The halide ions as reducing agents

Reducing ability of the halide ions

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:

2X- X2 + 2e
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.

Trends in the reducing power of the halide ions

the reducing power of the halides increases as you descend group 7 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:

sodium chloride(s) + sulfuric acid(aq) sodium hydrogensulfate(s) + hydrogen chloride(g)
NaCl(s) + H2SO4(aq) NaHSO4(s) + HCl(g)
If you use sodium fluoride instead of sodium chloride then an almost identical reaction takes place except that hydrogen fluoride gas is produced instead of hydrogen chloride gas:

child writing lines on the board What type of reaction is occurring here? 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 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 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 swap the halide ion, that is we sodium bromide. The equation below is similar to the one using fluoride and chloride but then further reactions take place since the bromide ion is a much better reducing agent than either the chloride or fluoride ions:

 reaction of sodium bromide and conc sulfuric acid 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: concentrated sulfuric acid and hydrogen bromide reaction 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:

reduction reaction: SO42-(aq) + 4H+(aq) + 2e → SO2(g) + H2O(l)
2Br-(aq) Br2(l) + 2e
This additional reaction with bromide occurs simply because the bromide ion is a much better reducing agent than either the chloride or fluoride ions. What happens 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 addconcentrated 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 oxised since their oxidation numbers remain unchanged.

reaction of sodium iodide with concentrated sulfuric acid 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. equation for the oxidation of hydrogen iodide by concentrated sulfuric acid

The above equation was as far as we went with bromide ions, however the iodide ions are able to further reduce the sulfuric acidto form solid sulfur. 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, an equation for this reaction is shown below. 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: equation to show the reduction of concentrated sulfuric acid by hydrogen iodide

Howver 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:

reduction of sulfuric acid using HI to form hydrogen sulfide gas

We can write 2 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 equation to get the required number of electrons needed for the reduction reaction taking place.

reduction reaction to produce sulfur: SO42-(aq) + 4H+(aq) + 2e → SO(g) + 2H2O(l)
reduction reaction : SO42-(aq) + 8H+(aq) + 6e → S(s) + 4H2O(l)
reduction reaction: SO42-(aq) + 10H+(aq) + 8e → H2S(g) + 4H2O(l)
oxidation reaction: 2Br-(aq) Br2(l) + 2e

Testing for the products of reactions

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)
  • sulfur dioxide gas has an eggy smell.
  • it will turn damp blue litmus paper red.
  • sulfur dioxide gas is a reducing gas, it will turn filter paper soaked in acidified potassium dichromate solution from orange to green.
Will turn strips of filter paper soaked in ornage acidified potassium dichromate solution from orange to green. This happens because the SO2 gas will reduce the orange Cr6+ion to the form the green Cr6+ ion.
hydrogen chloride gas (HCl)
  • will form misty fumes in mosit air.
  • If hydrogen chloride gas comes into contact with conc ammonia dense white fumes of ammonium chloride are formed.
  • Hydrogen chloride is an acidic gas, it will turn moist blue litmus paper red.
These test also work for hydrogen bromide gas, though with concentrated ammonia solution dense fumes of ammonium bromide form.

Key Points

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!

Practice questions

Check your understanding - Questions halides as reducing agents.