5 E T R

In a reaction between an acid and a base, a proton is transferred from acid to base. When
an electron (or a number of electrons) is transferred from one species to another, the
reaction is known as an oxidation-reduction or redox reaction. Such reactions are very
common, both in the laboratory for analysis of various species, or industrially, in the
manufacture of chemicals, or economically, as in the undesirable process of corrosion. All redox
reactions are, in theory, reversible, ie. they are equilibrium processes.
The name oxidation is derived from the observations by chemists of centuries ago that certain
materials gained oxygen during a reaction, e.g. rusting of iron to iron oxide. They didn't know
anything about electrons, but could observe this common type of reaction. The opposite reaction,
where species lost oxygen during a reaction, e.g. heating up copper oxide to form copper metal, they
called reduction because the amount of oxygen in the substance had been reduced. Later studies more
closely defined the two processes (in terms of electrons), and showed that there were plenty of redox
reactions where there wasn't a single oxygen atom in sight.
The species that gains the electron(s) is reduced, while the species that loses the
electron(s) is oxidised. It should be emphasised that the processes are indivisible: the
electrons lost in the oxidation must be gained by another species which is then reduced.
A simple, non-analytical example of oxidation and reduction is the reaction of metals with acids.
When dropped into a HCl, zinc begins to ‘dissolve’, and hydrogen gas is given off. The zinc does not
really dissolve, but in fact loses electrons (is oxidised) and becomes the ion, Zn 2+ (which is soluble),
while the protons in solution gain those electrons (are reduced) and forms hydrogen gas.
In acid-base reactions, the terms acid and base were used to describe the two species involved in
the proton transfer. Whereas in redox, the terms you have met so far - oxidation and reduction - refer
to the process. So what do we call the species involved? This is where it becomes a bit confusing!
Referring back to our original example, the zinc, by losing electrons, causes the
hydrogen ions to be reduced. Therefore, the zinc is called a reducing agent or
reductant: a species that causes another to be reduced. On the other hand, the hydrogen
ions, by gaining the electrons, cause the zinc to be oxidised. Thus, the hydrogen ion is
called an oxidising agent or oxidant: a species that causes another to be oxidised.
Why are these definitions possibly confusing? Consider what happens to the reducing agent - it
causes something else to be reduced, by getting itself oxidised. The reverse applies for the oxidising
Recognising an oxidation-reduction reaction is not necessarily as easy as one involving an acid-base
reaction, but there are some general rules which help. The following changes definitely mean
oxidation or reduction:
elements changing into compounds or ions, or vice versa, e.g. Zn  Zn2+, H+  H2, C  CO2,
a change in valency by a monatomic species, e.g. Fe 2+  Fe3+
But there are plenty of other reactions which are reduction or oxidation that don't fit these two rules,
e.g. MnO4-  Mn2+, and others, e.g. CaCO3  CaCl2, that might look candidates that aren't. So we
need a better system. This is known as oxidation numbers.
Chemistry 2
5. Electron Transfer Reactions
Which of the following processes are definitely redox reactions (without use of oxidation
(a) Sn4+ + 2Fe2+  Sn2+ + 2Fe3+
(b) HCl + NaOH  H2O + NaCl
(c) I- + Fe3+  I2 + Fe2+
(d) 2MnO4- + 6H+ + 5C6H8O6  2Mn2+ + 8H2O + 5C6H6O6
(e) ClO4- + C2H6  CO2 + H2O + Cl(f) O2 + H2O + Fe  Fe(OH)2
Calculating Oxidation Numbers
The oxidation state of each element (a bit like the valency, but it applies to each element within a
compound or ion) is the means of distinguishing redox reactions from others: if the oxidation state of
an atom changes during a reaction, then it has been reduced or oxidised.
The oxidation state is determined by the oxidation number system, which follows a set of
simple rules, as below:
The oxidation number of an atom when present in the elemental form is zero.
The normal oxidation number of hydrogen in compounds is +1, except when it is present as
hydride ion (H-), when it is -1.
The normal oxidation number of oxygen in compounds is -2, except when it is present as
peroxide ion (O22-), when it is -1.
The sum of oxidation numbers of the atoms in a compound or ion is equal to the charge on the
Consider the following examples of a series of sulfur species: (a) sulfur (S 8), (b) sulfide, (c)
sulfur dioxide, (d) sodium sulfate, (e) sulfuric acid, (f) thiosulfate (S 2O32-). What is the
oxidation number (ON) of sulfur in each of these species?
Sulfur is present as the elemental form, and therefore the oxidation number of sulfur is
In sulfide, the one sulfur atom has a charge of -2. Therefore, the ON. is -2.
In sulfur dioxide, the overall charge is 0. Therefore, the sum of the ON of the one S and
two O is zero. Since oxygen has an ON of -2 and there are two of them, the sulfur has
an ON of +4.
In sulfate, the overall charge is -2. Therefore, the sum of the ON of sulfur and 4
oxygens (4 x -2) = -2. The ON of sulfur is +6.
In sulfuric acid, the sum is zero. Therefore, 2 x +1 + ON + 4 x -2 = 0. The ON of sulfur
is + 6.
In thiosulfate, 2 x ON + 3 x -2 = -2. Therefore, ON of sulfur is +2.
Sulfur is clearly capable of existing in many different oxidation states. The higher (more positive)
oxidation number of an element, the more oxidised it is. Therefore, sulfur is most oxidised in sulfate
and sulfuric acid (ON = +6), and most reduced in sulfide (ON = -2).
How then do we use the oxidation number system in recognising redox reactions, and
more importantly, what has been oxidised and reduced? The following simple rules help
in the decision:
If the ON of two atoms in a reaction change, then a redox reaction is indicated.
If the ON of an atom within a species increases, the species has been oxidised.
If the ON of an atom within a species decreases, the species has been reduced.
Chemistry 2
5. Electron Transfer Reactions
Determine the oxidation number of the non-oxygen element in each of the following: (a) ClO3-,
(b) P2O72-, (c) N2O3, (d) MnO42-, (e)Fe2O3, (f)Cu2O, (g)C2H6O.
Calculate the oxidation number of nitrogen in the following species, and determine which is the
most oxidised and most reduced species: NO, NO3-, N2, N2O, NO2, N2O4.
Determine the oxidation numbers of all elements in the reactions in Q1. Identify any other redox
reactions, and the reducing and oxidising agents in each case.
Clearly, a redox reaction occurs in two stages: the reductant gives up its electron and then the oxidant
accepts it. These two ‘halves’ of the reaction could be represented separately in equation, as below
using the Zn/HCl example. Electrons become reactants or products in these equations
Zn 
Zn2+ + 2e
2H+ + 2e  H2
Oxidation = Electrons as products
Reduction = Electrons as reactants
Breaking up the reaction into its two halves is useful for a variety of reasons: firstly, we
can see clearly what is happening, and secondly, we can use the information about what
is happening when zinc is oxidised in any reaction where it is being oxidised. These
equations are called half-equations, because they describe exactly that: either the
reduction or oxidation part of the overall process.
TABLE 5.1 Common half equations
Ag+ + e  Ag
Cr2O72- + 14H+ + 6e  2Cr3+ + 7H2O
Cu2+ + 2e  Cu
Fe2++ 2e  Fe
Fe3+ + e  Fe2+
2H+ + 2e  H2
H2O2 + 2e + 2H+  2H2O
I2 + 2e  2IMnO4- + 4H+ + 3e  MnO2 + 2H2O
MnO4- + 8H+ + 5e  Mn2+ + 4H2O
O2 + 2H2O + 2e  4OHPb2+ + 2e  Pb
SO42- + 4H+ + 2e  SO2 + 2H2O
S4O62- + 2e  2S2O32Sn4+ + 2e  Sn2+
Zn2++ 2e  Zn
2CO2 + 2e   C2O42-
Chemistry 2
5. Electron Transfer Reactions
You might notice that all these half-equations are written in the reduction form, i.e. with electrons on
the left-hand side. However, this is simply a convention, and the double-headed arrows mean that any
of these can be reversed. Obviously, when using these to describe an overall redox reaction, one of the
two half-reactions must be an oxidation, i.e. reversed.
Using Half Equations
There is nothing special about a redox equation. It can be balanced in exactly the same way as any
equation; it is just that some are more complicated than others. You wouldn't need half equations to
write the balanced equation for the reaction of good old zinc and hydrochloric acid, but if balancing
equations isn't your strong point, you might struggle to balance some of the more complex examples.
The tables of half equations also mean that the other reactants and products involved don’t have to be
The steps to be followed in using half equations to produce a balanced overall equation are:
Identify the oxidant and reductant involved.
Locate half equations that have these species as reactants (on the left hand side).
Work out the correct half equation for each species, by knowledge of products and ensuring that
one reduction and one oxidation reaction is involved (two reductions or two oxidations can’t be
Add the two equations together, ensuring that no electrons remain in the overall equation. To
achieve this, use the cross-multiplication method: multiply one equation completely by the
number of electrons in the other half equation, and then do likewise for the opposite
Cancel out common species on both sides.
The reaction conditions (e.g. acidic, alkaline) can affect the reaction. For example, permanganate can
form at least three species when it is reduced, depending on the solution.
Remember that reactants have to end up on the left hand side of the final equation, but won't
necessarily be there in the half-equations from the table, since they are all in reduction form. If this is
the case, you must reverse the equation, taking all species on the left hand side and moving to the right
hand side, and vice versa.
Write the balanced equation for the reaction of permanganate reacting with iron (II) in acidic
Looking at Table 5.1, there are two occurrences of permanganate and two of iron (II), as
Fe2+ + 2e  Fe
Fe2+  Fe3+ + e (reversed to make iron (II) a reactant)
MnO4- + 8H+ + 5e  Mn2+ + 4H2O
MnO4- + 4H+ + 3e  MnO2 + 2H2O
In the case of permanganate, in acidic solution, it is more likely to form Mn 2+, not MnO2.
Therefore, the appropriate manganese half-equation is MnO4 - + 8H+ + 5e  Mn2+ +
Which iron (II) equation is the right one? Reduction and oxidation go hand in hand. If
permanganate is reduced, the iron (II) must be oxidised. Therefore, the correct half equation
is Fe2+  Fe3+ + e.
Chemistry 2
5. Electron Transfer Reactions
To produce the overall equation, cross multiply.
MnO4- + 8H+ + 5e
MnO4- + 8H+ + 5Fe2+
Mn2+ + 4H2O
5Fe + 5e
Mn2+ + 4H2O + 5Fe3+
Write the balanced overall equations for the following reactions.
(a) iodide and dichromate
(b) oxalate and permanganate in acidic solution
(c) hydrogen peroxide and iodide ion
(d) thiosulfate and iodine
(e) cerium (IV) ion and iron (II)
(f) oxalate and sulfate
While redox processes are reversible, they only occur spontaneously in one direction. If an iron nail is
dropped into a solution of copper ions, a copper coating appears instantly on the nail, and invisible to
us, iron atoms are oxidised to ions and dissolve into the solution. if, however, a piece of copper is
dropped into a solution of iron (II), nothing happens.
Clearly, some metals are more readily oxidised (e.g. iron) than others, and the ions of others
more readily reduced (e.g. copper). Placing the metals in order of ease of oxidation is known as the
activity series of metals, as shown in Figure 5.1.
ease of
of M+
of M
FIGURE 5.1 The activity series of metals
The higher up the activity series, the less likely a metal is to be oxidised. Said another way, if metal X
is placed in contact with a solution of the ions of metal Y, a reaction will only occur if metal X is
lower on the table.
Hydrogen is often added to this list, even though it is not a metal. It fits in this list between
copper and lead. We will see shortly a less obvious reason for its inclusion, but for the moment,
suffice to say, that we can tell whether a metal will react with HCl or H 2SO4 (be oxidised by H+) by
whether it is above or below hydrogen: those above will not, those below will.
Chemistry 2
5. Electron Transfer Reactions
What will happen when an iron nail is dipped into a copper (II) solution?
Since iron is lower than copper in this activity series, iron will be oxidised to iron (II) and
the electrons will reduce copper ions to copper metal. A copper coating will appear on
the nail.
Can HCl dissolve copper?
Copper is higher on the table than hydrogen. Therefore, copper will not be dissolved in
You are required to test the reactions of silver, copper, zinc, iron and lead in solutions of the
ions of these metals (use iron (II), not iron (III)).
This need only be done in test tubes. Ensure that the metals are clean by polishing with
emery paper.
In some cases, changes will be immediately apparent while still in the test tube. Where there
is no visible change, the piece of metal will have removed, and wiped with paper towel to see
whether any metallic residue is removed, which will be a sign of a reaction.
Q1. Draw up a table for your results, with the metals in columns and the ions in rows.
Indicate positive or negative reactions clearly (e.g. + and -).
Q2. Can see find a sequence of reactivity for the metal ions, i.e. did one metal ion react
with all the other metals, did one metal ion react with no metals?
Compare the sequence of reduction half-reactions for metal ions in Table 5.1 with those in the
activity series.
(a) Where would gold and platinum fit in Table 5.1?
(b) Where would the dividing line for metals affected or not affected by oxygen be drawn?
(c) Suggest why platinum and gold are called noble metals.
Predict what would happen if an iron nail was exposed to a copper salt in the presence of
When a species is oxidised, it transfers electrons to the oxidant. This, at a molecular level, is like a
flow of electrical current. After all, what is electrical current but a flow of electrons from one place to
another. In a redox reaction in a beaker, the electrons go directly from molecule to another. If,
however, we could separate the two processes - reduction and oxidation - so that electrons had to flow
between one beaker and another via a wire, then we have the capacity to generate a measurable (and
useful) electrical current. In a modified form, all our batteries (in your calculator, car etc) fit this
description: a chemical reaction producing electrical current.
In the laboratory, how would such a setup work? Figure 5.2 shows a typical apparatus, using
copper, zinc and solutions of their ions. Table 5.2 defines some key terminology.
Chemistry 2
5. Electron Transfer Reactions
connecting wire
FIGURE 5.2 A typical electrochemical cell
Thus, electrons flow from the zinc electrode through the external wire to the platinum where they are
used to reduce H+ to H2. Over time, the following observations could be made:
 the zinc will lose mass,
 the concentration of zinc ions in solution will increase,
 hydrogen gas will bubble off, and
 the concentration of hydrogen ions in solution will decrease.
TABLE 5.2 Important terminology
Electrochemical cell
Half cell
A redox reaction where the half-reactions are separated and the electrons
are forced to travel through external connections in the form of current
One of the halves of an electrochemical cell
a conducting material through electrons flow in electrochemical reactions
– there must always be two electrodes
10. Identify the following components of the apparatus in Figure 5.2:
- electrochemical cell
- each half-cell
- each electrode
Voltage Measurements
Electrons flow in an electrochemical cell because of a voltage (or potential) difference between one
electrode and another. Electrons flow from the electrode with the greater ability to lose them. It is like
water in a bucket half way up a hill. If you tip the bucket over, the water runs downhill, not uphill,
because of the laws of gravity.
Some species have a greater need to lose electrons than others. The potential (or ability) of a
half-cell to lose or gain electrons can be measured, but not by itself. Another half-cell must be present
to complete the circuit. This half-cell has its own potential, so any measurements of potential (voltage)
can only be relative between the two half-cells.
If this causes you a problem, imagine yourself standing on a wall between two bodies of water
(the two half-cells). You have a measuring stick (a voltmeter), which can reach the surface of each
body of water. You want to measure the depth of each body of water, but your measuring stick isn’t
long enough. However, you can measure the difference in their depths (potential difference). Figure
5.3 illustrates this idea.
Chemistry 2
5. Electron Transfer Reactions
FIGURE 5.3 The ‘depth of water’ analogy for potential difference
Electrode potentials are dependent on concentration, which means their measurement can be used to
tell something about the composition of a solution. Voltage measurements are among the most
important methods used in analytical instrumentation to determine the composition of solutions.
Chemical analysis makes considerable use of redox reactions and their associated electrochemical
measurements. Potassium permanganate and sodium thiosulfate are important redox titrants.
Permanganate, an oxidant, is used to analyse iron levels in samples, once the iron has been prereduced to iron (II). Thiosulfate, a reductant with some peculiar properties, only reacts usefully with
elemental iodine. However, a number of species can oxidise iodide ion to elemental iodine, allowing
the titration with thiosulfate.
You have met a pH meter and electrode in this and other modules. Simply put, it and the
solution being tested form an electrochemical cell. It produces a reading by an electrochemical
response at the electrode surface. Other so-called ion-selective electrodes have been built to perform
the same function, but which are selective for different ions, such as fluoride and sodium.
Industrially, electrolysis (the forced reversal of non-spontaneous electrochemical processes) is
an important method of producing a number of otherwise difficult chemicals. For example, aluminium
metal is obtained from the reduction in molten conditions of an aluminium fluoride compound.
Sodium metal and chlorine gas can be produced by the electrolysis of molten sodium chloride. Water
cannot be present in either of these processes because it will be preferentially used up before the
desired materials.
However, the most important commercial application of electrochemical processes is the
battery. Whether it is a battery that runs your watch, your calculator or your car, it produces its power
from chemical reactions between two electrodes. The only differences between a commercial battery
and the zinc/hydrogen cell used to illustrate the key points in Section 5.5 are the chemicals involved
and the design (to minimise the amount of solution slopping around).
You may have heard of the terms aerobic and anaerobic. They describe conditions (especially in
water) where oxygen is plentiful or absent, respectively. Aerobic conditions are oxidising because of
the oxygen, while anaerobic conditions are reducing. This will affect the fate of chemicals that find
their way into the different environments. For example, carbon-containing materials (eg dead plants)
are oxidised to carbon dioxide in aerobic conditions, but reduced to methane and other hydrocarbons
in anaerobic environments – this is how the oil deposits formed.
Not all redox processes are desirable. Corrosion is the oxidation of metals by atmospheric
oxygen gas or by other oxidising agents: rusting of iron is the most common example. Large amounts
of money are invested in attempting to minimise the effects of this process. Space does not allow
further discussion of this process here.
The redox potential of a solution or soil (known as the ORP) is an important measure of its
potential corrosiveness, and is simply measured with a voltmeter.
Chemistry 2
5. Electron Transfer Reactions
define terminology associated with redox reactions
distinguish the reactive species in redox reactions
write balanced half and overall equations for redox processes
draw a complete electrochemical cell and explain the function of its components
explain how a voltage difference applies between half-cells
outline important applications of redox reactions
Chemistry 2