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INORGANIC
Part 10 3d block TRANSITION METALS sub-index: 10.1-10.2
Introduction 3d-block Transition Metals * 10.3
Scandium
* 10.4 Titanium * 10.5
Vanadium * 10.6 Chromium
* 10.7 Manganese * 10.8
Iron * 10.9 Cobalt
* 10.10 Nickel
* 10.11 Copper * 10.12
Zinc
* 10.13 Other Transition Metals e.g. Ag and Pt * Appendix 1.
Hydrated salts, acidity of
hexa-aqua ions * Appendix 2. Complexes
& ligands * Appendix 3. Complexes and isomerism * Appendix 4.
Electron configuration & colour theory *
Appendix 5. Redox
equations, feasibility, Eø * Appendix 6.
Catalysis * Appendix 7.
Redox
equations
* Appendix 8. Stability Constants and entropy
changes *
Appendix 9. Colorimetric analysis
and complex ion formula * Appendix 10 3d block - extended data
* Appendix 11 Some 3d-block compounds, complexes, oxidation states
& electrode potentials * Appendix 12
Hydroxide complex precipitate 'pictures',
formulae and equations
Advanced
Level Inorganic Chemistry Periodic Table Index *
Part 1
Periodic Table history
* Part 2
Electron configurations, spectroscopy,
hydrogen spectrum,
ionisation energies *
Part 3
Period 1 survey H to He *
Part 4
Period 2 survey Li to Ne * Part
5 Period 3 survey Na to Ar *
Part 6
Period 4 survey K to Kr and important trends down a
group *
Part 7
s-block Groups 1/2 Alkali Metals/Alkaline Earth Metals *
Part 8
p-block Groups 3/13 to 0/18 *
Part 9
Group 7/17 The Halogens *
Part 10
3d block elements & Transition Metal Series
*
Part 11
Group & Series data & periodicity plots * All
11 Parts have
their own sub-indexes near the top of the pages
10.5. Chemistry
of Vanadium
V, Z=23, 1s22s22p63s23p63d34s2
-
V
data table 1 summary *
extended vanadium data table 2 *
Vanadium & electrode potential
chart of 3d-block
-
Summary of some
complexes-compounds & oxidation states of vanadium compared to other
3d-block elements
-
Vanadium is one of
many transition metals alloyed with iron to make specialist steels.
-
Vanadium(V) oxide, V2O5,
is used as a catalyst in the 'Contact Process'
in the production of sulphur
trioxide for the manufacture
of sulphuric acid.
-
The catalysing of the
conversion of sulphur dioxide into sulphur trioxide is explained via change in oxidation state
changes i.e. some classic transition metal chemistry.
-
2SO2(g)
+ O2(g) ==> 2SO3(g)
-
The mechanism,
somewhat simplified, goes
via the catalytic cycle ...
-
(i) SO2 + V2O5 ==> SO3
+ V2O4, then (ii) V2O4
+ 1/2O2 ==> V2O5
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The vanadium
changes oxidation state from +5 to +4 and back to +5 in the
catalytic cycle, a classic combination of two characteristics of
transition metals - variable oxidation state and catalytic
properties.
-
This is an
example of heterogeneous catalysis - reactants (g) and catalyst (s) in
different phases.
-
Vanadium shows a
'classic' display of variable oxidation states
of varying colours
when a solution of e.g. ammonium vanadate(V), is reduced by a
zinc/dilute sulphuric acid mixture. do picture
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Acidification changes
the vanadate(V) ion into the pale yellow oxo-cation VO2+
(oxovanadium(V) ion)
-
VO43-(aq) + 4H+(aq)
VO2+(aq) + 2H2O(l) [an
acid-base reaction, NOT a redox change]
-
Note: Highly charged cations >3+ rarely exist as the
simple 'hydrated' tetra or hexa-aqua ion.
-
The theoretical polarising power
of the 'central metal ion' is so strong that
they form oxocations (see above) or oxyanions e.g.
-
orange
dichromate(VI) Cr2O72-, yellow
chromate(VI) CrO42-, purple manganate(VII)
MnO4- etc.
-
For transition metals they may be coloured
even if electronically the theoretical 'central metal ion'
has a noble gas structure e.g. [Ar] in its maximum
oxidation state like V(V), Cr(VI) and Mn(VII).
-
These oxyanions are
called charge transfer complexes and the theory is beyond
pre-university chemistry.
-
Three
successive reduction
steps then follow to eventually give V2+ ions,
shown as half-cell equations:
-
(i) V(V, +5) ==> V(IV,
+4): VO2+(aq) + 2H+(aq) +
e-
VO2+(aq) + H2O(l)
-
(ii) V(IV, +4) ==> V(III,
+3): VO2+(aq) + 2H+(aq) +
e-
V3+(aq) + H2O(l)
-
EØhalf-cell
potential = +0.34V,
blue to
the
green vanadium(III) ion
-
Here the vanadium(III) ion, V3+, is actually the
green hexaaquavanadium(III) ion, [V(H2O)6]3+
-
Both V(IV) and V(III) species
are slowly oxidised by dissolved oxygen back to the V(V) compound in
acid solution (see electrode potential comments later).
-
(iii) V(III,
+3) ==> V(II, +2): V3+(aq) + e-
V2+(aq)
-
EØhalf-cell
potential = -0.26V,
green to the purple-violet vanadium(II) ion.
-
V2+(aq)
is powerful reducing agent and is unstable in the presence of
air. Any dissolved oxygen will oxidise V2+(aq)
back to the vanadium(III) cation.
-
V2+ is actually the
purple-violet hexaaquavanadium(II) ion, [V(H2O)6]2+
-
Note
-
The standard
electrode potential EØZn(s)/Zn2+(aq)
is -0.76V, so the reducing power of zinc is sufficient
to effect any of the three vanadium oxidation state reduction
changes described above.
-
The reduction occurs on the surface of the zinc
metal i.e. the
site of electron transfer and you can write the above reductions
as a fully balanced complete equations ...
-
(i) 2VO2+(aq) +
4H+(aq) + Zn(s)
==> 2VO2+(aq) + 2H2O(l)
+ Zn2+(aq)
-
(ii)
2VO2+(aq) + 4H+(aq)
+ Zn(s) ==> 2V3+(aq) +
2H2O(l) + Zn2+(aq)
-
(iii)
2V3+(aq) + Zn(s ==>
2V2+(aq) + Zn2+(aq)
-
EØreaction
= -0.26 - (-0.76) = +0.50V
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BUT
the vanadium(II) cation is unstable in the presence of
dissolve oxygen in air.
-
1/2O2(g) + 2H+(aq)
+ 2e-
H2O(l) has a standard electrode
potential of +1.23V,
-
so,
for the vanadium(II) oxidation reaction ...
-
1/2O2(g) + 2H+(aq)
+ 2V2+(aq) ==> 2V3+(aq)
+
H2O(l)
-
EØreaction
= EØreduction - EØoxidation
= +1.23 - (-0.26) = +1.49V
-
hence the if left standing open to air, the violet V2+(aq)
solution will gradually change to a green V3+(aq)
solution and in turn V3+(aq) will
revert back to VO2+(aq) in the
presence of air because of oxidation by dissolve oxygen
unless protected by an inert atmosphere. (see
Redox Electrode
Potential Chart, V2+/V3+
and V3+/VO2+ potentials are less
positive (below)
that for O2/H2O/H+
potentials).
-
You will see
hydrogen formed simultaneously from the unavoidable metal-acid
reaction.
-
Does vanadium
chemistry show an example of disproportionation?
-
This is just a little academic
exercise using standard electrode potential data.
-
A disproportionation
reaction is where a species in one oxidation state spontaneously and
simultaneously changes into two species of different oxidation states -
one higher and one lower in oxidation number.
-
Examples:
disproportionation in manganese(VI)
chemistry and
disproportionation in copper(I) chemistry
-
Question: In
terms of aqueous ions, is the disproportionation of vanadium(III) into
vanadium(II) and vanadium (IV) feasible?
-
(i) VO2+(aq) + 2H+(aq) + 2e-
V3+(aq) + H2O(l)
(EØVO2+/V3+ = +0.34V)
-
(ii) V3+(aq) + e-
V2+(aq) (EØV3+/V2+
= -0.26V)
-
The
disproportionation equation would be (iii) 2V3+(aq)
+ H2O(l)
V2+(aq) + VO2+(aq) + 2H+(aq)
-
For equation (iii),
(ii) will be the reduction half-cell equation and (i) reversed will be
the oxidation half-cell reaction.
-
EØreaction
= EØreduction - EØoxidation =
= EØV3+/V2+ - EØVO2+/V3+ =
(-0.26) - (+0.34) = -0.60V
-
showing the
disproportionation is thermodynamically NOT feasible i.e. EØreaction
is less than zero.
-
In fact what can
actually happen is if you mix salt solutions of vanadium(IV) and
vanadium(II) on an equimolar basis, you end up with a solution of
vanadium(III) salts, a sort of 'anti-disproportionation' reaction!
Scandium
* Titanium * Vanadium
* Chromium
* Manganese * Iron * Cobalt
* Nickel
* Copper *
Zinc
* Silver & Platinum
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