Common oxidising agents
An oxidising agent is an electron acceptor,
it will oxidise a substance when it reacts and
it will be reduced by gaining electron.
Some common oxidising agents you are likely to meet are listed
below with half equations to show the reduction of these oxidising agents:
| Oxidising agent |
Half-equation |
| chlorine |
Cl2(g) + 2e → 2Cl-(aq) |
| bromine |
Br2(g) + 2e → 2Br-(aq) |
| iodine |
I2(g) + 2e → 2I-(aq) |
| oxygen |
O2(g) + 4e → 2O2-(aq) |
There are other common oxidising agents such as acidified potassium permanganate,
acidified potassium dichromate,
hydrogen peroxide and concentrated sulfuric acid.
The half equations for these reduction reactions are a little more
complicated but easy to work out if you follow some simple rules; as set out below:
Example 1 - potassium dichromate
Acidified potassium dichromate is a very common oxidising agent used
in chemistry. Potassium dichromate is a bright orange solid which
being a salt of a group I metal is soluble in water; it dissolves to form a bright orange solution as shown in the image below.
The formula for potassium dichromate is K2Cr2O72-.
The chromium ion in the dichromate ion has an oxidation state of +6; the highest possible for a chromium ion, this makes
it an excellent oxidising agent. The ability of the dichromate ion to act as an
oxidising agent requires the
presence of a dilute acid, sulfuric acid is a suitable acid to use here since the dichromate ion is not a powerful enough oxidising agent to oxidise it, as it would for example if you used hydrochloric acid instead.
When the dichromate ion oxidises a substance the orange dichromate ion
is reduced to form the green Cr3+ ion as shown in the image below:
The equation for reduction of the dichromate ion could be written as:
reduction reaction: Cr2O72- + 6e → 2Cr3+
The oxidation state of the chromium in dichromate is +6 and it
gains 3 electrons to form the Cr3+
ion when it is reduced, now the Cr3+ obviously has an oxidation state of +3. However the reduction equation above is not balanced in terms of
the atoms present or the charges present on both sides of the equation. There are a number of simple steps we can take
to balance this equation:
- Step1 - add water to balance off the number of oxygen atoms present.
In this case there are 7 oxygen atoms
present in the dichromate ion but none on the product side of the above equation. So by adding
7H2O
molecules to the product side of the equation will balance off the oxygen atoms present on both sides
of the equation, this now gives:
reduction reaction: Cr2O72- + 6e → 2Cr3+ + 7H2O
- Step2 - However by adding 7H2O molecules to the product side of the equation we may have balanced
off the
oxygen atoms but there are now 14 hydrogen atoms on the product side of the equation and none on the
reactant side of the equation. So to balance off the 14 hydrogen atoms add
14H+ ions to the reactants
side of the equation. These 14H+ ions would represent the addition of the sulfuric acid, as all acids are
solutions which contain an excess of hydrogen ions (H+ ions).
This is why the potassium dichromate solution has to be acidified with sulfuric acid,
once acidified the
solution is an excellent oxidising agent. This gives the half-equation below:
reduction reaction: Cr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O
The number of each type atom on both sides of the reduction equation balance. However you must also check that
the total charge on each side of the equation also balances. So we have:
- Total charges on the reactant side of the equation= (2-) + (14+) + (6-)= +6.
- Total charges on the product side of the equation= (+3 x2) = +6.
So the equation balances for charges present and also for the number of atoms present.
Example 2- Potassium permanganate
Potassium permanganate is a dark purple solid which when it dissolves in a
dilute sulfuric acid solution forms a
strong oxidising agent. The Permanganate ion has the formula MnO4-. When the
permanganate ion
is reduced it forms the almost colourless Mn2+(aq) ion. In the
permanganate ion (MnO4-)
the metal manganese has an oxidation state of +7, so when it is
reduced to form the Mn2+(aq)
it will gain 5 electrons.
The colour change when the permanganate ion is reduced
can be shown as:
An equation to show how the permanganate ion, MnO4- is
reduced to form the colourless ion Mn2+(aq) could be:
reduction reaction: MnO4-(aq) + 5e → Mn2+(aq)
However this equation is not balanced in terms of both the number of atoms of each type present on both sides of the equation or the total charge present on each side of the equation, so to balance it simply
follow the procedure we used above:
-
Step1 - balance the oxygen atoms on each side of the equation by adding
water to the oxygen deficient side. In this
case add 4 water molecules to the product side of the equation to give:
reduction reaction: MnO4-(aq) + 5e → Mn2+(aq) + 4H20(l)
- Step2 - balance the hydrogen atoms
now by adding H+(aq) to the hydrogen deficient side, in
this case to the reactant side of the equation. This gives:
reduction reaction: MnO4 (aq)- + 8H+(aq)+ 5e → Mn2+(aq) + 4H20(l)
This equation now balances in terms of atoms and also the charges on both sides of the equation.
A redox titration using potassium permanganate
As an example of using an acidified potassium permanganate solution as an oxidising agent in a redox reaction consider the oxidation of Fe2+(aq) ions to Fe3+(aq) ions
by
an acidified potassium permanganate solution.
Iron (II) sulfate is present in many lawn care and lawn conditioners because it kills
moss, which can spoil the look of a lawn. The concentration of Fe2+(aq) ions in a sample of lawn conditioner
can be determined by carrying out a titration using an acidified potassium permanganate solution. The method used is shown opposite and described below:
The permanganate solution is placed in a burette, a solution of 0.02M is often used; as shown in the
image opposite while a known volume, usually 25ml of the solution of the Fe2+(aq) solution is pipetted into a conical flask which contains 25ml of 1M sulfuric acid, as shown in the image.
When the purple permanganate ion is added slowly into to the Fe2+(aq) ions in the conical flask it oxidises the
Fe2+(aq) ions to form
Fe3+(aq) ions and the purple MnO4- ions are reduced to form the almost colourless Mn2+ ions. The permanganate solution will be added until the solution in the conical flask is colourless or a pale pink. As with any titration the titration should be repeated until concurrent results are obtained and an average titre can then be calculated.
We already have a balanced half-equation for the reduction of the purple acidified permanganate
ion (MnO4-) to the almost colourless Mn2+(aq) ion:
reduction reaction: MnO4 (aq)- + 8H+(aq)+ 5e → Mn2+(aq) + 4H20(l)
The oxidation reaction here is simply the oxidation of the pale green Fe2+(aq) being oxidised to the pale yellow
Fe3+(aq) ion.
oxidation reaction: Fe2+(aq) → Fe3+(aq) + e
However you will obviously note that the oxidation reaction only releases 1 mole of electrons while the reduction of the permanganate ion
requires 5 moles of electrons. So to get the overall equation the oxidation half-equation will need to be multiplied by x5.
oxidation reaction: 5Fe2+(aq) → 5Fe3+(aq) + 5e
The overall equation for this redox reaction is obtained by simply adding together the two half-equation for the oxidation and reduction reactions taking place and cancelling out the electrons; as shown below:
reduction reaction: MnO4-(aq) + 8H+(aq)+ 5e → Mn2+(aq) + 4H20(l)
oxidation reaction: 5Fe2+(aq) → 5Fe3+(aq) + 5e
overall reaction: MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H20(l)
Key Points
- Oxidising agents such as dichromate and permanganate need to be
acidified by the addition of sulfuric acid before they can act as oxidising agents.
- To balance half-equations involving these oxyanions simply add water to the oxygen deficient side of the equation.
This should balance out the number of atoms of oxygen on the product and reactant sides of the half-equation.
- Next add hydrogen ions (H+(aq)) to the hydrogen deficient side of the equation.
- Check that the charges balance on each side of the overall equation.
- Finally check that the number of electrons on each side of the
oxidation and reduction half-equations balance.
The practice questions contain additional examples using these oxyanions acting as oxidising agents; why not have a go at them to quickly check your understanding.
Practice questions
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