Redox potential chart
comments and relative stability of oxidation states:
All data quoted is for
standard conditions i.e. 298K, 1 atm. pressure and 1 mol dm–3
solutions of ions.
Other than the solid
metals, MnO2 and FeO42–, hydrogen gas, you
can assume all ions are in aqueous media.
Unless an oxyanion,
oxocation or another ligand in a complex is indicated, you assume you are
dealing with hexaaqua–metal ions (H2O ligand only).
Further comments below
draw out some general patterns and other points of interest.
Some trends on the nature and
stability of transition metal ions:
The lower oxidation states of
transition metals are usually found as simple ionic compounds e.g.
containing ions such as Cr3+,
Mn2+, Fe2+, Co2+, Ni2+,
The transition metal compounds of
their higher oxidation states are usually bound to an relatively highly
electronegative element such as oxygen or fluorine in an anion
e.g. vanadate(V) VO4-,
manganate(VII) MnO4- and ferrate(VI) FeO42-.
These are potentially
strong oxidising agents e.g. potassium manganate(VII).
You do not usually get simple
ions like V4+, Fe6+ or Mn7+.
For the 3d block in general in terms
of redox reactions:
From left to right, the higher
oxidation states become less stable relative to lower oxidation
Compounds of transition metals in
a high oxidation state tend to be oxidising agents (comment above).
Compounds of transition metals in
a low oxidation state tend to be reducing agents e.g. Ti2+
The relative stability of the +2
state relative to the +3 state increases from left to right e.g. the
hydrated aqueous ions of Ti2+ and Fe2+.are
easily oxidised to Ti3+ and Fe3+, but
difficult to oxidise Co2+, Ni2+ and Cu2+
to Co3+, Ni3+ and Cu3+.
BUT, take care, complexing
the central ion with ligands other than water and
considerably change the relative stability of the complex i.e.
can cause the half-cell potential to be significantly changed.
All except scandium
(Sc/Sc3+), which is not that reactive towards acids despite the
relatively negative M/M3+ potential, form a hydrated M2+
ion either by reaction of the metal with acid or reduction of a higher
oxidation state complex–compound.
oxidation states in aqueous solution containing dissolved oxygen
from air tend to be the 'simple' hydrated ions such as ...
Sc3+ (only ion),
[TiO]2+, VO2+, Cr3+,
Mn2+, Fe3+, Co2+, Ni2+,
Cu2+ and Zn2+ (only ion), NOT Ti2+,
Cr2+ or Fe2+.
On the basis of the
electrode potential chart above, the argument is simple. In neutral or
acid solution the oxidising potential of the oxygen–proton–water system
is +1.23V. Therefore any e.g. M3+/M2+ potential
less positive than +1.23V will result in the oxidation of the lower
oxidation state species to the higher oxidation state species in the
presence of dissolved oxygen which is reduced to water.
higher than the stable ones tend to oxidise water liberating oxygen
and as mentioned above, lower oxidation states tend to be reducing and
liberate hydrogen from water.
So the Mn3+/Mn2+ and
Co3+/Co2+ potentials lie above +1.23V so Mn3+
and Co3+ will oxidise water and cannot be stable in acid
Note that the +4
oxidation states of Ti and V exist as hydrated oxo–cations because the
high polarising power of the highly charged central metal ion causes
Appendix 1. Acidity
of hexa–aqua ions).
The rest are [M(H2O)n]2+/3+
where n is usually 6, can be 4 for Cu and Zn.
Other comments on the relative
stability of transition metal ions
Apart from iron,
there is a tendency for the lower oxidation state to become increasingly
more stable with increasing atomic number.
states which are normally oxidising in aqueous solution can be
stabilised by complexing e.g. compare the Co(II)/Co(III) potential when
complexed with water (+1.82V) and with the ligand ammonia (+0.10V).
There are classic
examples of disproportionation where an intermediate oxidation state
species spontaneously changes into a higher and lower' oxidation state
species e.g. the disproportionation reactions
Cu(I) ==> Cu(0) +
Cu(II) and Mn(VI) ==> Mn(II) + Mn(VII).
These are described
in detail, complete with electrode potential arguments for thermodynamic
feasibility, under the respective metal.
How do you work out
what will oxidise what? or what will reduce what?
Using an electrode
potential chart like the one above or a list of redox potentials the
following rules apply.
To facilitate an
oxidation, the half–cell potential of the oxidising agent must be less
negative or more positive than the redox potential of the 'system' you
wish to oxidise.
So using at the
redox potential chart for example:
will oxidise Co2+ to Co3+ in presence of ammonia –
forms the amine complexes, but the hexaaqua complex ion of Co2+
is stable in the presence of oxygen if no ammonia present.
(H2O ligand, EØ = +1.82V), O2/H2O
(EØ = +1.23V in neutral solution)
(EØ = +1.23 is less than +1.82 but more than +0.10V), Co3+/Co2+
(NH3 ligand, EØ = +0.10V)
is stable in the presence of oxygen, but [Co(NH3)6]2+
will be oxidised to [Co(NH3)6]3+.
You can then
further predict that [Co(H2O)6]3+ will
oxidise water to oxygen.
To facilitate a
reduction, the half–cell potential of the reducing agent must be more
negative or less positive than the redox potential of the 'system' you
wish to reduce.
So using the redox
potential chart and the half–cell redox potential for I2/I–
ions will be reduced by iodide ions because EØ for Fe3+/Fe2+
(H2O ligand) is +0.77V
i.e. the Fe3+ will oxidise
the iodide ions rather than iodine oxidising the Fe2+ ions.
[Fe(H2O)6]3+ is reduced to [Fe(H2O)6]2+,
iron(III) to iron(II).
However if the
ligand is the cyanide ion, then iodide ions will not reduce the Fe3+
cyanide ion complex but iodine would oxidise [Fe(CN)6]4–
to [Fe(CN)6]3–, iron(II) to iron(III).
(H2O ligand, EØ = +0.77V), I2/I–
((EØ = +0.54)
(EØ = +0.54, less than +0.77 but more than +0.36V), Fe3+/Fe2+
(CN– ligand, EØ = +0.36V)
How to work out the
feasibility of reaction from electrode
potential data is described in Appendix 5.