Common oxidising agents
An oxidising agent is an electron acceptor,
it will oxidise a substance when it reacts and
it will be redu 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:
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 reactions are a little more
complicated by easy to work out if you follow some simple rules.
||Cl2(g) + 2e → 2Cl-(aq)
|| Br2(g) + 2e → 2Br-(aq)
|| I2(g) + 2e → 2I-(aq)
|| O2(g) + 4e → 2O2-(aq)
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 dichromate 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.
When it oxidises a substance the orange dichromate ion
is reduced to form the green Cr3+ ion.
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 3e to form the Cr3+
ion, which obviously has an oxidation state of +3. However this equation is not balanced in terms of
the atoms present or the charges 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
molecules to the products side of the equation we will balance off the oxygen atoms present on both sides
of the equation:
reduction reaction: Cr2O72- + 6e → 2Cr3+ + 7H2O
- step2 - However by adding 7H2O molecules to the product side of the equation we may have balanced
oxygen atoms but there are now 14 hydrogen atoms on the products side of the equation and none on the
reactants side of the equation. So balance off the hydrogen atoms by adding
14H+ ions to the reactants
side of the equation. These 14H+ ions would represent the addition of an acid, as all acids are
solutions of H+ ions.
This is why the potassium dichromate solution has to be acidified with sulfuric acid,
once acidified the
solution is a good oxidising agent. This gives the half-equation below:
reduction reaction: Cr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O
The number of atoms 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:
So the equation balances for charges present and also for the number of atoms present.
- 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.
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 agen. Permanganate has the formula MnO4-. When the
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 needs to gain 5 electrons.
The colour change when the permanagante 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 as above this equation is not balanced in terms of atoms or charge, so to balance it simply
follow the procedure we used above:
Step1 - balance the oxygen atoms on each side by adding
water to the oxygen deficient side. In this
case add 4 water molecules to the products 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 reactants side of the equation. This gives:
Reduction reaction: MnO4-(aq) + 8H+(aq)+ 5e → Mn2+(aq) + 4H20(l)
This now balances the equation in terms of atoms but also the charges on both sides of the equation.
A redox titration using potassium permanagante
As an example consider the oxidation of Fe2+(aq) ions to Fe3+(aq) ions
an acidified potassium permanganate solution.
Iron (II) sulfate is present in many lawncare 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 potassium permanagante.
The permanagante is placed in a burette, as shown in the
image opposite while a solution of the Fe2+(aq) ions is placed in a concial flask as shown.
When the purple permanagate ion is added to the Fe2+(aq) ions in the conical flask it oxidises the
Fe2+(aq) ions to form
Fe+(aq) ions and the purple MnO4- ions are reduced to form the almost colourless Mn2+ ions.
We already have a balanced half-equation for the reduction of the purple acidified permanagante
ion (MnO4-) to the almost colourless Mn2+(aq) ion:
Reduction reaction: MnO4-(aq) + 8H+(aq)+ 5e → Mn2+(aq) + 4H20(l)
The oxidation reaction is simply the pale green Fe2+(aq) being oxidised to the pale yellow
oxidation reaction: Fe2+(aq) → Fe3+(aq) + e
However you will obviously note that the oxidation reaction only releases 1 electron while the reduction of the permanaganate ion
requires 5 electrons. So to get the overall equation the oxidation half-equation needs to be multiplied by x5.
oxidation reaction: 5Fe2+(aq) → 5Fe3+(aq) + 5e
Overall equation for this redox reaction is obtianed by simply adding together the two half-equation and cancelling out the electrons:
Reduction reaction: MnO4-(aq) + 8H+(aq)+
5e → Mn2+(aq) + 4H20(l)
oxidation reaction: 5Fe2+(aq) → 5Fe3+(aq) +
overall reaction: MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H20(l)
- Oxidising agents such as dichromate and permanagnate require to be
acidified by the addition of sulfuric acid.
- 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 products and reactants side of the half-equation.
- Next add hydrogen ions (H+(aq)) to the hydrogen deficient side of the equation.
- Finally check that the number of electrons in each of the
oxidation and reduction half-equations balance.
The practice questions contain additional examples using these oxyanions as oxidising agents, have a go at them to quickly check your understanding.