AS Chemistry 9701: Disproportionation of Chlorine with Sodium Hydroxide (Group 17)

Chlorine gas is bubbled into cold, dilute sodium hydroxide solution to make a solution used as household bleach. In a separate experiment, chlorine gas is bubbled into hot, concentrated sodium hydroxide solution.

(a) Write a balanced symbol equation, including state symbols, for the reaction between chlorine and cold, dilute sodium hydroxide. [2]

(b) State the oxidation number of chlorine in each of the three chlorine-containing species in your equation from (a), and use these values to explain why this reaction is described as disproportionation. [2]

(c) Write a balanced symbol equation for the reaction between chlorine and hot, concentrated sodium hydroxide, and state the oxidation number of chlorine in the new chlorine-containing product that is not formed under cold conditions. [3]

(d) A 250 cm $^3$ sample of the cold sodium hydroxide bleach solution made in (a) is found to have a sodium chlorate(I), $\text{NaOCl}$ , concentration of 0.160 mol dm $^{-3}$ . Calculate the minimum mass of chlorine gas, $\text{Cl}_2$ , that must have reacted to produce this concentration of $\text{NaOCl}$ , giving your answer to 3 significant figures. [3]

Part (a): Chlorine with cold, dilute sodium hydroxide

Chlorine reacts with cold, dilute aqueous sodium hydroxide to form a mixture of sodium chloride and sodium chlorate(I). The active ingredient of household bleach:

$\text{Cl}_2\text{(g)} + 2\text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{NaOCl(aq)} + \text{H}_2\text{O(l)}$

Two moles of $\text{NaOH}$ are needed per mole of $\text{Cl}_2$ : one sodium ion ends up paired with $\text{Cl}^-$ , the other with $\text{OCl}^-$ , and the two hydroxide oxygens/hydrogens combine to form water.

Part (b): Oxidation numbers and disproportionation

Working out the oxidation number of chlorine in each species:

In $\text{Cl}_2$ , an element in its standard state, chlorine has oxidation number $\boxed{0}$ .
In $\text{NaCl}$ , sodium is $+1$ and the compound is neutral overall, so chlorine is $\boxed{-1}$ .
In $\text{NaOCl}$ , sodium is $+1$ and oxygen is $-2$ ; for the compound to be neutral, chlorine must be $\boxed{+1}$ .

Chlorine therefore starts at oxidation number $0$ in $\text{Cl}_2$ and, in the very same reaction, some chlorine atoms are reduced to $-1$ (in $\text{Cl}^-$ ) while others are oxidised to $+1$ (in $\text{OCl}^-$ ). A reaction in which the same element is simultaneously oxidised and reduced is called a disproportionation reaction.

Part (c): Chlorine with hot, concentrated sodium hydroxide

Under hot, concentrated conditions, chlorine disproportionates further, forming sodium chlorate(V) instead of sodium chlorate(I):

$3\text{Cl}_2\text{(g)} + 6\text{NaOH(aq)} \rightarrow 5\text{NaCl(aq)} + \text{NaClO}_3\text{(aq)} + 3\text{H}_2\text{O(l)}$

The new chlorine-containing product not seen under cold conditions is $\text{NaClO}_3$ (sodium chlorate(V)). Its oxidation number can be checked: sodium is $+1$ , each of the three oxygens is $-2$ (total $-6$ ), so chlorine must be $\boxed{+5}$ for the compound to be neutral overall.

Part (d): Mass of chlorine reacted

First find the moles of $\text{NaOCl}$ present in the 250 cm $^3$ sample, converting the volume to dm $^3$ ( $250\ \text{cm}^3 = 0.250\ \text{dm}^3$ ):

$n(\text{NaOCl}) = 0.160\ \text{mol dm}^{-3} \times 0.250\ \text{dm}^3 = 0.0400\ \text{mol}$

From the equation in part (a), one mole of $\text{Cl}_2$ produces one mole of $\text{NaOCl}$ (a 1:1 ratio), so:

$n(\text{Cl}_2) = 0.0400\ \text{mol}$

Using $M_r(\text{Cl}_2) = 2 \times 35.5 = 71.0$ :

$\text{mass} = 0.0400 \times 71.0 = 2.84\ \text{g}$

So the minimum mass of chlorine that must have reacted is $\boxed{2.84}$ g (3 s.f.).

Final answers

(a) $\text{Cl}_2\text{(g)} + 2\text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{NaOCl(aq)} + \text{H}_2\text{O(l)}$
(b) $\text{Cl}_2 = 0$ , $\text{Cl}^-$ (in NaCl) $= -1$ , $\text{Cl}$ (in NaOCl) $= +1$ ; chlorine is simultaneously oxidised and reduced, so this is disproportionation.
(c) $3\text{Cl}_2\text{(g)} + 6\text{NaOH(aq)} \rightarrow 5\text{NaCl(aq)} + \text{NaClO}_3\text{(aq)} + 3\text{H}_2\text{O(l)}$ ; chlorine in $\text{NaClO}_3$ has oxidation number $+5$ .
(d) Mass of $\text{Cl}_2$ = 2.84 g