Brown's A Level Chemistry Revision Notes
Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 7.2
7.2 Simple cell
notation and construction, electrode potential chart
You should know that redox reactions take place in
electrochemical cells where electrons are transferred from the reducing
agent to the oxidising agent indirectly via an external circuit. How do you write electrochemical cell
notation? how do you make a simple cell from two half–cells to make a
simple battery? What is an electrode potential chart? How do you work
out the voltage produced by a simple cell from the two half–cell
electrode potentials involved?
Chemical Equilibrium Notes Part 7 Index
Constructing a simple cell or
and electrode potential chart
below shows the full explanatory diagram of
how a zinc–copper Daniel cell works
A simple cell like the Daniel cell
('battery') is made by setting up a situation in which the two half–cell
reactions are kept physically separated.
To complete the electrical
circuit, the zinc and copper(II) ion solutions are connected by a salt
bridge (a supported ion solution).
Wires connect strips of zinc and
copper metal to a voltmeter or other device to complete the circuit.
The energy release from
the exothermic reaction is not in the form of heat energy
but in the form of electrical energy because the electrons from the
two electron transfers of the two half–cell reactions are 'forced'
around the circuit due to the potential difference between the two
shows how to set up
a simple electrochemical cell
(galvanic/voltaic cell or battery), and relates the direction of chemical changes
(on electrodes) to the +ve and –ve terminals and the direction of
electron flow. The left and right metal–metal ion solution 'beakers' constitute
the two half–cells which together make up the full cell.
(matching diagram, NOT IUPAC convention?)
Note a single vertical line |
signifies an electrode interface or phase boundary and a double vertical
|| signifies a salt bridge
of a suitable electrolyte such as potassium chloride or ammonium nitrate.
When two half–cells are
connected together with the high resistance voltmeter to make a complete
working cell, a salt bridge, a conducting electrolyte solution/gel,
is needed to complete the electrical circuit and allows the slow drift of
ions in both directions, NOT electrons – they only flow between the
metal electrodes via the external circuit wire and high resistance
Cell notation: See Ecell
calculation below and
The hydrogen electrode and standard conditions
If a half-cell contains a mixture of
aqueous ions involved in a redox change, both ions must be shown,
separated by a comma.
Introduction to Ecell
calculations. As explained in
7.1 Half cell equilibria, electrode potential the chemical
potential of zinc to form a zinc(II) ion is much greater than the
chemical potential of copper to form a copper(II) ion and the
difference in 'tendency' or chemical potential can be exploited in the construction of a
simple cell (battery) which demonstrates that half–cell redox
equilibria exist and when connected to form a circuit the equilibria
head in the direction dictated by the chemical potentials. In electrolysis these reversible
equilibria are forced in a particular direction by an external applied
potential difference (p.d., volts).
A simple cell or
battery is made from combining two half–cells and the system can be
used to determine electrode potentials and the data provided can be
used to theoretically 'test' the feasibility of a redox reaction.
potential is a measure of the tendency or chemical potential (in
a thermodynamic sense) of a species to lose/gain electrons in the
context of a half–cell equation.
Half–cell potentials are measured in volts compared to
the standard hydrogen electrode (EθH+(aq)/H2(g)
= 0.00V, fully described and explained in
The hydrogen electrode and standard conditions.
The more positive or less negative the potential,
the greater the tendency of the half–cell to act as reduction change.
less positive or more negative the electrode potential, the greater
the tendency for the half–cell reaction to be one of oxidation.
+ 2e– (EθZn2+(aq)/Zn(s)*
= –0.76V (most –ve or least +ve, oxidation, electron loss)
+ 2e– ==> Cu(s) (EθCu2+(aq)/Cu(s)*
= +0.34V (most +ve or least –ve, reduction, electron gain)
standard measurement of Eθcell
the metal strips of the half–cells are wired in series with a high resistance voltmeter
To complete the
circuit a 'salt
bridge' connects the two solutions of the half–cells.
consists of a conducting electrolyte solution of 'inert' ions e.g. potassium chloride.
solution is supported e.g. crudely with soaked filter paper or a gel
held in a U tube.
The ions are free to move and can carry current in any direction, but
must not chemically react with anything.
We now need to consider how to work
out theoretically the Emf the cell in operation (in crude terms, the
theoretical 'battery voltage', but best to talk about cell Emf or Eθcell
in doing calculations from standard electrode potential date.
One way of working out Eθcell values
the IUPAC convention for cell notation would
require the cell notation to be
with the positive
pole on the right so that for a feasible reaction Eθright
– Eθleft gives a value of >0V.
Eθcell = Eθ(right)
Eθcell = Eθ(+ve/red)
which amounts to the difference between the half–cell potentials on an
electrode potential chart (see Fig 3.
is the half-reaction electrode potential for the half-cell reduction
i.e. Cu2+ + 2e- ==> Cu
is the half-reaction electrode potential for the half-cell oxidation
i.e. Zn ==> Zn2+ + 2e-
is the most positive or the least negative half–cell potential, the strongest oxidising agent or electron acceptor of the two
half–cell systems, and the +ve pole of the cell, e.g. Cu2+/Cu
compared to Zn2+/Zn,
so the reduction (red) Cu2+(aq)
+ 2e– ==> Cu(s), rather than
reduction of Zn2+ to Zn.
is the least positive or the most negative half–cell potential, the
strongest reducing agent or electron donor of the two half–cell
potentials, and the –ve battery pole e.g. Zn2+/Zn
compared to Cu2+/Cu,
so the oxidation (ox) Zn(s) – 2e–
==> Zn2+(aq) happens rather
than oxidation of Cu to Cu2+,
overall cell redox
+ Zn(s) ==> Cu(s) + Zn2+(aq)
voltage–Emf for the copper–zinc cell:
= EθCu2+(aq)/Cu(s) = +0.34V
= EθZn2+(aq)/Zn(s) = –0.76V
pole/red) – Eθ(–ve
(+0.34V) – (–0.76) = + 1.10 V
If the voltage
cell Emf is positive the cell reaction is feasible! as calculated
based on the cell reaction with the matching oxidation (–ve pole) and
reduction (+ve pole) potentials.
If in the
calculation a negative voltage results, the reaction is
NOT feasible, but will in fact go in the reverse direction.
A note on cell electrode
potential measurements (cell EMF) and polarisation.
Half–cell potentials are
dependent on the concentration of any ion involved in any of the
electrode reactions, whether a product or reactant.
caused by the build up of the cell reaction products e.g. Zn2+
ions from oxidation of zinc electrode, or, the diminution of reactants
e.g. copper ions Cu2+ on their reduction to copper.
Any changes in concentration
will affect the half–cell potential of any electrode i.e. creating
different equilibrium non–standard conditions.
This is reduced by using a
very high resistance voltmeter to give the most accurate
readings because if very little, if any, current flows, this will
minimise the extent of the cell reaction and hence minimise any
An example of a simple electrode potential chart (right).
The basis of the
calculation can be illustrated by way of a simple electrode potential
chart like the one shown on the right. The Emf produced by the cell is
the difference between the two half–cell potentials. Note the relative
reducing/oxidising power potential trends.
See also the list of
half–cell potentials in 7.4 which constitutes a 'chart' when set out
in relative potential order.
The most positive
(or least negative) half–cell potential has the stronger the oxidising
power of the two half–cells in a complete cell/reaction, so it will
constitute the reduction change in the overall cell reaction.
The more negative
(or least positive) half–cell has the stronger reducing power of the two
half–cells, so it will constitute the oxidation change in the full cell
reaction. The full cell potential is expressed as the difference
between the two half–cell potentials ...
= Eθhalf–cell of
positive pole = most +ve or
least –ve Eθ for
the reduction half–cell reaction
– Eθhalf–cell of negative pole = least +ve or most
–ve Eθ for the oxidation half–cell
which I've expressed
simply in the Daniel Zn–Cu cell above, and subsequent examples below,
= Eθ(+ve/red) – Eθ(–ve/ox)
= EθCu2+(aq)/Cu(s) – EθZn2+(aq)/Zn(s)
= (+0.34V) – (–0.76V) = +1.10V
For the effect of
changing concentration on the value of the electrode potential see
Appendix 1. The Nernst Equation
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Chemical Equilibrium Notes Part 7 Index