Download Electrode Potentials

THE ELECTRONIC THEORY OF
OXIDATION & REDUCTION
1. Oxidation & Reduction
Simple definitions of oxidation and reduction are based on the loss/gain of oxygen or
the loss/gain of hydrogen. Oxidation is the gain of oxygen or the loss of
hydrogen; reduction is the loss of oxygen or the gain of hydrogen. These
definitions can only be used when a chemical reaction involves hydrogen and
oxygen, and therefore their usefulness is limited.
A more basic and more useful definition of oxidation and
reduction is based on the loss/gain of electrons.
OXIDATION IS LOSS OF ELECTRONS
REDUCTION IS GAIN OF ELECTRONS
In reactions involving simple ions, it is usually easy to tell
whether electrons are lost or gained, but it is less easy to
tell when complex ions or covalent molecules are involved.
Oxidation number is a useful concept for helping to decide
in these more awkward cases.
2. Oxidation Number
The oxidation number is used to express the oxidation state of an element, whether
as the uncombined element or when combined in a compound; it consists of a + or –
sign followed by a number, or it is zero.
Atoms of elements have no overall charge and are therefore given an oxidation
number of zero. When two elements combine, the atoms or ions of the more
electropositive element have a positive oxidation state, and those of the more
electronegative element a negative oxidation state. Elements become more
electronegative the higher their Group number and the lower their Period number;
therefore, the most electronegative element is fluorine.
The oxidation number of an element in a compound is equal to the charge which a
particle of the element would carry in the compound, assuming the compound is
ionic. This is a purely theoretical idea, and it is does not matter whether the
compound in question is really ionic or covalent.
TOPIC 13.20: ELECTRODE POTENTIALS
1
e.g.
compound
oxidation numbers
NaCl
CCl4
HBr
H2S
Na
C
H
H
+1
+4
+1
+1
Cl
Cl
Br
S
-1
-1
-1
-2
The following general rules are useful:
All free elements (i.e. those not combined with another element) have an
oxidation number of 0.
e.g. Na, Mg, Br2
In simple ions, the charge on the ion is equal to the oxidation number.
e.g.
Na+
+1
ion
oxidation number
Fe3+
+3
Br-1
O2-2
Since fluorine is the most electronegative element, it always has an oxidation
number in its compounds of –1.
Combined oxygen always has an oxidation number of –2, except when in
combination with fluorine or in peroxides.
e.g.
compound
oxidation numbers
Fe2O3
Mn2O7
CrO3
Fe +3
Mn +7
Cr +3
O -2
O -2
O -2
H2O2
F2O
H +1
O +2
O -1
F -1
BUT ….
Group I elements in their compounds always have an oxidation number of +1.
Group II elements in their compounds always have an oxidation number of +2.
Hydrogen in its compounds always has an oxidation number of +1, except
when it has combined with a reactive metal.
e.g.
compound
oxidation numbers
H2O
HCl
CH4
H +1
H +1
H +1
O -2
Cl -1
C -4
NaH
Na +1
H -1
BUT ….
The sum of the oxidation numbers of all the atoms in an uncharged molecule
is zero: in an ion, the sum is equal to the charge on the ion.
TOPIC 13.20: ELECTRODE POTENTIALS
2
e.g.
compound/ion oxidation numbers
NH3
NH4+
H2SO4
H +1 N -3
H +1 N -3
H +1 S +6 O -2
(-3+1+1+1=0)
(-3+1+1+1+1=+1)
(+1+1+6-2-2-2-2=0)
3. Number of Oxidation States
Many elements have several oxidation states:
e.g.
sulphur
chlorine
H2S
S
SCl2
SO2
SO3
+1 -2
0
+2 -1
+4 -2
+6 -2
HCl
Cl2
HOCl ClF3 KClO3 KClO4
+1-1
0
+1-2+1 +3-1
+1+5-2 +1+7-2
4. Redox Reactions
If, during a chemical reaction, the oxidation number of an element increases (i.e.
becomes more positive or less negative), then the element has lost electrons and
has been OXIDISED.
Conversely, if the oxidation number of an element decreases (i.e. becomes less
positive or more negative), then the element has gained electrons and has been
REDUCED.
Oxidation
-5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5
Reduction
REDuction and OXidation occur together in what is called a REDOX reaction.
Example:



work out the oxidation numbers for all the elements in the reaction
identify the oxidation and reduction steps
work out the total number of electrons transferred in each step: they should be
equal
oxidation: loss of 2e- x 1
Mg + 2HCl
0
MgCl2 + H2
+1 -1
+2 -1
0
check: 2e- x 1 = 1e- x 2
reduction: gain of 1e- x 2
TOPIC 13.20: ELECTRODE POTENTIALS
3
OXIDATION & REDUCTION
QUESTION SHEET 3
Work out the oxidation number of each element in the following chemical formulae
1.
Ca
…………………………………………………………………..
2.
LiF
………………………………………………………………….
3.
MgO
………………………………………………………………….
4.
I2
………………………………………………………………….
5.
Cr3+
………………………………………………………………….
6.
Al2O3
………………………………………………………………….
7.
HF
………………………………………………………………….
8.
Ni(NO3)2
………………………………………………………………….
9.
CrO42-
………………………………………………………………….
10.
SrCO3
………………………………………………………………….
11.
NaClO4
………………………………………………………………….
12.
SO32-
………………………………………………………………….
13.
NaIO3
………………………………………………………………….
14.
XeF4
………………………………………………………………….
15.
Pb(OH)2
………………………………………………………………….
16.
K2MnO4
………………………………………………………………….
17.
Al2(SO4)3
………………………………………………………………….
18.
NaVO3
………………………………………………………………….
19.
H3PO3
………………………………………………………………….
20.
NH4+
………………………………………………………………….
TOPIC 13.20: ELECTRODE POTENTIALS
4
OXIDATION NUMBERS
Transition elements are able to form compounds in which they exhibit a variety of
oxidation states. They form complex ions with oxygen and various ligands.
A ligand is a lone pair donor, which forms a coordinate bond with a central metal
ion or atom. The ligand may be a neutral molecule or a negative ion. Examples of
ligands are:
Ligand
Formula
Overall charge
water
H2O
0
ammonia
NH3
0
cyanide ion
CN-
-1
carbon monoxide
CO
0
1,2-diaminoethane
NH2CH2CH2NH2 or en
0
chloride ion
Cl-
-1
hydroxide ion
OH-
-1
thiosulphate ion
S2O32-
-2
ethanedioate ion
C2O42-
-2
bis[di(carboxymethyl)amino]ethane
EDTA4-
-4
Work out the oxidation number of the transition metal in each of the following
complexes.
1.
Cr2O72-
…………………………………………………………………..
2.
Ni(CO)4
………………………………………………………………….
3.
[CoCl4]2-
………………………………………………………………….
4.
[Co(H2O)6]2+
………………………………………………………………….
5.
VO2+
………………………………………………………………….
6.
VO2+
………………………………………………………………….
7.
[Fe(CN)6]4-
………………………………………………………………….
8.
[Ag(S2O3)2]3-
………………………………………………………………….
9.
[Fe(H2O)4(OH)2]+
………………………………………………………………….
10.
[Cu(NH3)4(H2O)2]2+ ………………………………………………………………….
11.
[Ag(NH3)2]+
………………………………………………………………….
TOPIC 13.20: ELECTRODE POTENTIALS
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12.
[CuCl2]-
………………………………………………………………….
13.
Co(en)2Cl2
………………………………………………………………….
14.
[Fe(C2O4)3]3-
………………………………………………………………….
15.
[Cu(EDTA)]2-
………………………………………………………………….
16.
[Cr(OH)6]3-
………………………………………………………………….
17.
[Cr(H2O)3(OH)3]
………………………………………………………………….
18.
[Ag(CN)2]-
………………………………………………………………….
19.
[Zn(OH)4]2-
………………………………………………………………….
20.
[VCl2(H2O)4]+
………………………………………………………………….
TOPIC 13.20: ELECTRODE POTENTIALS
6
REDOX EQUATIONS
Constructing Half –Equations
The half-equation shows either the oxidation or the reduction step of a redox change.
In a half-equation:

only one element changes its oxidation state

the atoms of each element must balance

the total charge on both sides of the equation must be the same
When constructing half-equations for reactions which take place in aqueous solution:

Water can be used as a source of oxygen atoms on the reactant side of the
equation, the hydrogen appearing in the products as H+.

Any extra oxygen atoms on the reactant side of the equation can be
converted into water by reaction with hydrogen ions from an acid.

The oxidation numbers of hydrogen and oxygen do not change.
Examples:
1. Deduce the half-equation for the reduction of VO3- to V2+ in acid
solution.
Vanadium is changing its oxidation state from +5 in VO3- to +2 in V2+ and
therefore needs to gain 3e-.
VO3- + 3eV2+
The acid solution provides the six hydrogen ions needed to react with the
three extra oxygen atoms in VO3- to form water.
VO3- + 6H+ + 3e-
V2+ + 3H2O
Finally, a check shows that the total charge on each side of the equation
is the same.
l.h.s. -1 + 6 - 3 = +2
TOPIC 13.20: ELECTRODE POTENTIALS
r.h.s. +2
7
2. Deduce the half-equation for the reduction of MnO4- to Mn2+ in
acid solution.
Manganese is changing its oxidation state from +7 in MnO4- to +2 in Mn2+
and therefore needs to gain 5e-.
MnO4- + 5eMn2+
The acid solution provides the eight hydrogen ions needed to react with
the four extra oxygen atoms in MnO4- to form water.
MnO4- + 8H+ + 5e-
Mn2+ + 4H2O
Finally, a check shows that the total charge on each side of the equation
is the same.
l.h.s. -1 + 8 - 5 = +2
r.h.s. +2
3. Deduce the half-equation for the reduction of NO3- to NO in acid
solution.
Nitrogen is changing its oxidation state from +5 in NO3- to +2 in NO and
therefore needs to gain 3e-.
NO3- + 3eNO
The acid solution provides the four hydrogen ions needed to react with
the two extra oxygen atoms in NO3- to form water.
NO3- + 4H+ + 3e-
NO + 2H2O
Finally, a check shows that the total charge on each side of the equation
is the same.
l.h.s. -1 + 4 - 3 = 0
TOPIC 13.20: ELECTRODE POTENTIALS
r.h.s. 0
8
Combining Half –Equations
The overall equation for a redox change can be obtained by combining the halfequations for the oxidation and reduction steps in such a way that the number of
electrons donated by the reducing agent is equal to the number accepted by the
oxidising agent.
Any molecules or ions which appear on both sides of the overall equation, such as H +
and H2O, then need to be cancelled down.
Examples:
4. Manganate(VII) ions (MnO4-) oxidise iron(II) ions (Fe2+) in acid solution,
forming iron(III) ions (Fe3+), and are reduced to manganese(II) ions (Mn2+).
Deduce the two half-equations for this reaction, and hence derive an overall
equation.
Manganese is changing its oxidation state from +7 in MnO4- to +2 in Mn2+ and
therefore needs to gain 5e-.
MnO4- + 5eMn2+
The acid solution provides the eight hydrogen ions needed to react with the four
extra oxygen atoms in MnO4- to form water.
MnO4- + 8H+ + 5e-
Mn2+ + 4H2O
Iron is changing its oxidation state from +2 in Fe2+ to +3 in Fe3+ and therefore
needs to lose 1e-.
Fe2+
Fe3+ + eTo balance the electrons, the first half-equation needs to be multiplied x1:
MnO4- + 8H+ + 5e-
Mn2+ + 4H2O
and the second x5:
5Fe2+
5Fe3+ + 5e-
The two half-equations are now added together:
5Fe2+ + MnO4- + 8H+ + 5e-
Mn2+ + 4H2O + 5Fe3+ + 5e-
The electrons cancel out to give the overall equation:
5Fe2+ + MnO4- + 8H+
Mn2+ + 4H2O + 5Fe3+
Finally, a check shows that the total charge on each side of the equation is the
same.
l.h.s. +10 -1 + 8 = +17
TOPIC 13.20: ELECTRODE POTENTIALS
r.h.s. +2 +15 = +17
9
5. Dichromate(VI) ions (Cr2O72-) oxidise sulphate(IV) ions (SO32-) in acid
solution, forming sulphate(VI) ions (SO42-), and are reduced to chromium(III)
ions (Cr3+). Deduce the two half-equations for this reaction, and hence
derive an overall equation.
Chromium is changing its oxidation state from +6 in Cr2O72- to +3 in Cr3+ and
therefore needs to gain 6e-.
Cr2O72- + 6e-
2Cr3+
The acid solution provides the fourteen hydrogen ions needed to react with the
seven extra oxygen atoms in Cr2O72- to form water.
Cr2O72- + 14H+ + 6e-
2Cr3+ + 7H2O
Sulphur is changing its oxidation state from +4 in SO 32- to +6 in SO42- and
therefore needs to lose 2e-.
SO32-
SO42- + 2e-
One water molecule is needed to provide the extra oxygen atom:
SO32- + H2O
SO42- + 2H+ + 2e-
To balance the electrons, the first half-equation needs to be multiplied x1:
Cr2O72- + 14H+ + 6e-
2Cr3+ + 7H2O
and the second x3:
3SO32- + 3H2O
3SO42- + 6H+ + 6e-
The two half-equations are now added together:
3SO32-+ 3H2O + Cr2O72- + 14H+ + 6e-
2Cr3+ + 7H2O + 3SO42- + 6H+ + 6e-
The electrons cancel out, and the numbers of H+ ions and H2O molecules are
simplified to give the overall equation:
3SO32- + Cr2O72- + 8H+
2Cr3+ + 4H2O + 3SO42-
Finally, a check shows that the total charge on each side of the equation is the
same.
l.h.s. -6 -2 + 8 = 0
TOPIC 13.20: ELECTRODE POTENTIALS
r.h.s. +6 -6 = 0
10
Sometimes the same species oxidises and reduces itself simultaneously. This
process is called disproportionation.
6. Under alkaline conditions, chlorine disproportionates to form chloride
ions (Cl-) and chlorate(V) ions (ClO3-). Deduce the two half-equations for
this reaction, and hence derive an overall equation.
Chlorine is changing its oxidation state from 0 in Cl2 to -1 in Cl- and therefore
needs to gain 1e-.
½ Cl2 + eClChlorine is changing its oxidation state from 0 in Cl2 to +5 in ClO3- and therefore
needs to lose 5e-.
1/ Cl
ClO3- + 5e2
2
Six hydroxide ions in the alkaline solution provide the three extra oxygen atoms
needed to form ClO3-, and also form water.
1/
2
Cl2 + 6OH-
ClO3- + 3H2O + 5e-
To balance the electrons, the first half-equation needs to be multiplied x5:
21/2 Cl2 + 5e-
5Cl-
and the second x1:
1/
2
Cl2 + 6OH-
ClO3- + 3H2O + 5e-
The two half-equations are now added together:
3Cl2 + 6OH- + 5e-
5Cl- + ClO3- + 3H2O + 5e-
The electrons cancel out to give the overall equation:
3Cl2 + 6OH-
5Cl- + ClO3- + 3H2O
Finally, a check shows that the total charge on each side of the equation is
the same.
l.h.s. -6
TOPIC 13.20: ELECTRODE POTENTIALS
r.h.s. -1 -5 = -6
11
ELECTRODE POTENTIALS
If a metal is placed in water or a solution of one of its salts, the metal tends to throw
off positive ions into the liquid. For example:
Zn2+(aq) + 2e-
Zn(s) + aq.
When positive ions are given off into the liquid, an excess of electrons is left on the
surface of the metal, which therefore acquires a negative charge. Owing to the
attraction of opposite charges, the hydrated cations remain close to the metal and an
electric double layer is formed.
Zn
Zn2+
Zn2+
Zn2+
Zn2+
Zn2+
------
-
Zn2+
Zn2+
Zn2+
Zn2+
Zn2+
There is also a tendency for ions in the solution to be deposited on the metal:
Zn2+(aq) + 2e-
Zn(s)
so that eventually an equilibrium is established; the rate of formation of the ions is
equal to the rate of deposition:
Zn2+(aq) + 2eZn(s)
The formation of the electric double layer causes a potential difference between the
surface of the metal and the liquid; this is called the electrode potential.
The further to the left the above equilibrium lies, the greater the potential difference
will be.
For a given metal, the position of the equilibrium depends on the concentration of
metal ions already present in the solution; the more dilute the solution, the higher the
electrode potential will be.
For different metals placed in solutions containing equal concentrations of their ions
at the same temperature, the position of the equilibrium, and therefore the electrode
potential, is governed by the overall energy change associated with the formation of
hydrated ions from the metal. This can be broken down into a series of energy
changes:
TOPIC 13.20: ELECTRODE POTENTIALS
12
Mn+(g) + ne- + aq.
Hhyd
HI
Mn+(aq) + ne-
Energy
M(g) + aq.
H
Hsub
M(s) + aq.
Energy change
Hsub
Na
+109
HI
+494
Hhyd
H
-406
+197
Mg
+ 150
+ 736
+1450
-1920
+ 416
Zn
+ 130
+ 908
+1730
-2050
+ 718
Cu
+ 339
+ 745
+1960
-2100
+ 944
As a result of these energy changes, from Na to Cu, the dissociation into ions will be
less extensive. Therefore, the electrode potential will decrease from Na to Cu.
ELECTRODES
There are three types of electrodes:
Metal Electrode
A metal electrode consists of a metal dipping into a solution of its ions in which the
following type of equilibrium is established:
Zn2+(aq) + 2e-
Zn(s)
Metal electrodes are represented by convention as, for example
Zn2+(aq) Zn(s)
where the vertical line represents a phase boundary.
The redox couple (e.g. Zn2+/Zn), which is always written with the reduced species on
the right, is a convenient shorthand way of representing the half reaction.
Gas Electrode
A gas electrode consists of a gas in equilibrium with a solution of its ions in the
presence of an inert metal, usually platinum. An example of a gas electrode is the
hydrogen electrode (H+/H2).
Pt(s) H2(g) H+(aq)
By convention, the reduced species is written nearer to the Pt.
TOPIC 13.20: ELECTRODE POTENTIALS
13
Redox Electrode
A redox electrode consists of an element in two different oxidation states in the
presence of an inert metal, usually platinum. An example of a redox electrode is
Fe3+/Fe2+.
Pt(s) Fe2+(aq), Fe3+(aq)
By convention, the reduced species is written nearer to the Pt. The two different ions
of iron are in the same phase and are separated by a comma.
Half-Equation Convention
The reaction occurring in an electrode is represented by a half-equation. By IUPAC
convention, redox half-equations are written as reductions, for example:
Zn2+(aq) + 2e-
Zn(s)
Electrochemical Cells
An electrode is half a cell. If two different electrodes are connected together, an
electrochemical cell is formed. Usually, two electrode compartments are joined by a
salt bridge. A salt bridge allows ions to flow, thus completing the electrical circuit, but
prevents the ions in the two electrode compartments from mixing to any great extent.
A salt bridge consists usually of a saturated solution of KNO 3 which may or may not
be thickened with gelatine or agar. A salt bridge is written in a cell representation as a
double vertical line.
The diagram below shows an electrochemical cell constructed from two metal
electrodes.
salt bridge
zinc
zinc sulphate
copper
copper sulphate
One electrode will form the negative terminal of the cell, the other the positive
terminal. When the cell is operating, electrons flow in the external circuit from the
negative terminal to the positive terminal. Therefore an oxidation reaction occurs at
the negative terminal and a reduction reaction at the positive terminal. The reactions
occurring in this particular cell are:
Cu2+(aq) + 2eZn(s)
Cu(s)
Zn2+(aq) + 2e-
Measurement of Electrode Potentials
The electrode potential cannot be measured absolutely. Measurement of the
potential difference between, for example, a metal and the solution of its ions
necessitates making an electrical connection between the metal and its solution. The
connection to the solution could only be made by immersing some other metal in the
solution, which would simply introduce another electrode potential.
TOPIC 13.20: ELECTRODE POTENTIALS
14
Only potential difference can be measured, and it is therefore necessary to define a
standard electrode, relative to which all other potentials are measured. The electrode
used for this purpose is the STANDARD HYDROGEN ELECTRODE (S.H.E.).
Pt(s) H2(g,1bar) H+(aq, 1.0 mol.dm-3)
E
= 0V at 298K
In this gas electrode, hydrogen at a pressure of 1 bar is bubbled over a strip of
platinum coated in platinum black and immersed in an acid solution containing a
hydrogen ion concentration of 1.0 mol.dm-3. Temperature = 298K
The following equilibrium is established:
H+(aq) + e-
1/
2H2(g)
To measure the standard electrode potential of a particular electrode, it is connected
to a S.H.E. to form an electrochemical cell. The cell potential is then measured. If the
electrode under test forms the negative pole of the cell, its standard electrode
potential is negative; if it forms the positive pole, the standard electrode potential is
positive. The following standard conditions are required:
 Zero current. This is achieved by measuring the cell potential with a
high resistance (solid-state) voltmeter.
 Ion concentrations of 1.0 mol.dm-3
 Temperature 298K
 Pressure 100 kPa (1 bar)
high resistance voltmeter
-
H2 at 100kPa
+
V
salt bridge – KNO3 soln.
Zn
T = 298K
[H+] = 1.0 mol.dm-3
Pt coated in Pt black
[Zn2+] = 1.0 mol.dm-3
The S.H.E. is made the left hand electrode and the unknown the right hand
electrode. If the voltmeter is connected as indicated, the electrode potential and its
sign can be read directly from the voltmeter. In this case, the reading is -0.76V.
Therefore the standard electrode potential of Zn2+/Zn is -0.76V.
TOPIC 13.20: ELECTRODE POTENTIALS
15
If a current is allowed to flow in the above cell, electrons will flow from the negative
electrode (Zn2+/Zn) to the positive electrode (H+/H2). Therefore zinc is oxidised and
hydrogen reduced; the reactions occurring in the cell are:
Zn2+(aq) + 2e-
Zn(s)
2H+(aq) + 2eThe overall cell reaction is:
Zn(s) + 2H+(aq)
H2(g)
Zn2+(aq) + H2(g)
Secondary Standard Electrodes
The hydrogen electrode is difficult to set up and use. Therefore, it is often more
convenient to use a secondary standard electrode, whose potential with respect to
the S.H.E. is accurately known. The electrode most frequently used as a secondary
standard electrode is the calomel electrode:
Eo = +0.27V
Pt(s) Hg(l) Hg2Cl2(s), KCl(aq)
Zn2+/Zn
SHE
-0.76
calomel
0
+0.27
Ecell = +0.76V
Cu 2+/Cu
+0.34
V
Ecell = +0.07V
Ecell = +0.34V
Ecell = +1.03V
Cell Representations





The more negative electrode is drawn on the left
Phase boundaries are shown as single vertical lines
A salt bridge is shown as a double vertical line
Substances in the same phase are separated by a comma
In gas electrodes and redox electrodes, where an inert metal is present, the
component in the reduced state is written nearest to the inert metal
Example 1: A cell comprises the electrodes Mg2+/Mg and Sn4+Sn2+. Write the
representation of this cell.
The magnesium electrode is more negative and is shown on the left of the cell.
Mg(s) Mg2+(aq) Sn4+(aq), Sn2+(aq) Pt(s)
Example 2: A cell comprises the electrodes Fe3+/Fe2+ and H+/H2. Write the
representation of this cell.
The hydrogen electrode is more negative and is shown on the left of the cell.
Pt(s) H2(g) H+(aq) Fe3+(aq), Fe2+(aq) Pt(s)
TOPIC 13.20: ELECTRODE POTENTIALS
16
Calculating Cell e.m.f.
To calculate the e.m.f. of an electrochemical cell made from any two electrodes, the
cell is first arranged so that the electrode with the more positive potential is on the
right hand side. Then
Eocell = EoR - EoL
where ER and EL are the reduction potentials of the right and left electrodes.
Example:
Calculate the standard e.m.f. of a cell with copper (Cu2+/Cu; Eo = +0.34V) and
zinc (Zn2+/Zn; Eo = -0.76V) electrodes. Write the representation of the
standard cell.
The more positive electrode (Cu2+/Cu) is made the right hand electrode, then
Eocell = EoR - EoL
Eocell = +0.34 - (-0.76) V
Eocell = +1.10V
Cell representation:
The zinc electrode is more negative and is shown on the left of the cell.
Zn(s) Zn2+(aq, 1.0mol.dm-3) Cu2+(aq, 1.0mol.dm-3) Cu(s)
TOPIC 13.20: ELECTRODE POTENTIALS
17
Electrochemical Series
The electrochemical series is a list of standard electrode
potentials in order of their voltages.
The list is presented either in descending order, with the most positive potentials first,
or in ascending order, with the most negative potentials first. Reactions are written as
reductions. Part of the electrochemical series is given below.
Reduction half-equation
Eo /V
F2(g) + 2e2F-(aq)
MnO4-(aq) + 8H+(aq) + 5eCl2(g) + 2e2Cl-(aq)
+2.85
+1.51
+1.36
Mn2+(aq) + 4H2O(l)
Cr2O72-(aq) + 14H+(aq) + 6eMnO2(s) + 4H+(aq) + 2eBr2(g) + 2e2Br-(aq)
better
oxidising
agent
2Cr3+(aq) + 7H2O(l) +1.33
Mn2+(aq) + 2H2O(l)
+1.23
+1.09
Cu2+(aq) + I-(aq) + eCuIs)
Ag+(aq) + eAg(s)
3+
Fe (aq) + e
Fe2+(aq)
I2(g) + 2e2I-(aq)
Cu+(aq) + eCu(s)
Cu2+(aq) + 2eCu(s)
+0.86
+0.80
+0.77
+0.54
+0.52
+0.34
Hg2Cl2(aq) + 2eAgCl(s) + eCu2+(aq) + eSn4+(aq) + 2e2H+(aq) + 2ePb2+(aq) + 2eSn2+(aq) + 2eNi2+(aq) + 2eV3+(aq) + eCr3+(aq) + e-
+0.27
+0.22
+0.15
+0.15
-0.00
-0.13
-0.14
-0.25
-0.26
-0.41
Fe2+(aq) + 2eZn2+(aq) + 2eAl3+(aq) + 3eMg2+(aq) + 2eNa+(aq) + eCa2+(aq) + 2eK+(aq) + eLi+(aq) + e-
2Hg(l) + 2Cl-(aq)
Ag(s) + Cl-(aq)
Cu+(aq)
Sn2+(aq)
H2(g)
Pb(s)
Sn(s)
Ni(s)
2+
V (aq)
Cr2+(aq)
Fe(s)
Zn(s)
Al(s)
Mg(s)
Na(s)
Ca(s)
K(s)
Li(s)
TOPIC 13.20: ELECTRODE POTENTIALS
better
reducing
agent
-0.44
-0.76
-1.66
-2.36
-2.71
-2.87
-2.92
-3.05
18
Spontaneous Reactions
The feasibility of a reaction is governed by the sign of the free energy change (G)
for the reaction; a spontaneous change will occur only if G is negative. G is related
to the e.m.f. of a cell (E):
G = -nEF
F and n are both positive constants, therefore a cell reaction is spontaneous (G is
negative) only if Ecell is positive.
Ecell = ER - EL
Ecell will only be positive if ER is more positive than EL. In other words, the right-hand
electrode must be the site of reduction and the left-hand electrode the site of
oxidation.
Deducing Cell Reactions & Cell e.m.f.
Cell reactions and the cell e.m.f. can be deduced from the electrochemical series.
Example: Deduce the cell reactions and the cell e.m.f. for a cell comprising the
electrodes Cl2/Cl- and V3+/V2+
2Cl-(aq)
At the -ve (left) electrode:
V2+(aq)
V3+(aq) + eThe reduction equation given in the
table needs to be reversed.
Ecell = ER - EL
Ecell = +1.36 - (-0.26)
= 1.62V
19
+
+1.51
2V3+(aq) + 2Cl-(aq)
+1.36
+0.77
-
The two equations are then added
together:
2V2+(aq) + Cl2(g)
TOPIC 13.20: ELECTRODE POTENTIALS
-0.26
-0.76
To balance the electrons, the second
equation must be multiplied by 2:
2V2+(aq)
2V3+(aq) + 2e-
Mn2+(aq) + 4H2O(l)
RED
OX
MnO4-(aq) + 8H+(aq) + 5e-
Cl2(g) + 2e-
Fe2+(aq)
At the +ve (right) electrode:
Cl2(g) + 2e2Cl-(aq)
Fe3+(aq) + e-
Zn(s)
The more negative potential (V3+/V2+)
forms the negative pole of the cell;
oxidation takes place here.
V2+(aq)
The more positive potential (Cl2/Cl-)
forms the positive pole of the cell;
reduction takes place here.
V3+(aq) + e-
Zn2+(aq) + 2e-
The electrochemical series is first rotated
so that the more positive potentials are to
the r.h.s.
Predicting the Feasibility of a Reaction
Electrode potentials can be used to predict the feasibility of redox changes. The
change can be broken down into two half-reactions, one of which is a reduction, the
other an oxidation. These two electrodes can be used to construct a cell, the
reduction step occurring at the right-hand electrode. Ecell is then calculated. If Ecell
is positive, the change is feasible.
Example 1: Is the disproportionation of copper(I) to copper(II) and copper a
spontaneous reaction?
2Cu+(aq)
The reaction is:
Cu2+(aq) + Cu(s)
This redox change can be broken down into two half-reactions:
1.
Cu+(aq)
Cu2+(aq) + e-
2.
Cu+(aq) + e-
Eo = +0.15V
Eo = +0.52V
Cu(s)
Reaction 1. is an oxidation and must be written as the left-hand electrode.
Reaction 2. is a reduction and must be written as the right-hand electrode.
Ecell = ER - EL
Ecell = +0.52 - (+0.15)
Ecell = +0.37V
Since the cell e.m.f. is positive, the reaction will go spontaneously.
Example 2: Will copper react spontaneously with an aqueous acid to form
hydrogen?
Cu(s) + 2H+(aq)
The reaction is:
Cu2+(aq) + H2(g)
This redox change can be broken down into two half-reactions:
Cu2+(aq) + 2e-
1.
Cu(s)
2.
2H+(aq) + 2e-
H2(g)
Eo = +0.34V
Eo = +0.00V
Reaction 1. is an oxidation and must be written as the left-hand electrode.
Reaction 2. is a reduction and must be written as the right-hand electrode.
Ecell = ER - EL
Ecell = +0.00 - (+0.34)
Ecell = -0.34V
Since the cell e.m.f. is negative, the reaction will not go spontaneously.
TOPIC 13.20: ELECTRODE POTENTIALS
20
Example 3: Will dichromate(VI) ions in acid solution oxidise chloride ions
spontaneously under standard conditions?
The reaction is:
6Cl-(aq) + Cr2O72-(aq) + 14H+(aq)
2Cr3+(aq) + 7H2O(l) + 3Cl2(g)
This redox change can be broken down into two half-reactions:
1. Cr2O72-(aq) + 14H+(aq) + 6e2. 6Cl-(aq)
2Cr3+(aq) + 7H2O(l)
3Cl2(g) + 6e-
Eo = +1.33V
Eo = +1.36V
Reaction 1. is a reduction and must be written as the right-hand electrode.
Reaction 2. is an oxidation and must be written as the left-hand electrode.
Ecell = ER - EL
Ecell = +1.33 - (+1.36)
Ecell = -0.03V
Since the cell e.m.f. is negative, the reaction will not go spontaneously.
Example 4: Will dichromate(VI) ions in acid solution oxidise bromide ions
spontaneously under standard conditions?
The reaction is:
6Br-(aq) + Cr2O72-(aq) + 14H+(aq)
2Cr3+(aq) + 7H2O(l) + 3Br2(g)
This redox change can be broken down into two half-reactions:
1. Cr2O72-(aq) + 14H+(aq) + 6e2. 6Br-(aq)
2Cr3+(aq) + 7H2O(l)
3Br2(g) + 6e-
Eo = +1.33V
Eo = +1.09V
Reaction 1. is a reduction and must be written as the right-hand electrode.
Reaction 2. is an oxidation and must be written as the left-hand electrode.
Ecell = ER - EL
Ecell = +1.33 - (+1.09)
Ecell = +0.24V
Since the cell e.m.f. is positive, the reaction will go spontaneously.
TOPIC 13.20: ELECTRODE POTENTIALS
21
Commercial Cells
Batteries and cells
There are three main types of commercial electrochemical cell:
• Primary cells are not rechargeable and are thrown away after they run
down. These types of cell are commonly used in torches or remote
controls.
• Secondary cells can be recharged after they run down. These are
found in mobile phones and mp3 players.
• Fuel cells produce electricity from gaseous or liquid fuels.
1) Primary cells
Early designs of primary cells were wet cells. The Daniell cell was
invented by John Daniell in 1836. It consists of a piece of zinc dipped
into aqueous zinc sulphate. The zinc acts as the anode. Electrons are
lost and the zinc is oxidized:
Zn(s) → Zn2+ (aq) + 2eThe copper acts as the cathode. Electrons are gained and copper(II)
ions
are reduced:
Cu2+ (aq) + 2e-→ Cu(s)
The overall cell reaction is
Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s)
with an Ecell of 1.1 V. This is however not very portable because of the
liquid electrolytes it contains. Most modern primary cells are dry cells
instead like the zinc-carbon cells you are most familiar with.
TOPIC 13.20: ELECTRODE POTENTIALS
22
Zinc–carbon cells
Zinc–carbon cells are cheap to make but easily run down under heavy
use. They produce a potential difference of about 1.5 V, which gradually
reduces to around 0.8 V with use.
Anode reaction
Zinc reacts with ammonia formed from the ammonium chloride
electrolyte.
Zn(s) + 2NH3(aq) → [Zn(NH3)2]2+(aq) + 2eCathode reactions
The reaction involves hydrogen ions released from the ammonium ions:
2MnO2(s) + 2H+(aq) + 2e- → Mn2O3(s) + H2O(l)
The overall cell reaction is:
Zn(s) + 2MnO2(s) + 2NH4+(aq) → [Zn(NH3)2]2+ (aq) + Mn2O3(s) + H2O(l)
Alkaline dry cells
Alkaline cells produce the same potential difference as zinc–carbon cells
but they last longer. They contain potassium hydroxide as the electrolyte
rather than ammonium chloride.
Anode reactions
At the anode, hydroxide ions react with the zinc ions:
Zn(s) + 2OH-(aq) → ZnO(s) + H2O(l) + 2e−
TOPIC 13.20: ELECTRODE POTENTIALS
23
Cathode reactions
2MnO2(s) + H2O(l) + 2e− → Mn2O3(s) + 2OH-(aq)
This is the overall cell reaction:
Zn(s) + 2MnO2(s) → ZnO(s) + 2 Mn2O3 (s)
2) Rechargeable cells
Rechargeable cells are secondary cells. They must be charged before
use by connecting them to the electricity supply. The lead–acid battery,
the type used in cars, is the oldest design.
The lead–acid battery
A typical car battery comprises six cells in series, each producing 2V,
giving a total voltage of 12V. The electrolyte is 6M H2SO4.
Anode reactions
This is the reaction that happens at the anode when the battery
discharges:
Pb(s) + SO42-(aq) → PbSO4(s) + 2eThe reaction is reversed during charging:
Cathode reactions
This is the reaction that happens at the cathode when the battery
discharges:
PbO2(s) + 4H+(aq) + SO42-(aq) + 2e- → PbSO4(s) + 2H2O(l)
The reaction is reversed during charging:
TOPIC 13.20: ELECTRODE POTENTIALS
24
Nicads
The most common rechargeable cell in everyday use is the nickel–
cadmium cell, often just called a nicad. The anode is made from
cadmium and the cathode from nickel(III) hydroxide. Potassium
hydroxide is the electrolyte.
Anode reactions
This is the reaction that happens at the anode when the battery
discharges:
Cd(s) + 2OH-(aq) → Cd(OH)2(s) + 2eThe reaction is reversed during charging:
Cathode reactions
This is the reaction that happens at the cathode when the battery
discharges:
NiO(OH)(s) + e- + H2O → Ni(OH)2(s) + OH- (aq)
The reaction is reversed during charging:
Write the overall reaction in the space below
……………………………………………………………………………………
……………………………………………………………………………………
TOPIC 13.20: ELECTRODE POTENTIALS
25
Lithium-ion batteries
Lithium-ion batteries are used in portable devices such as mobile phones
and laptop computers. They are more complex than the other secondary
cells described here, and include a computer chip to control charging
and discharging. They produce a much higher potential difference, too,
typically 3.7 V. The anode is made of graphite and the cathode is made
of lithium cobalt oxide, LiCoO2.
Anode reactions
This is an example of the reaction that happens at the anode when the
battery discharges:
Li+ + CoO2 + + e- → Li+(CoO2)The reaction is reversed during charging:
Cathode reactions
This is an example of the reaction that happens at the cathode when the
battery discharges:
Li → Li+ + eThe reaction is reversed during charging:
TOPIC 13.20: ELECTRODE POTENTIALS
26
3) Fuel Cells
The alkaline hydrogen–oxygen fuel cell
Fuel cells transform the chemical energy in a fuel such as hydrogen or
methanol directly into electrical energy. The fuel is oxidized by oxygen from
the air using electrochemical reactions in the fuel cell. A typical alkaline
hydrogen–oxygen fuel cell comprises two flat electrodes, each coated on
one side by a thin layer of platinum catalyst. A semi-permeable membrane
is sandwiched between the two electrodes and the electrolyte between
them is sodium hydroxide solution. Hydroxide ions produced at the cathode
travel across the membrane to the anode.
Anode reaction
The oxidation of hydrogen is catalysed by the layer of platinum:
H2(g) + 2OH- → 2H2O(l) + 2eThe electrons released by this reaction flow through the external circuit.
The hydrogen ions pass through the proton exchange membrane to the
cathode.
Cathode reaction
Oxygen is reduced to water vapour. It reacts with the hydrogen ions that
pass through the proton exchange membrane and electrons from the
external circuit:
2H2O(aq) + 4e- + O2(g) → 4OH-(aq)
Overall reaction
The overall reaction is: 2H2(g) + O2(g) → 2H2O(g)
TOPIC 13.20: ELECTRODE POTENTIALS
27
Benefits and drawbacks of electrochemical cells
Electrochemical cells are a very convenient, portable sources of
electricity. They reduce the need for expensive cabling and bring
electricity supplies to remote places. Spacecraft use hydrogen fuel cells
to provide electricity and drinking water.
Non-rechargeable cells are cheap, but are usually thrown away. This
wastes the energy and resources needed to manufacture them and they
usually go to landfill.
The use of rechargeable cells reduces the total number of batteries
thrown away each year. Rechargeable cells also improve the
performance of solar cells. These convert sunlight directly into electricity
but they do not work at night. Rechargeable cells store electricity
generated during the day and release it at night. The cadmium in nickel–
cadmium cells is very toxic. These cells should be recycled rather than
disposed of in a landfill site.
Water vapour is the only waste product of hydrogen–oxygen fuel cells.
These cells may replace petrol and diesel engines in the future, reducing
the amount of carbon dioxide produced by vehicles. This will help to
reduce the release of greenhouse gases, if the hydrogen is produced
without fossil fuels. But most industrial hydrogen is produced using these
fuels at the moment. It could be produced by electrolysis of water but this
uses electricity which has been generated by burning of fossil fuels.
Hydrogen itself is highly flammable, and is difficult to store and handle
safely.
TOPIC 13.20: ELECTRODE POTENTIALS
28