redox reactions

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:

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 reactions are a little more complicated by easy to work out if you follow some simple rules.

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.

acidified potassium dichromate

When it oxidises a substance the orange dichromate ion is reduced to form the green Cr3+ ion.

dichromate ion colour change when it is reduced

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:
reduction reaction: Cr2O72- + 14H+ + 6e2Cr3+ + 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.

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 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 needs to gain 5 electrons.

potassium permanganate

The colour change when the permanagante ion is reduced can be shown as: colour change when permanganate is reduced

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:
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

potassium permanganate and fe2+ ions reacting As an example consider the oxidation of Fe2+(aq) ions to Fe3+(aq) ions by 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 Fe3+(aq) ion.
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) + 5e
overall reaction: MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H20(l)

Key Points

The practice questions contain additional examples using these oxyanions as oxidising agents, have a go at them to quickly check your understanding.

Practice questions

Check your understanding - Questions on redox equations.

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