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Chemistry Unit 5
Chapter 12 – Thermodynamics
Chapter 12.1 – Enthalpy Change
Hess’s Law
The Hess’s Law states that:
It basically says that if we convert
reactant A into product B, the
aggregate enthalpy change will be
exactly the same as whichever
reaction route we have used.
Definitions:
Enthalpy change
The measure of heat change at a constant pressure. It is abbreviated as ΔH.
Standard enthalpy change
It refers to the enthalpy change under standard conditions and for specified
amount of chemicals in moles.
Standard conditions
A term used to define standard enthalpy and free energy changes. It relates to a
condition where temperature is 298 kelvins, a pressure of 1 bar (100kPa) and if
solution is involved, a concentration of 1 mol dm-3.
Standard states
The stable physical state of an element or compound under standard conditions.
For example, the standard state for water is a liquid; for oxygen it would be in
gaseous state.
Standard molar enthalpy of formation (ΔH°f )
It is the enthalpy change when one mole of compound is formed from its
constituent elements, under standard state, and all products and reactants in
their standard state.
The enthalpy change accompanying a chemical change, is
independent of the route by which the chemical change
occurs.
2
Standard enthalpy of atomisation (H°a )
It is the enthalpy change which accompanies the formation of one mole of
gaseous atoms from the elements in its standard states under standard
conditions.
First ionisation energy (1st IE)
It is the standard enthalpy change when one mole of gaseous atoms is converted
into a mole of gaseous ions each with a single positive charge.
Second ionisation energy (2nd IE)
It relates to the standard enthalpy change to remove an electron from a mole of
singly-positively charged ions.
First electron affinity (EA)
The standard enthalpy change when one mole of gaseous atoms is converted to a
mole gaseous ions with each a single negative charge.
Second electron affinity (EA)
The standard enthalpy change when a mole of electrons are added to a mole of
gaseous ions, each with a single negative charge to form ions each with two
negative charges.
Lattice formation enthalpy
It is the standard enthalpy change when a mole of solid ionic compound is
formed from its gaseous ions.
Lattice dissociation enthalpy
The opposite process of lattice formation: it has the same value as the lattice
formation enthalpy, but always carry a positive value.
It is the enthalpy change needed to convert 1 mole of solid crystal into its
scattered gaseous ions.
Enthalpy of hydration
The standard enthalpy change when water molecules surround one mole of
gaseous ions.
Enthalpy of solution
The standard enthalpy change when one mole of solute dissolves completely in
sufficient solvent to form a solution in which the molecules or ions are far
enough apart not to interact with each other.
Mean bond enthalpy
The enthalpy change when one mole of gaseous molecules each breaks a
covalent bond, to form two free radicals, averaged over a range of compounds.
3
Chapter 12.2 – Born-Haber cycle
Lattice enthalpies cannot be measured directly. A way to find the lattice enthalpy
is via Born-Haber cycle, involving enthalpy changes which can be measured.
Example
Below shows a Born-Haber cycle for sodium chloride.
Note that exothermic reaction points downwards; and vice-versa, endothermic
reaction points upwards.
We start with the two thick lines: NaCl (s), and the elements in their standard
states, Na (s) + 0.5 Cl (g). The whole cycle is to find the formation of sodium
chloride via a series of many small enthalpy changes.
The +107 is the atomisation enthalpy of sodium. We have to produce gaseous atoms so that we can use the next stage in the cycle.
The +496 is the first ionisation energy of sodium. Remember that first ionisation energies go from gaseous atoms to gaseous singly charged positive ions.
The +122 is the atomisation enthalpy of chlorine. Again, we have to produce gaseous atoms so that we can use the next stage in the cycle.
4
The -349 is the first electron affinity of chlorine. Remember that first electron affinities go from gaseous atoms to gaseous singly charged negative ions.
And finally, we have the positive and negative gaseous ions that we can convert into the solid sodium chloride using the lattice formation enthalpy.
And using Hess’s Law:
-411 = +107 + 496 + 122 - 349 + LE
LE = -411 - 107 - 496 - 122 + 349
LE = -787 kJ mol-1
5
Trends in lattice enthalpies
There are two factors which affect the lattice enthalpy: the charges on the ions
and the ionic radii (which affects the distance between the ions).
Charge of ions
For example, both sodium chloride (NaCl) and magnesium oxide (MgO) has
exactly the same ionic arrangement in the lattice. However the latter lattice
enthalpy is much greater.
This is because in the case of magnesium oxide, magnesium has 2+ charge while
an oxygen has a 2- charge; whereas in sodium chloride, the attraction is only
between 1+ and 1- ions. As a conclusion, increasing charge difference result in a
greater lattice enthalpy.
6
The radius of ions
As we go down the Group 7 of the periodic table, we expect that the lattice
enthalpies fall as negative ions are larger. This is because of the increased
distance between the centres of the oppositely charged ions.
7
Chapter 12.3 – More enthalpy changes
Enthalpy of solution:
Ions are strongly attracted to water, due to the fact that water is a polar molecule.
Ions can be dissolved when the lattice is broken up, which requires some
energy – lattice enthalpies. The constituent ions are then surrounded by cluster
of solvent ions. Positive ions are surrounded by negative ends of the dipole
water molecule, and negative ions surrounded by the positive ends of the dipole
water molecule. This process is known as hydration.
Size of enthalpy of hydration:
The size of enthalpy of hydration is determined by the amount of attraction
between the ions and the water molecules.
Size of ions
When the ions are smaller, there will be a stronger attraction. For example,
smallest ions in each group, i.e. lithium in Group 1 and fluoride ion in group 7 has
the highest hydration enthalpy. As ions get bigger, due to increased distance
between the centre of ion and the water molecule, the enthalpy of hydration falls.
Charge of ions
The attraction is greater if the charge of the ion is greater. The hydration
enthalpies of group 2 ions and much greater than those of group 1 ions (Mg2+ >
Na+).
Dissolving as a energy cycle:
We can assume dissolving an ionic compound as a three-step process.
Crystal lattice is broken up into separate gaseous ions – lattice dissociation
enthalpy.
Hydrating the positive ions – enthalpy of hydration.
Hydrating the negative ions – enthalpy of hydration.
8
Lattice enthalpies and bonding:
Purely ionic model of a lattice assumes that all ions are perfectly spherical, and
have their charge evenly distributed around them.
Often the theoretical value (which we worked out) agrees with the experimental
value (which we found out via experiment). Sometimes there is a large
discrepancy between the two values between the experimental and theoretical
lattice formation enthalpy.
An explanation for this would be due to the fact that the bond contains some
covalent character, but the ionic bond nature still co-exist.
This is called ion distortion. The positively charged ion, cation, slowly
approaches the anion (negatively charged ion), and deform the electron cloud of
the anion. It attracts the electron cloud towards the cation. How easily could a
cation distort an anion depends on its size and charge.
If an anion is huge in size, then the distance between the electrons and the
nucleus increases. If the anion is rich in charge, too, then there are more
electrons involved in the distortion. For example, S 2- ion is more easily distorted
than the Cl- ion.
If the cation is small is size, then the distance between the positively-charged
nuclei and the negative ion would be reduced, thus increase the strength of
distortion. With high positive charge, it increases the power in which the
electron is attracted towards it. This type of positive ion has high concentration
of positive charge. In any metallic groups of the periodic table, the member that
forms the smallest cation has the biggest tendency to pull electrons from the
anion, which form a high degree of covalent character. For example, lithium in
group 1.
In this scenario, the anion is said to be polarised.
All ionic and covalent bonds can be ‘transitional’ – there is a continuous
spectrum from ionic to purely covalent, depending on the strength of the ions.
For example, Cs+F-, with large, singly charged positively; and a small singly
negatively charged ion, has no degree of polarisation and is almost completely
ionic. On the contrary, a bond formed of two identical atoms must be covalent.
9
Chapter 12.4 – Calculating enthalpy changes using mean bond enthalpies
In any chemical reactions, energy is absorbed to break bonds, and released when
bonds are formed. The overall difference between energy absorbed and released
is the overall enthalpy change of reaction.
Enthalpy change of reaction = Total energy absorbed – Total energy
released
In order to find out the overall enthalpy change of reaction, we work out the total
enthalpy required to break the bonds, and the total enthalpy released as the
bonds are formed. The enthalpy change of reaction would be subtracting total
energy released from sum of total energy absorbed.
For example:
Calculate the overall enthalpy change for the following reaction,
CH4 + 2O2 CO2 + 2H2O
Bonds broken:
4 C-H bonds broken: 4 x 413
2 O=O bonds broken: 2 x 498
Total: 2648 kJ mol-1
Bonds formed:
2 C=O bonds formed: 2 x 805
4 O-H bonds formed: 4 x 460
Total: 3450 kJ mol-1
Overall enthalpy change of reaction: 2648 – 3450 = -802 kJ mol-1.
10
Hess’s Law route:
We can also use the Hess’s Law approach to find the overall enthalpy change of a
reaction.
Steps:
1. Write down the enthalpy change you want to find as a simple equation,
then draw a ‘delta H’ above the arrow.
2. Draw and write a box of the intermediates.
3. Then fit the information we have onto the diagram, forming a Hess’s Law
cycle.
4. Now find the alternative route of the reaction (not the most direct arrow).
Go with the flow of the arrows, if the arrows are in the wrong direction,
flip the arrow and change the ‘+/-‘ sign of the value.
Mean bond enthalpies:
In the above situation we make use of the mean bond enthalpies to work out the
overall enthalpy change of reaction. Bond enthalpies of a specific bond may vary
slightly in different environment. For example, the C-F bond has bond enthalpy
452 kJ mol-1 in CH3F, but 485 kJ mol-1 in CF4.
11
Chapter 12.5 – Why do chemical reactions take place?
Often we refer to the fact that as a chemical reaction takes place, the substances
become more stable, i.e. losing their potential energy via an exothermic reaction.
This still does not explain why many endothermic reactions happen
spontaneously.
This means that there must be other factors, apart from enthalpy change, that
would cause a reaction to proceed spontaneously.
Spontaneous reaction:
It refers to a reaction that tends to go without being driven by any external factor.
Entropy:
The other factor that determines whether would a reaction to proceed is entropy.
Entropy: a thermochemical quantity
that measures randomness,
or degree of disorder,
based on the number of
ways which molecules can
be arranged, and the
number of ways which
energy can be shared out
amongst the molecules.
Symbol is ‘S’. The unit for
entropy is J K-1 mol-1. They are usually quoted in standard
conditions: pressure of 100kPa and temperature of 298 kelvins.
Entropy helps us to predict the direction
of changes: change happens in the
direction that leads to an increase of
entropy – increasing randomness that
makes molecule more spread out.
The kinetics theory of gas, tells us that
molecules are in rapid motion, and
constantly colliding with each other and
with the walls of the gas container. The
volume of the gas remains constant, but they constantly re-arrange themselves
and redistributing its enrgy. Hence the entropy value is huge for a gas.
12
Factors affecting entropy:
There are a few factors which entropy could be affected:-
Physical state:
Physical state of a substance
affects entropy. To explain this
we have to look back into the
particle model. In a solid,
particles vibrate about a fixed
point; in liquid move around a
small region, hence with some
degree of ‘disorder’; in gas
particles moves rapidly and has
the most ‘random’ arrangement.
Therefore, more chaotic the
particle arrangement is, higher
the value of entropy.
Dissolution:
In a solid, molecules vibrate about a fixed point. As we dissolve a solid, dissolved
particles can move freely. Hence increases entropy.
Number of particles:
As we increasing the amount of particles present in a given space, there are more
ways which they can be physically and energetically arranged. In a reaction
where we can increase the amount of moles present, we increase entropy.
For example, a reaction of N2O4 2NO2 increases entropy.
13
Spontaneous endothermic reactions:
We know that in a spontaneous reaction, we don’t require any external factor to
provoke a reaction to occur. Therefore, because endothermic reactions require
external energy to be put in, that must not be spontaneous reaction. However,
this is not the case!
In some scenarios, because the reaction involves a massive increase in entropy,
the reaction simply happens spontaneously.
For example
Evaporation of water
At room temperature, some water molecules turn into water vapour. This
requires energy to break the hydrogen bond between the molecules, so it is
endothermic. But because it involves a change of physical state, the entropy
value increases massively, as the arrangement of particles in gas is ‘more chaotic’
than that in liquid, the reaction occurs.
Reactions between NaHCO3 and HCl
The reaction between hydrogencarbonate and hydrochloric acid is a
spontaneous endothermic reaction. This is because there are more number of
particles in the product side, compared to the reactants. Thus the increase of
entropy causes the reaction to occur.
NaHCO3 (s) + H+ (aq) Na+ (aq) + CO2 (g) + H2O (l)
Entropy changes and feasibility:
The Second Law of Thermodynamics states that any spontaneous reaction
involves an increase in entropy of the system – energy is dissipated.
This means that a reaction would only be feasible if the total enthalpy change, ΔS
total, is positive.
The entropy change for a reaction can be calculated by adding all the entropies of
the products and subtracting the sum of all entropies of the products.
Formula:
ΔS system = ΔS products + ΔS reactants
ΔS total = S system + S surroundings
ΔS surroundings = - (ΔH) / T
where S is entropy, measured in J K-1 mol-1;
ΔS is entropy change;
ΔH is enthalpy change, measured in J mol-1;
T is temperature, measured in kelvins (K).
14
Gibb’s free energy change:
It is a quantity invented by an America chemist, Josiah Willard Gibbs. It is the
enthalpy change and entropy change combined. It has the symbol ‘G’.
The free energy change, ΔG, combines the enthalpy change (ΔH) and entropy
change (ΔS) factors as follows:
ΔG = ΔH - TΔS
ΔG is dependent on temperature due to TΔS. This means that some reaction is
feasible in certain temperature but not the other.
It is a measure used to predict whether a reaction is feasible. If ΔG equals to zero
or below zero, then a reaction may happen itself. Therefore, an endothermic
reaction can become plausible when the temperature is increased.
Notice that a positive entropy change will make ΔG more negative, due to the
negative sign before it.
Despite the fact that the ΔG may show that a reaction is feasible, the reaction may
require high activation energy, or the reaction may proceed at a very slow speed
that we don’t notice any visible effect.
For a reaction to occur, the Gibb’s free energy change, ΔG, must be negative.
For example
CaCO3 (s) CaO (s) + CO2 (g)
ΔH = +178 kJ mol-1, ΔS = +161 J K-1 mol-1
Because enthalpy is measured in kilojoules per mole (kJ mol-1), we need to
convert the entropy change (ΔS) back into kJ K-1 mol-1 by dividing it by a
thousand.
Hence, at room temperature 298K:
ΔG = 178 – (298*0.161) = +131 kJ mol-1
Since ΔG in this case is positive, the reaction is not possible.
However, if we raise the temperature to 1500K:
ΔG = 178 – (1500*0.161)
ΔG = 178 – 242
ΔG = -64 kJ mol-1
In this scenario, ΔG is negative, hence the reaction is possible.
15
ΔG = 0
Temperature
At the point where ΔG equals zero, this is the point where the reaction is just
possible. We can now work out the temperature at that point.
Using the example above again:
ΔG = ΔH - TΔS
0 = ΔH - TΔS
TΔS = ΔH
when ΔH = +178kJ mol-1, and ΔS = 0.161 kJ K-1
Thus, temperature T = 1105.6K
Entropy change
We can also calculate the entropy change at the point where the reaction is just
feasible (ΔG = 0). For example, ice melts to become water at 273K. Given that the
enthalpy change is 6kJ mol-1.
Using the equation again:
ΔG = ΔH - TΔS
0 = ΔH - TΔS
TΔS = ΔH
at temperature 273K and ΔH = 6kJ mol-1:
273ΔS = 6
ΔS = 6/273
ΔS = +22 J K-1 mol-1
We should expect that the change of entropy value to be positive, this is because
that as ice turns into water, particles are more disordered.
16
Unit 13 – Periodicity
Unit 13.1 – Reactions of Period 3 elements
Period 3 Elements:
In Period 3 elements, the 3s and 3p orbitals are filling with electrons. Their
electronic configurations are given below:
Na [Ne] 3s1
Mg [Ne] 3s2
Al [Ne] 3s2 3px1
Si [Ne] 3s2 3px1 3py1
P [Ne] 3s2 3px1 3py1 3pz1
S [Ne] 3s2 3px2 3py1 3pz1
Cl [Ne] 3s2 3px2 3py2 3pz1
Ar [Ne] 3s2 3px2 3py2 3pz2
where [Ne] denotes the complete electronic structure of a neon atom.
Reactions of Period 3 elements:
The reactions of Period 3 elements are all redox reactions, as all elements in
Period 3 have oxidation states of zero initially. It is then changed to either
positive or negative after the reaction.
Sodium and magnesium are the first two elements in Period 3. Sodium is a
Group 1 element, and magnesium is a Group 2 element – sodium would lose 1
electron to form a Na+ ion, and magnesium would lose 2 electrons to form Mg 2+
ion.
Since sodium only need to lose 1 electron, it is more reactive as it takes less
energy to lose one electron than two, compared to magnesium. Hence more
(heat) energy is required to allow magnesium to react with water.
17
Reactions of Period 3 elements with water:
Because Na, Mg, Al and Si are all more electropositive than hydrogen, they can
reduce water to hydrogen gas.
Sodium
Sodium reacts vigorously with cold water in a very exothermic reaction,
producing hydrogen gas and a colourless solution of sodium hydroxide. The
‘molten ball’ fizzes on the surface, forming hydrogen gas.
The numbers below the equation denotes changes in oxidation state.
0 +1 -2 +1-2+1 0
The resulting solution contains NaOH (aq), and thus is strongly alkaline with pH
14.
Magnesium
Magnesium reacts at an extremely slow rate with cold temperature. Eventually if
the reaction goes ahead, the few small bubbles of hydrogen float to the surface.
As a consequence a thin layer of magnesium hydroxide is formed on magnesium
and tends to stop the reaction. This explains why magnesium hydroxide is only
weakly alkaline, as the solution is sparingly soluble, with pH 9 as its oxide is
slightly basic.
However magnesium burns in steam, giving white flame to produce white
magnesium oxide and hydrogen gas.
0 +1 -2 +2 -2 0
All of these reactions are redox: the oxidation state of the metal increases and
that of hydrogen decreases.
Aluminium
Aluminium powder heated in steam will produce hydrogen gas and aluminium
oxide. The reaction is slow due to the existing strong aluminium oxide layer on
the metal, and the oxide layer will build up even more during the reaction.
Silicon
There are many explanations to this reaction, due to the fact that different form
of silicon gives different reaction. Commonly, the grey lumps of silicon with
metallic-like appearance are rather unreactive. It is suggested that it reacts with
steam at strong heat to produce silicon dioxide and hydrogen.
18
Phosphorous and sulphur
No reaction with water.
Chlorine
Chlorine dissolves in water to an extent to give a green solution. It is a reversible
reaction – chlorine will disproportionate and produce a mixture of hydrochloric
acid and chloric(I) acid (hypochlorous acid)
The resulting solution contains HCl (aq) and is thus acidic (pH = 2).
In the presence of sunlight, due to ultraviolet (UV) rays, the chloric(I) acid slowly
decomposes to produce more hydrochloric acid, releasing oxygen gas.
Argon
No reaction with water.
19
Reactions with oxygen
Period 3 elements (except argon) are all reactive, forming oxides when they
react with oxygen. They are usually oxidised to their highest oxidation state –
the same as their group numbers. The reactions are exothermic.
Sodium:
Sodium burns in oxygen with a bright yellow flame, forming white solid mixture
of sodium oxide and sodium peroxide.
Simple oxide:
0 0 +1 -2
Peroxide:
Na2O is an ionic oxide.
Magnesium:
A strip of magnesium ribbon burns in air it will give a white flame, producing
white powdered magnesium oxide. The flame will be more intense if the
reaction takes place within an oxygen jar.
0 0 +2 -2
Because of the presence of nitrogen in air, magnesium also reacts with nitrogen
in the air, getting a mixture of magnesium oxide and magnesium nitride.
MgO is an ionic oxide.
Aluminium:
Aluminium will burn in oxygen when it is in powdered form, otherwise the
strong oxide layer will prohibits the reaction. Due to its oxide layer, aluminium
appears to be an unreactive metal, and is used to many daily applications. When
the aluminium is scratch, the exposed layer of ‘pure’ aluminium reacts rapidly to
reform the protective layer of aluminium oxide.
The reaction will give a bright flame, and form a white powder of aluminium
oxide.
0 0 +3 -2
Al2O3 is mostly ionic, but also exhibits a significant covalent character.
Silicon
Silicon burns in oxygen if it is heated strongly. Silicon dioxide is formed.
SiO2 is a giant covalent oxide.
20
Phosphorous: Red phosphorous must be heated before it reacts with oxygen. White phosphorous spontaneously ignites in air, giving a white flame and produce clouds of white smoke, which are a mixture of phosphorous (III) oxide and phosphorous(V) oxide (pentoxide). Red and white phosphorous are all allotropes of phosphorous – same elements which atoms are arranged differently.
For the phosphorus(III) oxide:
For the phosphorus(V) oxide:
Sulphur: Sulphur burns in air or oxygen with gentle heat, giving a pale blue flame. It forms colourless sulphur dioxide gas.
Slight amount of sulphur dioxide, in the presence of oxygen, can be converted into sulphur trioxide, with a catalyst (V2O5) and a carefully controlled conditions. This reversible process is known as the Contact Process.
Chlorine: It won’t react directly with oxygen.
Argon: No reaction occur.
21
Unit 13.2 – The oxides of elements in Period 3 Unit 13.3 – Oxides of the elements of Period 3 and their reactions with acids and bases
The oxides that we’re studying are:
The top row oxides are known as the highest oxides – the Period 3 elements are
in their highest oxidation states. This means that all outer electrons in the Period
3 elements are involved in bonding.
The metal oxides
Sodium, magnesium and aluminium oxides are all compounds that a metallic
element is combined with a non-metallic compound. They are formed as a giant
ionic compound, where the bond extends throughout the compound,
contributing to its high melting point.
Despite the bonding in aluminium oxide is mostly ionic, it also exhibits covalent
character. This is due to the fact that aluminium forms a very small ion with a
large positive charge, so as it approach O2- closely it can distort its electron
cloud, adding some covalent character.
Therefore it is possible to predict the ionic character, by considering the
difference of electronegativity of the two atoms. Bigger the difference is, greater
the ionic character of the bond.
Non-metal oxide
Silicon oxide has a giant covalent (macromolecular) structure, which the bond
extends throughout the giant structure. These compounds have a high melting
point due to the fact that we need to break down the extensive network of
covalent bonds before we melt it.
Compared to metallic ionic oxides, they possess melting & boiling points. This is
because the intermolecular forces holding one molecule to its neighbouts will be
van der Waals or dipole-dipole interactions. Again strength of these interactions
depend on the sizes of the molecules.
Na2O MgO Al2O3 SiO2 P4O10 SO3 Cl2O7 P4O6 SO2 Cl2O
22
General properties of Period 3 oxides:
The structure
The trend in structure is from metallic oxides containing giant ionic structures at
the left of the period, to the middle (going to the right) with macromolecular
structure (silicon oxide), then molecular oxides to the right.
Silicon dioxide
The electronegativity of elements increase as we go across the period, and as we
get to silicon there is not much electronegativity difference between the silicon
and oxygen to form an ionic bond. Hence silicon dioxide is a giant covalent
structure.
Melting and boiling points
The giant structures (i.e. metallic ionic structure and macro-structure in silicon
dioxide) have a high melting & boiling points, due to the vast amount of energy
required to break the strong bonds (ionic and covalent, sometimes).
Na2O, Al2O3 and MgO are ionic oxides and hence all have a high melting point.
Because MgO and Al2O3 have a higher charges than Na2O, they possess a higher
melting point.
The attractive forces between these molecules would be van der Waals and
dipole-dipole interactions. These attractive forces vary in size, depending on
their molecular size, shape and its polarity.
Electrical conductivity
There are no free electrons present in these oxides, in a consequence it is
impossible to conduct electricity when the oxides are in solid state. However,
ionic oxides can undergo electrolysis when they’re molten, due to that they can
conduct electricity when ions move towards the electrodes and discharge
themselves there.
One important example would be electrolysis of aluminium oxide to manufacture
aluminium.
The table below summarises the general properties of the Period 3 oxides:
Na2O MgO Al2O3 SiO2 P4O10 SO3 SO2 Melting Point
1548 3125 2345 1883 573 290 200
Bonding
Ionic Ionic Ionic/ Covalent
Covalent Covalent Covalent Covalent
Structure
Giant ionic
Giant ionic
Giant Ionic
Giant covalent
Molecular Molecular Molecular
23
Acid-base character of oxides:
Ionic oxides contain O2- ion, which is a strongly basic ion as it has a high
tendency to combine with hydrogen ions, forming hydroxide (OH-) ions.
O2-(aq) + H2O(l) 2OH-(aq)
Thus all ionic oxides are basic.
Covalent oxides do not contain ions, but have a strongly positive dipole on the
atom which is not oxygen. This attracts the lone pair on water molecules,
releasing H+ ions.
MO(s) + H2O(l) MO(OH)-(aq)+ H+(aq)
Thus all covalent oxides are acidic.
The trend is from strongly basic oxides on the left hand side, to the strongly
acidic ones on the right, with an amphoteric (showing both acidic and basic
properties) aluminium oxide at the middle.
Reactions with water:
Basic oxides:
Both sodium and magnesium oxides are basic.
Both of them react with acid, producing salt and water only.
Sodium oxide
Reactions with water
Sodium oxide reacts with water to give sodium hydroxide solution – a strongly
alkaline solution. Typically it has a pH around 14.
Reactions with base
Because it is a strong base, sodium oxide also reacts with acids. For example, it
would react with dilute hydrochloric acid to produce sodium chloride solution.
24
Magnesium oxide
Reactions with water
Same as sodium oxide, it is a simple basic oxide, as it also contains oxide ions.
However it is a weak base as it is sparingly soluble in water, the oxide ions aren’t
free at all. The magnesium oxide is only slightly soluble in water, forming
magnesium hydroxide on the soluble. Not much hydroxide ions are produced,
due to the insoluble layer of magnesium hydroxide floating on the water.
Reactions with acid
Similar to sodium oxide, it reacts with warm dilute hydrochloric acid to give
magnesium chloride solution.
25
Amphoteric oxide:
Amphoteric oxide: oxides which react both as acidic oxides and basic
oxides. They are anhydrous, and are often relatively
inert to aqueous reagents so that it is easier to
demonstrate their amphoteric behaviour.
Aluminium oxide is one of the example of an amphoteric oxide.
Aluminium oxide:
Reactions with water
Aluminium oxide does not react simply with water, like how sodium and
magnesium oxides previously do. Despite it still contains oxide ions, they are
held too strongly in the solid lattice to react with water.
Basic character of aluminium oxide
It contains oxide ions and so it reacts with acid, in the same manner as sodium
and magnesium oxides. For example, aluminium oxide reacts with hot dilute
hydrochloric acid to give aluminium chloride solution. In this scenario, being
amphoteric, it is displaying its basic character.
Acidic character of aluminium oxide
The reaction between aluminium oxide and sodium hydroxide displays its acidic
side of its amphoteric nature. The reason is because as electronegativity is low
in sodium, thus the massive difference of electronegativity present in sodium
oxide makes it an ionic compound. In the case of aluminium oxide, the
electronegativity difference is smaller, thus permits the formation of covalent
bonds between the two.
With hot, concentrated sodium hydroxide solution, aluminium oxide reacts to
give a colourless solution of sodium (tetrahydroxo)aluminate.
There are a variety of combinations for the products of this reaction. This is
greatly dependent on the temperature and the concentration of sodium
hydroxide used. Below are the possible combinations:
Al2O3(s) + 3H2O(l) + 6OH-(aq) 2Al(OH)63-(aq)
Al2O3(s) + 3H2O(l) + 2OH-(aq) 2Al(OH)4-(aq)
26
Acidic oxides:
Non-metals on the right of the periodic table usually form acidic oxides.
Silicon dioxide
As we go across the period, the electronegativity increases. At silicon there is
insufficient difference in electronegativity between the silicon and oxygen to
form an ionic bond. Since it has no oxide ions and does not react with acids, it
has no basic properties – instead it is weakly acidic.
Reactions with water
Silicon dioxide does not react with water, due to the high energy required to
break up the giant covalent structure.
Reactions with bases
Silicon dioxide reacts with strong base as a weak acid. It reacts with hot and
concentrated sodium hydroxide solution, forming a colourless solution of sodium
silicate.
Phosphorous oxides
We are looking at two phosphorous oxides, phosphorous(III) oxide, P4O6, and
phosphorous(V) oxide, P4O10.
Phosphorous(III) oxide
Reactions with water
It reacts with cold water to give a solution of the weak acid, H3PO3
(phosphorous acid). It reacts with hot water.
Direct reaction of phosphorous(III) oxide with sodium hydroxide solution
When we react phosphorous(III) oxide with sodium hydroxide solution, we end
with a salt.
Reactions between phosphorous acid with sodium
hydroxide
The hydrogen ions are not easily released as phosphorous
acid is a weak acid. However it has a pKa value of 2.00,
making it stronger than common organic acids.
With phosphorous acid, the two hydrogen atoms in the –OH groups are acidic,
but not the other. This tells us that there are two possible reactions, depending
on the proportion of sodium hydroxide solution used.
27
Phosphorous(V) oxide
Reactions with water
It reacts violently with water to give a solution containing a mixture of acids,
depending on the condition of reaction. It produce a solution of phosphoric(V)
acid, H3PO4. The diagram below shows the structure of the pure, un-ionised
acid.
It has a pH value of 3.
Reactions with base
Phosphoric(V) acid is a weak acid, too, with a pKa of
2.15, which is slightly weaker than that of phosphorous
acid. When we look at the structure, it contains 3 –OH
groups, each contains an acidic hydrogen atom. This
tells us that it ionises in three stages.
When we use different amount of sodium hydroxide, we can control the extent
which the acid ionises. In each turn, each hydrogen reacts with a hydroxide ion
and replace it with a sodium ion.
Sulphur oxides
There are two common oxides, sulphur dioxide (sulphur(IV) oxide, SO2), and
sulphur trioxide (sulphur(VI) oxide, SO3).
Sulphur dioxide:
It is a colourless gas at room temperture with a recognisable smell. It exists as a
simple SO2 molecule.
The sulphur uses 4 of its outer electrons to form double bonds with oxygen,
leaving the remaining two as a lone pair on the sulphur. This also explains why it
has a bent shape.
Reactions with water
It is fairly soluble in water, reacting and gives a solution of sulphuric(IV) acid,
H2SO3. This acid only exist in solution, and isolating it from its solution causes
sulphur dioxide to be given off again.
28
Reactions with base
Structure of the un-ionised form of acid is given below.
It is a weak acid with a pKa of 1.8, which is just slightly stronger than the two
previous phosphorous-containing acids.
Sulphur dioxide can also react directly with bases such as sodium hydroxide
solution. It reacts violently, forming sodium sulphite solution first
If sulphur dioxide is in excess, it further reacts to form sodium hydrogensulphite
solution.
Acid rain
Sulphur dioxide is a waste gas from many industrial processes. It is hazardous as
it dissolves in rainwater to give acid rain. It can be removed from waste gas by
dissolving it in an alkaline solution in the waste gas outlet, to minimise the
impact to the atmosphere.
Sulphur trioxide
It reacts violently with water to produce fog of concentrated sulphuric acid
droplets.
It also reacts with bases to form sulphates. For instance, it reacts with calcium
oxide to form calcium sulphate.
29
Unit 14 – Redox equilibria
Unit 14.1 – Redox equations
Redox reaction
Oxidation and reduction
The ‘OIL RIG’ acronym will help us to remember that
oxidation is the loss of electrons;
reduction is the gain of electrons.
When a species loses electrons, it is said to be oxidised.
Fe2+ Fe3+ + e
When a species gains electons, it is said to be reduced.
MnO4- + 8H+ + 5e Mn2+ + 4H2O
Combining half-equations to overall equations
An oxidation half-equation can be combined with reduction half-equation,
forming a full equation.
Firstly we multiply all the coefficients in one of the equation (or two), to ensure
that the number of electrons gained is equal to the number of electrons lost.
Fe2+ Fe3+ + e oxidation
MnO4- + 8H+ + 5e Mn2+ + 4H2O reduction
Multiplying all coefficients in the oxidation reaction by 5:
5Fe2+ 5Fe3+ + 5e
means that 5 electrons are gained and five are lost
overall equation:
MnO4- + 8H+ + 5Fe2+ Mn2+ + 4H2O + 5Fe3+
Secondly to balance the amount of oxygen atoms, we place an appropiate
number of water molecules on one side.
Thirdly to balance the amount of hydrogen atoms (by the water molecule), we
place an appropiate number of H+ ions on one side.
Lastly to balance the charge, we then put suitable amount of electrons to make
the overall charge of the reaction neutral.
Oxidation and reducing agents
Oxidation agent is a species which accept electrons from another speicies.
It is reduced in the redox reaction.
Eg MnO4- is the oxidizing agent in the above reaction.
Reducing agent is the one that donates electrons to the another species.
It is oxidised during the redox reaction.
Eg Fe2+ is the reducing agent in the above reaction.
30
Oxidation states
Also known as oxidation number. It is the charge that would exist on the atom
if the bonding were completely ionic, as it tells us the number of electrons that
the species have gained/lost when the bond is formed. It also helps us to keep
track on what species has been oxidised and which has been reduced.
Below are the rules for oxidation states:
1.
2. Uncombined elements have oxidation state of 0.
3. Elements bonded with idential atoms habe oxidation state of 0, i.e. O2.
4. The sum of oxidation states of a neutral compound is 0.
5. The sum of oxidation states of a complex ion, i.e. NH4 + & SO4 2-, is equal
to the charge of the ion.
6. The oxidation state of a simple ion, is equal to the charge of the ion. For
example, Na+, K+ all have oxidation number of +1; O2-, S2- all have
oxidation number of -2.
Elements Oxidation state in compound Example Hydrogen (H)
+1 (Except in metal hydrides, i.e. NaH, where it is -1) (Except in hydrogen gas form, H2, it is 0.)
HCl
Group 1
Always +1 NaCl
Group 2
Always +2 CaCl2
Aluminium (Al)
Always +3 AlCl3
Oxygen (O)
-2 (Except in peroxides where it is -1.) (Except when it is compounds with F, i.e. OF2, it is +2.)
Na2O
Fluorine (F)
Always -1 NaF
Chlorine (Cl)
-1 (Except in compounds with F and O, where it has positive values depending on the oxidation numbers of F and O.)
NaCl
31
Changes in oxidation state:
Oxidation states can increase or decrease, depending on how many electrons are
lost or gained. To figure out which species have been oxidised/reduced, we
lookat the changes of oxidation states.
For each electron lost, the oxidation state increases by one, hence it is oxidised;
for each electron gained, the oxidation state decreases by one, hence it is reduced.
Disproportionation
A disproportionation reaction refers to a reaction which the same element both
increases and decreases its oxidation number. They are special examples for
redox reactions.
There are many d-block species that readily undergo both oxidation and
reduction, and hence behave as both oxidising and reducing agents. For example,
we can look at Cu+, Mn 3+ and MnO4 2-:
Eg Cu+ Cu2+ + e oxidation
Cu+ + e Cu reduction
Eg Mn3+ + 2H2O MnO2 + 4H+ + e oxidation
Mn3+ + e Mn2+ reduction
Eg MnO42- MnO4- + e oxidation
MnO42- + 2H+ + 2e MnO2 + 2H2O reduction
32
Unit 14.2 – Electrode potentials and the electrochemial series
Electrochemical cells:
a series that shows half-reactions for oxidation and reduction.
A electrochemical cell can be made of two different metal dipped into the salt
solution of their own ions and connected by a wire. Two process happens in the
cell: oxidation and reduction, hence it is a redox process.
Electrode potentials
If we consider a zinc rod submersed in a solution with Zn2+ ions, i.e. ZnSO4.
zinc electrode
Zn2+ ions in solution
The Zn atoms on the rod can deposit two electrons on the rod and move in the
solution as Zn2+ ions.
Hence the equation can be written as:
Zn(s) Zn2+(aq) + 2e
The process will result in the accumulation of negative charge on the zinc rod.
And vice versa, the Zn2+ ions in solution would accept two electrons from the
rod, and move into the rod as Zn atoms.
Hence the equation can be written as:
Zn2+(aq) + 2e Zn(s)
The process will result in the accumulation of positive charge on the zinc rod.
In two cases, a potential difference is established between the rod and the
solution, and is known as an electrode potential. Note that a chemical reaction is
not taking place – there is simply a potential difference between the rod and the
solution.
Factors affecting the potential difference would depend on the nature of: nature
of ions in solution, the concentration of ions in solution, the type of electrode
used, and temperature.
33
Creating emf (electromotive force)
V
CuZn
Zn2+ Cu2+
+ve-ve
We cannot measure electric potential directly, however we can measure
potential difference across two electrodes.
When we connect the two electrodes, the potential differences between the two
electrodes provoke a current to flow between them. This means that the system
is generating electricity, hence an electromotive force (emf) is established.
To make a current flow we must complete the circuit to let ions flow from one
solution to the another. The salt bridge, usually a strip of filter paper saturated
with a solution of inert electrolyte (i.e. KNO3 (aq)), is used . This is allow the
charge to be balanced in the two solution by allowing them to flow.
The e.m.f. can then be measured using a voltmeter.
The combination of the two electrodes in this way is known as a
electrochemical cell, and can be used to generate electricity. The two
components which make up the cell are known as half-cells.
In the above graphical example…
The positive electrode is the one that most favours reduction: in the above
case the copper electrode is positive.
The negative electrode is the one that most favours oxidation: in the above
case the zinc electrode is negative.
Remember that electrons flow are in opposite direction to the conventional
current flow. Thus electrons flow from the zinc electrode to the copper
electrode.
34
Reduction thus takes place at the copper electrode: Cu 2+ (aq) + 2e- Cu (s)
Oxidation thus takes place at the zinc electrode: Zn (s) Zn 2+ (aq) + 2e-
The overall cell reaction can be expressed as:
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
The sulphate ions flow through the salt bridge from the Cu 2+ (aq) solution to
the Zn 2+ (aq) solution, to complete the electrical circuit and compensate for the
reduced Cu 2+ concentration and increased Zn 2+ concentration.
If we include the spectator ions, the cell reaction can be written as:
CuSO4(aq) + Zn(s) Cu(s) + ZnSO4(aq).
(The electrons are cancelled.)
To allow the electrons to flow in a circuit, the external circuit must be made of
metallic wire; and the salt bridge between the two cells must be made of an
aqueous electrolyte.
Electrochemical cells
Any half reactions can be used to make a half-cell.
If the electrode in the half-metal does not contain a metal in its elemental state,
then an inert electrode (i.e. platinum) must be used. If a gas is involved in the
reaction it must be bubbled through the solution, so that it is in contact with the
electrode.
For example:
1) Fe 3+ (aq) + e- ⇌ Fe 2+ (aq)
A platinum electrode is used, immersed in a solution containing both Fe
2+ and Fe 3+ ions.
Pt
mixture ofFe2+ and Fe3+
2) Cr2O7 2- (aq) + 14H+ (aq) + 6e- ⇌ 2Cr3+ (aq) + 7H2O (l)
A platinum electrode is used, immersed in a solution containing Cr2O7 2-,
H+ and Cr 3+ ions. Pt
mixture ofCr2O7
2-, H+
and Cr3+
35
3) Cl2 (g) + 2e- ⇌ 2Cl- (aq)
A platinum electrode is used, immersed in a solution containing Cl- ions.
Chlorine gas is bubbled through the solution, in contact with the electrode.
Pt
Cl2
Cl-
4) 2H+ (aq) + 2e- ⇌ H2 (g)
A platinum electrode is used, immersed in a solution containing H+ ions.
Hydrogen gas is bubbled through the solution, in contact with the
electrode.
Pt
H2
H+
Standard (Hydrogen reference) electrode
Because we are unable to measure the absolute voltage (potential difference)
between the metal and the solution, we compare the voltage with a standardised
system known as a reference electrode.
The another system is called the standard hydrogen electrode.
Hydrogen electrode: a half-cell of great theoretical importance but lack of
practical significance. By the definition of a standard
electrode potential of the hydrogen electrode, is zero
(0.00 volts).
A standard hydrogen electrode is set up in an
equilibrium between hydrogen ions in solution (1
mol dm-3) and hydrogen gas (1 bar) all at 298
kelvins on a platinum electrode.
36
Standard conditions
The electrode potential depends on the following conditions used:
temperature, pressure & concentration of reactants;
therefore it is imperative that we specify the conditions used when measuring
electrode potentials. The conditions set up are at temperature of 298K, gas
pressure of 1 bar (100kPa) and with all species in solution having a
concentration of 1.0 mol dm-3. Electrode potentials measured under these
conditions are known as the standard electrode potentials, and is denoted by
the symbol Eo.
Using the Le Chatelier’s principle, we can predict the effects on the electrode
potentials if non-standard condition is used.
If the oxidising agent has concentration more than 1.0 mol dm-3, it is more likely
to favour reduction and the electrode potential will be more positive than the
standard electrode potential. And vice-versa, if it has concentration of less than
1.0 mol dm-3, it is more likely to favour oxidation and the electrode potential will
be more negative than the standard electrode potential. The reverse theory is
also valid for reducing agents.
Eg: Fe2+(aq) + 2e ⇌ Fe(s)
Standard electrode potential = -0.44 V
If [Fe2+] = 0.1 moldm-3 the electrode potential = -0.50 V
The concentration is lower than standard so reduction is less likely to
take place, and hence the electrode potential is more negative than
expected.
If the temperature is greater than 298 kelvins, then the system will move in the
endothermic direction and the electrode potential will also change.
If the pressure is greater than 1 bar, then the system will move to the direction
that tends to reduce the system pressure, thus the electrode potential will
change accordingly.
In general, a change which favours the reduction reaction will make the
electrode more positive, and a change which favours the oxidation reaction
will make the electrode potential more negative.
37
Standard hydrogen electrode
We are able to measure the potential difference of the electrochemical cells (or
when there are two half-cells), but for the case of individual electrodes, electrode
potentials cannot be measured at all. Even if we immerse two electrodes into
the same solution, we are only measuring the potential difference across the two
electrodes.
The only possible way to overcome this issue is to arbitrarily assign a value to a
half-cell when they are allocated a value, and all other electrodes are measured
relative to it. This is known as a reference electrode. In this section, the
standard hydrogen electrode will be investigated.
The diagram illustrates the standard hydrogen electrode.
The standard hydrogen electrode is fixed at the standard condition as mentioned
above.
This electrode is arbitrarily assigned a value of 0.00 volts.
Using this reference electrode, we are able to assign an electrode potential to all
other half-cells.
The voltmeter measures the potential difference on the right hand side of the cell,
and subtracts it from the potential on the left hand side of the cell.
E.M.F = E RHS - E LHS
If the standard hydrogen electrode is placed at the left hand side, then E LHS will
be zero, and the e.m.f. displayed on the voltmeter will be the electrode potential
of the right hand side electrode.
38
In this case, for example:
Two equilibria are set up on the two electrodes (the magnesium and the porous
platinum).
Since magnesium has a much greater tendency to form ions than hydrogen, the
position of magnesium equilibrium will be well left, compared to that of the
hydrogen equilibrium. This means that there will be a much greater build up of
electrons on the magnesium, than that on the platinum strip.
The major difference between the charge (due to accumulation of electrons) of
the two electrodes, provoke a potential difference across the two electrodes
which can be measured using a voltmeter. In this example the voltmeter reading
is 2.37 volts, and it shows magnesium as the negative electrode and the
platinum/hydrogen electrode as ‘positive’. The platinum/hydrogen electrode is
not ‘positive’, in reality there are some slight amount of electrons built up on it.
Because voltmeter measures the difference of electric potentials, the platinum
electrode ‘relatively’ is positive compared to the magnesium electrode.
39
Reference electrode: electrodes used to measure electrode potentials
instead of the standard hydrogen electrode, as it is
inconvenient. It is easier to use a secondary
standard, i.e. silver/silver chloride electrode. They
are calibrated against a standard hydrogen electrode.
Conventional cell representations:
It is hectic and time-consuming to draw a full diagram to represent a
electrochemical cell. A simpler system of notation is used to describe the cell in
full.
Half-cell
Half-cells can be written as follows:
the electrode is placed on one side of the vertical line;
the species in solution, whether solid, liquid, aqueous or gaseous, are
placed together on the other side of the vertical line;
if there are more than one species in solution, and species are on the
different sides of the half-equation, the different species are separated by
a comma.
The vertical solid line denotes phase boundary – i.e. between a solid and a
liquid.
Eg Zn2+(aq) + 2e- ⇌ Zn(s)
Zn Zn2+
Eg Fe3+(aq) + e- ⇌ Fe2+(aq)
Pt Fe2+ Fe3+
,
Eg Cl2(g) + 2e- ⇌ 2Cl-(aq)
Pt Cl2, Cl-
40
Electrochemical cell
When two half-cells are connected to make a full electrochemical cell, then the
cell is written as follows:
the more positive electrode is always placed on the right;
the two half-cells are placed at the either side of the double vertical line;
the electrodes are placed on the far left and far right, and other species
are placed in between them, adjacent to the double vertical lines at the
centre;
on the left (oxidation), the species are written in increasing oxidation
states;
on the right (reduction), the species are written in decreasing oxidation
states.
The double vertical line denotes a salt bridge.
Attaching cell voltage sign
When giving a value of the emf, we state the polarity (positive or negative) of the
right hand electrode. For example:
If we reverse the direction of the reaction, the emf reading would be +2.37 volts
and +0.34 volts respectively. The second example tells us that the electron flow
from the platinum/hydrogen electrode to the copper electrode.
41
The electrochemical series
The electrochemical series is set up by arranging different redox equilibria in
order of their standard electrode potential. They are in the order that it starts at
the top with the most negative E° values, and the most positive at the bottom.
equilibrium E° (volts)
-3.03
-2.92
-2.87
-2.71
-2.37
-1.66
-0.76
-0.44
-0.13
0
+0.34
+0.77
+0.80
+1.33
+1.36
+1.50
Each E° value shows the relative position of the equilibrium, either to the left or
right, compared to the hydrogen equilibrium.
42
Metals at the top of the electrochemical series are good at giving away
electrons – they’re good reducing agents. Thus the reducing ability increases as
we ascent through the series.
Metals at the bottom of the series are good at receiving electrons – they’re good
oxidising agents. Thus the oxidising ability of the metal ions increases as we
descent through the series.
43
Unit 14.3 – Predicting the direction of redox reactions
Cell potential
When two half-cells are connected electrically, and a current is allowed to flow, it
will undergo a redox reaction: the more positive electrode undergo reduction
and the more negative electrode undergo oxidation. The oxidising agent at the
more positive electrode is reduced, thus oxidises the reducing agent at the more
negative electrode.
Hence it is possible to use the standard electrode potentials to predict the
feasibility of a reaction. A current only flows when the electron(s) move from
more negative electrode to a more positive electrode, not the other way round.
As a reminder:
More negative the E° value is, the further the position of equilibrium lies to the
left, relative to the hydrogen equilibrium.
For example:
The negative sign of the zinc E° value shows that it releases electrons more
readily than hydrogen does. The positive sign of the copper electrode shows that
it releases electrons less readily than hydrogen.
When the two equilibria are linked as a electric circuit, electrons flow from one
equilibrium to another. This will upset the equilibria, and the Le Chatelier’s
principle applies. This means that:
the equilibrium with more negative (less positive) E° value will shift to
the left;
the equilibrium with more positive (less negative) E° value will shift to
the right.
So when the electrodes are connected, the reaction takes place:
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
First we assume that if a reaction happens electrochemically, the same will apply
chemically. Hence if zinc metal is immersed into a solution of copper (II)
sulphate, the above reaction occurs. However, it is not the case for copper metal
immersing into a solution of zinc (II) sulphate, as for a reaction to occur in this
example, it would have to be:
Cu(s) + Zn2+(aq) Cu2+(aq) + Zn(s)
This reaction does not take place when the two half-cells are connected, and also
would not take place chemically, too.
44
Which reaction is feasible?
There are two reactions below. Which of them is feasible?
1. Magnesium reacting with dilute sulphuric acid;
2. Copper reacting with dilute sulphur acid.
Answer:
Reaction 1 is feasible; reaction 2 is impossible.
Reaction 1
We start with the magnesium metal and hydrogen ions in the acid. The sulphate
ions are spectator ions and have no role in the reaction.
The E° values are:
The two reactions are at opposite direction, due to the different polarity of the E°
values. Thus the magnesium can freely turn into magnesium ions and give
electrons to the hydrogen ions producing hydrogen gas. The reaction, therefore
is feasible.
45
Reaction 2
The E° values are:
When start from the copper metal, the copper equilibrium is well completely to
the right. Shall the reaction to occur, the equilibrium will need to shift to the left,
which does not agree with the E° values.
In similar logic, when we start from hydrogen ions (from dilute acid), the
hydrogen equilibrium is already far to the left as possible. Shall it react it would
move to the right, which again disagree with the polarity of the E° values.
Therefore, the reaction is infeasible.
46
Oxidising and reducing agents
Closer to the bottom of the electrochemical series, the electrodes will be more
positive. The oxidising agents in these systems will oxidise any reducing agent
which lies above it in the electrochemical series. For example:
Eg H+(aq) will oxidise Pb(s) to Pb2+(aq), and any other metal above it, but will
not oxidise Cu(s) to Cu2+(aq) or any metal below it.
Pb(s) + 2H+(aq) Pb2+(aq) + H2(g)
Acids such as nitric acid, however, which contains the more powerful oxidising
agent NO3-(aq), will oxidise any reducing agent with a standard electrode
potential more negative than +0.81V, eg Cu(s)
Cu(s) + 4H+(aq) + 2NO3-(aq) Cu2+(aq) + 2NO2(g) + 2H2O(l)
Vice versa, reducing agents will reduce any oxidising agent which lies below it in
the electrochemical series.
Eg Fe2+(aq) will reduce VO2+(aq) to VO2+(aq), but not VO2+(aq) to V3+(aq) or
V3+(aq) to V2+(aq)
VO2+(aq) + 2H+(aq) + Fe2+(aq) VO2+(aq) + H2O(l) + Fe3+(aq)
Disproportionaton
Using standard electrode potentials, we are able to predict whether a species will
disproportionate or not.
For example:
Predict whether or not Ag+ ions will disproportionate in aqueous solution.
Ag+ may be expected to disproportionate in the following half-reaction:-
Ag+(aq) + e- ⇌ Ag(s) reduction, Eo = + 0.80V
Ag+(aq) ⇌ Ag2+(aq) + e- oxidation, Eo = + 1.98V
ECELL = 0.80 - 1.98 = -1.18V
Therefore Ag+ will not disproportionate
47
Predict whether or not H2O2 will disproportionate in aqueous solution.
We may expect that H2O2 will disproportionate according to the
following half-reaction:-
H2O2(aq) + 2H+(aq) + 2e- ⇌ 2H2O(l) reduction, Eo = +1.77V
H2O2(aq) ⇌ 2H+(aq) + O2(g) + 2e- oxidation, Eo = +0.68V
ECELL = 1.77 - 0.68 = +1.09V
Therefore H2O2(aq) will disproportionate:
2H2O2(aq) + 2H+(aq) 2H+(aq) + O2(g) + 2H2O(l)
2H2O2(aq) 2H2O(l) + O2(g)
Non-standard conditions
Despite the (second method) cell potential method is a reliable method to
predict whether a given reaction will take place or not, it only strictly applies to
standard conditions. If solutions used are either very concentrated/dilute, then
the electrode potentials are no longer standard electrode potentials, and the
potential values may have a different polarity (+/-).
Therefore, often reactions that previously are predicted infeasible occur if the
solution is hot and concentrated, or reactions that are previously predicted
feasible becomes infeasible when the solutions are too dilute.
For example:
The reaction between manganese dioxide and hydrochloric acid.
MnO2(s) + 4H+(aq) + 2Cl-(aq) Mn2+(aq) + Cl2(g) + 2H2O(l)
Reduction: MnO2(s) + 4H2+(aq) + 2e ⇌ Mn2+(aq) + 2H2O(l) Eo = +1.23V
Oxidation: 2Cl-(aq) Cl2(g) + 2e Eo = +1.36V
ECELL = Er - Eo = -0.13V
This reaction does not occur under standard conditions. However if hot
concentrated HCl is used, the high Cl- concentration favours oxidation, the
electrode potential becomes less positive and ECELL thus becomes positive
and the reaction occurs.
48
The reaction between potassium dichromate (VI) and hydrochloric acid.
Cr2O72-(aq) + 14H+(aq) + 6Cl-(aq) 2Cr3+(aq) + 3Cl2(g) + 7H2O(l)
Reduction: Cr2O72-(aq) + 14H+(aq) + 6e == 2Cr3+(aq) + 7H2O(l)
Eo = +1.33V
Oxidation: 2Cl-(aq) == Cl2(g) + 2e Eo = +1.36V
ECELL = Er - Eo = -0.03V
This reaction does not occur under standard conditions. However if solid
potassium dichromate is dissolved in hydrochloric acid, the high Cr2O72-
concentration favours reduction and makes the electrode potential more
positive. Thus ECELL becomes positive and the reaction occurs.
Kinetic Stability
Cell potentials can only predict whether if a reaction is feasible, it does not tell us
the rate which a reaction will proceed. Sometimes the ECELL value shows that it
is positive but there is no visible reaction occurring. This is because the
reactants are kinetically stable, giving a high activation energy so that the rate of
reaction is very slow at room temperature.
For example:
Mg(s) + 2H2O(l) Mg2+(aq) + 2OH-(aq) + H2(g)
E = -0.42V, E = -2.38V so ECELL = Er - Eo = +1.96V
So a reaction is expected but no reaction takes place.
This is because the activation energy is too high (magnesium will react
with steam and slowly with hot water).
As a conclusion:-
If a reaction is expected to be feasible but is found not taking place, the two
possible reasons are:
the solutions are too dilute (i.e. it is not at standard conditions)
the reaction is taking place very slowly (i.e. reactants are kinetically
stable)
If a reaction is not expected to be feasible but is found to be taking place,
the possible reason is:
the conditions are non-standard (i.e. the solutions are concentrated)
49
Unit 14.4 – Electrochemical cells
Electrochemical cells form a basis for batteries. Batteries have two separate half-
cells, connected by a salt bridge or a semi-permeable membrane that permits the
flow of ions without completely mixing the solutions. The electrodes are
connected to battery terminals, and when an external closed circuit is connected,
the current flows.
Non-rechargeable cells
If the reactions taking place in the half-cells are irreversible, then the cell is non-
rechargeable. There are a few types of non-rechargeable cells.
Zinc/copper cells
Zinc/carbon cells
The positive electrode is made of carbon which is inert, surrounded with a
mixture of manganese (IV) oxide. It forms a basis for the most ordinary
disposable batteries. The electrolyte is ammonium chloride paste in water,
rather than a liquid.
The half-equations are:
Zn (s) + 2e- ⇌ Zn 2+ (aq) E=-0.8V
2NH4 + (aq) + 2e- ⇌ 2NH3 (g) + H2 (g) E=+0.7V
Note that they are not E° values as they are not in standard conditions.
In half-cells, the reaction takes place as:-
At zinc electrode
Zn (s) Zn 2+ (aq) + 2e-
At carbon rod
2 NH4 + (aq) + 2e- 2NH3 (g) + H2 (g)
50
The overall equation is:
2NH4+ (aq) + Zn (s) 2NH3 (g) + H2 (g) + Zn2+ (aq)
emf = 1.5V with zinc as negative terminal
The hydrogen gas is oxidised to water by the manganese(IV) oxide, to prevent a
build up of pressure, while ammonia dissolves in the water of the paste. As the
cell discharges, the zinc is used up, making the zinc casing thinner and is prone to
leakage. Since the ammonium chloride is acidic and corrosive, we should
replace the cell once it is used up.
Zinc/chloride cell
A variant of the zinc/carbon cell, which is similar but uses zinc chloride as the
electrolyte. They enjoy a longer life and are usually marked ‘extra life’ on them.
Long-life alkaline battery
They are also based on the same system, but with an electrolyte of potassium
hydroxide. Zinc is in powdered form, to enhance its surface area to allow the cell
to supply a high current. The cell is contained within a steel container to
prevent any linkage.
Rechargeable batteries
If the reactions taking place in the half-cells are reversible, then the battery is
rechargeable. When we connect the battery to an external power supply with a
larger emf, electrons and ions are forced around the circuit in the opposite
direction, this will overcome the spontaneous reaction, hence recharges the
battery. There are a few examples of rechargeable batteries:
Lead-acid batteries
They are a type of rechargeable battery that can be used to operate starter motor
of cars. They consist of 6 2 volts cells connected in series, to give a total voltage
of 12 volts. The positive electrode is made of lead coated with lead(IV) oxide,
PbO2, and the negative plate made of lead.
51
When discharging, the
following reactions occur as
the battery drives electrons
from the negative lead
electrode to the positive
lead(IV) oxide electrode.
At lead electrode:
Pb (s) + SO4 2- (aq)
PbSO4 (s) + 2e-
At lead-dioxide coated plate:
PnO2 (s) + 4H+ (aq) + SO4
2- (aq) + 2e- PbSO4 (s) +
2H2O (l)
The overall reaction as the cell discharges is:
PbO2 (s) + 4H+ (aq) + 2SO4 2- (aq) + Pb (s) 2PbSO4 (s) + 2H2O (l)
emf = about 2 volts
The lead-acid battery can be recharged flow electrons flow in the reverse
direction, driven by the car’s dynamo.
Portable batteries
Nickel/cadium
They are available in various sizes to replace traditional zinc/carbon batteries.
They have an alkaline electrolye. The two half-equations are:
Cd(OH)2 (s) + 2e- ⇌ Cd(s) + 2OH- (aq)
NiO(OH) (s) + H2O (l) + e- ⇌ Ni(OH)2 (s) + OH- (aq)
Overall:
2NiO(OH) (s) + Cd(s) + 2H2O (l) ⇌ 2Ni(OH)2 (s) + Cd(OH)2 (s)
emf = about 1.2 volts
Lithium ion
They are widely used in portable devices. They are lightweight due to the fact
that lithium is a light metal, compared to traditional heavier metals. They have a
polymer electrolyte, compared to the ordinary gaseous/liquid electrolyte
meaning that they cannot leak.
52
Fuel cell
A fuel cell is one that chemical reaction happens between a fuel and oxygen, to
create a voltage. The fuel and oxygen flow into the cell continuously, and the
products flow out of the cell. Hence it does not need to be recharged.
Hydrogen-oxygen fuel cell
Like ordinary electrochemical cells it is consisted of two half-cells connected by a
semi-permeable membrane.
It consists of two electrodes with a very thin layer of platinum, with polymer-
based membrane is used as electrolyte, which allow ions (protons) to diffuse.
One electrode is exposed to hydrogen gas, and the other electrode to oxygen or
air.
The half-equations involved are:
2H+ (aq) + 2e- ⇌ H2 (g) E = 0.0 volts
4H+ (aq) + O2 (g) + 4e- ⇌ 2H2O (l) E = 1.2 volts
As before, the conditions are not standard.
The catalyst exposed to hydrogen gas catalyses the reaction:
On the electrode exposed to the oxygen:
53
Hence, overall, at the left electrode, the hydrogen molecules are ionised to give
hydrogen ions and electrons. Those hydrogen ions (protons) diffuses through
the membrane to the right electrode, where they react with oxygen to produce
water.
The overall equation can be found by combining the two half-equations together,
but due to the need for four electrons at the right electrode, the left-side reaction
need to happen twice to balance the overall equation.
To give:
emf = 1.2 volts
This is the exactly same reaction as burning hydrogen directly on oxygen, except
the fact that in the case of hydrogen-oxygen cell, most of the energy is converted
into electric energy, with some heat energy release as ‘wasted’ energy.
Notice that the temperature never exceeds 358 kelvins (85C), which is far below
the temperature that nitrogen oxides form.
Also notice that with this cell, the only waste product is water.
Pros and cons for fuel cells
Benefits
The hydrogen-oxygen fuel cell produces water as the only product. It
does not produce any greenhouse or polluting gas associated with
combustion engines. Despite the fact that during the process of
generating hydrogen gas produces small amount of carbon dioxide, it is
still much less than that produced by IC engines.
Fuel cells are more efficient than combustion engines. They usually have
an efficiency of about 50%, while IC engines only have 20%.
Drawbacks
Extraction of hydrogen gas is from crude oil, which is a non-renewable
resource.
Hydrogen itself is a highly flammable gas with a low boiling point. Hence
it is difficult and dangerous to store and transport. It can be stored as
liquid under pressure, or as solid absorbed into the surface of a solid.
However these techniques are expensive.
Fuel cells use toxic chemicals during the manufacuturing process.
Fuel cells have a limited lifetime.
54
Unit 15 – The transition metals
Unit 15.1 – The general properties of transition metals
Transitional elements
They are d-block elements that have partially filled d-energy levels in one or
more of their oxidation states. This includes all d-block elements except
scandium and zinc. Scandium is excluded because it only forms compounds in
+3 state and it loses all its outer electrons when it forms a 3+ ions. Zinc is
excluded because all of its compounds are in +2 state, as losing the two 4s outer
electrons give an ion Zn 2+ and with 3d level full of electrons.
Transition metals
Not all d-block elements are transition metals! The definition for a transition
metal is:
For example, copper is definitely a transition metal because the Cu 2+ ion has an
incomplete d level.
The d-block elements
The elements in the Periodic Table that correspond to the d-level electron level
filling are known as the d-block elements.
The electronic configuration of the d-block elements are shown below:
Sc [Ar] 3d14s2
Ti [Ar] 3d24s2
V [Ar] 3d34s2
Cr [Ar] 3d54s1
Mn [Ar] 3d54s2
Fe [Ar] 3d64s2
Co [Ar] 3d74s2
Ni [Ar] 3d84s2
Cu [Ar] 3d104s1
Zn [Ar] 3d104s2
A transition metal is one which forms one or more stable
ions which have incompletely filled d-orbitals.
55
Notice that there are some unusual structures present in chromium and copper.
The 4s and 3d sub-shells are very similar in energy, hence it is easy to promote
electrons from the 4s into the 3d orbitals.
In the case of chromium, the 4s1 3d5 structure is adopted due to the repulsion
between the two paired electrons in the 4s orbital is more than the energy
difference between the 4s and 3d subshells. Therefore it is more stable to have
unpaired electrons in the higher energy 3d orbital than the paired electrons in
the lower energy 4s orbital.
In the case of copper, the 3d sub-shell is actually lower in energy than the 4s sub-
shell. The 3d sub-orbitals are therefore, filled first before the 4s orbital. Hence
copper has a 4s1 3d10 configuration.
Why zinc and scandium are omitted from transition metals?
Zinc has the electronic configuration [Ar] 3d10 4s2. As it form ions, it always
loses two 4s electrons to give a 2+ ion with the electronic structure [Ar] 3d10.
Moreover, the zinc ion has full d-orbitals, which does not meet the definition of a
transition metal either.
Scandium has the electronic structure [Ar] 3d1 4s2. As it form ions, it always
loses 3 outer electrons and ends up with an argon structure. The Sc 3+ ion has
no d-shell electrons, which does not meet the definition of a transition metal
either.
4s 3d
Sc [Ar] ↑↓ ↑
Ti [Ar] ↑↓ ↑ ↑
V [Ar] ↑↓ ↑ ↑ ↑
Cr [Ar] ↑ ↑ ↑ ↑ ↑ ↑
Mn [Ar] ↑↓ ↑ ↑ ↑ ↑ ↑
Fe [Ar] ↑↓ ↑↓ ↑ ↑ ↑ ↑
Co [Ar] ↑↓ ↑↓ ↑↓ ↑ ↑ ↑
Ni [Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑
Cu [Ar] ↑ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
Zn [Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
56
Madelung/ Klechkowski Rule of building up electrons
This ordering rule only applies to neutral atoms in their ground state.
Oxidation states
The oxidation state formed by an element in its compound is governed by the
maximum number of electrons which it can lose, without the need of much energy
to remove electrons that the energy cannot be recovered when the bond is formed
again.
In a s-block element, it can only form one stable oxidation state in their
compounds. They lose all their valence electrons easily but cannot lose any
more since there is a strong electrostatic force attracting, giving rise to more
energy required to remove electrons from inner shell.
Sodium, Na, always adopt +1 oxidation state in its compounds due to the big
jump (see below) between the first and second ionisation energies.
First eight ionisation energies of sodium
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
1 2 3 4 5 6 7 8
number of electrons removed
log
(io
nis
ati
on
en
erg
y
Series2
57
Magnesium, Mg, always adopt +2 oxidation state. In the diagram below, there is
a small jump between the first and secondary ionisation energies, but a very
large jump between the second and third ionisation energies.
However, one of the key features portrayed in transition metals, is that the metal
can show a variety of oxidation states.
In d-block elements, there are a large number of valence electrons and removing
them all would require so much energy that is deemed unfeasible. Therefore it is
usually only possible to remove some of the valence electrons.
All d-block elements can give up their 4s electrons very easily, but some d-
electrons are harder remove. Moreover as the successive ionisation energies of
d-electrons increase steadily, it is hard to predict how many can be lost.
Hence, d-block metals can adopt a variety of oxidation states. The most
common oxidation states found by the first row d-block elements are as follows:
Sc: +3 only (d0)
Ti: +3 (d1), +4 (d0)
V: +2 (d3), +3 (d2), +4 (d1), +5 (d0)
Cr: +3 (d3), +6 (d0)
Mn: +2 (d5), +3 (d4), +4 (d3), +6 (d1), +7 (d0)
Fe: +2 (d6), +3 (d5)
Co: +2 (d7), +3 (d6)
Ni: +2 (d8)
Cu: +1 (d10), +2 (d9)
Zn: +2 only (d10)
First eight ionisation energies of magnesium
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
1 2 3 4 5 6 7 8
number of electrons removed
log
(io
nis
ati
on
en
erg
y
Series2
58
Note that:
Scandium only forms +3 ions (d0), due to the low effective nuclear charge
on scandium, allowing all three valence electrons to be removed easily.
Zinc only forms +2 ions (d10), due to the high effective nuclear charge on
zinc, forbidding any 3d electrons from being removed.
All other elements form at least one stable ion with partially filled d-
orbitals, hence makes them fit within the definition of a transition metal.
It is occasionally possible to remove all 3d electrons, to form a d0 configuration.
However for elements beyond manganese it is rare to find an oxidation state
containing fewer than 5 3d electrons. This is because the d5 configuration is in
fact rather stable, as the paired d-electrons can be removed to reach d5 but an
unpaired electron must be removed to reach d4. Because electrons are spaced
out, giving rise to less repulsive force between electrons, unpaired electrons are
more difficult to remove, compared to paired electrons.
Transition metal ions
Transition metal atoms form positive ions. When this happens, the electrons in
the s-orbital are removed first, then the d-orbital electrons.
For example:
Iron forms Fe2+ ions and Fe3+ ions.
As the 2+ ions are formed, it loses both 4s electrons. So:
Fe = [Ar] 3d6 4s2 Fe2+ = [Ar] 3d6
Only when the 4s electrons are removed that we can remove a 4s electron.
Fe2+ = [Ar] 3d6 Fe3+ = [Ar] 3d5
Below are the steps that can help when we write the electronic configuration for
the transition metal ion:
1) Write down the electronic configuration of the element.
2) Work out how many electrons need to be removed to form the ion.
3) Remove that amount of electrons from the electronic configuration.
Starting from the s-orbital first, then the d-orbital.
(A good tool that can help us work out how many electrons are required to be
removed is via the oxidation state.)
59
Properties of transition metals
Physical properties
The physical properties of transition metals won’t change much as we go across
the group. They all share similar properties.
1. They all have high density.
2. They all have high melting and boiling points.
3. Their ionic radii are similar.
Chemical properties
There are four common features of transition metals.
1. Variable oxidation states
Transition metals have more than one oxidation state in their compounds,
i.e. Cu(I) and Cu(II). Hence they can take part in many redox reactions.
2. Majority of transition metal ions are coloured
For example, Cu 2+ (aq) is blue.
3. Catalysts
Catalysts are those that speeds up the rate of reaction, without being used
up. Transition metals and its compounds are widely used in catalysts.
For example, iron is the catalyst in Haber Process, vanadium(V) oxide as a
catalyst in the Contact Process.
4. Complex formation
Transition elements form complex ions. These are formed when a
transition metal ion is surrounded by ions or other molecules, known as
‘ligands’, which will then bonded to them by co-ordinate bonds.
For example, [Cu(H2O)6]2+ is a complex ion that is formed as copper
sulphate dissolves in water.
60
Unit 15.2 – Complex formation and the shape of complex ions
Complex ions
They consist of a central metal ion linked to a number of molecules or ions with
lone pairs of electrons. The surrounding molecules and ions are ligands which
use lone pairs of electrons to form a co-ordinate bond with the metal ion. The
number of ligands in a complex ion is typically two, four or six.
The number of co-ordinate bonds to ligands that surround the d-block metal ion
is called the co-ordination number.
Ligands
Ligands are species that can use its lone pair of electrons to form a dative
covalent bond with a transition metal. They are molecules or ions surrounding
the central metal ion.
Examples for simple ligands include water, ammonia, chloride ion and hydroxide
ion.
All of these ligands have active lone pairs of electrons in the outer energy level,
which forms the basis of the co-ordinate bond with the metal ion.
All ligands are lone pair donors, meaning that they function as a Lewis base.
Cations
Cations which form complex ions must have the below two features:-
- they must have a high charge density, and thus be able to attract
electrons from ligands;
- they must have empty orbitals of low energy, so that they can
accept the lone pair of electrons from the ligands.
Cations of d-block metals are small, have a high charge and have available empty
3d and 4s orbitals of low energy. Hence they form complex ions readily,
61
Co-ordination number
The co-ordination number of a complex ion counts the number of co-ordinate
bonds being formed by the metal ion at its centre.
It is dependent on the size and the electronic configuration of the cation, and also
the size and charge of the ligand. 6 is the most common co-ordination number,
while 2 and 4 is also known.
It is NOT the number the ligands co-ordinately bonded to the central metallic ion!
Bonding in simple complex ions
Al(H2O)6 3+
First we will investigate the complex ion Al(H2O)6 3+.
This complex ion is formed when water molecules attach themselves to an
aluminium ion to give Al(H2O)6 3+.
First we think of the naked aluminium ion electronic structure before it is
bonded with the water molecules:
Aluminium has electronic structure:
1s22s22p63s23px1
When it is turned to an Al 3+ ion, it loses 3rd level electrons to give:
1s22s22p6
This means that all the 3rd level orbitals are empty. The aluminium then uses six
of these to accept lone pairs from six water molecules. It then hybridises (re-
arranges) the 3s, three 3p and two 3d orbitals to produce six new orbitals all
with the same energy.
So why only six water molecules? This is because six is the maximum number of
water molecules, size-wise, to be physically able to fit around the aluminium ion.
Only one lone pair is shown on each water molecule, as the other lone pair is
pointing away from the aluminium and is not involved in the co-ordinate
bonding.
62
Hence the ion looks like this:
Due to the movements of the electrons towards the centre of the ion, the 3+ is no
longer located only on the aluminium, but instead spread over the whole of the
ion.
Because now the aluminium formed 6 bonds, the co-ordination number is said to
be 6.
63
CuCl4 2-
This is another example of a complex ion with a negative charge.
Copper has electronic structure of:
1s22s22p63s23p63d104s1
As it turns into a Cu 2+ ion it loses a 4s electron and one of the 3d electrons to
give:
1s22s22p63s23p63d9
To allow the four chloride ions to be bonded as ligands, the empty 4s and 4p
orbitals are used, in a hybridised form, to accept a lone pair of electrons from
each chloride ion. As chloride ion has a larger size than water molecules, we can
only fit four of them around the copper ion, instead of 6 like water do.
Note that only one of the four lone pairs on each chloride ion is shown, this is
because the other three are pointing away from the copper ion and are not
involved in the bonding.
Below shows the shape of the complex ion:
The ion carries 2 negative charges overall.
This is the resultant charge from the 2 positive
charges on the copper ion, and the 4 negative
charges from the 4 chloride ions.
Hence, in this case, the co-ordination number
is 4.
64
Ni (NH2CH2CH2NH2)3 2+ (or abbreviated as [Ni(en)3] 2+)
The structure of the ion is as follows:
In this case, the lone pairs are on the nitrogen atoms, and the two nitrogen atoms
are linked by the –CH2CH2- group.
Note that the arrangement of bonds around the central metallic ion is the same
as that with 6 water molecules attached. The only difference is that, at this
scenario, each ligand uses two of the positions available, at right angle to each
other.
More importantly, since nickel formed 6 co-ordinate bonds, the co-ordination
number for [Ni(en)3] 2+ therefore is 6, despite the fact that there are only 3
joined ligands. That illustrates the point that co-ordination number counts the
number of co-ordinate bonds formed, NOT the number of ligands.
65
Types of complex ions
Unidentate
In the above examples there are only one bond between each ligand and the
central metal ion to give the complex ion, which the ligand is said to be
unidentate – having only one ‘tooth’.
It has only one lone pair of electrons which can be used to bond with the metal,
and any other lone pairs are pointing at the wrong direction such that it is not
involved in the co-ordinate bond.
Bidentate
Some ligands have two lone pair of electrons that both can bond to the central
metal ion. These ligands are said to be bidentate – as it has two ‘teeth’.
Commonly used examples include 1,2-diaminoethane and the ethanedioate ion.
Despite for the case of ethanedioate ions, there are more than 2 lone pairs of
electrons, however effectively only two of the many lone pairs are used in
bonding.
Multidentate/polydentate
Some ligands have more than one lone pair of electrons that can be used to bond
to the central metallic ion – they have more than one ‘teeth’.
66
Shapes of complex ions
6-co-ordinated complex ions:
These ions are all octahedral, and are formed of small ligands such as H2O and
NH3. Four of the ligands are in one plane, with the fifth one and the sixth one
above and below the plane respectively.
67
4-co-ordinated complex ions:
There are two generally-occuring shapes for 4-co-ordinated complex ions.
Below we consider two similar ions, [CuCl4]2- and [CoCl4]2-.
Tetrahedral ions
In both cases, each copper(II) and cobalt(II) ions have four surrounding chloride
ions, rather than six. This is because chloride ions are too big in size to fit any
more around the centre metallic ion.
Square planar complex
In some occasions a 4-co-ordinated complex is square planar. One of the
example for it is cisplatin – an anti-cancer drug.
Cisplatin, Pt(NH3)2Cl2, is a neutral complex, which is due to that 2+ charge of the
original platinum(II) ion is exactly cancelled by the two negative charges
supplied by the chloride ions.
68
Stereoisomerism in complex ions
Some complex ions can demonstrate either geometric or optical isomerism.
Stereoisomerism means that the molecules have the same structural formula, but
with a different spatial arrangement.
Geometric isomerism
Requirement 1:
Octahedral with 4 of ligand A and 2 of ligand B.
For example: [Co(NH3)4Cl2]+
Requirement 2:
Square planar with 2 of ligand A and 2 of ligand B.
For example: Pt(NH3)2Cl2 (cis/trans-platin)
Cisplatin works by binding to the fast-growing cancerous DNA to prevent them
replicating themselves. This forms the basis of chemotherapy, which has some
unpleasant side effects.
The isomeric compound, transplatin, has no anti-cancer effect.
69
Optical isomerism
The requirements for optical isomerism to occur are
1) 3 bidentate ligands;
2) 2 bidentate ligands or 2 monodentate ligands in cis-isomer only;
3) hexadentate ligand
Examples include [Ni(NH2CH2CH2NH2)3]2+ or [Cr(C2O4)3]3-.
If we visualise the bidentate ligand as a ‘headphone’, thse complex ions look like
this:
We can see that the ion has no plane of symmetry, three-dimensional wise. As
we previously know, any substance without a plane of symmetry will have
optical isomers, which each isomer is mirror image of the another one.
There are no way of rotating or moving the second optical isomer so that it looks
exactly like the first one. They are non-super imposable images to each other.
One of the optical isomers will rotate the plane of polarisation of plane polarised
light clockwise, and the other isomer rotate it anti-clockwise.
70
Unit 15.3 – Coloured ions
The electromagnetic spectrum
The visible light is only a small part of the electromagnetic spectrum which also
consists of gamma rays, infra-red, radio waves etc. Each of the ‘section’ on the
electromagnetic spectrum has a specific range of wavelengths.
The visible light spectrum looks something like this:
Formation of coloured ions
The middle part of the Periodic Table is known as the d-block elements (not
transition metals as we previously mentioned about partially-filled d-orbitals).
In the above diagram, the shaded part is the first row of the d-block elements,
with 3d orbitals being filled.
When the cation forms a complex ion, the ligands bonded to it repel the electrons
in the atom, thus raising its electron energy. Due to that d-orbitals are quite
small, some of them are repelled more than others, splitting the d-orbital into 2
even smaller groups of orbitals, with 3 orbitals having lower energy than the
other two.
71
For example, the arrangements of d-orbital electrons in a Cu 2+ ion before and
after 6 water molecules are co-ordinated bonded to it:
The ‘height’ between these groups of orbitals is the energy difference, and is
similar to visible light energy. The relationship between the frequency of light
and the light energy is shown by the equation E=hf, where E is the light energy, h
is Planck’s constant, and f denotes frequency of light.
hf
If a metal has a partially-filled d-orbital (a transitional metal ions), some of the
lower energy electrons can be excited into the higher energy orbitals by
absorbing certain amount of white light energy corresponding to its relative
frequency.
Hence the resultant light ray transmitted is deficient in the corresponding
frequency absorbed, and thus appears coloured.
72
Criteria for an ion to be coloured
There must be a split of the d-orbitals. This can only happen when
there are ligands – meaning that only complex ions are coloured.
Since anhydrous ions do not contain split d-orbitals so it cannot absorb
any light energy in the visible light spectrum, and thus are white.
For example:
anydrous CuSO4 (d9) is white but hydrated CuSO4.5H2O is blue.
The d-orbitals must be partially-filled.
If the d-orbitals are empty (e.g. Sc 3+, Al 3+) then there are no electrons
that can be excited into higher energy d-orbitals and thus the ions will
be colourless.
If the d-orbitals are full (e.g. Cu+, Zn 2+) then there are no empty
orbitals at which electrons can be excited to, hence the ions will appear
colourless.
Factors affecting the colour of complex ions
It is dependent on:
The nature of ligand (depending on the difference/splitting of energy
gaps, which varies between different type of ligands)
The co-ordination number
The oxidation state of the metal
The identity of the metal
The list below shows some common ligands, which at the top produce the
smallest splitting, and the bottom produce the largest splitting. The greater the
splitting is, more energy is required to promote an electron from lower to higher
orbital – this means shorter wavelength light ray is absorbed. This also means
that as splitting increases, the light absorbed will shift from red end of the
spectrum.
73
Identifying a transitional metal complex using colorimetry
As transition metal ions can show a variety of colours, it is plausible to identify a
complex ion by its colour.
Determining concentration
A technique, called ultraviolet and visible spectrometry, can be used to
determine the concentration of a solution by measuring how much light it has
absorbed.
White light is shone through a filter, which is designed to only let white light to
pass. The light then passes through a filter to a colorimeter, which shows how
much light was absorbed by the sample.
Therefore, more concentrated the sample is, more light it will absorb. As the
light absorbance of the sample is proportional to its concentration, the
concentration of any sample can be determined by comparing to the absorbance
of a solution of a known concentration.
The process is as below:-
1. A solution containing a known concentration of the ion is prepared, and a
suitable ligand is added in order to intensify the colour (i.e. usually
thiocyanate ions, SCN-).
2. A sample of the solution is placed into a curvette of the colorimeter.
3. The light absorbance of the sample is recorded.
4. This process is repeated using a range of other known concentration.
5. Plot a graph of absorbance against concentration.
6. The solution of unknown concentration is then mixed with an excess of
the same ligand (in step 1) to intensify the colour.
7. The absorbance of the solution of unknown concentration is then
recorded using the same colorimeter.
8. The concentration of the solution can then be determined using the
previously plotted concentration absorbance graph.
74
Determining the formula
We can also use the same method to determine the ratio of metal ions to ligand
in the complex.
We can mix two solutions – one with metal ions and one with ligands, in different
proportions. When they are mixed in the same ratio as they are in the complex,
this will be the maximum concentration of the complex in the solution. Hence
the solution will absorb the most light.
For example:
The example will be about the red blood complex formed with iron(III) (Fe 3+)
and the thiocyanate ions (SCN-).
When potassium thiocyanate is added to a solution of Fe 3+ (aq), one of the
water molecules is replaced by the thiocyanate ion, forming red blood complex:
[Fe(H2O)6]3+ (aq) + SCN- [Fe(SCN)(H2O)5 2+] (aq) + H2O
As the concentration of the red blood complex builds up, the light intensity
passing through the solution reduces.
We start with two solutions of the same concentration, with one Fe 3+ (aq) ions
(i.e. iron(III) sulphate), and the another one containing SCN- (aq) ions (i.e.
potassium thiocyanate). We then mix them in different proportions, and add
water so that the total volume of solution in each test tube is constant
throughout.
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The result table explains this.
Tube no. 1 2 3 4 5 6 7 8
Vol of Fe 3+
(aq)
solution
(cm3)
10 10 10 10 10 10 10 10
Vol of
SCN- (aq)
solution
(cm3)
2 4 6 8 10 12 14 16
Vol of
water
(cm3)
28 26 24 22 20 18 16 14
Light
absorption
(arbituary)
0.15 0.33 0.48 0.63 0.70 0.70 0.70 0.70
Each tube is placed in the colorimeter and data of light absorbance is recorded.
From the data shown above, the maximum absorbance starts at tube 5: which
after tube 5, any further increase of SCN- ions will contribute no more colour
change in the solution. This tells us the ratio of SCN- ions to Fe 3+ ions in the
complex. Therefore, in tube 5 the volume ratio between Fe 3+ and SCN- ions will
the ratio of ions in the complex formula. As it has equal volume of SCN- and Fe3+
ions, the ratio in the complex must be 1:1. Hence this confirms that the formula
is:
[FeSCN(H2O)5]2+.
Compared to [Fe(H2O)6]3+, this tells us that the SCN- has substituted one of the
water molecules in the complex ion.
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Unit 15.4 – Variable oxidation states of transition elements
One of the important properties of transition metal is its variable oxidation
states.
In Group 1 metals, they lose their only outer electron to form only +1 ions; while
Group 2 metals lose two outer electrons to form 2+ ions in their compounds.
A transition element can exist in multiple oxidation states, as it can use its 3d as
well as its 4s electrons for bonding.
The most common oxidation state is 2+ as the 4s electrons are usually removed
first. Since 4s and 3d electrons are very close in energy, the 3d electrons can be
removed easily too. This results in a combination of ions that they can form, by
losing different amount of electrons which all are stable.
Changes in oxidation states also mean that the transition metals undergo a
colour change during a reaction.
The below diagram shows the oxidation numbers shown by elements of the first
d-series in their compounds.
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Note that only lower oxidation states of a transition metal actually exist as simple
ions, i.e. Mn2+ exist but Mn7+ do not. Mn(VII) as in MnO4-, manganese is
covalently bonded.
Redox reactions in transition metal chemistry
Majority of the reactions involving transition metal compounds are redox
reactions, in which metals are either oxidised or reduced. For example, iron
shows two oxidation states, Fe 3+ and Fe 2+.
Since Fe 2+ is a less stable state, it can be oxidised into Fe 3+ using oxygen in the
air and also by chlorine, as shown below:
2 Fe 2+ + Cl2 2Fe3+ + 2Cl-
(+2) (+3) (-1)
In this example, chlorine is the oxidising agent – as its oxidation state decreased
from 0 to -1 (due to gain of an electron); while iron has oxidation number
increased from +2 to +3 (due to lose of an electron), so it is oxidised.
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Redox titrations
A way to figure out the concentration of both reducing and oxidation agent is via
redox titrations. This is similar to the acid-base titration.
When we know the concentration of either the oxidising agent or the reducing
agent, we can use the titration results to work out the concentration of the other.
Steps for redox titration:
1. Firstly, we measure the quantity of a reducing agent, i.e. aqueous Fe 2+
ions using a pipette, and then transfer them into a conical flask.
2. Using a measuring cylinder, we add about 20cm3 of dilute sulphuric acid
to the flask. This amount is in excess to ensure that all oxidising agent to
be reduced.
3. Then we add the oxidising agent, i.e. aqueous potassium manganate(VII)
to the reducing agent using a burette. As it is being added we swirl the
conical flask.
4. The added oxidising agent reacts with the reducing agent. The reaction
will continue until all of the reducing agent is used up.
5. Stop adding oxidising agent into the mixture when the colour changes
into the colour of the oxidising agent. This is the end point of the titration.
6. Record the volume of oxidising agent added.
7. The first titration is a rough titration.
The latter titrations are accurate titrations, with results concordant to
each other.
Apparatus:
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We need to know two titration reactions of Fe 2+ ions, involving two different
oxidising agents.
1. Manganate(VII) ions (MnO4-) in aqueous potassium manganate(VII)
(KMno4). The solution turns into purple colour at the end point. The
equation is given by:
MnO4- (aq) + 8H+ (aq) + 5Fe2+(aq) Mn2+(aq) + 4H2O (l) + 5Fe3+ (aq)
2. Dichromate(VI) ions (Cr2O7 2-) in aqueous potassium dichromate(VI)
(K2Cr2O7). The solution turns orange at the end point. The equation is
given by:
Cr2O7 2- (aq) + 14H+ (aq) + 6Fe2+(aq) Cr3+(aq) + 7H2O (l) + 6Fe3+ (aq)
Oxidation of transition metal ions in alkaline solutions
Oxidation states of chromium
Chromium most commonly exists in the +3 and +6 oxidation state. Although it
can also exist in the +2 oxidation state, it is much less stable. The common ions
that chromium form is as below:
Oxidation state Formula of ion Colour of ion +6 Cr2O7 2- (aq) Orange +6 CrO4 2- (aq) Yellow +3 Cr 3+ (aq) Green / Violet +2 Cr 2+ (aq) Blue
In the +6 oxidation state, chromium can form chromate(VI) ions (CrO4 2-)and
dichromate(VI) (Cr2O7 2- ) ions. Both of them are good oxidising agents, as they
can be easily reduced to Cr 3+.
When Cr 3+ ions are surrounded by 6 water ligands, it appears violet. But since
water ligands are often substituted, the solution looks green instead.
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Chromate(VI) and dichromate(VI) ions in equilibrium
When an alkali (OH- ions) is added to aqueous dichromate(VI) ions (Cr2O7 2- ),
the orange colour turns yellow, because aqueous chromate(VI) (CrO4 2-) ions are
formed.
Cr2O7 2- (aq) + OH- (aq) 2CrO4 2- (aq) + H+ (aq)
Orange Yellow
When an acid (H+ ions) is added to aqueous chromate(VI) solutions, the yellow
colour turns orange, due to the formation of aqueous dichromate(VI) ions.
2CrO4 2- (aq) + H+ (aq) Cr2O7 2- (aq) + OH-(aq)
Yellow Orange
These two are opposite reactions, and both ions exist in equilibrium:
Cr2O7 2- (aq) + H2O (l) ⇌ 2CrO4 2- (aq) + 2H+ (aq)
The equilibrium position is dependent on pH value – as the Le Chatelier’s
principle explains. When H+ ions are added, the equilibrium shifts to the left
that orange Cr2O7 2- ions are formed. If OH- ions are added, H+ ions are removed
and the equilibrium shifts to the right, forming yellow CrO4 2- ions.