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Brown's Chemistry Clinic
Equilibria Part 7
"Redox equilibria, half-cell electrode potentials and electrolysis"
GCE-AS-A2-IB Advanced Level Theoretical-Physical Chemistry revision notes
Notes for e.g. GCE-AS-A2-IB
*
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GCSE
Notes on reversible reactions-equilibrium *
GCSE Notes on
Electrochemistry *
Advanced Part 1. Equilibrium,
Le Chatelier's Principle-rules * Part 2. Kc and Kp equilibrium expressions and
calculations * Part 3.
Equilibria and industrial processes * Part 4.
Partition,
solubility product and ion-exchange * Part 5.
pH, weak-strong acid-base theory and
calculations * Part 6. Salt hydrolysis,
Acid-base titrations-indicators, pH curves and buffers * Part
7 sub-index: 7.1 Half cell equilibria, electrode potential
* 7.2 Simple cells notation and construction *
7.3
The hydrogen electrode and standard conditions *
7.4
Half-cell potentials, Electrochemical Series and
using EØcell for reaction feasibility *
7.5 Electrochemical cells ('batteries') and fuel cell systems
*
7.6 Electrolysis
and the electrochemical series
* 7.7 Exemplar Questions * Appendix
1. The Nernst Equation *
Part 8. Phase equilibria-vapour
pressure, boiling point and intermolecular forces
* The K and ΔS-ΔG connection with EØcell
will be dealt with via new thermodynamics pages later, but an
example of a ΔS-ΔG calculation is given at the end of
the advanced kinetics pages and ΔG for cells is
mentioned in Equilibria Part 7.
M = old fashioned shorthand for mol dm-3
*
An advanced chemistry notes on REDOX reactions in
general
is presented on other pages and
prior study will help here.
7.1
Examination of some redox equilibrium situations
-
(i) A metal
atom and metal ion
-
When a piece of
metal (M) is dipped into an aqueous solution of its ions (Mn+(aq))
an equilibrium exists between the two chemical states of the metal,
and importantly, in different
oxidation states (0 for the element and +n for the cation).
-
(i)
Mn+(aq)
+ ne-
 M(s)
-
n is the
numerical value of
the positive charge on the cation and the number of electrons involved in
the oxidation state change.
-
This is an example
of a half-cell equation which represents the particular
oxidation (electron loss) or reduction (electron gain) of an atom/ion
related by electron transfer connecting two different oxidation states
of the element.
-
The same
idea of an equilibrium between two species of the same element in
two different oxidation states expressed as a half-cell equation also applies to many other
situations e.g.
-
(ii)
A non-metal atom/molecule and associated anion e.g.
-
(iii) Two different cations of the
same metal e.g.
-
(iv) An anionic and
cationic species of the same metal e.g.
-
MnO4-(aq)
+ 8H+(aq) + 5e-
Mn2+(aq)
+ 4H2O(l)
-
Interchange of
manganate(VII) and manganese(II) ions in acid solution, +7 and +2
ox. states.
-
Note that half-cell equations are
often, but not always, presented as
a reduction (electron gain). For
examples (ii) to (iv), a chemically inert platinum electrode
is dipped into the solution when used in electrochemical cell
construction (see later).
-
For an example of
situation (i) and its consequences consider the diagram
Fig.1 below
comparing the relative positions of the metal-metal ion equilibrium
for copper and zinc.
-
 Fig.1
-
In the case of the
copper atom/copper(II) ion
equilibrium, you get a positive charge on a piece of copper when it is
dipped into an aqueous copper(II) ion solution. This is due to an
electron deficiency on the copper because the right-hand side of the
equilibrium, the solid copper, is favoured, and copper(II) ions remove
negative electrons to form copper atoms on the surface (reduction
process). The +'s represent the electron deficiency in the copper metal,
which in cells would make it the positive pole, and the
-'s
represent surplus anion charge in the solution.
-
However, in the
case of the zinc atom/zinc(II) ion equilibrium, you get a negative charge
on a piece of zinc when it is dipped into an aqueous zinc(II) ion
solution. This is due to an electron surplus on the zinc metal because the
left-hand side of the equilibrium, the aqueous zinc(II) ion, is
favoured, and zinc atoms lose negative electrons to form zinc(II) ions
in the solution (oxidation process). For Zn/Zn2+:
The +'s represent the extra zinc ion charge in solution, the -'s
represent the surplus electrons in the zinc metal which in cells would
be the negative pole.
-
Quite simply,
thermodynamically, zinc atoms have a much greater tendency or
potential to lose electrons than copper atoms and copper(II) ions
have a much greater tendency or potential to accept electrons
than zinc ions. Therefore the two half-cell reactions would be
-
Zn(s)
==> Zn2+(aq)
+ 2e- and
Cu2+(aq)
+ 2e- ==> Cu(s)
-
and the overall
reaction is Zn(s)
+ Cu2+(aq) ==> Zn2+(aq) + Cu(s)
-
This overall
reaction can be accomplished in two very different ways.
-
(a) by adding zinc
metal to blue copper(II) sulphate solution when the blue colour fades
and a red-brown deposit of copper forms. The reaction is exothermic
which can be observed as a significant temperature rise if zinc powder
is added to 1M copper(II) sulphate solution.
-
(b) By setting up
a simple chemical cell ('battery') 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. The energy release from
the exothermic reaction is not in the form of heat energy as in (a)
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
half-cells. This, so-called Daniel cell is described in detail in the next section
7.2.
7.2
Constructing a simple cell or
battery
Fig.2
below shows the full explanatory diagram of
how a zinc-copper Daniel cell works
 Fig.2
-
The diagram
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.
-
Cell notation
(matching diagram, NOT IUPAC convention)
-
Cu(s)|Cu2+(aq)¦¦Zn2+(aq)|Zn(s)
-
Note |
signifies an electrode interface or phase boundary and ¦¦ signifies a salt bridge
of a suitable electrolyte such as potassium chloride or ammonium nitrate.
-
Cell notation: See Ecell
calculation below and standard conditions section
7.3)
-
Introduction to Ecell
calculations. As explained in
section 7.1 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. The
Daniell Cell,
shown above in Fig.2, is a simple 'battery' system for producing electrical
energy from chemical potential energy.
-
The half-cell
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 (EØH+(aq)/H2(g)
= 0.00V, fully described and explained in
section 7.3).
-
The more positive or less negative the potential,
the greater the tendency of the half-cell to act as reduction change.
The
less positive or more negative the electrode potential, the greater
the tendency for the half-cell reaction to be one of oxidation.
-
Zn(s)
==> Zn2+(aq)
+ 2e- (EØZn2+(aq)/Zn(s)*
= -0.76V (most -ve or least +ve, oxidation, electron loss)
-
and
Cu2+(aq)
+ 2e- ==> Cu(s) (EØCu2+(aq)/Cu(s)*
= +0.34V (most +ve or least -ve, reduction, electron gain)
-
To obtain
standard measurement of EØcell
(cell EmfØ)
the metal strips of the half-cells are wired in series with a high resistance voltmeter
for accuracy. This avoids significant current flow causing
polarisation, that is the build up of products or the diminution of
reactants, either of which will cause a change in the cell Emf.
-
To complete the
circuit a 'salt
bridge' connects the two solutions of the half-cells. This
consists of a conducting electrolyte solution of 'inert' ions e.g. potassium chloride. The
solution is supported e.g. crudely with soaked filter paper or a gel
held in a U tube. The ions can carry current in any direction, but
must not chemically react with anything.
-
One way of working out Eøcell values
-
Strictly speaking
the IUPAC convention for cell notation would
require the cell notation to be
-
Zn(s)|Zn2+(aq)¦¦Cu2+(aq)|Cu(s)
-
with the positive
pole on the right so that for a feasible reaction EØright
- EØleft gives a value of >0V.
-
EØcell = EØ(+ve/red)
– EØ(-ve/ox)
-
which amounts to the difference between the half-cell potentials on an
electrode potential chart (see Fig 3.
below).
-
EØ(+ve/red)
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.
-
EØ(-ve/ox)
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
reaction: Cu2+(aq)
+ Zn(s) ==> Cu(s) + Zn2+(aq)
-
Calculating the
voltage-Emf for the copper-zinc cell:
-
EØ(+ve
pole/red)
= EØCu2+(aq)/Cu(s) = +0.34V
-
EØ(-ve pole/ox)
= EØZn2+(aq)/Zn(s) = -0.76V
-
EØcell
= EØ(+ve
pole/red) – EØ(-ve
pole/ox) =
(+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.
 Fig.3
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 ...
EØcell
= EØhalf-cell of
positive pole = most +ve or
least -ve EØ for
the reduction half-cell reaction
- EØhalf-cell of negative pole = least +ve or most -ve EØ for the oxidation half-cell
reaction
which I've expressed
simply in the Daniell Zn-Cu cell above, and subsequent examples below,
as
EØcell
= EØ(+ve/red) – EØ(-ve/ox)
e.g.
EØcell
= EØCu2+(aq)/Cu(s) – EØZn2+(aq)/Zn(s)
so EØcell
= (+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.
7.3
The Hydrogen Electrode, standard
electrode potential and
standard conditions
-
You cannot measure
or calculate absolute half-cell potentials, so, as in the case of ΔH
values, where the elements in their normal stable states at 298K are
arbitrarily given the enthalpy value of zero, in electrode potentials,
the aqueous hydrogen ion/hydrogen gas half-cell is given the arbitrary
standard value of EØH2(g)/H+(aq) = 0.00V
under standard conditions. It is illustrated as the right-hand side of
Fig.4 and Fig.5.
-
The half-cell
equation is: 2H+(aq)
+ 2e-
H2(g) for
the standard hydrogen electrode.
-
Ø
Means standard conditions
-
298 K for
temperature (25oC),
-
1 atmosphere pressure
(101kPa) of hydrogen gas (H2(g) for hydrogen electrode),
-
1 mol dm-3
concentration of H+(aq) (hydrogen ions for
hydrogen electrode), e.g. 1M HCl(aq) or 0.5M H2SO4(aq),
-
1M of the metal ion in the
other half-cells mentioned above (1.0 mol dm-3
concentration of Zn2+(aq) or Cu2+(aq)
ions).
-
The standard
electrode potential of system is defined as the e.m.f. of an
electrochemical cell in which at 298K, the hydrogen gas (1
atm)/aqueous hydrogen ion (1M [H+(aq)] half-cell
electrode is coupled with the other half-cell electrode system
i.e. the half-cell potential of a system is its e.m.f. compared to the
standard hydrogen electrode
-
In principle, any
accurately known half-cell potential can be used in a cell system to
obtain an unknown half-cell potential by measuring the voltage of
the complete cell using a high resistance voltmeter at virtually zero
current flow .
-
The diagram
Fig.4 below shows the measurement of the
standard copper atom/copper(II) ion half-cell potential using a
standard hydrogen electrode.
-
The hydrogen
electrode consists of an 'inlet' system for hydrogen gas to come into
contact with hydrogen ions into which is dipped a platinum electrode.
The Pt electrode allows electron transfer between the aqueous
hydrogen ions and hydrogen gas.
 Fig.4
-
The copper/aqueous
copper(II) ion half-cell registers +0.34V with respect to the hydrogen
electrode. The electron flow is from the most negative/least positive
potential half-cell (-ve pole) to the most positive/least negative
potential half-cell (-ve pole), i.e. H2/H+ to Cu/Cu2+.
-
Cell notation
(matching diagram and IUPAC convention)
-
Pt|H2(g)|H+(aq)¦¦Cu2+(aq)|Cu(s)
-
On setting up the
cell it is found that the copper strip is the positive electrode i.e.
where the reduction occurs.
-
EØcell
= +0.34V = EØ(+ve/red) - EØ(-ve/ox)
= EØ(Cu2+/Cu) - EØ(H+/H2)
-
so EØ(Cu2+/Cu)
= EØcell + EØ(H+/H2) =
+0.34 + 0.00 = +0.34V
-
for the feasible
cell reaction: Cu2+(aq) + H2(g)
==> Cu(s) + 2H+(aq)
-
The diagram
Fig.5 below shows the measurement of the
standard zinc
atom/zinc ion half-cell potential using a standard hydrogen electrode.
The hydrogen electrode is kept on the left to compare the zinc/zinc
ion with the copper/copper(II) ion half-cell above.
 Fig.5
-
The zinc/aqueous
zinc ion half-cell registers -0.76V with respect to the hydrogen
electrode. In this case the electron flow is the opposite of the
copper-hydrogen cell in Fig.4. the electron flow is always from the
most negative/least positive potential half-cell (-ve pole) to the
most positive/least negative potential (+ve pole), i.e. here it is Zn/Zn2+
to H2/H+. This is the complete opposite of the
effect of the Cu/Cu2+ half-cell (refer
back to Fig.1).
-
Cell notation
(to match the diagram)
-
Pt|H2(g)|H+(aq)¦¦Zn2+(aq)|Zn(s)
-
On setting up the
cell it is found that the zinc strip is negative electrode i.e. where
the oxidation occurs.
-
EØcell
= -0.76V, so EØ(Zn2+/Zn)
= -0.76V
-
for the cell
reaction: Zn2+(aq) + H2(g) ==>
Zn(s) + 2H+(aq)
-
but since the cell
voltage is negative, this is NOT the feasible, so
-
EØcell
= -0.76V +0.76V for the feasible cell reaction
-
Zn(s)
+ 2H+(aq) ==> Zn2+(aq)
+ H2(g)
-
IUPAC cell
notation Zn(s)|Zn2+(aq)¦¦H+(aq)|H2(g)|Pt
-
Fig.6 shows how to measure the
standard potential of a half-cell consisting of an element in two
different oxidation states in aqueous solution.
Fig. 6
-
The diagram above
illustrates in principle how the half-cell potential for the
inter-conversion of iron(II)/iron(III) ions can be measured. The value
obtained is +0.77 so the electrons move from negative pole (Pt) of the
H2/H+ half-cell to the positive pole (Pt) of the
Fe2+/Fe3+ half-cell.
-
Fe3+
(aq) + e-
Fe2+ (aq)
[Fe(III) => Fe(II)]
-
Cell notation
(to match the diagram and IUPAC convention)
-
Pt|H2(g)|H+(aq)¦¦Fe3+(aq),Fe2+(aq)|Pt
-
You can do the
same for many other systems mixing solutions of the two species in the
'half-cell beaker e.g.
-
manganese(II) and
manganate(VII) ions in acidified solution
-
aqueous chlorine
molecule and the chloride ion
-
oxo-vanadium(V)
cation and the oxo-vanadium(IV) cation
-
aqueous iodine
molecule and the iodide ion
-
For the
effect of changing concentration on the value of the electrode
potential see Appendix 1. The Nernst Equation.
7.4
Half-cell potentials,
Electrochemical Series and
using EØcell for reaction feasibility
-
From the
experiments described in 7.2 and 7.3 you can measure a half-cell
potential for many redox systems.
-
From half-cell
potential data you can theoretically calculate the
EØfull
cell reaction and hence the thermodynamic feasibility of the reaction.
-
The electrode
potential is defined as the emf of a cell in which the electrode on
the left is a standard hydrogen electrode (0.00V under standard
conditions, see 7.3), and that on the right is
the electrode in question.
-
Therefore in IUPAC
cell notation for the definition and measurement of a half-cell
electrode potential
-
Pt|H2(g)|H+(aq)¦¦Zn2+(aq)|Zn(s)
gives an EØcell of -0.76V for the zinc
electrode, and for the copper electrode
-
Pt|H2(g)|H+(aq)¦¦Cu2+(aq)|Cu(s)
gives an EØcell of +0.34V, so the arithmetical
sign associated with the electrode potential of the electrode in
question is the polarity of this electrode when constituting the
right-hand electrode
-
Any half-cell
with a negative potential of less than 0.00V can theoretically reduce
aqueous hydrogen ions.
-
Any half-cell with
a positive potential of over 0.00V can theoretically oxidise hydrogen
molecules.
-
HALF-CELL
POTENTIALS are listed below in relative order and the half-cell
reactions shown as
reductions (with oxidation state change and this constitutes
what is called The Electrochemical Series.
-
-2.71
for Na+(aq)
+ e-
Na(s)
-
-2.37
for Mg2+(aq)
+ 2e-
Mg(s)
-
-1.66
for
Al3+(aq) +
3e-
Al(s)
-
-0.76
for
Zn2+(aq) + 2e-
Zn(s)
-
-0.56 for
Fe(OH)3(s) + e-
Fe(OH)2(s) + OH-(aq)
-
-0.44 for
Fe2+(aq) + 2e-
Fe(s)
-
-0.10 for
[Co(NH3)6]3+(aq) + e-
[Co(NH3)6]2+(aq)
-
0.00
for 2H+(aq) + 2e-
H2(g)
the arbitrary assumed standard
-
+0.34 for
Cu2+(aq)
+ 2e- Cu(s)
-
+0.40 for
1/2O2(g) + H2O(l)
+ 2e-
2OH-(aq)
-
+0.54 for
I2(aq) + 2e-
2I-(aq)
-
+0.77 for
Fe3+(aq) + e-
Fe2+(aq)
-
+1.09
for
Br2(aq) + 2e-
2Br-(aq)
-
+1.23 for
1/2O2(g)
+ 2H+(aq)
+ 2e-
H2O(l)
-
+1.33 for
Cr2O72-(aq) + 14H+(aq)
+ 6e-
2Cr3+(aq) + 7H2O(l)
-
+1.36 for
Cl2(aq) + 2e-
2Cl-(aq)
-
+1.51 for
MnO4-(aq) + 8H+(aq)
+ 5e-
Mn2+(aq) + 4H2O(l)
-
+1.77 for
H2O2(aq) + 2H+(aq) + 2e-
2H2O(l)
-
+1.82 for
Co3+(aq) + e-
Co2+(aq)
-
+2.01 for
S2O82-(aq) + 2e-
2SO42-(aq)
-
+2.87
for
F2(aq) + 2e-
2F-(aq)
-
Relative
oxidising and reducing power
-
Down the
list the more
positive/less negative the electrode potential the stronger the
oxidising power of the half-cell system.
-
Up the list the more
negative/less positive the electrode potential the stronger the
reducing power of the half-cell system.
-
So at the top of
the list above you get the powerful reducing reactive metals like
sodium, (-2.71V), and magnesium (-2.37V) with very negative
half-cell potentials.
-
At the bottom of
the list you get the powerful oxidising agents like potassium
manganate(VII), (+1.51V), hydrogen peroxide, (+1.77),
peroxodisulphate ion (+2.01V) and fluorine (+2.87).
-
The
Electrochemical Series
-
The list of
half-cell reactions and half-cell potentials involving the elements
is often referred to as the 'Electrochemical Series', though in
reality, it is the whole list of all of them whether an element in
its elemental state is involved at all.
-
Has as been
mentioned already, it gives an accurate prediction of (i)
oxidising/reducing power, (ii) reactivity trends for metals or
non-metals and (iii) which ions are likely to be preferentially
discharged in electrolysis.
-
Electrode
potential and patterns of 'reactivity' e.g.
-
The reactivity
series of metals: 'Up' the metal reactivity series the half-cell
potential voltages becomes more negative and the metal becomes 'more
reactive' e.g. Na (-2.71V) > Mg (-2.37) > Zn (-0.76V) > Fe (-0.44) > Cu (+0.34) etc. Metallic elements react by electron loss (ox.
state increase) to form a positive cation (e.g. magnesium ion Mg2+),
so, as the electron loss potential increases, so the metallic
element's reactivity increases. A metallic element more -ve/least
+ve potential
-
The Group 7 Halogen
reactivity series: Down group 7 the reactivity decreases as the
oxidising power decreases. The half-cell potential decreases down
the group e.g. F (+2.87) > Cl (+1.36) > Br (+1.09) > I (+0.54V). Halogen
elements react by electron gain (ox. state decrease) to form a
single covalent bond (e.g. HCl) or the negative anion (e.g. chloride
ion Cl- in NaCl), so, as the electron accepting capacity
power decreases, so does the element's reactivity.
-
Other half-cells,
they don’t have to simple metal/ metal ions, all you need is two
interchangeable oxidation states
-
e.g. Cl2(aq)/Cl-(aq)
or Mn2+(aq)/MnO4-(aq)
etc. but both components of the half-cell must be in the same solution
and in contact with a platinum electrode that connects to the rest of
the circuit.
-
Relating EØcell to the direction of overall chemical change
and feasibility of reaction. If you
calculate a -ve cell voltage for a given reaction, that is not the way
cell reaction goes! please reverse the equation re-calculate!
-
The Ebattery-cell
must be >0.00V for the cell, and any
other redox reaction, to be theoretically feasible.
-
The free
energy change must be negative <0 for the cell reaction to be
feasible (matching the Ecell rule of >0V for
feasibility).
-
ΔGØcell = -nEØF J
(ΔG not needed by all A2 syllabuses)
-
n =
number of moles of electrons transferred in the reaction per mol of
reactants involved,
-
EØ is the
Emf for the overall reaction in volts,
-
F = the Faraday
constant (96500 C mol-1).
-
e.g. for the
zinc-copper Daniel cell producing +1.10V,
-
2 electrons
transferred (M2+(aq) + 2e- <=> M(s))
-
cell reaction: Cu2+(aq)
+ Zn(s) ==> Cu(s) + Zn2+(aq)
-
ΔGØcell
= -2 x 1.10 x 96500 = 212300 Jmol-1 or
212.3 kJmol-1
-
These theoretical
calculations can be used for any redox reaction BUT there are
limitations:
-
You can’t say
the reaction will definitely spontaneously happen (go without
help!) because there may be rate
limits especially if the reaction has a high activation energy or a
very low
concentration of an essential reactant.
-
However you can
employ a catalyst, raise reactant concentrations or raise the temperature to get the reaction
going! There is usually a way of getting most, but not all,
feasible reactions to actually occur.
-
For the
effect of changing concentration on the value of the electrode
potential see Appendix 1. The Nernst Equation.
7.5
Electrochemical cells
('batteries') and fuel cell systems
-
Primary Cells
-
The first
primary cells were galvanic cells in which the reactants are sealed
in when manufactured and ready for immediate use i.e. the chemicals
are capable of spontaneously reacting and the redox changes released
energy as an electron flow (rather than heat energy). They cannot be
recharged, and when they run down, that is the chemical reactants
are completely depleted, they stop working and are discarded!
-
The common ones
such as the zinc-carbon batteries are used in
torches, radios, cameras, flashlights, cameras etc.
-
Hopefully
recycling of the materials will be increasingly possible as well as
being worthwhile from the point of view of conserving valuable
resources and minimising environmental pollution from poisonous
metals or their compounds.
-
Dry cell
zinc-carbon battery, 1.5V falling to 0.8V as reaction
products build up.
-
A rod of carbon
cathode (+) is set into a paste of zinc and ammonium chloride (weakly acid
electrolyte) and fine particles of manganese(IV) oxide and carbon
contained in a zinc anode (-) 'compartment'. Although called a 'dry'
cell, the paste must contain water, which is thickened with e.g.
starch.
-
Zn(s)|ZnCl2(aq),NH4Cl(aq)|MnO(OH)(s)|MnO2(s)|Cgraphite
-
anode
discharging reaction (i) Zn(s) + 4NH3(aq)
==> [Zn(NH3)4]2+(aq) + 2e-
-
cathode
discharging reaction (ii) MnO2(s) + NH4+(aq)
+ e- ==> MnO(OH)(s) + NH3(aq)
-
overall working
cell reaction (iii) Zn(s) + 4NH3(aq)
+ 2MnO2(s) + 2NH4+(aq)
-
oxidation state
changes: (i) oxidation Zn(0) ==> Zn(+2), (ii) reduction Mn(IV) ==> Mn(III)
-
Advantages: Low
cost and non-toxic materials.
-
Disadvantages:
Cannot be recycled, can leak (weak acid electrolyte reacts with
zinc), short shelf-life, unstable voltage and current (as battery
'runs down') and low power.
-
The dry cell
alkaline battery, 1.5-1.9V depending on constituents.
-
The electrolyte
is the strong base sodium/potassium hydroxide contained in
'typically' zinc anode (-) compartment and a cathode of
manganese(IV) oxide. Metals like cadmium or aluminium can be used as
the anode, and copper, iron, lead, mercury, nickel and silver oxide
can be used as cathode materials.
-
Zn(s)|ZnO(s)|OH-(aq)|Mn(OH)2(s)|MnO2(s)|Cgraphite
-
anode
discharging reaction (i) Zn(s) + 2OH-(aq)
==> ZnO(s) + H2O(l) + 2e-
-
cathode
discharging reaction (ii) MnO2(s) + 2H2O(l)
+ 2e- ==> Mn(OH)2(s) + 2OH-(aq)
-
overall cell
reaction (iii) Zn(s) + MnO2(s) + H2O(l)
==> ZnO(s) + Mn(OH)2(s)
-
oxidation state
changes: (i) oxidation Zn(0) ==> Zn(+2), (ii) reduction Mn(IV) ==> Mn(II)
-
Advantages: Low
cost and non-toxic materials. The alkaline electrolyte does not
readily react with zinc (compare Zn-C cell above) giving a much
longer shelf-life (5 years) and the current and voltage are steady
(handy in smoke alarms!) due to the strong base/alkali electrolyte
having a smaller resistance the ammonium chloride-carbon paste.
-
Disadvantages:
Cannot be recycled, more expensive due to extra sealing and low
power.
-
Fuels cells
are a development of primary cells but with one
significant difference from their predecessors, the chemical
potential energy source or 'fuel' can be continually fed in to give
the cell a long active life.
-
Secondary Cells
(electrical 'accumulators')
-
Secondary
cells are galvanic cells that must be charged before they can be
used and rechargeable many times. In the charging process, the
spontaneous-feasible cell reaction that produces electrical energy
is reversed, so building up the chemical potential of the cell
system.
-
Lead-acid
storage battery, 2 V. (usually 6 in series to give 12V
supply).
-
The electrodes
are initially hard lead-antimony alloy plates coated in a paste of
lead(II) sulphate encased in dilute sulphuric acid. During the first
charging some of the lead(II) sulphate is reduced lead(0) on one of
the electrodes (this will acts as the (-) anode in discharging).
Simultaneously in charging, lead(II) sulphate is oxidised to lead(IV)
oxide on the other electrode which acts as the cathode (+) in
discharging.
-
Pb(s)|PbSO4(s)|H+(aq),HSO4-(aq)|PbO2(s)|PbSO4(s)|Pb(s)
-
anode
discharging reaction (i) Pb(s) + HSO4-(aq)
==> PbSO4(s) + H+(aq) + 2e-
-
cathode
discharging reaction (ii) PbO2(s) + 3H+(aq)
+ HSO4-(aq) + 2e-
==> PbSO4(s) + 2H2O(l)
-
working cell
reaction (iii) PbO2(s) + 2H+(aq)
+ 2HSO4-(aq) + Pb(s)
==> 2PbSO4(s) + 2H2O(l)
-
oxidation state
changes: (i) oxidation Pb(0) ==> Pb(II) : (ii) reduction Pb(IV) ==> Pb(II)
-
The charging
reactions will be the opposite of (i) and (ii)
-
Advantages:
Inexpensive, high power density (can car starter motor as well as
lights), long shelf life, readily recharges, so has a long working
life of many years.
-
Disadvantages:
Lead needs to be recycled to avoid environmental contamination,
sometimes generates hydrogen gas at the cathode when charging
(explosive in air + spark) - though seem to be made of a high
standard these days in completely sealed units.
-
Uses: Car
batteries.
-
The NiCad
Cell, 1.25 V.
-
diagram?
-
Cd(s)|Cd(OH)2(s)|KOH(aq)|Ni(OH)3(s)|Ni(OH)2(s)|Ni(s)
-
anode
discharging reaction (i) Cd(s) + 2OH-(aq)
==> Cd(OH)2(s) + 2e-
-
cathode
discharging reaction (ii) 2Ni(OH)3(s) + 2e-
==> 2Ni(OH)2(s) + 2OH-(aq)
-
overall cell
reaction (iii) Cd(s) + 2Ni(OH)3(s) ==>
Cd(OH)2(s) + 2Ni(OH)2(s)
-
oxidation state
changes: (i) oxidation Cd(0) ==> Cd(II), (ii) reduction Ni(III) ==> Ni(II)
-
The charging
reactions will be the opposite of (i) and (ii)
-
Advantages:
-
Disadvantages:
Cadmium is a toxic metal.
-
Uses: Portable
computers
-
The voltage and power
available from a battery or cell
-
The voltage
depends primarily on the materials used in the chemical process
generating the electrical energy.
-
Since the
voltage is small from an individual cell, (typically 0.4 to 2V),
several cells can be assembled in parallel to increase the voltage.
-
The power
primarily depends on the amount of material and how fast the
chemicals can react. For a single cell the voltage will depend on
the half-cell potentials of chemicals employed, but the current flow
depends on the bulk reaction rate of the chemicals.
7.6
Electrolysis
and the electrochemical series
-
Right at the
start, in 7.2, it was pointed out that electrode potentials are
based on equilibria such as ...
-
Cu2+(aq)
+ 2e- Cu(s)
-
Since these
reactions are reversible, they can be used 'spontaneously' in cells to
generate electrical energy via the overall redox reaction BUT the
reverse process can be 'enforced' in electrolysis by applying a
potential difference ('voltage') across a suitable aqueous solution or
molten compound.
-
The GCSE notes on
the Extra Electrochemistry
page contains most of the electrolysis details you need for advanced
level, but there are some other points to make, which are outlined
below.
-
The
electrochemical series of half-cell reaction potentials, can be used
to predict which ions are likely to be preferentially discharged to
form electrolysis products on the cathode(+ pole in electrolysis) or
anode (+ pole in electrolysis).
-
At the negative
(-) cathode electrode (reduction half reaction)
-
The more
positive/less negative the half-cell potential, the more easily the
cation is discharged by reduction.
-
e.g. copper metal
from copper(II) ions (+0.34V) will be discharged deposited on a
cathode preferentially from iron from iron(II) ions (-0.44V) from a
solution containing both ions.
-
Since the process
involves electron gain, the cation with the greatest potential to gain
electrons is the one that is preferentially discharged
-
i.e.
Cu2+(aq)
+ 2e- ==> Cu(s) occurs
more readily than Fe2+(aq)
+ 2e- ==>
Fe(s)
-
At the positive
(+) anode electrode (oxidation half reaction)
-
The less positive
the half-cell potential, the more easily the anion is discharged by
oxidation.
-
e.g. in an aqueous
mixture of bromide and chloride ions, bromide forms bromine (+1.09V)
more readily than chloride ion forms chlorine (+1.36V).
-
Since the process
involves electron loss, the ion which is the most readily formed will
be the ion which is least readily discharged.
-
i.e.
2Br-(aq)
==> Br2(aq) + 2e- occurs more
readily than 2Cl-(aq)
==> Cl2(aq) + 2e-
-
But sometimes
other factors come into consideration e.g.
-
Concentrated or
dilute sodium chloride solution (brine)
-
In concentrated
NaCl(aq) evolution of chlorine predominates from
2Cl-(aq)
==> Cl2(aq) + 2e-
-
but in very dilute NaCl(aq)
evolution of oxygen predominates from
4OH-(aq)
==> O2(g) + 2H2O(l)
+ 4e-
-
Theoretically
oxygen should be discharged first, but the hydroxide ion concentration
is so low compared to the chloride ion that little oxygen is produced
on anode-ion collision probability. Also, oxygen has a high 'overpotential'
(which you can equate to a high activation energy giving a very slow
rate of reaction
-
) which also
inhibits its formation.
-
There will be
differences in electrolysis products between molten salts and aqueous
solutions due to the presence of water.
-
There will be
differences in electrolysis products between inert and non-inert
electrodes.
-
See the list of
electrode reactions-products on the
GCSE Extra Electrochemistry
page section 2a.
-
Also on the GCSE notes
Extra Electrochemistry
page is ...
-
All the
basics of
electrolysis, e.g. theory, methods, electrode equations etc.
including the explanation and discussion of the electrolysis of acidified water and
of dilute/concentrated sodium
chloride solution (brine) are on this GCSE-GCE notes page.
-
There
are links to
descriptions of industrial electrolysis processes including
'reactive' metal extraction e.g. aluminium, copper
purification/plating and electrolysis of brine (aqueous sodium
chloride).
-
Electrolysis calculations
e.g. example calculations/explanations of the relative amounts of products formed depends on the ion
charge, current flowing and how long the electrolysis is done for
etc. is dealt with on this GCSE-GCE notes page.
7.7
Exemplar questions
Appendix 1. The
Nernst Equation
-
The Nernst
equation allows the calculation-prediction of what a half-cell
potential will be for a different ion concentration (or temperature) from
that of the standard conditions i.e. 1.0 mol dm-3 and
298K.
-
The Nernst
equation is not needed (as far as I know?) for UK GCE-A2 and IB
courses, but it may be needed in coursework projects! You do NOT
have to present the thermodynamic derivation of the equation!
-
For a reduction
half-cell equilibrium i.e.
oxidised state + electrons
reduced state, the Nernst equation for the variation of
the half-cell potential is
-
E = EØ
+ (RT/nF) ln {[ox]/[red]}
-
E =
half-cell potential in V which varies with the molar concentrations
[mol dm-3] of the oxidised and reduced species.
-
EØ
is the standard half-cell potential e.g. at 298K when the molarities
are 1.0 mol dm-3.
-
in the case of
two aqueous ions like Fe2+/Fe3+, [red] = [ox]
i.e. equal concentrations, so that E varies with the ratio of the
two ion concentrations.
-
R = 8.314
J mol-1 K-1 (the ideal gas constant).
-
T the
absolute temperature in K (Kelvin = oC + 273).
-
n = moles
of electrons transferred per mole of reactants.
-
F =96500
C mol-1 (the Faraday constant).
-
ln =
natural/Napierian logarithm.
-
[ox/red]
= molar concentration of the oxidised or reduced species. University
students might well be using activity values as well as molarity
values, in which case the activity of the metal is considered to be
unity, i.e. aM(s) = 1.000 (see next paragraph).
-
Note the equivalent
equations (lg = log to base 10, log or log10)
-
E = EØ - (RT/nF) ln {[red]/[ox]}
-
E = EØ - (2.303RT/nF) lg {[red]/[ox]}
-
E = EØ + (2.303RT/nF) lg {[ox]/[red]}
-
For a
metal-metal half-cell equilibrium:
Mn+(aq)
+ ne-
M(s)
-
the Nernst
equation is EMn+/M = EØMn+/M + (RT/nF) ln
{[Mn+(aq)]/[M(s)]}
-
BUT the
concentration of the metal cannot change so its molarity or
'thermodynamic activity' is considered to be unity, in which case
the Nernst equation for the simple metal-metal ion equilibrium is
-
(i)
EMn+/M = EØMn+/M
+ (RT/nF) ln [Mn+(aq)]
-
(using natural
logarithm ln, watch on calculator!)
-
or (ii) EMn+/M = EØMn+/M
+ (2.303RT/nF) log10 [Mn+(aq)]
-
(log to base 10, lg, log or log10)
-
at 298K for ln
expression: RT/nF = (8.314 x 298)/(n x 96500) = 0.0257/n
-
for lg or log10:
RT/nF = (2.303 x 8.314 x 298)/(n x 96500) = 0.0591/n
-
Note: (i) ,
(ii)
-
Five examples of
calculations using the Nernst equation are outlined below.
-
What is the
half-cell potential for copper when dipped into a 2.0 mol dm-3
solution of copper(II) sulphate?
-
Cu2+(aq)
+ 2e-
Cu(s) (EØ =
+0.342V)
-
ECu2+/Cu = EØCu2+/Cu
+ (RT/nF) ln [Cu2+(aq)]
-
E =
+0.342 + {(8.314 x 298)/(2 x 96500)} ln [2.0]
-
E =
+0.342 + 0.01284 ln(2)
-
ECu2+/Cu
= +0.342 + 0.009 = 0.351 V (3sf, 0.35V 2sf)
-
Le
Chatelier comment: The increase in Cu2+
concentration increases the oxidising potential of the
half-cell i.e. a more positive half-cell potential acting from left to
right in the equilibrium.
-
What is the
half-cell potential of zinc dipped into a 0.1 molar zinc
sulphate solution?
-
Zn2+(aq)
+ 2e-
Zn(s) (EØ =
-0.763V)
-
EZn2+/Zn = EØZn2+/Zn
+ (2.303RT/nF) lg [Zn2+(aq)]
-
E =
-0.763 + {(2.303 x 8.314 x 298)/(2 x 96500)} lg [0.1]
-
EZn2+/Zn
= -0.763 + (-0.029) = -0.792 V (3sf, -0.79V
2sf)
-
Le
Chatelier comment: The decrease in Zn2+
concentration increases the reducing potential of the
half-cell i.e. a more negative potential half-cell potential
acting from right to left in the equilibrium.
-
What is the
theoretical half-cell potential of aluminium dipped into a 0.001
molar aluminium sulphate solution?
-
Al3+(aq)
+ 3e-
Al(s) (EØ =
-1.662V)
-
EAl3+/Al = EØAl3+/Al
+ (RT/nF) ln [Al3+(aq)]
-
E =
-1.662 + {(8.314 x 298)/(3 x 96500)} ln [0.001]
-
EAl3+/Al
= -1.662 + (-0.059) = -1.721 V (4sf, -1.72V
3sf)
-
Le
Chatelier comment: The decrease in Al3+
concentration increases the reducing potential of the
half-cell i.e. a more negative potential half-cell potential
acting from right to left in the equilibrium.
-
What is the
half-cell potential for a solution of iron(II) and iron(III)
ions, containing 0.20 mol dm-3 of Fe2+
and 0.10 mol dm-3 of Fe3+?
-
Fe3+(aq) + e-
Fe2+(aq) (EØ = +0.771V)
-
EFe3+/Fe2+ = EØFe3+/Fe2+
+ (RT/nF) ln [ox]/[red]
-
EFe3+/Fe2+ = EØFe3+/Fe2+
+ (2.303RT/nF) lg [Fe3+(aq)]/[Fe2+(aq)]
-
E =
+0.771 + (0.0591/1) log10 (0.1/0.2)
-
EFe3+/Fe2+
= +0.771 + (-0.018) = +0.753 V (3sf, 0.75V
2sf)
-
Le
Chatelier comment: The lower Fe3+
concentration compared to Fe2+, decreases the
oxidising potential of the
half-cell i.e. a less positive potential half-cell potential
acting from left to right in the equilibrium.
-
Very
accurate measurement
half-cell electrode potentials can be used to determine the
concentration of an ion (e.g. pH with glass electrode). When a copper strip
(+ve pole) is dipped
into a copper(II) ion solution, and combined with a saturated
KCl-calomel electrode (-ve pole, E = 0.244V at 298K), the cell
voltage measured 0.088 V at 298K. From the information deduce
the concentration of copper(II) ions in the solution.
-
(i)
You first need to calculate the copper half-cell
potential.
-
Ecell
= E+pole - E-pole
-
Ecell
= ECu/Cu2+ - Ecalomel
-
Ecell
= 0.088 = ECu/Cu2+ - (+0.244) = ECu/Cu2+
- 0.244
-
therefore ECu/Cu2+ = Ecell + 0.244 =
0.088 + 0.244 = +0.332 V
-
(ii)
You then need to rearrange the Nernst equation to calculate
the concentration.
-
EØCu2+/Cu
= +0.342V, and from example 1.
-
ECu2+/Cu = EØCu2+/Cu
+ (RT/2F) ln [Cu2+(aq)]
-
rearranging (with care!) gives
-
(RT/2F)
ln [Cu2+(aq)] = ECu2+/Cu
- EØCu2+/Cu
-
ln [Cu2+(aq)]
= (ECu2+/Cu - EØCu2+/Cu)/(RT/2F)
-
[Cu2+(aq)]
= e{(ECu2+/Cu - EØCu2+/Cu)/(RT/2F)}
-
[Cu2+(aq)]
= e{(0.332 - 0.342)/0.01284}
-
|
