Lecture Notes 3
rd
Series: Electrochemistry
Oxidation number or states
When atoms gain or lose electrons they are said to change their oxidation number or oxidation state.
If an element has gained electrons in a chemical reaction it is said to be reduced and its oxidation number
becomes more negative (or less positive).
If an element loses electrons in a chemical reaction it is said to be oxidized and its oxidation number
becomes more positive (or less negative).
Consider the following chemical reaction :
2+
Ca(s) → Ca (aq)
2+
Ca(s) has an effective charge of zero. Therefore its oxidation number is set at zero. Ca (aq) has an
effective charge of two plus. Its oxidation number is given as +2. Thus the oxidation number has
increased (become more positive) during the reaction. This is called an oxidation.
Oxidation refers to any process in which the oxidation number of an atom becomes more positive
For the above reaction we can write a reaction half-reaction.
2+
–
Ca(s) → Ca (aq) + 2e
For a reaction ½- equation it is important to note that the overall charge on the LHS of the equation (zero)
is equal to the overall charge on the RHS of the equation.
+
[(2 ) + (2 x 1-)] = 0
from an electron
If we take another example.
Cl2(g) → 2Cl (aq)
–
Here the oxidation number of Cl2(g) is zero and the oxidation number of the chloride ion is minus one.
Therefore the oxidation number has become more negative during the reaction and therefore chlorine
must have gained electrons and become reduced. We can therefore write a balanced ½- reaction as
Cl2(g) + 2e → 2Cl (aq)
–
–
Reduction refers to any process that leads to a decrease in the oxidation number of an atom
There is one most important thing to remember about ½- equations. You cannot have a reduction taking
place unless some other species is being oxidised. ½- reactions therefore are only theoretical
representations of what is happening to one of the reactants. Electrons are actually transferred from one
species to another in a chemical reaction. Free electrons do not exist in chemistry. Therefore oxidation is
always accompanied by reduction. These reactions are called redox reactions.
A redox reaction is one in which electrons are transferred between species and thus changes in oxidation
number occur
38
Balancing Redox Equations
A redox reaction, as we have seen, consists of two ½-reactions; a REDuction and an OXidation. It is vitally
important that you learn how to balance these equations as they are one of the most common, and most
important, classes of reactions in chemistry. As well as balancing the number of electrons involved in
both ½-reactions, one must also balance the elements which are not directly involved in the redox
reaction. Remember the Law of Conservation of Mass!
A further consideration is whether the redox reaction is taking place in acidic media or basic media.
+
Acidic media contains H /H2O
Basic media contains OH /H2O
Example 3.1
Question:
Balance the following redox equation in acidic media
2+
MnO4 + Fe
→
Answer:
39
3+
Fe
+ Mn
2+
You might well be asking yourself “ How can we have a charged solution when all matter is electrically
neutral?” The answer is quite straightforward. We don’t! What chemists tend to do is miss out so-called
SPECTATOR IONS, i.e. those that remain unchanged in the reaction. Therefore for the above equation we
could easily add these spectator ions and the charge would be zero on both sides of the equation. For
instance we could react KMnO4 with FeCl2 in the presence of H2SO4.
2+
2+
3+
KMnO4 + 5 FeCl2 + 4 H2SO4 → K + Mn + 10 Cl + 5 Fe + 4 H2O + 4 SO4
The charge is now zero on both sides but the equation looks clumsy as there are now far more species
involved which do not add to our understanding of the chemistry. We will continue to omit the spectator
ions.
Example 3.2
Question:
Balance the following reaction in basic media:
MnO4 + Br → BrO3
Answer:
40
+ MnO2
Exercise 3.1
Balance the following redox equation in acidic media.
22+
Cr2O7 + Sn
→
Cr
3+
+ Sn
4+
Exercise 3.2
Balance the following ½-reaction equations in acidic media and identify each as either an oxidation or a
reduction.
(a)
I → IO3
(b)
NO → NO3
2+
3+
(c)
VO → V
2(d)
PbSO4 → PbO2 + SO4
Exercise 3.3
Balance the following ½-reaction equations in basic media and identify each as either an oxidation or a
reduction.
(a)
ClO → Cl
(b)
IO3 → IO
(c)
NO3 → NO2
2(d)
CrO4 → Cr(OH)3
Exercise 3.4
Balance the following redox reaction equations.
(a)
Al + Cr2O3 → Al2O3 + Cr
22+
3+
3+
(b)
Cr2O7
+ Fe → Cr + Fe
(c)
MnO2 + BrO → MnO4 + Br
(acidic)
(acidic)
(basic)
Oxidising and Reducing Agents
In a redox reaction one (or more) species is reduced and one (or more) species is oxidised. If we look at a
simple redox equation:
2+
4+
3+
3+
Fe + Ce
→ Ce + Fe
where Ce is the element cerium (atomic number 58).
2+
3+
4+
Fe is oxidised to Fe by Ce
4+
3+
2+
Ce is reduced to Ce by Fe
This means that the species that is oxidised is acting as a reducing agent and the species that is reduced is
acting as an oxidizing agent.
An oxidising agent is itself reduced
A reducing agent is itself oxidized
The Relative Strength of Oxidising and Reducing Agents
0
A very useful concept in chemistry is the Standard Electrode Potential. This is given the symbol E and is
0
measured in volts. E values give us an idea as to how much one species wants to gain electrons relative
0
to another species. You can think of it as a tug-of-war for the electrons. The more positive the E value is,
41
the more a species wants to gain electrons and therefore the stronger the oxidising power of the species.
If we look at the following two redox ½- reactions:
3+
2+
Fe + e → Fe
+
2+
MnO4 + 8H + 5e → Mn + 4H2O
0
we can immediately tell which is the stronger oxidising agent by looking at the table of E values on the
3+
2+
0
2+
next page. The Fe /Fe redox couple has an E value of + 0.77 V while the MnO4 /Mn redox couple has
0
3+
an E value of + 1.51 V. Thus MnO4 is a stronger oxidising agent than Fe .
0
We can also use the E values to determine how strong the reducing agents are relative to each other.
0
You will notice that the E values in the table are all quoted for the ½- reactions written as reductions.
This means that the oxidising agents are on the LHS of the equation and the reducing agents are on the
RHS.
e.g.
+
Na + e → Na
+
Na - oxidizing agent and Na – reducing agent
o
The general rule is that the more negative the E value, the stronger is the reducing agent on the RHS of
the equation.
Standard Electrode Potentials
Reduction half reactions
F2 + 2e → 2F
3+
2+
Co + e → Co
+
H2O2 + 2H + 2e → 2H2O
+
Au + e → Au
4+
2+
Pb + 2e → Pb
4+
3+
Ce + e → Ce
+
2+
MnO4 + 8H + 5e → Mn + 4H2O
3+
2+
Mn + e → Mn
3+
Au + 3e → Au
Cl2 + 2e → 2Cl
2+
3+
Cr2O7 + 14H + 6e → Cr + 7H2O
+
O2 + 4H + 4e → 2H2O
+
ClO4 + 2H + 2e → ClO3 + H2O
+
2+
MnO2 + 4H + 3e → Mn + 2H2O
Br2 + 2e → 2Br
+
NO3 + 4H + 3e → NO + 2H2O
2+
2+
2Hg + 2e → Hg2
ClO + H2O + 2e → Cl + 2OH
2+
Hg + 2e → Hg
+
NO3 + 2H + e → NO2 + H2O
+
Ag + e → Ag
2+
2Hg2 + 2e → 2Hg
3+
2+
Fe + e → Fe
MnO4 + 2H2O + 2e → MnO2 + 4OH
o
E /V
+2.87
+1.81
+1.78
+1.69
+1.67
+1.61
+1.51
+1.51
+1.40
+1.36
+1.33
+1.23
+1.23
+1.23
+1.09
+0.96
+0.92
+0.89
+0.86
+0.80
+0.80
+0.79
+0.77
+0.60
Reduction half reactions
2+
Cu + 2e → Cu
Hg2Cl2 + 2e → 2Hg + 2Cl
AgCl + e → Ag + Cl
2+
Cu + e →
4+
2+
Sn +2e → Sn
+
2H + 2e → H2
3+
Fe + 3e → Fe
2+
Pb + 2e → Pb
2+
Sn + 2e → Sn
2+
Ni +2e → Ni
2+
Co + 2e → Co
+
Tl + e → Tl
2+
Cd + 2e → Cd
3+
2+
Cr + e → Cr
2+
Fe + 2e → Fe
2S + 2e → S
3+
Cr + 3e → Cr
2+
Zn + 2e → Zn
2H2O + 2e → H2 + 2OH
2+
Cr + 2e → Cr
2+
Mn + 2e → Mn
3+
Al + 3e → Al
2+
Mg + 2e → Mg
3+
Ce + 3e → Ce
42
o
E /V
+0.34
+0.27
+0.22
+0.16
+0.15
0, by definition
-0.04
-0.13
-0.14
-0.23
-0.28
-0.34
-0.40
-0.41
-0.44
-0.48
-0.74
-0.76
-0.83
-0.91
-1.18
-1.66
-2.36
-2.48
MnO4 + e → MnO4
I2 + 2e → 2I
+
Cu + e → Cu
I3 + 2e → 3I
ClO4 + 2H2O + 2e → ClO3 + 2OH
-
-
2-
La + 3e → La
+
Na + e → Na
2+
Ca + 2e → Ca
+
K +e →K
+
Li + e → Li
3+
+0.56
+0.54
+0.52
+0.53
+0.36
-
-2.52
-2.71
-2.87
-2.93
-3.05
Question 3.1
+
+
(a)
Which is the stronger oxidising agent, Na or K ?
(b)
Which is the stronger reducing agent, Na or K?
Predicting the Direction of Spontaneous Change
o
Knowing the relative values of E will also allow us to predict whether certain reactions will occur or not.
Take for instance the following two reactions.
2+
2+
Cu + Zn → Zn + Cu
2+
2+
Zn + Cu → Cu + Zn
Only one of these reactions can occur spontaneously. A spontaneous reaction is one which goes in the
0
energetically favourable direction. We can work out which one from the E values.
2+
o
Zn + 2e → Zn
E = -0.76 V
2+
o
Cu + 2e → Cu
E = +0.34 V
2+
2+
We can see that Cu is a much stronger oxidising agent than Zn and therefore the copper ion “grabs”
2+
2+
the electrons before zinc can. Thus Cu will oxidise zinc metal to Zn and
2+
2+
Cu + Zn → Zn + Cu
will be the spontaneous reaction.
Another way of looking at the above reaction is to recognise Zn is a stronger reducing agent than Cu (i.e. it
0
2+
has a more negative E value) and will thus reduce Cu to Cu.
0
0
For the reaction above we can calculate the difference in E values, ΔE .
By definition
o
o
o
ΔE = E (for the reduction ½-reaction) – E (for the oxidation ½-reaction).
o
o
o
ΔE = E red – E ox
Thus for the above reaction
o
o
2+
o
2+
ΔE
= E (Cu /Cu) – E (Zn /Zn)
2+
because Cu is being reduced and Zn is being oxidised.
o
∴ ∆E = +0.34 – (-0.76) V = +1.10 V
It is a general rule that ∆E will always be positive for a spontaneous reaction.
o
Example 3.3
Question
Which is the direction of spontaneous change for the following reaction – to the left or to the right?
3+
3+
2+
2+
2 Cr + 6 Fe + 7 H2O ↔ Cr2O7 + 14 H + 6 Fe
43
Answer
Exercise 3.5
o
Calculate the ∆E value for the following reactions and predict whether they are spontaneous or not.
2+
+
(a)
2 Li + Mg → 2 Li + Mg
4+
2+
2+
3+
(b)
Sn + 2 Fe → Sn + 2 Fe
The following two ½-reactions
Zn + 2e → Zn
E = - 0.76 V
2+
o
Cu + 2e → Cu
E = + 0.34 V
represents half-cells where a strip of metal, Zn or Cu called electrodes are immersed in solutions
containing ions of the same metal. Electrodes are classified according to whether oxidation or reduction
takes place there. If oxidation takes place, the electrode is called an anode, if reduction takes place, the
electrode is called the cathode. Charge is carried through solutions by the migration of ions through a Utube called a salt bridge and the connected combination of two half-cells is called an electrochemical cell.
2+
2+
The half-cells are represented as Zn(s)/Zn (aq) ane Cu (aq)/Cu(s). The spontaneous reaction in the
electrochemical cell as shown in the figure below is
2+
Oxidation:
Zn(s) → Zn (aq) + 2 e
2+
Reduction:
Cu (aq) + 2 e → Cu(s)
2+
2+
Overall:
Zn(s) + Cu (aq) → Zn (aq) + Cu(s)
2+
-
o
44
The cell diagram corresponding to the above figure and reaction is written as:
2+
2+
anode → Zn(s)Zn (aq)  Cu (aq) Cu(s) ← cathode Ecell = 1.103 V
Standard Electrode Potential (Ecell), Free Energy (ΔG) and Equilibrium Constant (K)
o
ΔG = -nFE
where
o
cell
n is the number of electrons transferred in the reaction
F is Faraday’s constant, the charge of 1 mole of electrons which can be found on the data sheet
-1
as 96485 C mol
o
We must of course remember that when we use E values to predict the direction of spontaneous change,
o
we are not saying that the reaction goes to completion. E values are a measure of the relative ‘demand’
between two species for elections.
This “tug-of-war” for electrons results in an equilibrium being established. If we have another look at
example 3 then we have the following equilibrium.
2-
+
Cr2O7 + 14 H + 6Fe
2+
3+
3+
⇌ 2 Cr + 6 Fe + 7 H2O
∆E = +0.56 V
o
The positive sign of the ∆E value tells us the forward reaction is spontaneous. This in turn means that the
o
position of the equilibrium is to the RHS of the equation. If the ∆E value tells us something about the
equilibrium state then it must, somehow, be linked to the equilibrium constant, K, for the reaction.
o
If ∆E is large and positive then the reaction favours the products and therefore we would expect the
equilibrium constant to be a large number. The mathematical relationship between the two is given by
the following equation:
o
where:
n is the number of electrons transferred in the reaction
F is Faraday’s constant, the charge of 1 mole of electrons which can be found
-1
on the data sheet as 96485 C mol
-1 -1
R is the gas constant, 8.315 J mol K
o
T is the temperature, which for standard conditions is 298 K (25 C)
45
Example 3.4
Question:
2+
Calculate the equilibrium constant for the reaction between MnO4 and Fe in acidic aqueous solution at
o
25 C
Answer:
Exercise 5.6
o
Calculate the KC value for the following reactions at 25 C
(a)
2-
2+
3+
-
Cr2O7 ions with Mn ions to give Cr ions and MnO4 ions in acidic media.
46
3+
–
2+
(b)
Co ions with I ions to give Co ions and I2 molecules in acidic solution.
Exercise 5.7
Which of the equilibria equations in Exercise 6 are spontaneous and which favour the products? Explain
your answer.
Ecell as a function of concentrations
Nernst equation:
Example 3.5
Question:
Consider the electrochemical cell represented by :
Zn(s)  Zn
2+
 Fe  Fe(s)
3+
3+
2+
-3
Determine E for the cell when the concentration of Fe is 10.0 M and that of Zn is 1.00 x 10 M.
Answer:
47