Reduction

Chapter 8
Reduction and oxidation
TOPICS
 Redox reactions and oxidation states (an overview)
 Reduction potentials and Gibbs energy
 Disproportionation
 Potential diagrams
 Frost--Ebsworth diagrams
 The effect of complex formation or precipitation on
Mz+/M reduction potentials
 Applications of redox reactions to industrial processes
Oxidation and reduction
Oxidation refers to gaining oxygen, losing hydrogen or losing one or
more electrons.
Reduction refers to losing oxygen, gaining hydrogen or gaining one or
more electrons
Magnesium acts as the reducing agent or reductant
, while O2 acts as the oxidizing agent or oxidant.
Redox is an abbreviation for reduction–oxidation.
In an electrolytic cell, the passage of an electrical current initiates
a redox reaction
In a galvanic cell, a spontaneous redox reaction occurs and generates
an electrical current
Oxidation states
An oxidation process is accompanied by an increase in the oxidation
state of the element involved; conversely, a decrease in the oxidation
state corresponds to a reduction step.
8.2 Standard reduction potentials, Eo, and relationships between
Eo, DGo and K
Half-cells and galvanic cells
The reaction is written as an equilibrium
The Daniell cell is an
example of a galvanic
cell.
Eocell is the standard cell potential.
The standard cell potential (at 298 K) for the Daniell cell is 1.10 V. Calculate the
corresponding values of DGo and K and comment on the thermodynamic
viability of the cell reaction:
Defining and using standard reduction potentials, Eo
For example, if Zn metal is placed into dilute acid, H2 is evolved. Thus, when
the standard hydrogen electrode is connected in a galvanic cell with a Zn2+/Zn
electrode, The following reaction is the spontaneous cell process.
The following two half-reactions correspond to two half-cells that are combined
to form an electrochemical cell:
(a) What is the spontaneous cell reaction? (b) Calculate Eocell.
(a) First, look up values of Eo for the half-reactions.
(b) The cell potential difference is the difference between the standard reduction
potentials of the two half-cells:
For example, to investigate the reaction between Fe and aqueous Cl2,
we consider redox couples 8.16–8.18.
For reaction 8.19 (where z = 2), DGo = 347 kJ per mole of reaction, while for
reaction 8.20 (z = 6), DGo =  810 kJ per mole of reaction. Per mole of Fe, the
values of DGo =  347 and  810 kJ , revealing that reaction 8.20 is
thermodynamically favoured over reaction 8.19.
Dependence of reduction potentials on cell conditions
Application of the Nernst equation to the Zn2+/Zn half-cell ( Eo =  0.76 V) gives
E =  0.79 V for [Zn2+] = 0.10 mol dm3
The more negative value of E ,
corresponding to a more positive
value of DG, signifies that it is more
difficult to reduce Zn2+ at the lower
concentration
Now consider the effect of pH (pH=  log[H+]) on the oxidizing ability of [MnO4] in
aqueous solution at 298 K.
E = Eo when [H+] =1 mol dm3, and [Mn2+] = [MnO4] =1 mol dm3. As [H+] increases (i.e.
the pH of the solution is lowered), the value of E becomes more positive. The fact that
the oxidizing power of [MnO4] is lower in dilute acid than in concentrated acid
explains why, for example, [MnO4] will not oxidize Cl in neutral solution, but liberates
Cl2 from concentrated HCl.
Worked example 8.4 Oxidation of Cr2+ ions in O2-free, acidic, aqueous solution
Explain why an acidic, aqueous solution of Cr2+ ions liberates H2 from solution
(assume standard conditions). What will be the effect of raising the pH of the
solution?
Raising the pH of the solution lowers the concentration of H+ ions. Let us consider a
value of pH 3.0 with the ratio [Cr3+] : [Cr2+] remaining equal to 1.
The 2H+ = H2 electrode now has a new reduction potential.
Now we must consider the following combination of half-cells, taking Cr3+ =Cr2+ still
to be under standard conditions
Thus, although the reaction still
has a negative value of DG , the
increase in pH has made the
oxidation of Cr2+ less
thermodynamically favourable.
8.3 The effect of complex formation or precipitation on Mz+/M
reduction potentials
Half-cells involving silver halides
Under standard conditions, Ag+ ions are reduced to Ag(equation 8.29), but if the
concentration of Ag+ is lowered, application of the Nernst equation shows that
the reduction potential becomes less positive (i.e. DG is less negative).
In practice, a lower concentration of Ag+ ions can be achieved by dilution of the
aqueous solution, but it can also be brought about by removal of Ag+ ions from
solution by the formation of a stable complex or by precipitation of a sparingly
soluble salt.
Reduction of Ag(I) when it is in the form of solid AgCl occurs according to
reaction 8.31, and the relationship between equilibria 8.29–8.31 allows us to
find, by difference, DGo for reaction 8.31. This leads to a value of Eo = + 0.22
V for this half-cell
The difference in values of Eo for half-reactions 8.29 and 8.31 indicates that it is
less easy to reduce Ag(I) in the form of solid AgCl than as hydrated Ag+.
Silver iodide ( Ksp = 8.51x10 17) is less soluble than AgCl in aqueous solution,
and so reduction of Ag(I) in the form of solid AgI is thermodynamically less
favourable than reduction of AgCl.
However, AgI is much more soluble in aqueous KI than AgCl is in aqueous KCl
solution. The species present in the iodide solution is the complex [AgI3]2,
Modifying the relative stabilities of different oxidation states of a metal
consider the Mn3+/Mn2+ couple, for which equation 8.34 is appropriate for aqua
species.
In alkaline solution, both metal ions are precipitated, but Mn(III) much more
completely than Mn(II) since values of Ksp for Mn(OH)3 and Mn(OH)2 are =1036
and = 2x10 13, respectively.
Precipitation has the effect of significantly changing the half-cell potential for the
reduction of Mn(III).
In solutions in which [OH]= 1M, Mn(III) is stabilized with respect to reduction to
Mn(II) as the value of Eo[OH]=1 for equation 8.35 illustrates. Compare this with
equation 8.34.
First, find the half-equations that are relevant to the question; note that
pH 0 corresponds to standard conditions in which = [H+] 1 mol dm3.
Most d- block metals resemble Mn in that higher oxidation states are more
stable (with respect to reduction) in alkaline rather than acidic solutions. This
follows from the fact that the hydroxide of the metal in its higher oxidation state
is much less soluble than the hydroxide of the metal in its lower oxidation state.
The following equations show the reduction of Co3+
The overall formation constant for [Co(NH3)6]3+ =1030
A similar comparison can be made for the reduction of the hexaaqua ion of Fe3+
and the cyano complex (equations 8.38 and 8.39), and leads to the conclusion
that the overall formation constant for [Fe(CN)6]3 is =107 times greater
8.4 Disproportionation reactions
Worked example 8.6 Disproportionation of copper(I)
Using appropriate data from Table 8.1, determine K (at 298 K) for the equilibrium:
Three redox couples in Table 8.1 involve Cu(I), Cu(II) and Cu metal:
The disproportionation of Cu(I) is the result of combining half-reactions (1) and (3). Thus
The value indicates that
disproportionation is
thermodynamically
favourable.
8.5 Potential diagrams
Consider manganese as an example. Aqueous solution species may contain
manganese in oxidation states ranging from Mn(II) to Mn(VII), and equations
8.42–8.46 give half-reactions for which standard reduction potentials can be
determined experimentally.
These potentials may be used to derive values of Eo for other half-reactions
such as 8.47.
Potential Diagrams – Latimer Diagrams
The diagram corresponding to the two half-reactions
Combining these two half-cells give reaction 8.48 for which Eocell = 1.20 V and
DGo (298K) =  231 kJmol1
This indicates that reaction 8.48 is spontaneous. Similarly, at pH 0, Mn3+ is
unstable with respect to disproportionation to MnO2 and Mn2+
Worked example 8.7 Potential diagrams
The following potential diagram summarizes some of the redox chemistry of
iron in aqueous solution. Calculate the value of Eo for the reduction of Fe3+(aq)
to iron metal.
Although there are short cuts to this problem, the most rigorous method is to
determine DGo (298K) for each step. Fe3+ to Fe2+is a one-electron reduction.
8.6 Frost-Ebsworth diagrams
Frost--Ebsworth diagrams and their relationship to potential diagrams
In a Frost-Ebsworth diagram, values of DGo or, more commonly, DGo/F for
formation M(N) from M(0), where N is the oxidation state, are plotted againest
increasing N, From the relationship:
Fig. 8.4 Frost–Ebsworth diagrams in aqueous solution at pH 0, i.e. [H+] =1 mol
dm3, for (a) chromium
Fig. 8.4 Frost–Ebsworth diagrams in aqueous solution at pH 0, i.e. [H+] =1 mol
dm3, for (b) phosphorus
Fig. 8.4 Frost–Ebsworth diagrams in aqueous solution at pH 0, i.e. [H+] =1 mol
dm3, for (c) nitrogen
8.7 The relationships between standard reduction potentials and some
other quantities
Worked example 8.9 Determination of DsolGo for an ionic salt
Ellingham
Diagram