Unit 12: Chemical Properties of Halogens & Halides I
Understanding the redox chemistry of Group 17 elements and their ions.
12.4 Halogens as Oxidising Agents
Halogen atoms have seven valence electrons. They have a strong tendency to gain one electron to achieve a stable octet, forming a halide ion ($X^-$). A substance that gains electrons is being reduced, and in the process, it causes another substance to be oxidized. Therefore, the halogens are strong oxidising agents.
The general reduction half-equation for a halogen is: $$ X_2 + 2e^- \rightarrow 2X^- $$
The oxidising power of the halogens decreases down the group from fluorine to iodine.
Fluorine > Chlorine > Bromine > Iodine
This is because as you go down the group, the atomic radius increases and there is more electron shielding from the inner shells. The nucleus has a weaker attraction for an incoming electron, making the atom less effective at gaining one. Fluorine is the most electronegative element and the strongest elemental oxidising agent.
Solved Examples:
-
Which is a stronger oxidising agent, chlorine or bromine? Explain
why.
Solution: Chlorine is a stronger oxidising agent. A chlorine atom is smaller than a bromine atom and has less electron shielding. This means its nucleus can exert a stronger pull on an incoming electron, making it more readily reduced than bromine. -
Write the half-equation for the reduction of an iodine
molecule.
Solution: $I_2(aq) + 2e^- \rightarrow 2I^-(aq)$
12.5 Halide Ions as Reducing Agents
Conversely, halide ions ($F^-, Cl^-, Br^-, I^-$) can lose an electron to be converted back into a halogen atom. A substance that loses electrons is being oxidized, and in doing so, it causes another substance to be reduced. Therefore, the halide ions can act as reducing agents.
The general oxidation half-equation for a halide ion is: $$ 2X^- \rightarrow X_2 + 2e^- $$
The reducing power of the halide ions increases down the group from fluoride to iodide.
Iodide > Bromide > Chloride > Fluoride
This is because as you go down the group, the outermost electron is in a higher energy level, is further from the nucleus, and is more shielded. It is held less tightly and is therefore easier to lose. The iodide ion is the largest and is the strongest reducing agent among the halides.
Solved Examples:
-
Which is a stronger reducing agent, a bromide ion or a chloride ion? Explain
why.
Solution: A bromide ion ($Br^-$) is a stronger reducing agent. The outermost electron in a bromide ion is further from the nucleus and experiences more shielding than in a chloride ion. Therefore, it is more easily lost (oxidized). -
Write the half-equation for the oxidation of fluoride ions. Why is this
reaction very difficult to achieve?
Solution: $2F^-(aq) \rightarrow F_2(g) + 2e^-$. This reaction is extremely difficult because the fluoride ion is very small and the electron is held very strongly by the nucleus. Fluoride is a very poor reducing agent.
12.6 Halogen Displacement Reactions
The trends in redox power can be clearly demonstrated by halogen displacement reactions. In these reactions, a more reactive halogen (a stronger oxidising agent) will displace a less reactive halide ion (from a stronger reducing agent) from its aqueous solution.
Rule: A halogen will displace any halide ion that is below it in the periodic table.
Examples:
- Chlorine + Bromide Ions: Chlorine is more reactive than bromine. It
will oxidize bromide ions to bromine, and the solution will turn from colorless to
orange/brown.
$$ Cl_2(aq) + 2Br^-(aq) \rightarrow 2Cl^-(aq) + Br_2(aq) $$ - Chlorine + Iodide Ions: Chlorine is more reactive than iodine. It will
oxidize iodide ions to iodine, and the solution will turn brown (or purple if a
non-polar solvent like hexane is added).
$$ Cl_2(aq) + 2I^-(aq) \rightarrow 2Cl^-(aq) + I_2(aq) $$ - Bromine + Iodide Ions: Bromine is more reactive than iodine. It will
oxidize iodide ions to iodine, turning the solution brown.
$$ Br_2(aq) + 2I^-(aq) \rightarrow 2Br^-(aq) + I_2(aq) $$ - Bromine + Chloride Ions: No reaction occurs because bromine is less reactive than chlorine and cannot displace it.
Solved Examples:
-
What would you observe if you bubbled chlorine gas through a solution of
potassium iodide? Write the ionic equation.
Solution: You would observe the colorless solution turning a dark brown color. This is because the more reactive chlorine oxidizes the iodide ions to form iodine.
Ionic Equation: $Cl_2(aq) + 2I^-(aq) \rightarrow 2Cl^-(aq) + I_2(aq)$ -
Will iodine displace bromide ions from a solution of sodium bromide?
Explain.
Solution: No. Iodine is below bromine in Group 17, making it a weaker oxidising agent. It is not reactive enough to take electrons from the bromide ions. -
In the reaction between bromine water and potassium iodide, what is oxidized
and what is reduced?
Solution: The iodide ions ($I^-$) lose electrons to form iodine ($I_2$), so the iodide ions are oxidized. The bromine molecules ($Br_2$) gain electrons to form bromide ions ($Br^-$), so the bromine is reduced. - A student mixes chlorine water with sodium chloride solution. What is
observed?
Solution: Nothing is observed. A halogen cannot displace itself. - How could you confirm that the brown substance formed when chlorine reacts
with potassium iodide is iodine?
Solution: Add a few drops of starch indicator. If iodine is present, a deep blue-black color will form. - Arrange the halide ions ($Cl^-, Br^-, I^-$) in order of increasing strength
as reducing agents.
Solution: The reducing strength increases down the group. The order is $Cl^- < Br^- < I^-$.