Every oxidation or reduction process can be expressed by a half equation
showing only the species being oxidised (or reduced) and any water or
hydrogen ions needed in the process.
Constructing half equations
The atom or species being oxidised (or reduced) is written in its two
different species on each side of the equation. The appropriate nuber
of electrons is placed to balance the change in oxidation number of the
particular atom being oxidised (or reduced). If necessary use water or
hydrogen ions to balance the equation.
Example 1: The
oxalate ion C2O42- changes to carbon
dioxide when oxidised. Construct the half equation.
Step 1: place the two species on either side
Step 2: Examine the oxidation number of the
carbon on each side and add the necessary electrons. In this case
carbon is changing from the III oxidation state to the IV oxidation
state and therefore each carbon atom needs to lose 1 electron. There
are two carbon atoms and so 2 electrons must be lost.
2CO2 + 2e
Step 3: The equation is now checked for balancing
and is balanced in terms of both atoms and charges
Example 2: Sulphur dioxide is a good reducing
agent and turns to sulphate ions in the process.
Construct the half equation.
Step 1: Write out the two species on each
Step 2: Sulphur is in the form of sulphur
(IV) at first and becomes Sulphur (VI), losing two electorns in
Step 3: Here we can see that the atoms are
not balanced and we must provide oxygen on the left hand side by
adding a suitable number of water molecules. This will leave hydrogen
ions on the right hand side
SO2 + 2H2O
+ 2e + 4H+
On inspection the equation is now balanced
in terms of both electrical charges and atoms.
Combination of half equations to give balanced redox equations
Each half equation represents one half of the redox process - either
the reduction or the oxidation. A reduction process can be combined with
an oxidation process to give a redox equation by simply ensuring that
the number of electrons is the same for each equation and then adding
them and cancelling out any common factors.
Construct the equation for the reaction between manganate (VII)
ions and iron (II) ions in aqueous acidic solution
+ 8H+ + 5e
Mn2+ + 4H2O
In this example the Manganate (VII) ion is
being reduced to Manganese (II) ions. This process cannot happen
in isolation- it needs something to provide the 5 electrons. A reducing
A typical reducing agent is iron (II) sulphate
as it contains the iron (II) ions that can release an electron to
give iron (III) ions
Fe3+ + 1e
These two equations can be combines by simple
addition to give the ionic redox reaction that occurs when manganate
(VII) ions react with Iron (II) ions. But before they can be added
the electrons must be equalised by multiplying the second equation
5Fe3+ + 5e
Now add the reduction and the oxidation half
+ 8H+ + 5Fe2+
Mn2+ + 4H2O + 5Fe3+
the redox process
In general the procedure is as follows:
Write down the two half equations, one reduction and the other oxidation.
Equalise the number of electrons by suitable multiplication of one
or both half equations
Add the two equations together and cancel out the electrons on both
Standard electrode potential is the potential difference between a given
half cell (at 1 mol dm-3 conc) and the standard
The standard hydrogen
electrode consists of a solution of H3O+ ions
at 1 mol dm-3 in a beaker. Placed into this is a platinum electrode
surrounded by a gas tube submerged in the solution, with hydrogen at 1
atm inside. The circuit to the other half cell is then attached to the
platinum electrode, and a salt bridge saturated in potassium chloride.
The entire process should take place at 298K and 1 atm pressure.
Measuring cell potentials
Cell potential is the potential difference between two half cells. They
can be calculated by comparing the standard electrode potential of each
half cell and taking the difference.
Example: What is the emf of a cell made from
a zinc|zinc sulphate half cell connected to a copper|copper sulphate
zinc|zinc sulphate half cell Eº = -0,76
copper|copper sulphate half cell Eº
= +0,34 V
E cell = the difference between
the two values = 1,10 V
Direction of electron flow will be from the
better reducing agent i.e. in this case from the zinc (the half
cell with the greater negative value) to the copper half cell
Electrode potentials can be used to predict the feasibility of a reaction.
If the electrode potentials are compared for two processes and the Eº
for a hypothetical cell determined from the relationship
Eº = E(red) - E(ox)
If the value of Eº calculated is positive and greater than +0,3V
(approximately) then the reaction is likely to occur.
If the value calculated is between 0 and +0,3V then its likely to
be an equilibrium.
If the value is negative then it is unlikely to occur.
hydrogen peroxide react with iron (II) ions in acidic solution?
Electrolysis is the situation when redox cells are forced to run in reverse
by attaching an electricity source to overcome the potential difference.
In aqueous solutions, the ions present in water compete at the electrodes,
and will sometimes be oxidised/reduced in preference to the dissolved
It is possible to use the standard electrode potentials to predict this,
in that species above water (when it is on the left) will not be oxidised,
and species below water (on the right) will not be reduced in an aqueous
At the cathode:
Species more reducing than the H+|H2 (half cell = 0V) will
not be released at the cathode in aqueous solution. Instead the reaction
is: 2H+(aq) + 2e
At the anode:
Species with a lower oxidising power than the O2|H2O
couple will not be released from the anode. The usual reaction is 4OH-
Highly concentrated solutions may overcome this to some degree. It is
possible for Cl2 to be released from a strong solution of chloride
the products of electrolysis of strong calcium chloride solution.
At the cathode
Species present Ca2+ and H+. Ca2+ is higher
in the reactivity seriers than hydrogen and therefore cannot be
released. The reaction is therefore: 2H+(aq)
At the anode
Species present OH- and Cl-
. The chloride concentration is strong and so it is preferentially
oxidised and the reaction is: 2Cl-(aq)
Cl2(g) + 2e
Species remaining in solution: Calcium ions
and hydroxide ions
Quantities produced by electrolysis
Faraday's law states that the mass of product produced will be proportional
to the charge passed.
The charge (Coulombs)= current (amps) x time (seconds)
Faraday's law may also be restated as...the number of faradays required
to discharge 1 mol of an ion at an electrode equals the number of charges
on that ion.
1 Faraday = 96500 Coulombs
Calculate the mass of copper
released by a current of 10A passing for 200 seconds through a Copper
II sulphate solution.
Charge = Amps x seconds =
10 x 200 = 2000 coulombs
Number of Faradays = 2000/96500
= 0,0207 Faradays
The reaction occurring at the cathode is:
Cu2+(aq) + 2e-
Therefore: 2 Faradays will produce 1 mole
Hence 0,0207 Faradays will produce 0,0207/2
moles of copper = 0,0104 moles
Therefore the mass of copper produced = 0,0104
x 63,5 = 0,658g
Copper ions leave the impure copper anode and copper metal is deposited
pure on the metal cathode.