Reactions of Chlorine and Water Purification

AS · 11 min

Chlorine is unusual in that, when it reacts with water or with sodium hydroxide, the same element is oxidised and reduced at the same time. This is called disproportionation. This note covers the reactions of chlorine with cold and with hot aqueous sodium hydroxide, interpreted using oxidation numbers, and the use of chlorine to purify drinking water, including the active species that kill bacteria. These reactions are a favourite source of oxidation-number questions in Papers 1 and 2, and the analysis of bleach by titration links them to Paper 3.

Oxidation numbers of chlorine

Chlorine forms compounds in many oxidation states. The ones you need are:

species	name	oxidation number of Cl
$\ce{Cl-}$	chloride	$-1$
$\ce{Cl2}$	chlorine	0
$\ce{ClO-}$	chlorate(I) (hypochlorite)	$+1$
$\ce{HOCl}$ (also written $\ce{HClO}$ )	chloric(I) acid (hypochlorous acid)	$+1$
$\ce{ClO3-}$	chlorate(V)	$+5$

Work out $\ce{ClO3-}$ like this: $x + 3(-2) = -1$ , so $x = +5$ . The Roman numeral in the name is the oxidation number of chlorine.

Disproportionation

Definition

Disproportionation is a redox reaction in which the same element is both oxidised and reduced.

To show that a reaction is disproportionation, find one element whose oxidation number both increases (in one product) and decreases (in another product), starting from a single oxidation state in the reactant. Chlorine in $\ce{Cl2}$ is at 0, between $-1$ and its positive oxidation states, so it can go both ways.

Chlorine with cold aqueous sodium hydroxide

At room temperature or below (cold, dilute alkali), chlorine reacts to form sodium chloride and sodium chlorate(I):

\ce{Cl2(g) + 2NaOH(aq) -> NaCl(aq) + NaClO(aq) + H2O(l)}

The ionic equation is

\ce{Cl2(aq) + 2OH-(aq) -> Cl-(aq) + ClO-(aq) + H2O(l)}

	in $\ce{Cl2}$	in $\ce{Cl-}$	in $\ce{ClO-}$
oxidation number	0	$-1$	$+1$
change		decrease by 1: reduced	increase by 1: oxidised

One chlorine atom of each $\ce{Cl2}$ molecule is reduced and the other is oxidised: disproportionation. The total decrease (1) equals the total increase (1), as it must.

The solution of sodium chlorate(I), $\ce{NaClO}$ , is household bleach. The mixture is also how bleach is manufactured.

Chlorine with hot aqueous sodium hydroxide

With hot, concentrated sodium hydroxide (about 70 °C), the products are sodium chloride and sodium chlorate(V):

\ce{3Cl2(g) + 6NaOH(aq) -> 5NaCl(aq) + NaClO3(aq) + 3H2O(l)}

Ionic equation:

\ce{3Cl2(aq) + 6OH-(aq) -> 5Cl-(aq) + ClO3-(aq) + 3H2O(l)}

	in $\ce{Cl2}$	in $\ce{Cl-}$	in $\ce{ClO3-}$
oxidation number	0	$-1$	$+5$
number of Cl atoms	6	5	1
total change		$5 \times (-1) = -5$ : reduced	$1 \times (+5) = +5$ : oxidised

Again disproportionation. Notice why the coefficients are 5 and 1: five chlorine atoms each gain one electron to balance the five electrons lost by the one chlorine atom that goes to $+5$ .

Tip

The hot reaction can be seen as the chlorate(I) formed in the cold reaction disproportionating further when heated: $\ce{3ClO- -> 2Cl- + ClO3-}$ (chlorine $+1 \to -1$ and $+1 \to +5$ ). Adding three times the cold equation to this gives the hot equation. This is why bleach should be stored cool.

Method

Balancing a disproportionation equation using oxidation numbers

Write the formulas of the reactant and both products containing the element.
Find the change in oxidation number for each product.
Choose numbers of each product so that the total increase equals the total decrease (for example, 5 $\ce{Cl-}$ for every 1 $\ce{ClO3-}$ ).
Balance the element on the left ( $\ce{Cl2}$ ), then oxygen with $\ce{OH-}$ / $\ce{H2O}$ , then check hydrogen and charge.

Chlorine in water purification

Chlorine and water

Chlorine dissolves in water and reacts reversibly, again by disproportionation:

\ce{Cl2(aq) + H2O(l) <=> HCl(aq) + HOCl(aq)}

Chlorine goes from 0 to $-1$ (in HCl) and to $+1$ (in HOCl, chloric(I) acid). The chloric(I) acid is a weak acid and partly dissociates to give the chlorate(I) ion:

\ce{HOCl(aq) <=> H+(aq) + ClO-(aq)}

Key result

The active species that kill bacteria in chlorinated water are chloric(I) acid, $\ce{HOCl}$ , and the chlorate(I) ion, $\ce{ClO-}$ . Both are strong oxidising agents that destroy bacteria and other micro-organisms. $\ce{HOCl}$ is the more effective of the two, so water treatment works best when the water is not too alkaline.

How chlorination is used

Chlorine (or sodium chlorate(I)) is added to drinking water supplies and swimming pools in small, controlled amounts, roughly 1 mg per $\text{dm}^3$ in drinking water.
It keeps working in the pipes after treatment, because some $\ce{HOCl}$ / $\ce{ClO-}$ remains (a residual concentration), preventing re-contamination.
It has hugely reduced deaths from waterborne diseases such as cholera and typhoid.

Risks

Chlorine gas is toxic and must be handled carefully at treatment works.
Chlorine can react with organic matter in water to form small amounts of chlorinated organic compounds, some of which may be harmful in large doses over long periods.
These risks are judged to be far smaller than the risk of untreated water, so chlorination remains standard.

In sunlight, chloric(I) acid decomposes: $\ce{2HOCl(aq) -> 2HCl(aq) + O2(g)}$ . This is why outdoor pools need chlorine topping up and why chlorine water is stored in dark bottles.

Chlorine water with indicator

Universal indicator added to chlorine water first turns red (because HCl and HOCl are acids) and is then bleached (decolourised) by the oxidising chloric(I) acid. Damp blue litmus paper behaves the same way in chlorine gas: it turns red, then white.

Worked examples

Identifying disproportionation

For the reaction $\ce{Cl2 + 2NaOH -> NaCl + NaClO + H2O}$ , give the oxidation number of chlorine in each chlorine-containing species and explain why this is a disproportionation reaction.

Solution

$\ce{Cl2}$ : 0. $\ce{NaCl}$ : $-1$ . $\ce{NaClO}$ : $+1$ ( $+1 + x - 2 = 0$ ).

Chlorine is reduced from 0 to $-1$ in NaCl and oxidised from 0 to $+1$ in NaClO. The same element is simultaneously oxidised and reduced, so the reaction is disproportionation.

Hot alkali: balancing and naming

Chlorine is passed into hot concentrated aqueous sodium hydroxide. Name the chlorine-containing products, write the ionic equation, and use oxidation numbers to explain the ratio in which they form.

Solution

Products: sodium chloride and sodium chlorate(V).

\ce{3Cl2 + 6OH- -> 5Cl- + ClO3- + 3H2O}

Chlorine to chloride: $0 \to -1$ , a decrease of 1 per atom. Chlorine to chlorate(V): $0 \to +5$ , an increase of 5 per atom. For the electrons to balance, five chlorine atoms must be reduced for every one oxidised, so chloride and chlorate(V) form in the ratio 5 : 1.

Check: Cl $6 = 5 + 1$ ; O $6 = 3 + 3$ ; H $6 = 6$ ; charge $-6 = -5 - 1$ .

Reacting quantities with cold alkali

$1.42\ \text{g}$ of chlorine reacts completely with cold aqueous sodium hydroxide. Calculate the mass of sodium chlorate(I) formed and the minimum volume of $2.00\ \text{mol dm}^{-3}$ sodium hydroxide required. ( $A_r$ : Na 23.0, Cl 35.5, O 16.0)

Solution

$n(\ce{Cl2}) = \dfrac{1.42}{71.0} = 0.0200\ \text{mol}$

$\ce{Cl2 + 2NaOH -> NaCl + NaClO + H2O}$

$n(\ce{NaClO}) = 0.0200\ \text{mol}$ ; $M(\ce{NaClO}) = 74.5\ \text{g mol}^{-1}$ ; mass $= 1.49\ \text{g}$ .

$n(\ce{NaOH}) = 0.0400\ \text{mol}$ ; volume $= \dfrac{0.0400}{2.00} = 0.0200\ \text{dm}^3 = 20.0\ \text{cm}^3$ .

Exam-style: chlorine in water treatment

(a) Write the equation for the reaction of chlorine with water, and use oxidation numbers to explain why it is a disproportionation reaction.

(b) Name the two species formed that kill bacteria, and write an equation linking them.

Solution

(a) $\ce{Cl2(aq) + H2O(l) <=> HCl(aq) + HOCl(aq)}$ . Chlorine goes from 0 in $\ce{Cl2}$ to $-1$ in HCl (reduced) and to $+1$ in HOCl (oxidised). Chlorine is simultaneously oxidised and reduced.

(b) Chloric(I) acid, $\ce{HOCl}$ , and the chlorate(I) ion, $\ce{ClO-}$ : $\ce{HOCl(aq) <=> H+(aq) + ClO-(aq)}$ .

(c) Any one: chlorine is toxic, so leaks are dangerous; it can form chlorinated organic compounds with organic matter in the water, which may be harmful; excess chlorine gives an unpleasant taste and smell.

Exam-hard: analysing bleach by titration

$10.0\ \text{cm}^3$ of household bleach is diluted to $250\ \text{cm}^3$ in a volumetric flask. A $25.0\ \text{cm}^3$ portion is added to excess acidified potassium iodide. The chlorate(I) ions oxidise iodide to iodine:

\ce{ClO-(aq) + 2I-(aq) + 2H+(aq) -> Cl-(aq) + I2(aq) + H2O(l)}

The iodine is titrated with $0.100\ \text{mol dm}^{-3}$ sodium thiosulfate, using starch indicator near the end-point; the mean titre is $22.40\ \text{cm}^3$ :

\ce{I2(aq) + 2S2O3^2-(aq) -> 2I-(aq) + S4O6^2-(aq)}

Calculate the concentration of $\ce{ClO-}$ in the original bleach in $\text{mol dm}^{-3}$ and the mass of $\ce{NaClO}$ per $\text{dm}^3$ . ( $M(\ce{NaClO}) = 74.5\ \text{g mol}^{-1}$ )

Solution

$n(\ce{S2O3^2-}) = \dfrac{0.100 \times 22.40}{1000} = 2.240 \times 10^{-3}\ \text{mol}$

$n(\ce{I2}) = \dfrac{2.240 \times 10^{-3}}{2} = 1.120 \times 10^{-3}\ \text{mol}$

$n(\ce{ClO-})$ in $25.0\ \text{cm}^3 = 1.120 \times 10^{-3}\ \text{mol}$ (1 : 1 with $\ce{I2}$ )

In $250\ \text{cm}^3$ (which contains all of the $10.0\ \text{cm}^3$ of bleach): $1.120 \times 10^{-2}\ \text{mol}$

c(\ce{ClO-}) = \frac{1.120 \times 10^{-2}}{0.0100} = 1.12\ \text{mol dm}^{-3}

Mass concentration of NaClO $= 1.12 \times 74.5 = 83.4\ \text{g dm}^{-3}$ .

Overall ratio: 1 $\ce{ClO-}$ $\equiv$ 1 $\ce{I2}$ $\equiv$ 2 $\ce{S2O3^2-}$ .

Watch out

Cold alkali gives chlorate(I), $\ce{ClO-}$ ; hot alkali gives chlorate(V), $\ce{ClO3-}$ . Swapping them is the most common error.
In the hot reaction the ratio of $\ce{Cl-}$ to $\ce{ClO3-}$ is 5 : 1, not 1 : 1.
Chlorine with water is reversible: use $\ce{<=>}$ . It forms HCl and HOCl, not $\ce{HClO3}$ .
"Chlorine kills bacteria" is not enough when the question asks for the active species: name $\ce{HOCl}$ and $\ce{ClO-}$ .
Disproportionation needs one element both oxidised and reduced. A reaction where chlorine is reduced and a different element is oxidised (for example $\ce{Cl2 + 2Br-}$ ) is not disproportionation.

Exam tip

"Describe and interpret in terms of changes in oxidation number" means: give the equation, state the oxidation numbers of chlorine in every species, say which is oxidised and which is reduced, and name the reaction disproportionation.
Learn the names with Roman numerals: sodium chlorate(I), sodium chlorate(V), chloric(I) acid. Both "chloric(I) acid" and "hypochlorous acid" are accepted.
State the conditions: "cold, dilute $\ce{NaOH(aq)}$ " and "hot, concentrated $\ce{NaOH(aq)}$ ".
Water purification questions: equation, active species $\ce{HOCl}$ and $\ce{ClO-}$ , and the fact that they kill bacteria (micro-organisms). A sensible risk earns a further mark.

Practical skills

The concentration of bleach is found by an iodine–thiosulfate titration (see Titrations). Bleach is diluted accurately (pipette into a volumetric flask) because commercial bleach is too concentrated to titrate directly. Excess potassium iodide and acid are added; the brown iodine is titrated with sodium thiosulfate until pale yellow, starch is added (blue-black), and the titration continues until the blue-black colour just disappears. Adding starch too early can trap iodine in the starch complex and give an inaccurate end-point. Bleach should be handled with gloves and eye protection, and must never be mixed with acid outside the titration flask, since this releases chlorine.

Summary

Disproportionation: the same element is simultaneously oxidised and reduced.
Cold $\ce{NaOH}$ : $\ce{Cl2 + 2OH- -> Cl- + ClO- + H2O}$ (Cl: $0 \to -1$ and $+1$ ); the product is bleach.
Hot $\ce{NaOH}$ : $\ce{3Cl2 + 6OH- -> 5Cl- + ClO3- + 3H2O}$ (Cl: $0 \to -1$ and $+5$ , ratio 5 : 1).
Water: $\ce{Cl2 + H2O <=> HCl + HOCl}$ , then $\ce{HOCl <=> H+ + ClO-}$ .
$\ce{HOCl}$ and $\ce{ClO-}$ are the active species that kill bacteria in drinking water.
Benefits (safe drinking water) far outweigh the risks (toxic gas, chlorinated organic by-products).

Practice

Question

Give the oxidation number of chlorine in $\ce{HOCl}$ , $\ce{NaClO3}$ , $\ce{Cl2}$ and $\ce{MgCl2}$ .
Define disproportionation.
Write the ionic equation for the reaction of chlorine with cold aqueous sodium hydroxide, and name the products.
Explain why the reaction $\ce{Cl2 + 2KI -> 2KCl + I2}$ is a redox reaction but not a disproportionation reaction.
Write the equation for the reaction of chlorine with water and name the two chlorine-containing products.
Explain why universal indicator added to chlorine water first turns red and then becomes colourless.
$480\ \text{cm}^3$ of chlorine at room conditions is passed into excess hot concentrated sodium hydroxide. Calculate the mass of sodium chlorate(V) formed. ( $A_r$ : Na 23.0, Cl 35.5, O 16.0)
A town's water supply treats $2.0 \times 10^{6}\ \text{dm}^3$ of water per day with chlorine at a dose of $1.0\ \text{mg dm}^{-3}$ . Calculate the mass of chlorine used per day in kg.
When bleach (sodium chlorate(I) solution) is warmed, the chlorate(I) ions disproportionate into chloride and chlorate(V) ions. Write the ionic equation and use oxidation numbers to explain the ratio of products.
$5.00\ \text{cm}^3$ of a bleach is diluted to $100\ \text{cm}^3$ . A $20.0\ \text{cm}^3$ portion is added to excess acidified KI, and the iodine released needs $18.60\ \text{cm}^3$ of $0.0500\ \text{mol dm}^{-3}$ sodium thiosulfate. Calculate the concentration of chlorate(I) ions in the bleach. Equations as in the worked example.

Answers

$\ce{HOCl}$ $+1$ ; $\ce{NaClO3}$ $+5$ ; $\ce{Cl2}$ 0; $\ce{MgCl2}$ $-1$ .
A redox reaction in which the same element is simultaneously oxidised and reduced.
$\ce{Cl2(aq) + 2OH-(aq) -> Cl-(aq) + ClO-(aq) + H2O(l)}$ : chloride ions and chlorate(I) ions (with water).
Chlorine is reduced (0 to $-1$ ) and iodine is oxidised ( $-1$ in $\ce{I-}$ to 0 in $\ce{I2}$ ), so it is redox. But the element oxidised (iodine) is different from the element reduced (chlorine), so it is not disproportionation.
$\ce{Cl2(aq) + H2O(l) <=> HCl(aq) + HOCl(aq)}$ : hydrochloric acid and chloric(I) acid.
Both HCl and HOCl are acids, so the indicator first shows an acidic colour (red). Chloric(I) acid is a strong oxidising agent that then bleaches (oxidises) the indicator dye, so the colour disappears.
$n(\ce{Cl2}) = 0.480 / 24.0 = 0.0200\ \text{mol}$ . 3 mol $\ce{Cl2}$ give 1 mol $\ce{NaClO3}$ : $n(\ce{NaClO3}) = 0.00667\ \text{mol}$ . $M(\ce{NaClO3}) = 106.5$ ; mass $= 0.710\ \text{g}$ .
Mass $= 2.0 \times 10^{6} \times 1.0\ \text{mg} = 2.0 \times 10^{6}\ \text{mg} = 2.0 \times 10^{3}\ \text{g} = 2.0\ \text{kg}$ .
$\ce{3ClO-(aq) -> 2Cl-(aq) + ClO3-(aq)}$ . Chlorine in $\ce{ClO-}$ is $+1$ . Going to $\ce{Cl-}$ is a decrease of 2 per atom; going to $\ce{ClO3-}$ is an increase of 4. To balance, two chlorine atoms are reduced ( $2 \times 2 = 4$ ) for each one oxidised (4), giving $\ce{Cl-}$ : $\ce{ClO3-}$ = 2 : 1. Check: Cl 3 = 3; O 3 = 3; charge $-3 = -2 - 1$ .
$n(\ce{S2O3^2-}) = 0.0500 \times 18.60 / 1000 = 9.300 \times 10^{-4}\ \text{mol}$ . $n(\ce{I2}) = 4.650 \times 10^{-4}\ \text{mol} = n(\ce{ClO-})$ in $20.0\ \text{cm}^3$ . In $100\ \text{cm}^3$ : $2.325 \times 10^{-3}\ \text{mol}$ , which came from $5.00\ \text{cm}^3$ of bleach. $c = 2.325 \times 10^{-3} / 0.00500 = 0.465\ \text{mol dm}^{-3}$ .

Oxidation numbers of chlorine

Disproportionation

Chlorine with cold aqueous sodium hydroxide

Chlorine with hot aqueous sodium hydroxide

Chlorine in water purification

Chlorine and water

How chlorination is used

Risks

Chlorine water with indicator

Worked examples

Practice

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