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Brown's Chemistry Advanced A Level Notes - Theoretical–Physical
Advanced Level
Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 7
Part 7.3 The Hydrogen
Electrode, standard electrode potential, standard conditions and cell notation
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ALL my advanced A
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What is, and how is, a standard hydrogen
electrode is constructed?
The standard electrode potential of a
half–cell is based on an arbitrary zero for the hydrogen ion/hydrogen
gas half–cell.
What is a standard electrode potential? What do we mean
by standard conditions?
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 EθH2(g)/H+(aq) = 0.00V
under standard conditions.
-
The half–cell
equation is: 2H+(aq)
+ 2e–
H2(g) for
the standard hydrogen electrode.
-
The hydrogen
electrode is quite difficult to set up accurately with a
specially prepared inert platinum electrode, and in practice, other
accurate reference electrodes are used e.g. the calomel electrode.
-
You should also
remember that half–cell potentials are dependent on the
concentration of any ion involved in any of the electrode reactions,
whether a product or reactant.
-
θ
Means standard conditions, here, for any half–cell electrode
system
-
298 K for
temperature (25oC), for any standard electrode
-
1 atmosphere pressure
(101kPa) of hydrogen (or any other) gas H2(g) for hydrogen electrode,
-
1 mol dm–3
concentration of H+(aq) (hydrogen ions for
hydrogen electrode),
-
e.g. 1 mol dm–3 HCl(aq)
maybe quoted with caution, BUT they represent an unfortunate
over–simplification, read on!
-
... meaning the choice
of acid requires some comment.
-
Hydrochloric acid
is the most widely quoted (exam board mark schemes, textbooks,
internet technical references etc),
and at school/college pre–university level, it seems perfectly
acceptable in answers to quote the electrolyte for the hydrogen
electrode as 1.00 mol dm–3 HCl(aq) i.e. equal to 1.00 mol
dm–3 H+(aq)
-
Hydrochloric acid is
an uncomplicated monobasic/monoprotic acid, non–oxidising and is
fully ionised in water.
-
With a pKa
= –7, Ka = 1.0 x 107 mol dm–3, no problem
on ionisation here! and a 1 molar solution theoretically gives a
1 molar solution of hydrogen ions.
-
However if you
research this aspect of the hydrogen electrode on the internet, you find that a
1.8 molar solution of hydrochloric acid may be used to give an
activity' of [H+(aq)] of 1.00 mol dm–3,
but please do not worry about this, this is undergraduate
stuff!!
-
I'm not going to the
details of why, but unfortunately, in any equilibrium situation,
solutions do NOT behave ideally and I'm afraid 1 molar HCl ain't
1 molar H+(aq).
-
However, in pH
calculations at pre–university level, its acceptable to assume
this, so why not in an answer to a hydrogen electrode question?
-
There is an
electrode potential limit to the use of hydrochloric acid.
-
The half–cell
potential for Cl2/Cl– is +1.36V, therefore
if the other cell has an electrode potential of > +1.36V, you
may get oxidation of the chloride ion at the hydrogen
electrode.
-
This is NOT the
standard reaction required, i.e. you will not measure the
correct cell voltage from which to calculate the standard
electrode potential of the half–cell under investigation!
-
Sulfuric acid is
sometimes quoted i.e. 0.50 mol dm–3 H2SO4(aq),
should on complete ionisation give a 1 molar solution of hydrogen
ions.
-
However, it is a
dibasic/diprotic acid, the first ionisation is ~100% complete (pKa1
very negative, Ka extremely high), therefore ~100%
ionisation.
-
However, the 2nd
ionisation/dissociation is far from complete,
-
pKa2 =
1.92, Ka = 1.20 x 10–2 mol dm–3, so
sulfuric acid is not fully ionised and a 0.50 molar solution of
sulphuric acid will NOT give a 1.00 molar solution of hydrogen ions.
-
However, in pH
calculations at pre–university level, it seems acceptable to assume
this, so why not in an answer to a hydrogen electrode question?
-
BUT I've also done a
calculation to show that a 0.500 molar solution of sulfuric acid
only contains 0.511 mol dm–3 of hydrogen ions, which
means technically, 0.50 molar sulphuric acid is not a good answer!
-
See bottom of
equilibrium page 5b on strong acid
calculations
for more details on this.
-
Nitric acid may
seem a good monobasic/monoprotic alternative to hydrochloric acid.
-
It's pKa = –1.4, Ka = 40
mol dm–3, so is pretty well fully ionised (~97.5%), but
it is a strong oxidising agent and may theoretically oxidise the hydrogen gas
and there are other possible complications too.
-
Phosphoric acid
would be useless, its a tribasic/triprotic acid, too complicated,
and the first ionisation is far from complete (see below), quoting 0.333 molar H3PO4
is not a good idea!
-
H3PO4(aq)
H+(aq)
+ H2PO4–(aq)
-
for H3PO4(aq)
pK1 = 2.15, Ka1 = 7.08 x 10–3 mol
dm–3, and the other two Ka values are much smaller!
-
Final comment and
conclusion ...
-
As far as I can tell, 1.00 mol dm–3 HCl(aq)
will be the more accurate to use rather than 0.5 mol dm–3
H2SO4(aq), but don't quote anything else,
and there is an alternative and absolutely correct and universal
answer!
-
The fact of the
matter is, at this pre–university academic level, a simplified
answer may suffice, as to which acid, and its concentration, to
use in a hydrogen electrode, but my advice is, make sure quote
in any answer ...
-
.. an acid
solution equal to a 1 molar solution of hydrogen ions
-
i.e. [H+(aq)]
= 1.00 mol dm–3, and that aught to satisfy
any answer scheme, and if it doesn't, then chief examiners
need to do some research for themselves!
-
1 mol dm–3
concentration of metal ion in the other half–cells mentioned
in these pages.
-
e.g. 1.0 mol dm–3
concentration of Zn2+(aq) or Cu2+(aq)
ions.
-
Bearing in mind the
above discussion on the hydrogen ion concentration in standard
cells, it should be absolutely correct in an exam to quote
molarities of all ions as 1 molar for any standard cell at 298K, 1
atm.
-
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.
-
Why use a high
resistance voltmeter?
-
By minimising the
current flow to ~zero, inaccurate readings due to polarisation
are avoided.
-
Polarisation is caused
by the build up of the cell reaction products or the diminution of
reactants.
-
Any change in
concentrations will affect the half–cell potential, giving
non–standard readings.
-
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 platinum electrode
must specially prepared and beyond the scope of these notes, but
more importantly, the Pt electrode allows electron transfer between the aqueous
hydrogen ions and hydrogen gas, ideally creating true equilibrium
conditions.
-
When connected to another
half–cell and high resistance voltmeter a salt bridge
(electrolyte solution/gel) is needed to complete the electrical circuit and allows the slow drift of
ions in both directions (NOT electrons). The best salt bridge
contains a gel of potassium chloride solution, but even a bit of filter
paper soaked in saturated potassium chloride solution will work for
simple school/college experiments.
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)
-
Note the standard
conventions in common use
-
the | notation indicates
a phase boundary
-
the
||
notation 'divides' the two half cells and
-
On setting up the
cell it is found that the copper strip is the positive electrode i.e.
where the reduction occurs.
-
Eθcell
= +0.34V = Eθ(+ve/red) – Eθ(–ve/ox)
= Eθ(Cu2+/Cu) – Eθ(H+/H2)
-
Eθ(Cu2+/Cu)
= Eθcell + Eθ(H+/H2) =
+0.34 + 0.00 = +0.34V
-
for the feasible
cell reaction: Cu2+(aq) + H2(g)
==> Cu(s) + 2H+(aq)
-
Here, polarisation
is caused by the build up of the cell reaction products e.g. H+
ions from the oxidation of the hydrogen gas and the diminution of the
copper ions Cu2+ on their reduction to copper.
-
This is reduced by using a
very high resistance voltmeter.
-
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
Fig.1
-
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 (see Fig.1 above).
-
Cell notation
(to match the diagram)
-
Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt
-
Note the standard
conventions in common use
-
the | notation indicates
a phase boundary
-
the
||
notation 'divides' the two half cells (salt bridge connection).
-
On setting up the
cell it is found that the zinc strip is negative electrode i.e. where
the oxidation occurs.
-
Eθcell
= –0.76V, so
Eθ(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
-
Eθcell
= –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
-
Here, polarisation
is caused by the build up of the cell reaction products e.g. Zn2+
ions from the oxidation of the zinc electrode and the diminution of
hydrogen ions H+ on their reduction to hydrogen gas.
-
This is reduced by using a
very high resistance voltmeter.
-
The best salt bridge
contains a gel of conducting potassium chloride solution, but even
filter paper soaked in saturated potassium chloride solution will work
for simple school/college experiments.
-
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
-
Note the standard
conventions in common use
-
the | notation indicates
a phase boundary
-
the
||
notation 'divides' the two half cells and
-
The best salt bridge
contains a gel of conducting potassium chloride solution, but even
filter paper soaked in saturated potassium chloride solution will
work for simple school/college experiments.
-
You can do the
same for many other systems mixing solutions of the two species in the
same
'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,
WHAT NEXT?
INDEX 7.
Redox equilibria, half–cells, electrode potentials,
electrolysis, electrochemical series
Index of ALL my chemical equilibrium
context revision notes Index
Advanced Equilibrium Chemistry Notes 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 between two
phases, solubility product Ksp, common ion effect,
ion–exchange systems *
Part 5. pH, weak–strong acid–base theory and
calculations * Part 6. Salt hydrolysis,
acid–base titrations–indicators, pH curves and buffers * Part 7.
Redox equilibria, half–cell electrode potentials,
electrolysis and electrochemical series
*
Part 8.
Phase equilibria–vapour
pressure, boiling point and intermolecular forces watch out for sub-indexes
to multiple sections or pages
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