This page describes reactions of the halogens that do not fall under the other categories in other pages in this section. All the reactions described here are redox reactions.
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Reactions with iron
With the exception of iodine, iron burns in halogen vapor, forming iron(III) halides. Iodine is less reactive, and produces iron(II) iodide.
Reactions with solutions containing iron(II) ions
Only the reactions of chlorine, bromine, and iodine can be considered. Aqueous fluorine is very reactive with water. Chlorine and bromine are strong enough oxidizing agents to oxidize iron(II) ions to iron(III) ions. In the process, chlorine is reduced to chloride ions, bromine to bromide ions.
This process is easiest to visualize with ionic equations:
For the bromine equation, Br is substituted for Cl.
The pale green solution containing the iron(II) ions turns into a yellow or orange solution containing iron(III) ions. Iodine is not a strong enough oxidizing agent to oxidize iron(II) ions, so there is no reaction. In fact, the reverse reaction proceeds. Iron(III) ions are strong enough oxidizing agents to oxidize iodide ions to iodine as shown:
\< 2Fe_3+ + 2I^- \rightarrow 2Fe^{2+} + I_2\>
The reaction of chlorine with cold sodium hydroxide solution
Chlorine and cold, dilute sodium hydroxide react as follows:
\< 2NaOH + Cl_2 \rightarrow NaCl + NaClO + H_2O\>
NaClO (sometimes written as NaOCl) symbolizes sodium chlorate(I). The traditional name for this compound is sodium hypochlorite; the solution on the product side of the equation is commonly sold as bleach.
Consider this reaction in terms of oxidation states. Chlorine displays an obvious state change from its elemental form to ionic compounds. The oxidation numbers for each element are shown below:
Chlorine is the only element that changes oxidation state—it is both oxidized and reduced. One atom is reduced because its oxidation state has decreased; the other is oxidized. This is a good example of a disproportionation reaction, a reaction in which a single substance is both oxidized and reduced.
The reaction of chlorine with hot sodium hydroxide solution
Chlorine reacts with hot, concentrated sodium hydroxide as follows:
\< 6NaOH + 3Cl_2 \rightarrow 5NaCl + NaClO_3 + 3H_2O\>
The product formed is sodium chlorate(V) - NaClO3. As before, the oxidation states of each element are calculated. Once again, the only change is in chlorine, from 0 in the chlorine molecules on the reactant side to -1 (in the NaCl) and +5 (in the NaClO3). This is another example of a disproportionation reaction.
Balancing equations for these reactions
The first equation is simple to balance. The second one is more difficult; oxidation states are used to derive it.
The two main products of the reaction are NaCl and NaCIO3, so the reaction can be tentatively written as follows:
\< NaOH + Cl_2 \rightarrow NaCl + NaClO_3 + ? \>
In its conversion to NaCl, the oxidation state of the chlorine decreases from 0 to -1. When converted to NaClO3, it increases from 0 to +5. The positive and negative oxidation state changes must cancel out, so for every NaClO3 formed, there must be 5 NaCl:
\< NaOH + Cl_2 \rightarrow 5NaCl + NaClO_3 + ? \>
Now it is a simple task to balance the sodium and the chlorine atoms, after which there are enough hydrogen and oxygen atoms to make 3H2O.
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Bromine and sodium hydroxide solution
For bromine, the formation of the sodium bromate(V) happens at around room temperature. Sodium bromate(I) must be formed at about 0°C.
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