Electrochemical cells & Thermodynamics.pdf
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Transcript of Electrochemical cells & Thermodynamics.pdf
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CHAPTER 1
ELECTROCHEMICAL NATURE OF AQUEOUS
CORROSION
CMT555 Electrochemical & Corrosion Science
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ELECTROCHEMISTRY What is electrochemistry? yElectrochemistry is a branch of chemistry that
deals with the interconversion between electrical energy and chemical energy.
Electrical ↔ Chemical yThe conversion takes place in an electrochemical
cell.
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REDOX REACTIONS y In an electrochemical cell, there are two half-
reactions (redox processes) occur at two different electrodes:
Reaction Electrode Reduction (Gain of electron)
Oxidation (Loss of electron)
Anode
Cathode
* Remember: OIL RIG
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Example 1:
Half-reaction Anode: Cathode:
Overall:
)(2)( 22 gaq HeH o� ��
)(22)()()( 2 gaqaqs HZnHZn �o� ��
�� �o eZnZn aqs 22)()(
(reduction)
(oxidation)
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A summary of redox terminology Process Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) OXIDATION ¾ One reactant loses electrons. ¾ Reducing agent is oxidized. ¾Oxidation number increases.
9 Zinc loses electrons. 9 Zinc is the reducing agent and become oxidized. 9 The oxidation no. of Zn increases from 0 to +2.
REDUCTION ¾ Other reactant gains electrons. ¾ Oxidising agent is reduced. ¾ Oxidation number decreases.
9 Hydrogen ion gains electrons. 9 Hydrogen ion is the oxidizing agent and becomes reduced. 9 The oxidation no. of H decreases from +1 to 0.
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• Oxidising agent: ? • Reducing agent:?
• When a piece of zinc is placed in an aqueous
solution of copper(II) sulphate, brown solids (copper metal) will form and collect at the bottom of the container. The blue colour of the solution slowly fades, and the piece of zinc gets smaller.
Electron acceptor
Electron donor
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• In the reaction, the zinc metal gives up electrons and slowly dissolves:
Zn → Zn2+ + 2e-
• The electrons are taken up by the Cu2+ ions in the solution, and get deposited as copper:
Cu2+ + 2e- → Cu • In the process, zinc gets oxidised while the Cu2+
ions get reduced. Zn + Cu2+ → Zn2+ + Cu
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• In the reaction, zinc supplies electrons to Cu2+ ions. Hence, zinc is the reducing agent. Cu2+ ions accept electrons from zinc. Hence, the Cu2+ ions is the oxidising agent.
• In a redox reaction, the reducing agent undergoes oxidation, while the oxidising agent undergoes reduction.
Zn + Cu2+ → Zn2+ + Cu oxidation
reduction
reducing agent
oxidising agent
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Balancing Redox Equations using Half-reactions
• This method divides the overall redox reaction into oxidation and reduction half-reactions.
• Since neither oxidation nor reduction can
actually occur without the other, we refer to the separate equations as half-reactions.
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The general rule involves the following: 1. Divide the reaction into two half-reactions. 2. Balance each half reaction separately.
a) Balance atoms other than O & H in each half-reaction separately.
b) Balance O by adding H2O to the opposite side. c) Balance H by adding H+ as appropriate. d) Balance the charge by adding e-. For example, if the
reactant side of the equation has a total charge of +3, the product side must also equal +3.
e) Balance the charges of the two half-reactions by multiplying appropriately.
3. Add two half-reactions together and balance the final equation by inspection.
4. Cancel the electrons on both sides.
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Example 2: • Consider the following reaction:
22
4 ClMnClMnO �o� ���
Reduction Oxidation 1)
2)
3)
4) 5)
1)
2) ..........
3) ..........
4)
5)
�� o 24 MnMnO 2ClCl o�
�� o 24 MnMnO
OHMnMnO 22
4 4�o ��
OHMnMnOH 22
4 48 �o� ���
OHMnMnOHe 22
4 485 �o�� ����
OHMnMnOHe 22
4 8221610 �o�� ����
22 ClCl o�
�� �o eClCl 22 2
�� �o eClCl 10510 2
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• Finally, combine half-reactions and cancel terms:
��
����
���
o���
eClOHMnClMnOHe
10582
1021610
222
4
222
4 58210216 ClOHMnClMnOH ��o�� ����
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Exercise 1 Balance the following equations:
���� �o� 332272 FeCrFeOCr
�� �o� 2422 SOClSOCl
2242 SOCOSOHC �o�
(a)
(b)
(c)
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ELECTROCHEMICAL CELL • Oxidation-reduction or redox reactions take
place in electrochemical cells.
• There are several types of electrochemical cells: 1. Galvanic/voltaic/chemical cell/daniel cell 2. Electrolytic cell 3. Concentration cell
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There are 4 essential components in a cell: 1. Anode 2. Cathode 3. Ionic conductor (electrolyte) 4. Metallic conductor (electrical connection)
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Galvanic Cells • Spontaneous chemical reaction which
generates electrical energy Chemical energy → Electrical energy In the cell reaction the difference in chemical
potential energy between higher energy reactants and lower energy products is converted into electrical energy.
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• A common galvanic cell is shown below:
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Electrolytic Cells �Electrical energy is used to bring about a non-
spontaneous reaction. Electrical energy → Chemical energy �This type of cell is formed when an external
source of electrical energy is introduced into the system.
In the cell reaction electrical energy from an
external power supply converts lower energy reactants into higher energy products.
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• A common electrolytic cell is shown below:
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Similarities between Galvanic Cell and Electrolytic Cell
1. Both involve redox reactions. 2. Anode is the site of oxidation. 3. Cathode is the site of reduction. 4. Involve the flow of electrons from the Anode to the Cathode. 5. Both can have a salt bridge (for the passage of
ions so as to maintain electrical neutrality).
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Differences between Electrolytic Cell and Galvanic Cell
Galvanic cell Electrolytic cell
1. Chemical energy → Electrical energy
1. Electrical energy → Chemical energy
2. Spontaneous reaction 2. Non-spontaneous reaction
3. Positive terminal of a cell is cathode
Negative terminal of a cell is anode
3. Positive terminal of a cell is anode
Negative terminal of a cell is cathode
4. 'G < 0 4. 'G > 0
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Cell Notation
• Cell notation in chemistry is a shorthand way of expressing a certain reaction in an electrochemical cell.
Components of anode compartment (oxidation half-cell)
Components of cathode compartment
(reduction half-cell) ║
LEFT SIDE RIGHT SIDE
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Cell Notation
CuCuZnZn �� 22
Electrode (s)
Electrolyte (aq)
Electrolyte (aq)
Electrode (s)
Reduction Oxidation
Phase boundary (aqueous/solid)
Salt bridge or porous partition
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Concentration Cells • A galvanic cell in which both compartments have
the same material but at different concenration. • The electrons flow in the direction that tends to
equalize concentrations.
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Concentration Cell
Diluted solution Concentrated solution
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• [Zn2+] in the two compartments are different, 1.0 M and 0.1 M, respectively.
• The cell will try to equalize the [Zn2+] in the two
compartments – by transfering electrons from: ¾Compartment containing 0.1 M Zn2+ to the one
containing 1.0 M Zn2+ (left to right). ¾The electron transfer will produce more Zn2+ in the
left compartment and consume Zn2+ (to form Zn) in the right compartment.
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CELL POTENTIAL • Electron flows from the anode to the cathode
because there is a difference in electrical potential energy between the electrodes.
• The difference in electrical potential between the anode and the cathode is measured by a voltmeter.
• The voltage across the electrodes of a cell is called the cell voltage or cell potential or electromotive force (emf) → (E)
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STANDARD ELECTRODE POTENTIAL • According to the IUPAC convection, all half-cell
reactions are written as REDUCTION.
• The standard electrode potential is sometimes called as standard reduction potential or standard redox potential.
• The electrode potential measured under standard-state conditions is called standard electrode potential, E° measured in volts.
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• Conditions for the measurement of the electrode potential have to be standardised.
• The standard-state conditions are:
Temperature (T) fixed at 25°C or 298 K Pressure (P) fixed at 1 atm or 101 kPa
Concentration of aqueous ions fixed at 1.0 M
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• The absolute value of the standard electrode potential of an electrode system cannot be measured.
• The value can be determined if a particular
electrode is chosen as the standard or reference electrode and all other electrode systems are measured against this standard electrode.
• The standard hydrogen electrode (SHE) was
chosen by IUPAC as the reference electrode for all measurements of E°.
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• The standard reference half-cell is a standard hygrogen electrode, which consists of a specially prepared platinum electrode immersed in a 1 M aqueous solution of a strong acid, H+ (aq) [or H3O+ (aq) ], through which H2 gas at 1 atm is bubbled. )1;(2)1;(2 2 atmgHeMaqH l� �� E°reference = 0.00 V
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• E° for the hydrogen electrode is arbitrarily fixed at 0.00 V.
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• Tables of Standard Electrode Potentials for Half-Reactions allow us to determine the voltage of electrochemical cells.
• These tables compare the ability of different half-reactions to compete for electrons (become reduced).
• Since the values are given in their ability to be reduced, the bigger the E°, the easier they are to be reduced.
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Selected Standard Electrode Potentials (298K)
Half-Reaction Eo(V)
2H+(aq) + 2e− H2(g)
F2(g) + 2e− 2F−(aq) Cl2(g) + 2e− 2Cl−(aq) MnO2(g) + 4H+(aq) + 2e− Mn2+(aq) + 2H2O(l) NO3
-(aq) + 4H+(aq) + 3e− NO(g) + 2H2O(l) Ag+(aq) + e− Ag(s) Fe3+(g) + e− Fe2+(aq) O2(g) + 2H2O(l) + 4e− 4OH−(aq) Cu2+(aq) + 2e− Cu(s)
N2(g) + 5H+(aq) + 4e− N2H5+(aq)
Fe2+(aq) + 2e− Fe(s) 2H2O(l) + 2e− H2(g) + 2OH−(aq) Na+(aq) + e− Na(s) Li+(aq) + e− Li(s)
+2.87
−3.05
+1.36 +1.23 +0.96 +0.80 +0.77 +0.40 +0.34
0.00 −0.23 −0.44 −0.83 −2.71
strength of reducing agent
stren
gth
of o
xidi
zing
agen
t
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Example 3: If copper and hydrogen half-cells are joined
together, we find that the copper half-cell will gain electrons from the hydrogen half-cell:
� � � �saq CueCu o� �� 22 VE 34.0 $
� � � �aqaq HeH 222 o� �� VE 00.0 $
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• Since both half-reactions cannot undergo
reduction, we must reverse the equation of the reaction that will undergo OXIDATION.
� � � �saq CueCu o� �� 22
� � � ��� �o eHH aqg 222
VE 34.0 $
VE 00.0 $
Reduction
Oxidation
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Calculating Voltages of Electrochemical Cells
• Before calculating the voltage of a cell you must first determine which half-cell will be oxidized and which one will be reduced.
anodecathodecell EEE $$$ �
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Example 4 Zinc – copper electrochemical cell
• From Table of Standard Electrode Potential:
• In the Table, all reactions are written as reduction reactions.
• The E° values indicate which half-reaction is better at competing for electrons.
� � � �saq ZneZn o� �� 22
� � � �saq CueCu o� �� 22 VE 34.0 $
VE 76.0� $
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• Since the copper half-reaction has a larger value for E° than the zinc half-reaction, copper will be reduced, forcing zinc to be oxidized.
• So we reverse the zinc equation.
• We can then add the two equations together to get the full redox reaction and determine the voltage of the cell.
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� ��� �o eZnZn s 22
� � � �saq CueCu o� �� 22 V34.0
� �V76.0��
� � � � � � � �ssaq CuZnZnCu aq �o� �� 22 V10.1
A positive value of E° indicates a spontaneous reaction
A negative value of E° indicates a non-spontaneous reaction
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Exercise 2 Predict the spontaneity of the following
reactions: a) Reduction of Sn2+ to Sn by Mg b) Oxidation of Cl- to Cl2 by acidified Cr2O7
2-
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SPONTANEITY OF REDOX REACTIONS
• In an electrochemical cell, chemical energy is converted to electrical energy.
• Electrical energy in this case is the product of the emf of the cell and the total electrical charge (in coulombs) that passes through cell:
W = Q E unit: W = Joules (J) Q = Coulombs (C) ΔE = Volts (V)
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• The total charge is determined by the number of moles of electrons (n) that pass through the circuit.
• By definition: total charge = nF (Q = nF) 1F = 96500 C/mol • The measured emf is the max. voltage that the cell can
achieve. This value is used to calculate the maximum amount of electrical energy that can be obtained from the chemical reaction.
• This energy is used to do electrical work (wele), so wmax = wele
= -nFEcell
• wmax = is the max. amount of work that can be done.
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Gibb’s Free Energy (ΔG) •The Gibb’s free energy (ΔG) represent the max. amount of useful work that can be obtained from a reaction. •ΔG is the negative value of the max. electrical work: ΔG = Wmax
= -nFEcell
For reactions in which reactants and products are in their standard states,
ΔG° = -nFE°cell
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• Spontaneous reaction: ΔG° = negative E°cell = positive
• Non-spontaneous reaction: ΔG° = positive E°cell = negative
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Applications of Thermodynamics to Corrosion
• A definite relation between the free energy change and the cell potential of an electrochemical reaction.
• ΔG is used to predict the spontaneous direction of any electrochemical reaction.
• The sign of ΔG is the most important factor to indicate whether or not the reaction is spontaneous.
ΔG° = -nFE°cell
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• In any electrochemical reaction, the most negative (active) half-cell tends to be oxidized and the most positive (noble) half-cell tends to be reduced.
• From the rule: All metals with potentials more negative
(active) than hydrogen will tend to be corroded by acid solutions.
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• However Cu, Ag, Hg, Pt and Au are not corroded in acid solutions.
• But if dissolved oxygen is present, there is a possibility of corrosion to occur for Cu and Ag due to oxygen reduction.
• Cu and Ag tend to corrode spontaneously in the presence of oxygen.
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Example 5 i) Cu + H2SO4 → No reaction (Cu/Cu2+ > positive than H2/H+) ii) 2Cu + 2H2SO4 + O2 → 2CuSO4 + 2H2O (O2/H2O > positive than Cu/Cu2+)
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Exercise 3 1. Would it be possible to store a silver spoon in
a zinc nitrate solution? 2. Would it be possible to store a silver nitrate
solution in a copper container? 3. Calculate the standard free-energy change
for the following reaction at 25°C:
Is these reaction spontaneous?
)(32
)( 3)1(2)1(32 ss CaMAuMCaAu �o� ��
)(22)()( 2 gaqs HZnClHClZn �o�
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Nernst Equation y Used when all or some of the components are under
non-standard-state conditions.
y Shows how the cell potential depends on the concentrations of the cell components. Q
nFRTEE cellcell ln� $
QnF
RTEE cellcell log3.2� $
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Qn
EE cellcell log0592.0� $
Where R = universal gas constant (8.314472 J K-1 mol-1) T = temperature (K) F = Faraday constant (96500 C mol-1) n = no. of electron Q = reaction quotient
> @> @treac
productQtan
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Cell Potential and the Nernst Equation
• The Nernst equation can be used to calculate the cell potential of electrochemical cells which are not under standard conditions.
• Consider the galvanic cell operate at 25°C: Zn (s) │Zn2+ (1.8 M) ║ Cu2+ (0.2 M) │ Cu (s)
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Exercise 4 • Predict whether the following reaction would
proceed spontaneously as written at 298 K: Co(s) + Fe2+
(aq) → Co2+ (aq) + Fe(s)
given that [Co2+] = 0.15 M and [Fe2+] = 0.68 M.
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Equilibrium Constants • For a general cell reaction:
where a, b, c, and d are the stoichiometric coefficients for the reacting species A, B, C, and D.
• For the reaction at a particular temperature:
where K is the equilibrium constant.
dDcCbBaA �o�
ba
dc
BADCK][][][][
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• At equilibrium, there is no net transfer of electrons, so E = 0 and Q = K.
• Applying these conditions to the Nernst equation (valid at 25°C)
• Therefore,
Qn
EE cellcell log0592.0� $
Kn
E cell log0592.00 � $
0592.0)(log cellEnK
$
? K
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Example 6 Calculate the equilibrium constant, K, for the
following reaction:
AgSnAgSn 22 42 �o� ���
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Relationship of Gibb’s Free Energy and Nernst Equation
• Consider a redox reaction of type:
• The free energy change, ΔG, is given by thermodynamic equation:
Where Q = reaction quotient
dDcCbBaA �o�
QRTGG ln�' ' $
> @ > @> @ > @ba
dc
BADC
= [Product] [Reactant]
![Page 59: Electrochemical cells & Thermodynamics.pdf](https://reader036.fdocuments.in/reader036/viewer/2022082215/5695d0091a28ab9b0290ad55/html5/thumbnails/59.jpg)
• Because ΔG = -nFE and ΔG° = -nFE°, the equation can be expressed as
• Dividing the equation through by –nF, we get
• Using the base-10 logarithm of Q
QRTnFEnFE ln�� � $
QnFRTEE ln� $ (Nernst Equation)
Qn
EE log0592.0� $
![Page 60: Electrochemical cells & Thermodynamics.pdf](https://reader036.fdocuments.in/reader036/viewer/2022082215/5695d0091a28ab9b0290ad55/html5/thumbnails/60.jpg)
'Go
Eocell K
'Go K Reaction at standard-state
conditions
Eocell
The interrelationship of 'Go, Eo, and K.
< 0 spontaneous
at equilibrium
nonspontaneous
0
> 0
> 0
0
< 0
> 1
1
< 1
'Go = -RT lnK 'Go = -nFEocell
Eocell = RT lnK
nF
Kn
Ecell log0592.0 $ (at 298 K)
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HOMEWORK
1. Calculate E° and Ecell for the following cell reactions.
a.
b.
2. What is the emf of a cell consisting of a Pb2+/Pb
half-cell and a Pt/H+/H2 half-cell if [Pb2+] = 0.10 M, [H+] = 0.050 M and PH2 = 1.0 atm?
� � � � � �
> @ > @ MSnMMg
SnMgSnMg saqaqs
035.0,045.0 22
2)(
2
�o���
��
� � � � � �
> @ > @ MZnMCr
CrZnCrZn saqaqs
0085.0,010.0
232323
2)(
3
�o���
��
![Page 62: Electrochemical cells & Thermodynamics.pdf](https://reader036.fdocuments.in/reader036/viewer/2022082215/5695d0091a28ab9b0290ad55/html5/thumbnails/62.jpg)
3. Calculate the emf of the following concentration cell:
READ on batteries topic -dry cell battery -mercury battery -lead storage battery -solid state lithium battery
� � � � � � � �ss MgMMgMMgMg 53.024.0 22 ��
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Solubility Product Constant • When a sparingly soluble salt (a salt which has
very low solubility in water) is added a little at a time to water, a saturated solution is eventually formed.
• That is, the solution is in equilibrium with excess undissolved solid.
• When a saturated solution of silver chloride is in contact with solid silver chloride, the following equilibrium is established.
)()()( aqClaqAgsAgCl �� ��
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• The product of the concentration of silver ions and chloride ions is known as the solubility product (Ksp) of silver chloride.
• The solubility product of a sparingly soluble salt is defined as the product of the concentration of the ions in a saturated solution of the salt, raised to the power of the stoichiometric coefficient of the respective ions.
]][[ �� ClAgKsp
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Example 7 1. 2. 3.
)()()( 22 aqSaqNisNiS �� �� ]][[ 22 �� SNiKsp
)(2)()( 22 aqClaqPbsPbCl �� �� 22 ]][[ �� ClPbKsp
�� �� 34
3243 2)(3)()( POaqCasPOCa 23
432 ][][ �� POCaKsp
![Page 66: Electrochemical cells & Thermodynamics.pdf](https://reader036.fdocuments.in/reader036/viewer/2022082215/5695d0091a28ab9b0290ad55/html5/thumbnails/66.jpg)
• For a sparingly soluble salt AxBy, the Ksp is given by:
• The value of Ksp changes when there is a change in temperature.
• Generally, the higher the value of Ksp, the higher the solubility.
)()()( aqyBaqxAsBA xyyx
�� ��
yxxysp BAK ][][ ��
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Example 8 Write down the expression for the solubility
product for the following salts. a) Pb(OH)2
b) Ag2CrO4
c) Al(OH)3
d) MgF2
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Example 9 Calculate the solubility products of the salts from
the data below: a) The solubility of silver chloride in water is
1.453 x 10-3 g dm-3. b) The solubility of silver sulphate in water is 1.5 x
10-2 mol dm-3.
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Solubility Product and Precipitation • When an aqueous solution of calcium nitrate is
added to an aqueous solution of sodium sulphate, one of the products is calcium sulphate, a sparingly soluble salt.
• Whether a precipitate of CaSO4 will be formed or not depends on the concentration of the Ca2+
ions and SO4- ions in the mixture.
)(2)()()()( 344223 aqNaNOsCaSOaqSONaaqNOCa �o�
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• If, [Ca2+][SO4
2-] > Ksp : precipitation will occur [Ca2+][SO4
2-] < Ksp : no precipitation takes place
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Example 10 25.0 cm3 of 1.0 x 10-5 mol dm-3 silver nitrate is
added to 25.0 cm3 of 1.0 x 10-5 mol dm-3 sodium chloride.
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Common Ion Effect • Consider a saturated solution of silver chloride:
• If a little sodium chloride is added to the above solution, according to Le Chatelier’s Principle, the equilibrium will shift to the left to get rid of some of the Cl- ions added.
• Hence, the amount of solid silver chloride will increase.
• In other words, the solubility of silver chloride is decreased by the addition of sodium chloride (which contains the Cl- ions).
• This phenomenon is known as the common ion effect.
)()()( aqClaqAgsAgCl �� ��