both reduced and oxidised from a single starting state, so this is disproportionation.

The solution of NaClO and NaCl is household bleach. Its ClO⁻ ions kill bacteria, as they do in chlorinated water (below).

Chlorine with hot aqueous sodium hydroxide

With hot, concentrated NaOH(aq), the oxidised product is different:

3Cl₂(g) + 6NaOH(aq) → 5NaCl(aq) + NaClO₃(aq) + 3H₂O(l)

Ionic equation: 3Cl₂(g) + 6OH⁻(aq) → 5Cl⁻(aq) + ClO₃⁻(aq) + 3H₂O(l)

The products are sodium chloride and sodium chlorate(V), NaClO₃. In ClO₃⁻, $x + 3(-2) = -1$, so chlorine is +5.

Chlorine in	Oxidation number	Change per atom
Cl₂	0	start
Cl⁻ (NaCl)	−1	reduced, gains 1 e⁻
ClO₃⁻ (NaClO₃)	+5	oxidised, loses 5 e⁻

Chlorine again goes both up and down from 0, so this is also disproportionation.

Where the 5 : 1 ratio comes from. Use the rule from 6.1: the total increase in oxidation number equals the total decrease.

1. One chlorine atom rising from 0 to +5 is an increase of 5.
2. Each atom falling from 0 to −1 is a decrease of 1, so five must fall to balance it.
3. That is six chlorine atoms, 3Cl₂, giving 5Cl⁻ and one ClO₃⁻.
4. The right side has a total charge of −6, so add 6OH⁻ on the left. Their 6 hydrogen atoms form 3H₂O.

Check: each side has 6 Cl, 6 O, 6 H and a total charge of −6.

Why the temperature matters. Chlorate(I) ions are unstable when heated: they disproportionate further, to chloride (−1) and chlorate(V) (+5). So hot alkali ends up with ClO₃⁻ instead of ClO⁻.

	Cold NaOH(aq)	Hot NaOH(aq)
Oxidised product	NaClO, Cl +1	NaClO₃, Cl +5
NaCl : oxidised product	1 : 1	5 : 1
NaOH : Cl₂	2 : 1	2 : 1

Worked example 1: the same chlorine, cold or hot

Question: 1.44 dm³ of chlorine, measured at room conditions, is passed into an excess of aqueous sodium hydroxide. Calculate the mass of the chlorine-containing salt, other than sodium chloride, that forms when the alkali is (a) cold and (b) hot. (c) Use oxidation numbers to explain why the mass in (b) is smaller. Use Na = 23.0, Cl = 35.5 and O = 16.0, and 24.0 dm³ mol⁻¹ as the molar volume at room conditions.

Step 1: amount of chlorine. $$n(\text{Cl}_2) = \frac{1.44}{24.0} = 0.0600\ \text{mol}$$

(a) Cold. One Cl₂ gives one NaClO, so $n(\text{NaClO}) = 0.0600$ mol. The molar mass is 23.0 + 35.5 + 16.0 = 74.5 g mol⁻¹. $$m = 0.0600 \times 74.5 = 4.47\ \text{g}$$

(b) Hot. Three Cl₂ give one NaClO₃, so $n(\text{NaClO}_3) = 0.0600 \div 3 = 0.0200$ mol. The molar mass is 23.0 + 35.5 + 3(16.0) = 106.5 g mol⁻¹. $$m = 0.0200 \times 106.5 = 2.13\ \text{g}$$

(c) Explain. The 0.0600 mol of Cl₂ contains 0.120 mol of chlorine atoms.

• Cold: each oxidised atom loses 1 electron, taken by one reduced atom, so half the atoms, 0.0600 mol, are oxidised.
• Hot: each oxidised atom loses 5 electrons, taken by five reduced atoms, so only one atom in six, 0.0200 mol, is oxidised.

NaClO₃ has the larger molar mass, 106.5 against 74.5 g mol⁻¹, but that is less than three times as much, so it cannot make up for a third of the amount.

Answer: (a) 4.47 g of NaClO; (b) 2.13 g of NaClO₃; (c) each oxidised chlorine atom rises by 5 in hot alkali but by 1 in cold, so only 1 atom in 6 ends up in the oxidised salt instead of 1 in 2.

Chlorine in water purification

Chlorine is added to drinking water to kill bacteria. It dissolves in water and reacts with it:

Cl₂(aq) + H₂O(l) ⇌ HCl(aq) + HOCl(aq)

HOCl is chloric(I) acid, also written HClO; the formula HOCl shows that hydrogen is bonded to oxygen. HCl is a strong acid, fully dissociated in water, so you can also write the products as H⁺(aq) + Cl⁻(aq) + HOCl(aq).

This is disproportionation too. Chlorine goes from 0 in Cl₂ to −1 in HCl (reduced) and to +1 in HOCl (oxidised). In HOCl, $(+1) + (-2) + x = 0$, so chlorine is +1.

HOCl is a weak acid, so some of it dissociates:

HOCl(aq) ⇌ H⁺(aq) + ClO⁻(aq)

Chlorinated water therefore contains both HOCl and ClO⁻. These are the active species. In both, chlorine is +1 and is easily reduced to −1, so both are oxidising agents. They oxidise and damage substances in bacterial cells, which kills the bacteria. Chloride ions cannot do this: chlorine in Cl⁻ is already at −1, its lowest oxidation number, so Cl⁻ is not an oxidising agent.

Chlorine is toxic, so only small, controlled amounts are added.

Worked example 2: chlorinating a water tank

Question: The water in an 800 dm³ storage tank is treated with chlorine at a dose of 2.0 mg of Cl₂ per dm³ of water. (a) Calculate the amount, in mol, of Cl₂ added. (b) Assume that all the chlorine reacts with the water and none is used up in other ways. Calculate the total concentration of the active species, HOCl and ClO⁻. Use Cl = 35.5.

(a) Mass of Cl₂ = 800 × 2.0 = 1600 mg = 1.6 g, and $M(\text{Cl}_2)$ = 2 × 35.5 = 71.0 g mol⁻¹. $$n(\text{Cl}_2) = \frac{1.6}{71.0} = 0.0225\ \text{mol}$$ The dose is given to 2 s.f., so this is 0.023 mol.

(b) In Cl₂(aq) + H₂O(l) ⇌ HCl(aq) + HOCl(aq), each Cl₂ gives one HOCl. Each HOCl that dissociates gives one ClO⁻, so the total amount of HOCl and ClO⁻ equals the amount of Cl₂. $$c = \frac{0.0225}{800} = 2.8 \times 10^{-5}\ \text{mol dm}^{-3}$$

Answer: (a) 0.023 mol; (b) 2.8 × 10⁻⁵ mol dm⁻³.

Only half of the chlorine atoms added become active species; the other half become chloride ions, which do not kill bacteria.

Key Equations

Cold aqueous NaOH (chlorine: 0 to −1 in NaCl and to +1 in NaClO):

Cl₂(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l)

Ionic: Cl₂(g) + 2OH⁻(aq) → Cl⁻(aq) + ClO⁻(aq) + H₂O(l)

Hot aqueous NaOH (chlorine: 0 to −1 in NaCl and to +5 in NaClO₃):

3Cl₂(g) + 6NaOH(aq) → 5NaCl(aq) + NaClO₃(aq) + 3H₂O(l)

Ionic: 3Cl₂(g) + 6OH⁻(aq) → 5Cl⁻(aq) + ClO₃⁻(aq) + 3H₂O(l)

Chlorine with water (chlorine: 0 to −1 in HCl and to +1 in HOCl): Cl₂(aq) + H₂O(l) ⇌ HCl(aq) + HOCl(aq)

Chloric(I) acid, a weak acid (chlorine stays +1): HOCl(aq) ⇌ H⁺(aq) + ClO⁻(aq)

Balancing rule (from 6.1): total increase in oxidation number = total decrease. In the hot reaction, one atom up by 5 balances five atoms down by 1.

Common Mistakes to Avoid

• Mistake: leaving out the conditions, or writing just "NaOH". Fix: write "cold aqueous NaOH" for NaClO and "hot aqueous NaOH" for NaClO₃.
• Mistake: giving an acid or oxygen as a product in alkali, or leaving out water, as in Cl₂ + NaOH → NaCl + HOCl. Fix: the alkali neutralises any acid, so the products are two sodium salts and water: Cl₂ + 2NaOH → NaCl + NaClO + H₂O.
• Mistake: writing the hot equation as Cl₂ + 2NaOH → NaCl + NaClO₃ + H₂O. Fix: one chlorine atom rising to +5 loses 5 electrons, so five atoms must each gain one: 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O. The wrong equation does not even balance for oxygen (2 on the left, 4 on the right).
• Mistake: mixing up chlorate(I) and chlorate(V), or giving +1 for chlorine in NaClO₃. Fix: the Roman numeral is chlorine's oxidation number. In NaClO₃, $(+1) + x + 3(-2) = 0$ gives $x = +5$, so it is sodium chlorate(V); NaClO is sodium chlorate(I).
• Mistake: saying "NaOH" or "the compound" is oxidised and reduced, or giving only one change. Fix: name the element and both changes: "chlorine is oxidised from 0 in Cl₂ to +1 in ClO⁻ and reduced from 0 in Cl₂ to −1 in Cl⁻".
• Mistake: for water treatment, writing Cl instead of Cl₂, an equation with no HOCl (such as 2Cl₂ + 2H₂O → 4HCl + O₂), or saying that chloride ions or just "chlorine" kill the bacteria. Fix: write Cl₂(aq) + H₂O(l) ⇌ HCl(aq) + HOCl(aq) and name HOCl and ClO⁻ as the species that kill bacteria.

Exam Tips

• Describe a reaction of chlorine with sodium hydroxide by giving the reagent and conditions (aqueous; cold or hot), the names and formulae of both salts, and a balanced equation with state symbols.
• Explain or show that a reaction is disproportionation by naming the element (chlorine), its starting oxidation number (0 in Cl₂) and both final ones, each with its species. A definition on its own does not show it.
• Construct the hot equation from the electron balance: one chlorine at +5 needs five at −1, so 5Cl⁻ to one ClO₃⁻. Balance the charge with OH⁻ and the hydrogen with H₂O, then check atoms and charges.
• Explain, with an equation, the use of chlorine in water purification in three parts: the equation Cl₂ + H₂O ⇌ HCl + HOCl (and HOCl ⇌ H⁺ + ClO⁻), the active species (HOCl and ClO⁻), and their role (they kill bacteria).