Oxidation and reduction
Reactive metals will react with oxygen to produce metal oxides.
metal(s) + oxygen(g) → metal oxide(s)
No doubt at some time in your science lessons you have held a piece of magnesium ribbon in a Bunsen flame and
cautiously observed the bright flash from the burning magnesium.
The product of this reaction is magnesium oxide. No doubt from your gcse chemistry you will recall that we
as the addition of oxygen to a substance and reduction as the removal
of oxygen from a compound, while these definitions of oxidation
and reduction are useful they are by no means the only way in which
we can talk about substances being oxidised and reduced.
In the example above the magnesium metal has had oxygen added to it to form the
compound magnesium oxide, we would say that the
magnesium has been oxidised. We can write a word equation to show this simple reaction:
oxygen(g) → magnesium oxide(s)
and a symbolic equation for this reaction is:
O2(g) → 2MgO(s)
Now magnesium oxide is an ionic compound which contains
Mg2+ and O2- ions. If we consider how these ions are formed:
Mg -2e → Mg2+
The magnesium atoms are losing 2 electrons to form
magnesium ions (Mg2+). The addition of oxygen or the
loss of electrons is also called oxidation. So in this example
magnesium atoms have been oxidised to form
(Mg2+). So if losing electrons is another way in
which to define oxidation then if a substance gains electrons
it will be reduced.
or the preferred way to show oxidation is with the electrons on the right hand side of the equation:
Mg → Mg2+ + 2e
The oxygen atoms in this reaction have gained 2 electrons
from the magnesium atoms to form an oxide ion (O2-). We can show this as:
O + 2e → 2O2-
or since oxygen is a diatomic gas we should really multiply this equation by x2 to give:
O2 + 4e → O2-
The gain of electrons is called reduction
and when a substance gains electrons we say it has been reduced.
Burning copper metal
If a square of copper metal is held with a pair of tongs in a hot Bunsen flame for about 30 seconds the shiny
bronze coloured copper turns black. No flames or bright flashes are produced. The copper
metal is reacting
with the oxygen in the air, it is being oxidised. An equation for this reaction is:
copper(s) + oxygen(g) → copper oxide(s)
Copper oxide is an ionic compound which contains Cu2+ ions and O2- ions.
The copper ions are formed when the
copper atoms are oxidised, that is they lose 2
electrons to form copper ions:
Cu → Cu2++ 2e
and the oxide ions are formed when the oxygen atoms gain 4 electrons from the copper atoms:
O2 + 4e → 2O2-
The oxygen atoms have been reduced (gain electrons) by the copper atoms
to form oxide ions (O2-). A substance which
donates electrons is called a reducing agent. In this example the
copper atoms are the reducing agent, they supply or donate
2 electrons to the oxygen atoms. Confusing the
reducing agent is oxidised when it reacts, that is when
it loses or gives away electrons.
An oxidising agent is an electron acceptor, it will accept
electrons from another
substance and it in turn will be reduced. It is easy to see why some people can easily get these terms mixed up but with practice you will not make that mistake!
The equation for the oxidation of copper atoms:
Cu → Cu2++ 2e
and the reduction of oxygen atom into oxide ions:
O2 + 4e → 2O2-
These equations are called ion-electron half equations or simply half-equations for obvious reasons, they represent only half of the
reaction taking place. To get the overall equation for the reaction we simply add together the two half equations.
However the 2 equations need to balance in terms of atoms and also electrons lost and gained.
The copper atoms lose 2 electrons but the
oxygen atoms need to gain 4 electrons,
so we need to x2 the half-equation for the oxidation of copper to balance off
2Cu → 2Cu2++
4e → 2O2-   reduction
2Cu + O2 → 2Cu2+O2-   overall equation
Example 2- The displacement reaction between copper sulfate solution and iron
When an iron nail is placed in a beaker containing a solution of copper(II) sulfate a displacement reaction occurs, an
equation for the reaction is shown below:
iron(s) + copper sulfate(aq) → iron sulfate(aq) + copper(s)
Fe(s) + CuSO4(aq) → Fe2+SO4(aq)2- + Cu
If we now write an ionic equation it may allow us to better understand what has been oxidised and what has been
reduced in this redox reaction:
Fe + Cu2+ SO42-→ Fe2+SO42- + Cu
the sulfate ion, SO42- is unchanged in the reaction, it appears on both the reactant and product
side of the equation and it is unchanged. It takes no part in the reaction and is only present to balance off the
present. Ions like this which take no part in the reaction are called spectator ions. To simplify the equation above we can
remove the spectator ions, this will give a clearer view of what is actually happening in the reaction:
Fe + Cu2+ → Fe2+ + Cu
We can break this down into 2 separate half equations. The iron atoms lose
2 electrons and are being oxidised to
form iron ions (Fe2+). The half equation for this is:
Fe → Fe2+ + 2e
While the copper ions (Cu2+) in the copper sulfate solution
gain 2 electrons and are reduced to form metallic copper.
is the brown substance that coats the iron nail and collects in the bottom of the beaker in the image above. The half equation for
this reduction reaction is:
Cu2+ + 2e → Cu
Example 3 - The reaction of sodium and chlorine
Sodium is an alkali metal with an electronic configuration of
1s22s22p63s1, so when it reacts it will lose its outer 3s1 electron
and in the process be oxidised. Chlorine is a reactive halogen in group 7 with an
electronic configuration of
1s22s22p63s23p5. To gain a np6 or noble gas electronic
configuration chlorine only has to gain 1 electron, this means it will be looking
to gain 1 electron in its reactions, this
will make it an excellent oxidising agent and by contrast sodium
will be looking to lose its outer 3s1 electron to
gain a noble gas electronic configuration. This means sodium,
like most metals, is looking to lose electrons. This means that
metals are good reducing agents.
If a small piece of sodium metal is placed on a pile of sand in a flask
filled with chlorine gas you might a violent reaction to
occur, however the reaction is slow. To start the reaction a few drops of
water from a pipette, as shown in
the image below, are added to the sodium metal, a violent reaction
between the sodium and chlorine gas then starts. The flask quickly fills with
white fumes of sodium chloride.
A half equation for the oxidation of sodium is:
Na → Na+ + e
and a half equation for the reduction of chlorine gas (Cl2) is:
Cl2 + 2e → 2Cl-
Since chlorine is a diatomic gas it will require 2 moles of electrons to
reduce it to form
2 moles of chloride ions (2Cl-) however the half equation for
the oxidation of sodium
only releases 1 mole of
electrons. This means that in order to balance these
two equations the half equation for the oxidation of sodium will have to
be multiplied by x2.
oxidation half equation: 2Na → 2Na+ +
reduction half equation: Cl2 +
We can simply cancel out the electrons, as they appear on both sides of the equation to get the
overall equation for the reaction:
2e → 2Cl-
overall equation: 2Na + Cl2 → 2NaCl
- Oxidation is the loss of electrons or the addition of oxygen or the removal of
hydrogen from a substance.
- Reduction is the gain of electrons
or the removal of oxygen or the addition of hydrogen to a substance.
- An oxidising agent is an electron acceptor.
An oxidising agent will oxidise a substance,
that is remove electrons from it
and in the process it is itself reduced.
- A reducing agent is an electron donor. A
reducing agent will reduce a substance by supplying electrons
to it and in the
process it will be oxidised.
- Remember the word OilRig - oxidation is the loss of electrons,
reduction is the gain of electrons.
- To get the overall equation for a reaction from 2 half-equations ensure
that the number of electrons lost and gained by
the reacting chemicals balance.