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Brown's A Level Chemistry Revision Notes
Theoretical–Physical
Advanced Level
Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 7.4
7.4 Half–cell
potentials, Electrochemical Series and calculating & using Eθcell for reaction feasibility
Using Half–cell
potentials , the electrochemical series and the relative oxidising power or
reducing power of a half–reaction. How to calculate and use Eθcell
(Emf of cell) to determine the feasibility of a redox reaction.
Chemical Equilibrium Notes Part 7 Index
7.4
Half–cell potentials,
Electrochemical Series and
using Eθcell 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 Eθfull
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
The hydrogen electrode and standard conditions), 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
-
Zn2+(aq)|Zn(s)||Pt|H2(g)|H+(aq)
gives an Eθcell of –0.76V for the zinc
electrode, and for the copper electrode
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Pt|H2(g)|H+(aq)||Cu2+(aq)|Cu(s)
gives an Eθcell 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, these can be represented in the form of an
electrode potential chart (above right).
-
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 changes in (), and this constitutes
what is called The Electrochemical Series i.e from the most
negative half–cell electrode potentials to the most positive
half–cell potentials.
-
–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)
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+1.51 for
MnO4–(aq) + 8H+(aq)
+ 5e–
Mn2+(aq) + 4H2O(l)
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+1.77 for
H2O2(aq) + 2H+(aq) + 2e–
2H2O(l)
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+1.82 for
Co3+(aq) + e–
Co2+(aq)
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+2.01 for
S2O82–(aq) + 2e–
2SO42–(aq)
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+2.87
for
F2(aq) + 2e–
2F–(aq)
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Relative
oxidising and reducing power
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Down the
list the more
positive/less negative the electrode potential the stronger the
oxidising power of the half–cell system.
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Up the list the more
negative/less positive the electrode potential the stronger the
reducing power of the half–cell system.
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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).
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The
Electrochemical Series
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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.
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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.
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Electrode
potential and patterns of 'reactivity' e.g.
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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
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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.
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Other half–cells,
they don’t have to simple metal/ metal ions, all you need is two
interchangeable oxidation states
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Relating Eθcell to the direction of overall chemical change
and feasibility of reaction.
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Examples of predicting
the feasibility of a reaction by doing a theoretical
calculation using the electrode potential data from above.
I've included the oxidation state changes too.
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(i) You probably know
that aqueous chlorine can readily oxidise iron(II) ions to
iron(III) ions, BUT can aqueous bromine also oxidise
iron(ii) ions to iron(III)?
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First set up the
equation is: 2Fe2+(aq) + Br2(aq)
==> 2Fe3+(aq) + 2Br-(aq)
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Reduction: Br2
==> Br- (0 to -1); Oxidation: Fe2+
==> Fe3+ (+2 to +3)
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Eθreaction = Eθ(reduced)
– Eθ(oxidation) = (+1.09V) - (+0.77V) =
+0.32V, >0V so reaction feasible
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I/I2
half-cell potential is only +0.54V and so iodine
cannot oxidise Fe2+ to Fe3+,
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The equation is:
2Fe2+(aq) + I2(aq) ==> 2Fe3+(aq)
+ 2I-(aq)
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Eθreaction
= Eθ(red) – Eθ(ox)
= (+0.54V) - (+0.77V) = -0.23V, so NOT feasible
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Without doing a
second calculation you can immediately deduce that
iron(III) ions can oxidise iodide ions, the reverse
reaction must be feasible!
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(ii) Can cheap iron
displace more expensive zinc from an aqueous solution of a
zinc salt?
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Reaction: Fe(s) + Zn2+(aq)
===> Zn(s) + Fe2+(aq)
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Reduction: Zn2+
==> Zn (+2 to 0); Oxidation: Fe ==> Fe2+ (0
to +2)
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Eθreaction
= Eθ(red) – Eθ(ox)
= (-0.76V) - (-0.44V) = -0.32V, so NOT feasible
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However, cheap (but
more 'reactive') iron CAN displace more valuable copper
from an aqueous solution of copper(II) sulfate.
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Reaction: Fe(s) + Cu2+(aq)
===> Cu(s) + Fe2+(aq)
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Reduction: Cu2+
==> Cu (+2 to 0); Oxidation: Fe ==> Fe2+ (0
to +2)
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Eθreaction
= Eθ(red) – Eθ(ox)
= (+0.34V) - (-0.44V) = -+0.78V, so feasible
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(iii) It doesn't matter
how complicated the balanced equation, the Eθreaction
calculation is quite simple as long as you use the correct
electrode potential data!
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e.g. can potassium
manganate(VII) oxidise hydrochloric acid?
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Balanced equation: 2MnO4–(aq) +
16H+(aq) + 10Cl–(aq)
==>
2Mn2+(aq)
+ 8H2O(l) + 5Cl2(aq)
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Reduction: MnO4–
==> Mn2+ (+7 to +2); Oxidation: Cl–
==> Cl2 (-1 to 0)
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Eθreaction
= Eθ(red) – Eθ(ox)
= (1.51V) - (+1.36V) = +0.15V, so feasible
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(iv)
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The Ebattery–cell
must be >0.00V for the cell, and any
other redox reaction, to be theoretically feasible.
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The Gibbs
Free Energy change ΔG is only on some A level syllabuses.
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The free
energy change must be negative <0 for the cell reaction to be
feasible (matching the Ecell rule of >0V for
feasibility).
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ΔGθcell =
–nEθF
(J)
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n =
number of moles of electrons transferred in the reaction per mol of
reactants involved,
-
Eθ is the
Emf for the overall reaction in volts i.e. for the theoretical cell,
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F = the Faraday
constant (96500 C mol–1).
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e.g. for the
zinc–copper Daniel cell producing +1.10V,
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2 electrons
transferred (M2+(aq) + 2e– <=> M(s))
-
cell reaction: Cu2+(aq)
+ Zn(s) ==> Cu(s) + Zn2+(aq)
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ΔGθcell
= –2 x 1.10 x 96500 = –212300 Jmol–1 or
–212.3 kJmol–1
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See also
Appendix 2 Free Energy
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These theoretical
calculations can be used for any redox reaction BUT there are
limitations:
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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
Chemical Equilibrium Notes Part 7 Index
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Chemical Equilibrium Notes Part 7 Index
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