Reducing agents are electron donors. To act as a reducing agent the halide ions (F-, Cl-, Br-, I- ) need to lose electrons and form halogen atoms. We can represent this oxidation reaction as:
Where X represents any of the halide ions. You should recall that all the halogens consist of diatomic molecules which is why the above equation is multiplied by x2.
The reducing power of the halide ions increases as we descend group 7.
This trend is easily explained; as we descend group 7
the halide ions increase in size and the outer electrons
are further from the nucleus. The outer electrons also experience
more shielding by the inner
electrons so it requires less energy
to remove an outer electron from a large halide anion than a smaller one.
One of the reactions which is often used to demonstrate the reducing power
of the halide ions is the reaction of concentrated
sulfuric acid (a reasonable oxidising agent) with sodium or potassium halides e.g. sodium chloride reacts with
conc sulfuric acid
according to the equation below:
What type of reaction is occurring in each of these two reaction? To try and help us decide we can look at the oxidation numbers
for the elements involved in the reaction.
The equation above shows the oxidation numbers for most of the reacting elements. You will see that
none of the elements changes its oxidation number during the reaction so we can at least say that this is NOT
a redox reaction. Nothing has been reduced or
oxidised since none of the oxidation numbers have changed.
The sodium ions are spectator ions, so from the point of view of the fluoride ion it has lost a sodium atom and gained a
hydrogen ion during the reaction;
that is the F- ion is acting as a proton (H+) acceptor or a base.
So this is an acid base reaction with the concentrated sulfuric acid acting as a
H+ donor; that is an acid and
the fluoride ion is acting as a base.
Concentrated sulfuric acid is not able to oxidise
the fluoride or the chloride ion in any of these reactions. The 2 reactions above
are acid-base reactions. However what
happens if change the halide ion; that is we sodium bromide. The equation below represents the initial reaction which takes place between sodium bromide and concentrated sulfuric acid and it
is very similar to the reactions using sodium fluoride and sodium chloride, however further reactions are able to take place since the bromide ion is a
much better reducing agent than either the chloride
or fluoride ions:
This reaction is exactly the same as the one with NaF and NaCl, however the conc sulfuric acid
is able to oxidise the
hydrogen bromide gas produced to give bromine, sulfur dioxide gas and
water, an equation for the reaction is shown below:
This time there has been a redox reaction. The oxidation state
of the sulfur in sulfuric acid has gone from +6 to +4 in
the sulfur dioxide gas produced; that is the sulfuric acid has gained 2 electrons
and been reduced. The electrons needed
for this reduction have come
from the HBr gas. One of the products of the reaction is elemental bromine;
which being an element has an oxidation number of 0. This
bromine has come from the hydrogen bromide gas which has been oxidised. Ion-electron half equations for these 2 reactions are shown below:
This additional reaction with bromide occurs simply because the bromide ion is a much better reducing agent
than either the chloride or fluoride ions. So what would happen if
we use a salt containing iodide (I-) ions and react it with concentrated sulfuric acid?
If we place sodium or potassium iodide in a boiling tube and add concentrated sulfuric acid then this time its chaos! Iodide is the best reducing agent in group 7 and a mixture of products is obtained including:
The reaction of sodium iodide with concentrated sulfuric acid starts off in exactly the same way as the other halide ions with the production of hydrogen iodide gas and solid sodium hydrogen carbonate. This is shown below, as you can see it is simply an acid-base reaction with the iodide ion (I-) acting as a base and accepting a hydrogen ion to form hydrogen iodide gas (HI(g)). If you study the equation you will see that none of the reactants or products is reduced or oxidised since their oxidation numbers remain unchanged.
However as before the HI(g) which is produced in this reaction can be oxidised
by the concentrated sulfuric acid to
produce iodine and the smelly gas sulfur dioxide. The equation below gives the
oxidation numbers to help you visualise what
has been reduced and what has been oxidised.
You should note that there are 2 moles of HI on the reactants side of the
equation and that the oxidation number of the sulfur has been reduced by 2.
The above equation was as far as we went with bromide ions, however the iodide ions are able to further
reduce the sulfuric acid to
form solid sulfur; as shown in the equation below. Here the oxidation number of the sulfur atom in
sulfuric acid has gone from +6 to 0, in the solid elemental sulfur. The 6
electrons needed
for this reduction are all provided by the hydrogen iodide. This time
there are 6 moles of HI on the reactants side of the
equation and that the oxidation number of the sulfur has been reduced by 6,
these six moles of hydrogen iodide provide the 6 electrons needed:
However this is not the end of the story! The hydrogen iodide is able to reduce the sulfuric acid to form hydrogen sulfide gas. Here the oxidation number of the sulfur has gone from +6 to -2 in the hydrogen sulfide gas. An equation for this redox reaction is shown below:
We can write ion-electron half equations for the production of sulfur and hydrogen sulfide gas. The equations below have all the spectator ions removed to make it simplier to see exactly what is happening here. The electrons needed for these reactions come from the oxidation of the iodide ions (I-) to form iodine, as shown in the last equation below. All that is needed is to multiply this final equation to get the required number of electrons needed for the reduction reaction taking place.
For these reactions of the hydrogen halides with concentrated sulfuric acid there is obviously a range of different products produced, the table below summaries a range of tests to identify these individual substances:
| product | Possible test | Observations |
|---|---|---|
| hydrogen sulfide (H2S) | Strips of filter paper are soaked in a saturated solution of lead ethanoate and allowed to dry. Dry lead ethanoate paper turns black in the presence of hydrogen sulfide gas. | The lead ions (Pb2+) react present in the lead ethanoate solution react with hydrogen sulfide gas to form the black solid lead sulfide (PbS) which causes the paper to blacken. |
| sulfur dioxide gas (SO2) |
|
Will turn strips of filter paper soaked in orange acidified potassium dichromate solution from orange to green. This happens because the SO2 gas will reduce the orange Cr6+ion in the dichromate to the form the green Cr3+ ion. |
| hydrogen chloride gas (HCl) |
|
These test also work for hydrogen bromide gas, though with concentrated ammonia solution dense fumes of ammonium bromide form. |
| Halide | Products of reaction with concentrated sulfuric acid. | Observations |
|---|---|---|
| sodium fluoride | hydrogen fluoride gas (HF) | misty white fumes in moist air |
| sodium chloride | hydrogen chloride gas (HCl) | misty white fumes in moist air |
| sodium bromide | hydrogen bromide gas (HBr) bromine gas (Br2) sulfur dioxide gas (SO2) |
dense white fumes in moist air. brown fumes of bromine vapour. colourless eggy smelling gas produced. |
| sodium iodide | hydrogen iodide gas (HI) iodine gas (I2) sulfur dioxide gas (SO2) solid sulfur (S) hydrogen sulfide gas (H2S) |
dense white fumes in moist air. violet fumes of iodine vapour. colourless eggy smelling gas produced. side walls of test-tube covered in yellow sulfur colourless vile smelling gas released, smells of really bad eggs! |