What is the difference between the products of chlorine reacting with cold vs. hot sodium hydroxide?

In cold $NaOH$ ($15^{\circ}C$), chlorine forms sodium chlorate(I) ($NaClO$), where Cl is in the $+1$ state. In hot $NaOH$ ($70^{\circ}C$), it forms sodium chlorate(V) ($NaClO_3$), where Cl is in the $+5$ state.

How does the reaction of chlorine with water differ from its reaction with sodium hydroxide?

The reaction with water is a reversible equilibrium producing $HCl$ and $HClO$, whereas the reaction with $NaOH$ is a full neutralization that goes to completion, forming salts ($NaCl$ and $NaClO/NaClO_3$) and water.

When comparing chlorate(I) and chlorate(V) ions, which is more stable at high temperatures?

Chlorate(V) ($ClO_3^-$) is more stable at high temperatures. Chlorate(I) ($ClO^-$) tends to disproportionate into chloride and chlorate(V) when heated.

What is a common mistake when assigning oxidation states to the chlorine in $NaClO_3$?

Students often assume chlorine is $+1$ because it is a halogen, but in $NaClO_3$, the three oxygens contribute $-6$ and sodium contributes $+1$, requiring chlorine to be $+5$ to make the compound neutral.

Why is it incorrect to say that chlorine gas ($Cl_2$) kills bacteria in water directly?

Chlorine gas reacts with water to form chloric(I) acid ($HClO$). It is the $HClO$ and the dissociated $ClO^-$ ions that actually act as the sterilizing agents by oxidizing biological molecules in bacteria.

What error occurs if you fail to balance the electrons in the hot alkali disproportionation equation?

If unbalanced, the ratio of $Cl^-$ to $ClO_3^-$ will be wrong. One chlorine atom loses $5$ electrons to become $Cl^{+5}$, so $5$ other chlorine atoms must each gain $1$ electron to become $Cl^-$, resulting in a $5:1$ product ratio.

Define 'Disproportionation' in the context of chlorine chemistry.

Disproportionation is a redox reaction where chlorine atoms from the same $Cl_2$ molecules are simultaneously oxidized to a positive state (like $+1$ or $+5$) and reduced to the $-1$ state.

What is the chemical formula and oxidation state of the chlorine in 'Chloric(I) acid'?

The formula is $HClO$, and the oxidation state of chlorine is $+1$. It is formed when chlorine reacts with water or cold alkalis.

State the ionic equation for the reaction of chlorine with hot aqueous hydroxide ions.

$$3Cl_2 + 6OH^- \rightarrow 5Cl^- + ClO_3^- + 3H_2O$$ This shows the $5:1$ ratio of chloride to chlorate(V) ions.

Why does the reaction of chlorine with $NaOH$ require more hydroxide ions at higher temperatures?

The formation of chlorate(V) involves a greater change in oxidation state ($0$ to $+5$), which requires more $OH^-$ ions to balance the stoichiometry and the resulting water molecules produced.

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Chemistry

11. Group 17

The Reactions of Chlorine

Summary

Chlorine is a highly reactive halogen that primarily undergoes disproportionation reactions, where it is simultaneously oxidized and reduced. Its behavior is significantly influenced by temperature and the medium of reaction, leading to the formation of various oxoanions used in sterilization and industrial bleaching.

1. Core Concept: Disproportionation

Disproportionation is a specific type of redox reaction where a single element in a single oxidation state is simultaneously oxidized and reduced to form two different products.

In the case of chlorine ( $Cl_2$ ), the initial oxidation state is $0$ . During reaction, some atoms lose electrons to reach positive oxidation states (like $+1$ or $+5$ ), while others gain electrons to reach the $-1$ state.

This dual behavior allows chlorine to act as both an oxidizing agent and a reducing agent within the same chemical environment.

2. Reaction with Cold, Dilute Alkali

Diagram showing the two different disproportionation paths of chlorine in NaOH based on temperature.

3. Reaction with Hot, Concentrated Alkali

4. Reaction with Water & Purification

5. Key Distinctions in Chlorine Oxoanions

6. Exam Strategy & Tips

The Reactions of Chlorine

Summary

1. Core Concept: Disproportionation

Disproportionation is a specific type of redox reaction where a single element in a single oxidation state is simultaneously oxidized and reduced to form two different products.

This dual behavior allows chlorine to act as both an oxidizing agent and a reducing agent within the same chemical environment.

2. Reaction with Cold, Dilute Alkali

When chlorine gas is reacted with cold, dilute sodium hydroxide (approximately $15^{\circ}C$ ), it forms a mixture of sodium chloride and sodium chlorate(I).

The chemical equation for this process is: $Cl_2(aq) + 2NaOH(aq) \rightarrow NaCl(aq) + NaClO(aq) + H_2O(l)$ or ionically: $Cl_2(aq) + 2OH^-(aq) \rightarrow Cl^-(aq) + ClO^-(aq) + H_2O(l)$

In this reaction, chlorine is reduced to the $-1$ state in $Cl^-$ and oxidized to the $+1$ state in the chlorate(I) ion ( $ClO^-$ ). This mixture is the primary component of household bleach.

Diagram showing the two different disproportionation paths of chlorine in NaOH based on temperature.

3. Reaction with Hot, Concentrated Alkali

If the temperature of the sodium hydroxide is increased (to approximately $70^{\circ}C$ ), the chlorine undergoes a more extensive oxidation process.

The resulting products are sodium chloride and sodium chlorate(V) ( $NaClO_3$ ), where the chlorine reaches a higher oxidation state of $+5$ .

The balanced equation is: $3Cl_2(aq) + 6NaOH(aq) \rightarrow 5NaCl(aq) + NaClO_3(aq) + 3H_2O(l)$ This demonstrates that higher thermal energy facilitates the loss of more electrons from the chlorine atoms.

4. Reaction with Water & Purification

Chlorine reacts reversibly with water to produce a mixture of hydrochloric acid ( $HCl$ ) and chloric(I) acid ( $HClO$ ).

The reaction is: $Cl_2(aq) + H_2O(l) \rightleftharpoons HCl(aq) + HClO(aq)$ This is also a disproportionation reaction ( $0 \rightarrow -1$ and $+1$ ).

Chloric(I) acid is the active species responsible for water sterilization; it penetrates bacterial cell walls and disrupts vital enzymes through oxidation.

5. Key Distinctions in Chlorine Oxoanions

It is vital to distinguish between the different chlorate ions formed under varying conditions to predict chemical behavior and oxidation states.

Ion Name	Formula	Oxidation State of Cl	Formation Condition
Chloride	$Cl^-$	$-1$	All aqueous reactions
Chlorate(I)	$ClO^-$	$+1$	Cold alkali / Water
Chlorate(V)	$ClO_3^-$	$+5$	Hot alkali

Note that Chlorate(I) is unstable at higher temperatures and will itself disproportionate into Chloride and Chlorate(V) if heated.

6. Exam Strategy & Tips

Check the Temperature: Always look for 'cold' vs 'hot' or specific temperatures like $15^{\circ}C$ vs $70^{\circ}C$ to determine if the product is $ClO^-$ or $ClO_3^-$ .
Balance by Oxidation State: In the hot alkali reaction, remember the $1:5$ ratio of oxidized to reduced chlorine atoms ( $1$ atom goes to $+5$ , while $5$ atoms go to $-1$ ).
Identify the Agent: In water treatment questions, specify that $HClO$ (chloric(I) acid) or the $ClO^-$ ion is the actual sterilizing agent, not the $Cl_2$ molecule itself.
Oxidation State Sums: Always verify that the sum of oxidation states in an ion equals its charge (e.g., in $ClO_3^-$ , $+5 + 3(-2) = -1$ ).