Some
chemistry of Iron (2):
Edexcel
24. be
able to record observations and write suitable equations for the reactions of Cr3+(aq),
Fe2+(aq), Fe3+(aq), Co2+(aq) and Cu2+(aq)
with aqueous sodium hydroxide and aqueous ammonia, including in excess.
25. be able
to write ionic equations to show the difference between ligand exchange and
amphoteric behaviour for the reactions in (24) above.
34.
understand the role of Fe2+ ions in catalysing the reaction
between I− and
S2O82— ions.
Manganate(VII)
with iron (II) titration self-indicating
catalyst in
Haber process
AQA
Exchange of
the ligand H2O by Cl– can involve a change of co-ordination number (e.g.
Fe3+(aq),
Co2+(aq) and Cu2+(aq).
Haem is an
iron(II) complex with a multidentate ligand.
Oxygen forms
a co-ordinate bond to Fe(II) in haemoglobin, enabling oxygen to be transported
in the blood.
Carbon
monoxide is toxic because it replaces oxygen co-ordinately bonded to Fe(II) in
haemoglobin.
The redox
titrations of Fe2+(aq)
and C2O42– with MnO4—
Students should be able to perform calculations for these titrations and similar
redox reactions.
Examples
include, finding:
•
the mass of iron in an iron tablet
•
the percentage of iron in steel
•
the Mr
of hydrated ammonium iron(II) sulfate
•
Fe is used as a heterogeneous catalyst in the Haber process.
In aqueous
solution, the following metal-aqua ions are formed:
[M(H2O)6]2+,
limited to M = Fe and Cu
[M(H2O)6]3+,
limited to M = Al and Fe
The acidity
of [M(H2O)6]3+ is greater than that of [M(H2O)6]2+
Students should be able to:
•
explain, in terms of the charge/size ratio of the metal ion, why the
acidity of [M(H2O)6]3+ is greater than that of [M(H2O)6]2+
•
describe and explain the simple test-tube reactions of M2+(aq)
ions, limited to M = Fe and Cu, and of M3+(aq) ions, limited to M =
Al and Fe, with the bases OH–, NH3 and CO32—
•
Students could carry out test-tube reactions of metal-aqua ions with NaOH,
NH3 and
Na2CO3
Halogen
carrier in arene organic chemistry
OCR
Redox reactions
(k) redox
reactions and accompanying colour changes for:
(i) interconversions
between Fe2+ and
Fe3+
Fe2+ can be
oxidised with H+/MnO4– and Fe3+ reduced
with I–
7. Iron as a catalyst
Haber
Process
a) The Haber
process is used to form ammonia from atmospheric nitrogen and hydrogen
extracted from methane (natural gas).
The
industrial process uses iron as a catalyst in a process in which the pressure
is around 200atm and the working temperature around 450oC.
N2(g) +
3H2(g) ⇌ 2NH3(g)
You can find
out more about the Haber Process if you follow this link:
Iron here
acts as a heterogenous catalyst because it is in a different state to the
gaseous reactants.
b) Redox
catalysis of the peroxodisulphate/iodide reaction
The reaction
between peroxodisulphate ions and iodide ions is typically used to illustrate
the action of iron ions and their catalytic effect in this reaction.
This is the
equation for the redox reaction between peroxodisulphate ions and iodide
ions.
2I— +
S2O82— ⟶ I2 +
2SO42—
As you can
see both ions are anions negatively charged and so will repel each other.
The result
is that this reaction has a very high activation energy.
Iron ions
have a catalytic effect because they lower the activation energy of the
reaction.
When a small
amount of iron(II) ions (Fe2+) are added to the iodide
-peroxodisulphate mixture, the peroxodisulphate ions oxidise them to iron(III)
ions.
2Fe2+ +
S2O82— ⟶ 2Fe3+ +
2SO42—
Then the
iron(III) ions formed oxidise the iodide ions to iodine and they are reduced
back to iron(II)
2I— +
2Fe3+ ⟶ I2 +
2Fe2+
These
iron(II) ions can now start the redox process all over again.
The reaction
is much faster because the two reactions catalysed with iron ions have much
lower activation energies.
In this
example the iron(II) ions act as a homogenous catalyst since they are in the
same state as the iodide and peroxodisulphate ions.
8. Titration of iron(II) with manganate(VII)
ions
Example 1:
Finding the mass of iron in an
iron tablet:
A typical
multivitamin with iron tablet weighs 0.364g and is said to contain 14mg of
iron.
Is this the
case?
To determine
the iron content of this tablet first weigh two tablets and then grind them up
and dissolve them in dilute (0.5M) sulphuric acid and make up the solution to
250ml in a volumetric flask.
The role of
the sulphuric acid is there to prevent oxidation of the iron (II) ions to
iron(III) ions. I’m here assuming that
the iron in the tablet is in the form of iron(II)sulphate.
Next titrate
a 25ml aliquot of the solution from the volumetric flask with potassium
manganate(VII) solution (0.0005M).
Manganate(VII)
ions oxidise the iron(II) ions according to the equation below.
5Fe2+ +
MnO4—
+ 8H+ 5Fe3+ +
Mn2+ + 4H2O
The
titration results show that it requires 20ml of the manganate(VII) solution to
completely oxidise the iron(II) ions.
The end
point of the titration is when the iron(II) solution turns a very pale purple
because manganate(VII) is the purple and it is therefore self-indicating.
Calculation:
a)
Calculate moles of manganate(VII) ions from the titration.
Use n=cV
Therefore
n= 0.0005 ×
20/1000 = 0.00001 moles manganate(VII) ions.
b)
Determine the number of moles iron(II) ions in the aliquot from the
equation since 5 moles iron(II) ions react with 1 moles manganate(VII) ions.
5Fe2+ +
MnO4—
0.00005 0.00001
c)
if there are 0.00005 moles iron(II) ions in the aliquot then there are 10×
this number of moles in the volumetric flask i.e. 0.0005 moles.
d)
Calculate the mass of iron that this number of moles represents.
Use m = n
× M therefore mass Fe2+ ions = 0.0005
× 55.9 =
0.02795g or 27.95mg.
This mass of iron corresponds closely to the mass of iron reportedly
present in 2 of these tablets 28mg.
Example 2:
Finding the percentage of iron
in a piece of steel weighing 0.034g.
First thing
to do is dissolve the steel in the minimum volume of 2M sulphuric acid. You may need to heat the flask in which you
carry out this part of the experiment and do so in fume cupboard because of the
hydrogen fumes that will be evolved.
The
resultant solution should look pale green in colour showing the presence of
iron(II) ions.
Next allow
the solution to cool and then transfer it all to a 250ml volumetric flask and
make up to the mark with distilled water.
Next prepare
a solution of potassium manganate(VII) 0.0005M and titrate a 25ml aliquot of
the iron(II) ions solution with the purple manganate(VII) solution.
If the
titration results show that 23.7ml of the manganate(VII) solution completely
oxidise the iron(II) ions, calculate the % of iron in the sample of steel.
a)
Calculate moles of manganate(VII) ions from the titration.
Use n=cV
Therefore
n= 0.0005 ×
23.7/1000 = 0.00001185 moles manganate(VII) ions.
b)
Determine the number of moles iron(II) ions in the aliquot from the
equation since 5 moles iron(II) ions react with 1 moles manganate(VII) ions.
5Fe2+ + MnO4—
0.00005925 0.00001185
c)
if there are 0.00005925 moles iron(II) ions in the aliquot then there are
10× this number of moles in the volumetric flask i.e. 0.0005925 moles.
d)
Calculate the mass of iron that this number of moles represents.
Use m = n ×
M therefore mass Fe2+
ions = 0.0005925 ×
55.9 = 0.03312g.
e)
Therefore the percentage of iron = 0.03312/0.034 × 100 = 97.4
Example 3:
Titration of iron(II) with
ethandioate ions
Follow this
link here to details about this titration.
9. Iron and Haemoglobin
Haem is an
iron(II) complex with a multidentate ligand.
Oxygen forms
a co-ordinate bond to Fe(II) in haemoglobin, enabling oxygen to be transported
in the blood.
Carbon
monoxide is toxic because it replaces oxygen co-ordinately bonded to Fe(II) in
haemoglobin.
You can find
out more about this area by following this
link: Haemoglobin.
No comments:
Post a Comment