2. Thermodynamics
• Law of thermodynamics deals with energy
changes of macroscopic systems involving a large
number of molecules
• Thermodynamics is not concerned about how
and at what rate these energy transformations
are carried out, but is based on initial and final
states of system undergoing the change
• Law of thermodynamics apply only when a
system is in equilibrium or moves from one
equilibrium state to another equilibrium states
2
3. Chemical thermodynamics
• Thermodynamics provide information regarding
the feasibility of a chemical reaction
• It tells us whether a particular reactions is
possible under given set of conditions and also
the direction in which the reaction tend to occur
• Such information and study of energy
transformation in chemical reactions constitute
the subject matter of chemical thermodynamics
3
4. Chemical thermodynamics
• How do we determine the energy change
involved in a chemical reaction/process? Will
it occur or not?
• What drives a chemical reaction/process?
• To what extent do the chemical reaction
proceed?
4
5. System is defined as any portion of matter under consideration, which is separated
from the surroundings by real or imaginary boundaries. System may be
homogeneous or heterogeneous, but has to be macroscopic(Au block in figure).
Surroundings is the rest of the universe around the system. A system and its
surrounding are always separated by real or imaginary boundaries, through which
matter and energy may be exchanged between the two(the water bath at constant
‘temperature’).
Universe = System + Surrounding (the part that is within the dotted line box in the figure below)
More practically, we can consider the ‘Surrounding’ as the immediate neighborhood
of the system (the part of the universe at large, with which the system ‘effectively’
interacts).
Things that matter for the surrounding: (i) T (ii) P (iii) ability to: do work, transfer
heat, transfer matter, etc. Parameters for the system: (i) Internal energy, (ii) Enthalpy,
(iii) T, (iv) P, (v) mass, etc.
In TD we usually do not worry about
the universe at large!
Terminology used in Thermodynamics
5
6. To a thermodynamic system two ‘things’ may be added/removed:
energy (heat, work) matter.
An open system is one to which you can add/remove matter (e.g. a open beaker to which
we can add water). When you add matter- you also end up adding heat (which is contained
in that matter).
A system to which you cannot add matter is called closed.
Though you cannot add/remove matter to a closed system, you can still add/remove heat
(you can cool a closed water bottle in fridge).
A system to which neither matter nor heat can be added/removed is called isolated.
A closed vacuum ‘thermos’ flask can be considered as isolated.
Open, closed and isolated systems
Type of boundary Interactions
Open All interactions possible (Mass, Work, Heat)
Closed Matter cannot enter or leave
Semi-permeable Only certain species can enter or leave
Insulated Heat cannot enter or leave
Rigid Mechanical work cannot be done*
Isolated No interactions are possible**
* By or on the system
** Mass, Heat or Work
Mass
Heat
Work
Interactions possible
6
7. Open Closed II
Open Isolated
System
Capable of
exchanging
both matter
and energy
with the
surroundings.
e.g. water in
an open
container
Capable of
exchanging
only energy
with the
surroundings.
e.g. water in a
closed
container
Capable of
exchanging
neither
matter nor
energy
with the
surroundings.
e.g. water in
thermos flask
7
8. Isothermal process → the process takes place at constant temperature
(e.g. freezing of water to ice at –10C)
Isobaric → constant pressure
(e.g. heating of water in open air→ under atmospheric pressure)
Isochoric → constant volume
(e.g. heating of gas in a sealed metal container)
Reversible process → the system is close to equilibrium at all times (and
infinitesimal alteration of the conditions can restore the universe (system +
surrounding) to the original state.
Cyclic process → the final and initial state are the same. However, q and w need
not be zero.
Adiabatic process → dq is zero during the process (no heat is added/removed
to/from the system)
A combination of the above are also possible: e.g. ‘reversible adiabatic process’.
Processes in thermodynamics
8
9. Reversible Process
• A process is said to be reversible, when the energy change in
each step of the process can be reversed in direction by a small
change in a variable (like temperature, pressure etc) acting on the
system.
10
10. Irreversible Process
A irreversible process is one in which the system or surroundings are
not restored to their initial state at the conclusion of their reverse
process.
All naturally occurring processes are irreversible. They always tend to
proceed in a definite direction.
They are also called Spontaneous process, which means direction of
the process is spontaneous.
11
11. Spontaneous Processes
• Spontaneous processes are
those that can proceed
without any outside
intervention.
• The gas in vessel B will
spontaneously effuse into
vessel A, but once the gas is
in both vessels, it will not
spontaneously
12
13. Extensive Property: Depends on the amount of substance
taken.
e.g. mass, volume, pressure, internal energy, heat capacity,
enthalpy.
Intensive Property: Does not depend on the amount of
substance taken.
e.g. density, viscosity, refractive index, specific heat capacity.
Property of a System
14
14. Internal Energy
• It is a function of temperature, chemical nature of substance, Pressure
and volume.
• The magnitude is determined by the kinetic, rotational and vibrational
movement
• It is denoted by the letter U or E.
• Impossible to determine absolute value. It is concerned with change
in internal energy. (ΔE = E2 – E1)
• Internal energy is a state function.
• It is a combination of heat content and work of the system
Final State, UF
Initial State, Ui
I II III IV
15
15. Consider the reaction between H2 and O2 to produce water
1) H2(g) + O2(g) H2O(l)
ΔE < 0, since Efinal < Einitial
Thus during the process, system loses energy.
2) H2O(l) H2(g) + O2(g)
ΔE > 0, since Efinal > Einitial
Thus during the process system gains energy
16
16. Changes in Internal Energy
ΔE < 0 ΔE > 0
H2(g) + O2(g) H2O(l) H2O(l) H2(g) + O2(g)
ΔE = - 483.6 KJ ΔE = + 483.6 KJ
Exergonic Process Endergonic Process
17
The endergonic and exergonic refer to changes in free energy (usually the Gibbs Free Energy)
ΔRG; while endothermic and exothermic refer to transfer of heat q or changes in enthalpy ΔRH.
17. First Law of Thermodynamics
First Law of thermodynamics is also known as the law of conservation
of energy.
Postulates of First Law:
1. The energy of an isolated system remains constant.
2. Energy can neither be created nor destroyed, although it can be
changed from one form to equivalent amount of another form.
3. Total energy (system + surroundings) is constant.
4. Energy is transferred back and forth between system and
surrounding in the form of work and heat.
18
18. Mathematical form of First Law
Suppose a system absorbs dq amount of heat, then its internal energy
increases by this amount.
If dw is the work done on the system, then internal energy also increases
by this amount. Hence, the net change in internal energy ΔU or ΔE,
ΔE = dq + dw
ΔE can have either negative or positive value, depending on q and w
values
19
19. Convention of signs regarding heat and
Work
The following sign conventions are accepted in thermodynamics:
1. Heat gained by the system is given positive sign (+q); while heat lost
by system is given by negative sign (-q)
2. The work done (w) by the system on the surroundings is given
negative sign (-w) and work done on the system is given positive sign
(+w)
3. As a system containing a definite mass at a definite temperature
possesses definite internal energy, E. Actual magnitude of internal
energy is not known, but the change in internal energy can be
measured and it is indicated by ΔE
20
20. System
ΔE > 0
System
ΔE < 0
Surroundings Surroundings
Heat, q> 0
Work, w > 0
Heat, q< 0
Work, w < 0
For q, + means system gains heat - means system loses heat
For w, + means work done on system - means work done by system
For ΔE, + means net energy gain by - means net energy lost by
system system
21
21. Implications of First Law
1. The energy of system will remain constant, if it is left undisturbed.
2. If the system interacts with the surroundings, then energy may
change but there will be equal and opposite change in the energy of
surroundings.
3. Work is a form of energy, so it is not possible for a machine to keep
on doing work indefinitely.
22
22. Problem 1
Calculate the change in the internal energy (ΔE) for a process in which
the system liberates 750 J of heat to the surroundings, while the
surroundings do 480 J of work on the system.
ΔE = Δq + Δw
Δq = -750 J Δw = + 480 J
ΔE = - 750 + 480
= - 270 J
23
23. Problem 2
Calculate the change in the internal energy (ΔE) for a process in which
the system absorbs 370 J of heat from the surroundings, and does
120 J of work on the surroundings.
ΔE = Δq + Δw
Δq = + 370 J Δw = - 120 J
ΔE = + 370 - 120
= 250 J
24
24. Chemical Change and Work
Many chemical reactions involve work done on
or by the system.
The work is either electrical or mechanical
energy.
Mechanical work is done in reactions involving
gaseous reagents.
Ex. In the reaction between Cu metal and
HNO3, NO2 gas is evolved.
Cu(s) + HNO3 (aq) Cu2+ + NO2(g) + H2O
25
25. Work at constant pressure
At constant pressure, the mechanical work done by the system involving
gaseous products is called pressure-volume work or P-V work.
W = - PΔV
where, P is the pressure of the system and ΔV is the change in volume
of the system. – ve sign indicates work done by the system.
At constant pressure, the mechanical work done on the system by
Compression of gas is also a P-V work.
W = P ΔV
Where, P is the pressure of the system, ΔV is the change in volume of
The system. +ve sign indicates work done on the system.
26
26. Work done at constant volume
In pressure-volume type of work, the work done is directly proportional
to the change in volume (ΔV) of the system.
Thus if there is no change in the volume of system:
(ΔV = 0), the w = 0, no work done.
In such cases the change in internal energy is equal to the change in
heat energy of the system.
ΔE = Δq + Δw
ΔE = Δq + 0
27
28. Enthalpy, H
Enthalpy is often referred to as “heat content”
• Again, we can only measure the change in enthalpy, ΔH
• Enthalpy equals the sum of the internal energy of the product
at the pressure and volume of the system
Thus, H = E + PV
• At constant pressure, the change in enthalpy is given as:
ΔH = ΔE + PΔV
29
29. Enthalpy (H): method for measuring energy changes during
chemical reactions
At constant pressure, the change in enthalpy of a system is the amount
of energy released or absorbed
Exothermic reactions
• release heat to their surroundings
• negative enthalpy change, ΔH < 0
Endothermic reactions
• take in heat from their surroundings
• positive enthalpy change, ΔH > 0
Heat given out
reactants
products
reactants
products
Heat taken in
30
Enthalpy
30. Exothermic Reactions
• When energy of products are less than the reactants, heat is liberated
(to surroundings) during the process.
• Such reactions are exothermic reactions.
• Example is combustion of fuels and hydrocarbons, sodium reacting
with water.
31
31. Endothermic Reactions
• When energy of products are more than the reactants, heat is
absorbed (from surroundings) during the process.
• Such reactions are endothermic reactions.
• Example is melting of ice, reaction of barium hydroxide and
ammonium nitrate
32
32. Sign of ΔH
When H is positive (i.e. qp is positive) the
system has gained heat from the surroundings
• Such a process is endothermic (ΔH > 0), ex.
melting of ice
• When H is negative (i.e. qp is negative) the
system has lost heat to the surroundings
• Such a process is exothermic (ΔH < 0), ex.
combustion of hydrocarbons
System
Heat
ΔH > 0, endothermic
System
Heat
ΔH < 0, exothermic
33
33. ΔH of a reaction
The enthalpy change that accompanies a reaction is called the
enthalpy of reaction or heat of reaction (Hrxn)
• Thus, for a reaction, Hrxn = Hproducts – Hreactants
• Consider the reaction involving production of water:
2H2(g) + O2(g) 2H2O(g) ΔH = – 483.6 kJ
• The reaction is exothermic since H is negative
• According to the equation 483.6 kJ of energy is released to the
surroundings when water is formed
34
36. ΔH- An extensive property
Enthalpy change is an extensive property
• Therefore, the magnitude of enthalpy is directly proportional to the
amounts of reactants consumed
• Ex. In the combustion of methane, if one mole of CH4 is burned in
oxygen to produce CO2 and water, 890 kJ of heat is released to the
surroundings
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = –890 kJ
• If two moles of CH4 are burned, then 1780 kJ of heat is released
2CH4(g) + 4O2(g) 2 CO2(g) + 4H2O(l) ΔH = –1780 kJ
37
37. Problem
Calculate the amount of heat released when 5.00 g of H2O2 is
decomposed (at constant pressure) in accordance with the
following reaction:
2H2O2(l) 2H2O(l) + O2(g) ΔH = − 196 kJ
Molar mass of H2O2 = (2 x 1) + (2 x 16) g = 2 + 32 g = 34 g
2 moles or 2 x 34 = 68 g gives out -196 KJ of heat
5 g gives out - 196 x 5 = -14.41 KJ of heat
68
38
38. By using a bomb calorimeter in the lab, we can determine the reaction
enthalpy
The equation to calculate the heat change is
q = m s ΔT
where q is the heat change
m is the mass of the sample
s is the specific heat of the sample
ΔT is the temperature change during reaction
The specific heat of a substance is the amount of heat required to raise
the temperature of one gram of that substance by one degree Celsius
To electrical
source
Thermometer
Mechanical stirrer
Material
combusted
in oxygen
39
Specific heat water =
S=q/mΔT
39. Example
A sample of 350g of water is heated from 10.5°C to 15.0°C. The specific
heat of water is 4.184 J g-1 °C-1. Calculate the heat change.
q = m s ΔT
m = 350g s = 4.184 J g-1 °C-1 ΔT = (15.0 – 10.5)°C =
4.5°C
q = (350 g)(4.184 J g-1 °C-1)(4.5 °C)
= 6589.8 J
= 6.59 kJ
Question
1)A 560g sample of mercury is heated from 40°C to 78°C. The
specific heat of mercury is 0.139 J g-1 °C-1. What is the heat change
for the reaction? Answer: 2.96 kJ
2)A 782g sample of water is cooled from 25°C to 1°C. The specific
heat of water is 4.184 J g-1 °C-1. What is the heat change for the
reaction? Answer: -78.5 kJ40
40. Standard enthalpy of reaction
The total enthalpy of a system cannot be measured
directly, the enthalpy change of a system is measured
instead. Enthalpy change is defined by the following
equation:
41
ΔH = Hf – Hi
ΔH is the "enthalpy change",
Hf is the final enthalpy of the system (in a chemical reaction,
the enthalpy of the products),
Hi is the initial enthalpy of the system (in a chemical reaction,
the enthalpy of the reactants).
41. Chemical properties:
• Enthalpy of reaction, defined as the enthalpy change observed in a constituent of
a thermodynamic system when one mole of substance reacts completely.
• Enthalpy of formation, defined as the enthalpy change observed in a constituent
of a thermodynamic system when one mole of a compound is formed from its
elementary antecedents.
• Enthalpy of combustion, defined as the enthalpy change observed in a constituent
of a thermodynamic system when one mole of a substance burns completely with
oxygen.
• Enthalpy of hydrogenation, defined as the enthalpy change observed in a
constituent of a thermodynamic system when one mole of an unsaturated
compound reacts completely with an excess of hydrogen to form a saturated
compound.
• Enthalpy of atomization, defined as the enthalpy change required to atomize one
mole of compound completely.
• Enthalpy of neutralization, defined as the enthalpy change observed in a
constituent of a thermodynamic system when one mole of water is formed when
an acid and a base react.
• Enthalpy of hydration, defined as the enthalpy change observed when one mole of
gaseous ions are completely dissolved in water forming one mole of aqueous ions.
42
42. Physical Properties
• Enthalpy of fusion, defined as the enthalpy change required to
completely change the state of one mole of substance between
solid and liquid states.
• Enthalpy of vaporization, defined as the enthalpy change required
to completely change the state of one mole of substance between
liquid and gaseous states.
• Enthalpy of sublimation, defined as the enthalpy change required
to completely change the state of one mole of substance between
solid and gaseous states.
• Lattice enthalpy, defined as the energy required to separate one
mole of an ionic compound into separated gaseous ions to an
infinite distance apart (meaning no force of attraction).
• Enthalpy of mixing, defined as the enthalpy change upon mixing
of two (non-reacting) chemical substances.
43
43. Enthalpy of reaction (ΔHrxn)
44
Using standard thermodynamic values, calculate the change in
enthalpy of reaction (ΔHrxn) in the formation of liquid water from
hydrogen and oxygen gas.
Solution
Chemical Equation:
H2(g)+½O2(g)→H2O(l)+heat
Product:
ΔHf H2O(l) = -285.83 kJ/mol
Reactants:
ΔHf H2(g) = 0 kJ/mol (the ΔHf of elements in their standard state is
defined to be 0 kJ)
ΔHf O2(g) = 0 kJ/mol x 2
Use ΔH of formation (Hf) for each of the chemicals involved in the
reaction found in a standard table or reference book.
[(ΔHf H2O = -285.83 kJ/mol)] - [(½)*(ΔHf O2 = 0 kJ/mol) + (ΔHf H2= 0
kJ/mol)]
ΔHrxn = SUM [(-285.83 kJ) – ((½)*0 kJ + 0 kJ)] = -285.83 kJ/mol
44. Enthalpy of the reaction
45
Using standard thermodynamic values, calculate the enthalpy of the reaction of
the combustion of methane gas with oxygen gas to form carbon dioxide and liquid
water.
Solution
Chemical Equation:
CH4(g) + 2 O2(g) => CO2(g) + 2 H2O(l) + heat
Products:
ΔHf H2O(l) = -285.83 kJ/mol x 2
ΔHf CO2(g) = -393.51 kJ/mol
Reactants:
ΔHf CH4(g) = -74.87 kJ/mol
ΔHf O2(g) = 0 kJ/mol x 2
Use values found in a standard table or reference book
[2*(ΔHf H2O(l) = -285.83 kJ/mol) + ΔHfCO2(g) = -393.51 kJ/mol]
- [2*(ΔHf O2 = 0 kJ/mol) + (ΔHf CH4= -74.87 kJ/mol)] =
ΔHrxn = [2*(-285.83 kJ) + (-393.51 kJ)] – [(2*0 kJ) + (-74.87 kJ)] = -890.3 kJ/mol
45. The enthalpy change accompanying a chemical change is
independent of the route by which the chemical change occurs.
HESS'S LAW AND ENTHALPY CHANGE CALCULATIONS
According to Hess's Law, if you convert reactants A into products
B, the overall enthalpy change will be exactly the same whether
it in one step or two steps or however many steps.
46
46. Hess’s law: if a reaction is carried out in a number of steps,
H for the overall reaction is the sum of H for each individual
step.
For example:
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ
2H2O(g) 2H2O(l) H= - 88 kJ
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890 kJ
Hess’s Law
47
47. The overall reaction enthalpy is equal to the sum of the individual
enthalpies for the reactions which make it up
Example
Given that for the combustion of glucose
i)C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) ΔH = -2816 kJ
and for the combustion of ethanol,
ii)C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g) ΔH = -1372 kJ
Calculate ΔH (in kJ) for the fermentation of glucose:
C6H12O6(s) 2C2H5OH(l) + 2CO2(g) ΔH = ?
C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) ΔH = -2816 kJ
4CO2(g) + 6H2O(g) 2C2H5OH(l) + 6O2(g) ΔH = +2744 kJ
C6H12O6(s) 2C2H5OH(l) + 2CO2(g) ΔH = -72 kJ
Exothermic reaction
48
48. Question
1. Given the following information:
2C(s) + 3H2(g) C2H6(g) ΔH = -84.68 kJ mol-1
C(s) + O2(g) CO2(g) ΔH = -393.51 kJ mol-1
H2(g) + ½O2(g) H2O(l) ΔH = -285.83 kJ mol-1
Calculate the standard enthalpy of combustion of ethane:
C2H6(g) + 3½O2 2CO2(g) + 3H2O(l)
Answer: -1559.8 kJ mol-1
2. Given the following information:
S(s) + O2(g) SO2(g) ΔH = -296.1 kJ mol-1
C(s) + O2(g) CO2(g) ΔH = -393.5 kJ mol-1
CS2(l) + 3O2(g) CO2(g) + 2SO2(g) ΔH = -1072 kJ mol-1
Calculate the enthalpy of formation of carbon disulfide, CS2:
C(s) + 2S(s) CS2(l)
Answer: +86.3 kJ mol-1
49
49. Entropy
• Entropy is a quantitative measure of the number of
microstates available to the molecules in a system.
• It is a measure of the number of ways in which energy or
molecules can be arranged.
• Entropy is the degree of randomness or disorder in a system
• The Entropy of all substances is positive
Ssolid < S liquid < Sgas
• ΔSsys is the Entropy Change of the system
• ΔSsur is the Entropy Change of the surroundings
• ΔSuni is the Entropy Change of the universe
• S has the units J K-1mol-1
50
50. • The sign of ΔSsur
depends on the
direction of the heat
flow.
• The magnitude of ΔSsur
depends on the
temperature
surr
H
S
T
If the reaction is exothermic, ΔH has a
negative sign and ΔSsurr is positive
If the reaction is endothermic, ΔH has a
positive sign and ΔSsurr is negative
Suniverse = Ssystem + Ssurroundings
This is ΔH of the system.
51
51. Entropy
• May be increased by increasing number of ways of arranging
components. Explained by Boltzmann equation:
S = k lnW
where S = entropy of system
k = Boltzmann constant=1.38 × 10−23 J/K.
W = number of possible arrangements
• Second Law of Thermodynamics: spontaneous processes (those
which occur naturally without any external influence) are
accompanied by an increase in entropy of the universe
• Absolute entropies may be determined from Third Law of
Thermodynamics: At zero degrees Kelvin, the entropy of a perfect
crystal is zero
• Because this starting point exists, can measure standard molar
entropies: entropy change for 1 mol of a pure substance at 1 atm
pressure (usually 25°C) 52
52. Standard molar Entropies
54
• Standard molar entropies for a reference temperature (like 298 K)
and 1 atm pressure (i.e. the entropy of a pure substance at 298 K
and 1 atm pressure). A table of standard molar entropies at 00 K
would be pretty useless because it would be 0 for every substance.
• When comparing standard molar entropies for a substance that is
either a solid, liquid or gas at 298 K and 1 atm pressure, the gas will
have more entropy than the liquid, and the liquid will have more
entropy than the solid.
• The entropy change in a chemical reaction is given by the sum of the
entropies of the products minus the sum of the entropies of the
reactants.
• Calculations related to balanced equations, the coefficients of each
component must be taken into account in the entropy calculation
(the n, and m, terms below are there to indicate that the
coefficients must be accounted for):
ΔS0=∑nS0(products)−∑mS0(reactants)
53. Predicting Entropy Changes
• An increase in temperature leads to greater kinetic energy of moving
particles, more motion and hence greater S°
• Going from solid to liquid to gas (i.e. to less ordered systems) leads to an
increase in S°
• For spontaneous change, ΔS must be greater than zero. For negative ΔS
values, the process is spontaneous in the reverse direction
Example
Predict whether the entropy change for the following reaction will be
positive or negative:
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
6 gas molecules 3 gas molecules + 4 liquid molecules
A decrease of the more disordered gas system indicates the entropy change
for the reaction should be negative
55
54. Given the following information, calculate ΔS° for the reaction
= [(3 mol)(213.7 J/molK) + (4 mol)(69.9 J/molK)] –
[(1 mol)(269.9 J/molK) + (5 mol)(205.0 J/molK)]
S° (J / mol K)
CO2 213.7
H2O 69.9
C3H8 269.9
O2 205.0
ΔSrxn = ΣSproducts - ΣSreactants
= [(3 mol CO2)(S° of CO2) + (4 mol H2O)(S° of H2O)] –
[(1 mol C3H8)(S° of C3H8) + (5 mol O2)(S° of O2)]
= -374.2 J/K
= 920.7 J/K – 1294.9 J/K
= 641.1 J/K + 279.6 J/K – (269.9 J/K + 1025 J/K)
56
55. 57
Calculate the change in entropy associated with the Haber process for
the production of ammonia from nitrogen and hydrogen gas.
N2(g)+3H2(g)⇌2NH3(g)
At 298K as a standard temperature:
•S0(NH3) = 192.5 J/mol K
•S0(H2) = 130.6 J/mol K
•S0(N2) = 191.5 J/mol K
Solution
ΔS0 = 2*S0(NH3) - [S0(N2) + (3*S0(H2))]
ΔS0 = 2*192.5 - [191.5 + (3*130.6)]
ΔS0 = -198.3 J/mol K
Problems
It would appear that the process results in a decrease in entropy - i.e. a
decrease in disorder. This is expected because we are decreasing the
number of gas molecules. In other words the N2(g) used to float around
independently of the H2 gas molecules. After the reaction, the two are
bonded together and can't float around freely from one another.
ΔSrxn = ΣSproducts - ΣSreactants
56. Question
Predict the sign of ΔS° and calculate its value from the following:
2NO(g) + O2(g) 2NO2(g)
S° (J / mol K)
NO 210.65
O2 205.0
NO2 239.9
Answer: should be negative
ΔS° = -146.5 J/K
58
57. 59
If 1.6g of CH4 reacts with oxygen gas to form water and carbon dioxide. what is the change in
entropy for the universe?
Solution
Reaction Equation:
CH4+2O2→CO2+2H2O
To solve this problem the following equations are also necessary:
ΔS System =ΣΔSProducts– ΣΔSReactants
ΔS System =[(.21374 kJ/mol)+(2* .06995 kJ/mol)]-[(2*.20507 kJ/mol)+( .18626 kJ/mol)] = -.24276
kJ/mol
ΔH System =ΣΔHProducts– ΣΔHReactants
ΔH System = [( -393.509 kJ/mol)+(2* -285.83 kJ/mol)]-[(2*0)+( -74.87 kJ/mol)] = -890.229 kJ/mol
ΔS Surroundings =-ΔHSystem /T
ΔS Surroundings = -890.229/298 = 2.9873 kJ/mol
ΔS Universe= ΔS Surroundings +ΔS System
ΔS Universe= 2.9873 kJ/mol + (-.24276 kJ/mol) = 2.745 kJ/mol
Entropy Numerical
59. Spontaneity and Gibbs Free Energy
• Gibbs Free energy is a measure of the spontaneity of a process
• ΔG is the free energy change for a reaction under standard state conditions
• At constant temperature and pressure:
ΔG = ΔH – TΔS
– an increase in ΔS leads to a decrease in ΔG
– if ΔG < 0, the forward reaction is spontaneous
– if ΔG > 0, the forward reaction is nonspontaneous
– if ΔG = 0, the process is in equilibrium
– if ΔH and ΔS are positive, then ΔG will be negative only at high T, i.e.
forward reaction spontaneous
– if ΔH is positive and ΔS is negative, ΔG will always be positive,
i.e. forward reaction nonspontaneous
– if ΔH is negative and ΔS is positive, ΔG will always be negative,
i.e. forward reaction spontaneous
– if ΔH and ΔS are negative, ΔG is negative only at low T, i.e. forward
reaction spontaneous
61
60. Example
At 27°C, a reaction has ΔH = +10 kJ mol-1 and ΔS = +30 J K-1 mol-1.
What is the value of ΔG?
ΔG = ΔH – TΔS T = 300 K
ΔG = (+10 kJ mol-1) – (300 K)(+30 J K-1 mol-1)
ΔG = +10 kJ mol-1 – 9000 J mol-1
ΔG = +10 kJ mol-1 – 9 kJ mol-1
ΔG = + 1.0 kJ mol-1
Question
For the reaction
4KClO3(s) 3KClO4(s) + KCl(s)
Calculate ΔG for the process at 298 K if ΔH = -144.3 kJ and ΔS = -36.8
J K-1 mol-1
Answer: -133.3 kJ 62
61. Examples
Predict if the reaction H2(g)+½O2(g)→H2O(l) is spontaneous at 25°C
using standard thermodynamic values.
Given: ΔHf H2O(l) = -285.83 kJ/mol. Elements in their standard state
have ΔHf as 0 kJ
ΔS0 (kJ/mol K) for H2, O2 and H2O are 130.6, 205.0, 188.7 respectively.
• Ans: ΔH = H products – H reactants = -285.83 kJ/mol - 1M
• ΔS = S products – S reactants = 188.7- (130.6 +205/2)
• = -44.4 kJ/ mol - 1M
• ΔG = ΔH - TΔS
• = -285.83 – (298 x -44.4) = -285.83 + 13231.2 = 12945.37 kJ - 1M
• ΔG is positive, so reaction is non-spontaneous
63
62. Find the change is Gibbs Free Energy for the reaction of hydrochloric
acid and sodium hydroxide to form liquid water and sodium chloride
at 310C.
Solution: Calculate ΔH and ΔS for the reaction.
HCl(aq) + NaOH(aq) H2O(l) + NaCl
( values for ΔH and ΔS taken from the table)
• ΔHRxn =ΣΔHProducts – ΣΔHReactants
ΔHRxn =[(-285.8+(-411.54)) kJ/mol]-[(-167.16 + (-470.1) kJ/mol] = -60.05 kJ/mol
• ΔSRxn =ΣΔSProducts – ΣΔSReactants
ΔSRxn=[(0.06991+ 0.07238) kJ/molK]-[0.0565 + 0.0482 kJ/molK] = 0.03759 kJ/molK
• convert temperature to Kelvins by 273.15 to its value in Celsius.
31 + 273.15 = 304.15 K
• Finally, all of these values are plugged into the original equation, ΔG=ΔH-TΔS.
ΔG= -60.05 KJ/mol – 304.15 K *(0.03759 KJ/molK) = -71.4830
64
63. Galvanic (Voltaic) and Electrolytic
Cells
• Cell reaction – Redox reaction involved in
electrochemical cell.
• Voltaic (galvanic) cell – reaction is
spontaneous and generates electrical
current.
• Electrolytic – non-spontaneous reaction
occurs due to passage of current from
external power source. E.g. charging of
batteries.
65
64. • An electrochemical cell is composed to two
compartments or half-cells, each composed of
an electrode dipped in a solution of
electrolyte. These half-cells are designed to
contain the oxidation half-reaction and
reduction half-reaction separately as shown
below.
What is electrochemical cell
67
65. English chemist John Daniell (left) and
physicist Michael Faraday (right), both credited
as founders of electrochemistry today.
Electrochemistry is defined as the branch of chemistry
that examines the phenomena resulting from combined
chemical and electrical effects.
Electrochemistry
68
66. • An electrochemical cell is a device that generates electricity from a
chemical reaction. Essentially, it can be defined as a device that
converts chemical energy into electrical energy. A chemical reaction
that involves the exchange of electrons is required for an
electrochemical cell to operate. Such reactions are called redox
reactions.
• A cell is characterized by its voltage. A particular kind of cell
generates the same voltage irrespective of the size of the cell. The
only thing that depends on the cell voltage is the chemical
composition of the cell, given the cell is operated at ideal
conditions.
• Normally, the cell voltage may be different from this ideal value,
due to several factors like temperature difference, change in
concentration, etc. Nernst equation formulated by Walther Nernst
can be used to calculate the EMF value of a given cell, provided the
standard cell potential of the cell.
What is an Electrochemical Cell?
69
67. Type of Electrochemical Cells
• Voltaic (galvanic) cell – reaction is
spontaneous and generates electrical
current.
• Electrolytic – non-spontaneous reaction
occurs due to passage of current from
external power source. E.g. charging of
batteries.
70
68. • A cell is characterized by its voltage. The cell
voltage depends on the 1)chemical composition
of the cell 2) given the cell is operated at ideal
conditions.
• An electrochemical cell is composed to two
compartments or half-cells, each composed of an
electrode dipped in a solution of electrolyte.
These half-cells are designed to contain the
oxidation half-reaction and reduction half-
reaction separately.
Working of Electrochemical cell
71
70. • The half-cell, called the anode, is the site at
which the oxidation of zinc occurs as shown
below.
• Zn (s) ----------> Zn+2 (aq) + 2e-
• During the oxidation of zinc, the zinc electrode
will slowly dissolve to produce zinc ions (Zn+2),
which enter into the solution containing Zn+2
(aq) and SO4-2 (aq) ions.
73
71. • The half-cell, called the cathode, is the site at
which reduction of copper occurs as shown
below.
• Cu+2 (aq) + 2e- -------> Cu (s)
• When the reduction of copper ions (Cu+2)
occurs, copper atoms accumulate on the
surface of the solid copper electrode.
74
72. Salt Bridge
• The reaction in each half-cell does not occur unless
the two half cells are connected to each other.
• It is an inverted U-tube containing an electrolyte e.g
KCL, KNO3 etc it act as bridge by connecting two half
cells
-----------Helps in
• To completing the electric circuit
• To prevent mixing of soln of two half cell.
• To help maintain electric neutrality
75
74. Cell notation
• Anode on left, cathode on right
• Electrons flow from left to right
• Oxidation on left, reduction on right
• Single vertical = electrode/electrolyte boundary
• Double vertical = salt bridge
Anode:
Zn →Zn2+ +
2e
Cathode:
Cu2+ + 2e
→Cu
77
75. Cell potentials, Ecell
• The cell potentials, Ecell is the measure of the
potential difference between two half cells in an
electrochemical cell.
• The potential difference is caused by the ability of
electrons to flow from one half cell to the other.
• It is measured by Ecell=Ecathode− Eanode
• Temperature, surface area, and concentration are
the main factors influencing chemical reactions
hence cell potential.
78
76. Standard Cell potentials, E°cell
• The Standard cell potentials, E°cell is the measure of the
potential difference between two half cells in an
electrochemical cell under standard condition.
• It is measured by E°cell=E°cathode− E°anode
E°cell=standard cell potential (under 1M, 1 Barr and 298 K).
E°cathode=is the standard reduction potential for the
reduction half reaction occurring at the cathode
E°anode=is the standard reduction potential for the
oxidation half reaction occurring at the anode
79
77. Electromotive force (emf)
• The electromotive force (EMF) is the maximum potential
difference between two electrodes of a galvanic or
voltaic cell.
• This is related to the tendency for an element, a
compound or an ion to acquire (i.e. gain) or release (lose)
electrons.
• For example, the maximum potential between Zn and
Cu of a well known cell
Zn(s)|Zn2+(1M)||Cu2+(1M)|Cu(s)
has been measured to be 1.10 V. A concentration of 1 M in
an ideal solution is defined as the standard condition, and
1.10 V is thus the standard electromotive force, Eo, or
standard cell potential for the Zn−Cu galvanic cell.
80
78. Cell
Cu
aq
u
C
aq
Zn
Zn )
(
)
( 2
2
V
E
e
aq
Zn
s
Zn o
76
.
0
,
2
)
(
)
( 2
V
E
s
Cu
e
aq
Cu o
34
.
0
),
(
2
)
(
2
zinc/copper cell
zinc/copper cell has an emf of about 1.1 volts
under standard conditions 81
Anode
Cathode
80. Applying standard reduction potentials
• Consider the reaction
• The half reactions are:
What is the cell potential?
• E° = 0.80 V – (-0.76 V) = 1.56 V
• NOTE: Although there are 2 moles of Ag reduced for each mole of Zn
oxidized, we do not multiply the potential by 2.
)
(
2
)
(
)
(
2
)
( 2
s
Ag
aq
Zn
aq
Ag
s
Zn
e
aq
Zn
s
Zn 2
)
(
)
( 2
)
(
)
( s
Ag
e
aq
Ag
83
E°cell=E°cathode− E°anode
E° = -0.76 V
E° = 0.80 V
81. Electrolytic Cell
• It is a device to convert Electric Energy into
Chemical Energy
• An electrolytic cell is an electrochemical cell in
which the energy from an applied voltage is
used to drive an otherwise nonspontaneous
reaction. Such a cell could be produced by
applying a reverse voltage to a voltaic cell like
the Daniell cell.
84
83. In galvanic cells the reactions are separated so electrons
can be used as they are transferred.
Electrolytic cells have two electrodes and an electrolyte.
Electrolytic cells differ from galvanic cells in that the
anode is the positive electrode while the cathode is the
negative electrode.
The polarities of the anode and cathode change because
in galvanic cells the polarity is imposed by the reactions
themselves.
Galvanic & Electrolytic Cells - A Comparison
86
85. Difference b/w Voltaic cell and
Eletrolytic Cell
Voltaic cell
• Chemical energy
converted into Electric
energy
• It is based on redox rxn
which is spontaneous
Eletrolytic Cell
• Electric energy
converted into Chemical
energy
• It is based on redox rxn
which is non-
spontaneous
88
86. Electrode potential
Working and counter electrodes
The electrode at which the reaction of interest occurs is called
the working electrode.
The electrode at which the other reaction occurs is called the
counter electrode.
A third electrode, called the reference electrode may also be
used.
For a given set of two reversible redox reactions,
Thermodynamics predicts which reaction proceeds as an
oxidation and which proceeds as a reduction.
Electrode potential
The electrode potential for a reaction is derived directly from the
free energy change for that reaction
ΔG = -nFE
89
87. The standard oxidation potential is equal in magnitude, but opposite in sign
to the standard reduction potential. The reaction with the lower standard
reduction potential gets oxidized -the other reaction proceeds as a
reduction.
Ex.:-
Zn - 2e = Zn2++ 2e --------- (1) E° red= -0.76 V
Cu2+++ 2e = Cu------------- (2) E° red= 0.34 V
Thus, in the above example, Zn is oxidized, and Cu is reduced.
E cell= E cathode–E anode
For a feasible reaction: E cell must be positive and ΔG cell is negative)
Therefore:
E cathode – E anode > 0 or
E cathode > E anode
Since oxidation occurs at the anode the species with the lower reduction
potential will get oxidized. This is to ensure that ΔG cell is negative. This is
why Zn got oxidized (and Cu reduced) in the above example.
In this case: E cell= 1.102., If the reverse were to occur, E cell would be: -
1.102, leading to a positive ΔG cell.
90
88. 91
The product of an electrolytic reaction depends on the
nature of the material being electrolysed and the type of
electrodes used.
In the case of an inert electrode such as platinum or gold,
the electrode does not participate in the chemical reaction
and acts only as a source or sink for electrons. While, in the
case of a reactive electrode, the electrode participates in
the reaction.
Hence, different products are obtained for electrolysis in
the case of reactive and inert electrodes.
Oxidizing and reducing species present in the electrolytic
cell and their standard electrode potential too, affect the
products of electrolysis.
89. Measuring Standard Electrode Potential
Potentials are measured against a hydrogen ion reduction
reaction, which is arbitrarily assigned a potential of zero volts.
We cannot measure the potential of an individual half-cell!
92
Standard Hydrogen Electrode (SHE)
• This is the reference to
which all other oxidations
are compared to
• This indicates standard
states of 25°C, 1
atmosphere pressure,
1M solutions.
Eo
H
+
/H2
half-cell = 0.000 V
90. Zn+2 SO4
-2
Anode
H+ Cl-
H2 in
Cathode
1 M HCl
1 M ZnSO4
0.76V
93
Thus, standard oxidation potential of Zn is 0.76V
Conversely, the standard reduction potential of Zn is ─0.76V
(Eo
Zn/Zn
2+ = 0.76 V)
(Eo
Zn
2+
/ Zn = ─ 0.76 V)
The Zinc-SHE Cell
91. The Zinc-SHE Cell
• For this cell the components are:
1. A Zn strip immersed in 1.0 M zinc (II) sulfate.
2. The other electrode is the Standard Hydrogen Electrode.
3. A wire and a salt bridge to complete the circuit.
• The initial cell voltage is 0.763 volts.
94
92. Relating free energy and cell potential
• The cell potential of an electrochemical cell is the potential difference
occurring between the two electrodes of the cell, and arises due to the
transfer of electrons through the external circuit of a cell that has not reached
equilibrium. This is related to the Gibbs free energy change for the cell
reaction given by: ΔG=−nFE
• where ∆G = Gibbs free energy change (J), n = number of electrons transferred
per unit overall reaction (mol), Ecell = cell potential (V), F = Faraday constant (96
485 C mol− 1).
• At standard conditions (1 M, 1 atm, 25°C)
• Furthermore, the cell potential is related to the composition of the reaction
mixture via the Nernst Equation.
nFE
G
95
93. Nernst Equation
• The equation was introduced by a German chemist named
Walther Hermann Nernst.
• The Nernst equation is often used to calculate the cell
potential of an electrochemical cell at any given temperature,
pressure, and reactant concentration.
• The Nernst equation provides a relation between the cell
potential of an electrochemical cell, the standard cell potential,
temperature, and the reaction quotient.
• Even under non-standard conditions, the cell potentials of
electrochemical cells can be determined with the help of the
Nernst equation.
• The Nernst Equation is derived from the emf and the Gibbs
energy under non-standard conditions. ΔG=−nFE.
• The equation above indicates that the electrical potential of a
cell depends upon the reaction quotient Q of the reaction
96
94. Nernst equation
• The general Nernst equation correlates the Gibb's Free Energy G and the EMF
of a chemical system known as the galvanic cell.
Ecell = E cell - (RT/nF)lnQ
Ecell = cell potential of the cell; E0 = cell potential under standard conditions
R = universal gas constant; T = temperature
n = number of electrons transferred in the redox reaction
F = Faraday constant; Q = reaction quotient
• Change in the Gibbs free energy (∆G = – nFEred) is an indication of the
spontaneity of process.
• During the cell reaction, concentration keeps changing and the potential also
will decrease with the rate of reaction. To get the maximum work or
maximum free energy change, the concentrations have to be maintained the
same. This is possible only by carrying out the reaction under a reversible
equilibrium condition.
• For a reversible equilibrium reaction, vant Hoff isotherm says:
• G = G + RT ln Q 97
95. Nernst Equation
• This means
−nFE = −nFE + RT ln Q
• Dividing both sides by −nF, we get the Nernst
equation:
or, using base-10 logarithms,
E = E −
RT
nF
ln Q
E = E −
2.303 RT
nF
ln Q
98
96. Nernst Equation
At room temperature (298 K), and
R = 8.314 J/mol K
F = 96,485 J/V-mol
The final form of the Nernst Equation becomes
E = E − 0.0592
n
ln Q
2.303 RT
F
= 0.0592 V
99
97. When the reactants and the products of the electrochemical
cell reach equilibrium, the value of ΔG becomes 0.
At this point, the reaction quotient and the equilibrium
constant (Kc) are the same.
Since ΔG = -nFE, the cell potential at equilibrium is also 0.
Substituting the values of Q and E into the Nernst equation,
the following equation is obtained.
0 = E0
cell – (RT/nF) ln Kc
The relationship between the Nernst equation, the
equilibrium constant, and Gibbs energy change is illustrated
below.
Relationships between ΔG°, K, and E°
cell
100
98. Effect of Concentration on Cell EMF:
The Nernst Equation
The Nernst Equation
E0
cell = (0.0592V/n) log Kc
Rearranging this equation, the following equation can be
obtained.
log Kc = (nE0
cell)/0.0592V
When Kc is greater than 1, the value of E0cell will be greater
than 0, implying that the equilibrium favours the forward
reaction.
Similarly, when Kc is less than 1, E0cell will hold a negative
value which suggests that the reverse reaction will be favoured.
101
99. Example
Calculate the electrode potential at a copper electrode dipped in a
0.1M solution of copper sulphate at 250C . The standard potential
of Cu2+/Cu system is 0.34 volt at 298 K.
Solution:
0 2
red
Putting the values of E 0.34 V,n 2 and [Cu ] 0.1 M
red 10
0.0591
E 0.34 log [0.1]
2
0.34 0.02955 ( 1)
0.31045 volt
Cu2+ + 2e- Cu
2 2
0 2
10
Cu /Cu Cu /Cu
0.0591
We know that E E log [Cu ]
n
102
Nernst equation: Ecell = E0cell – (0.0592/n) log10Q
100. The Nernst Equation
• n = number of electrons involved
• For the redox reaction:
• If [Cu2+] = 5.0 M, and [Zn2+] = 0.05 M, then at these concentrations the
emf of the cell using the Nernst equation is:
103
Zn (s) + Cu2+
(aq) Zn2+
(aq) + Cu (s) E˚cell = + 1.1 V
The electrode potential and the emf of the cell depend upon the nature of the
electrode, temperature and the activities (concentrations) of the ions in solution.
101. The Nernst Equation
(Non-standard conditions: Example)
• For the cell: Zn│Zn2+║Ag+│Ag . Calculate Ecell
• If [Ag+] = 5.2 × 10─6 M, and [Zn2+] = 1.3 × 10─3 M, calculate the emf of
the cell
• Using the Nernst equation:
104
Hint: Write the reactions at anode and cathode, and then the balanced
redox reaction, here ‘n’ = 2
Zn (s) + 2Ag+
(aq) Zn2+
(aq) + 2Ag(s)
Given: E˚cell = + 1.5 V
Ans. Ecell = 1.273 V
102. Numerical on Nernst equation
105
Question- What is the Cell Potential of the electrochemical cell in which the cell
reaction is: Pb2+ + Cd → Pb + Cd2+ ; Given that Eocell = 0.277 volts, temperature =
25oC, [Cd2+] = 0.02M, and [Pb2+] = 0.2M.
Solution:
Since the temperature is equal to 25oC, the Nernst equation can be written as follows;
Ecell = E0cell – (0.0592/n) log10Q
Here, two moles of electrons are transferred in the reaction. Therefore, n = 2. The
reaction quotient (Q) is given by [Cd2+]/[Pb2+] = (0.02M)/(0.2M) = 0.1.
The equation can now be rewritten as:
Ecell = 0.277 – (0.0592/2) × log10(0.1) = 0.277 – (0.0296)(-1) = 0.3066 Volts
Thus, the cell potential of this electrochemial cell at a temperature of 25oC is 0.3066
volts.
103. Limitations of Nernst Equation
• The activity of an ion in a very dilute solution is close to
infinity and can, therefore, be expressed in terms of the ion
concentration. However, for solutions having very high
concentrations, the ion concentration is not equal to the
ion activity. In order to use the Nernst equation in such
cases, experimental measurements must be conducted to
obtain the true activity of the ion.
• Another shortcoming of this equation is that it cannot be
used to measure cell potential when there is a current
flowing through the electrode. This is because the flow of
current affects the activity of the ions on the surface of the
electrode. Also, additional factors such as resistive loss and
over potential must be considered when there is a current
flowing through the electrode.
106
104. Nernst Equation Applications
The Nernst equation can be used to calculate:
Single electrode reduction or oxidation potential at any
conditions
Standard electrode potentials
Comparing the relative ability as a reductive or oxidative
agent.
Finding the feasibility of the combination of such single
electrodes to produce electric potential.
Emf of an electrochemical cell
Unknown ionic concentrations
The pH of solutions and solubility of sparingly soluble salts
can be measured with the help of the Nernst equation.
107
105. • The half-cell, called the anode, is the site at
which the oxidation of zinc occurs as shown
below.
• Zn (s) ----------> Zn+2 (aq) + 2e-
• During the oxidation of zinc, the zinc electrode
will slowly dissolve to produce zinc ions (Zn+2),
which enter into the solution containing Zn+2
(aq) and SO4-2 (aq) ions.
108
107. Cell notation
• Anode on left, cathode on right
• Electrons flow from left to right
• Oxidation on left, reduction on right
• Single vertical = electrode/electrolyte boundary
• Double vertical = salt bridge
Anode:
Zn →Zn2+ +
2e
Cathode:
Cu2+ + 2e
→Cu
110
108. Electromotive force (emf)
• The two half-cells are also connected externally. The
electrons provided by the oxidation reaction are
forced to travel via an external circuit to the site of
the reduction reaction. The fact that the reaction
occurs spontaneously once these half cells are
connected indicates that there is a difference in
potential energy. This difference in potential energy
is called an electromotive force (emf) and is
measured in terms of volts.
• The overall cell reaction is the sum of the two half-
reactions, but the cell potential is the difference
between the reduction potentials:
• E°cell=E°cathode− E°anode
111
109. )
(
)
(
)
(
)
( 2
2
s
Cu
aq
Zn
aq
Cu
s
Zn
Cu
aq
u
C
aq
Zn
Zn )
(
)
( 2
2
V
E
e
aq
Zn
s
Zn o
76
.
0
,
2
)
(
)
( 2
V
E
s
Cu
e
aq
Cu o
34
.
0
),
(
2
)
(
2
zinc/copper cell
zinc/copper cell has an emf of about 1.1 volts
under standard conditions 112
111. Applying standard reduction potentials
• Consider the reaction
• What is the cell potential?
• The half reactions are:
• E° = 0.80 V – (-0.76 V) = 1.56 V
• NOTE: Although there are 2 moles of Ag reduced for
each mole of Zn oxidized, we do not multiply the
potential by 2.
)
(
2
)
(
)
(
2
)
( 2
s
Ag
aq
Zn
aq
Ag
s
Zn
e
aq
Zn
s
Zn 2
)
(
)
( 2
)
(
)
( s
Ag
e
aq
Ag
114
E°cell=E°cathode− E°anode
112. Electrolytic Cell
• It is a device to convert Electric Energy into
Chemical Energy
• An electrolytic cell is an electrochemical cell in
which the energy from an applied voltage is
used to drive an otherwise nonspontaneous
reaction. Such a cell could be produced by
applying a reverse voltage to a voltaic cell like
the Daniell cell.
115
114. Difference b/w Electrochemical cell and
Eletrolytic Cell
Electrochemical cell
• Chemical energy
converted into Electric
energy
• It is based on redox rxn
which is spontaneous
Eletrolytic Cell
• Electric energy
converted into Chemical
energy
• It is based on redox rxn
which is non-
spontaneous
117
115. Relating free energy and cell potential
• The Faraday:
F = 96 485 C/mole
Standard conditions (1 M, 1 atm, 25°C)
nFE
G
nFE
G
120
116. Thus, if ΔG°, K, or E0 is known or can be calculated, the other
two quantities can be readily determined. The relationships
are shown graphically in Figure 1.
Figure 1. The relationships between ΔG°, K, and E°
cell. Given any one of the three
quantities, the other two can be calculated, so any of the quantities could be used to
determine whether a process was spontaneous.
Relationships between ΔG°, K, and E°
cell
121
117. Nernst Equation
Electrochemistry deals with cell potential as well
as energy of chemical reactions. The energy of a
chemical system drives the charges to move,
and the driving force give rise to the cell
potential of a system called galvanic cell. The
energy aspect is also related to the chemical
equilibrium. All these relationships are tied
together in the concept of Nernst equation.
122
118. • The general Nernst equation correlates the
Gibb's Free Energy G and the EMF of a
chemical system known as the galvanic cell.
• Ecell = E0
cell - (RT/nF)lnQ
• G = G + RT ln Q
123
119. Nernst Equation
• This means
−nFE = −nFE + RT ln Q
• Dividing both sides by −nF, we get the Nernst
equation:
or, using base-10 logarithms,
E = E −
RT
nF
ln Q
E = E −
2.303 RT
nF
ln Q
124
120. Nernst Equation
At room temperature (298 K), and
R = 8.314 J/mol K
F = 96,485 J/V-mol
The final form of the Nernst Equation becomes
E = E − 0.0592
n
ln Q
2.303 RT
F
= 0.0592 V
125
121. Effect of Concentration on Cell EMF:
The Nernst Equation
• Recall that G = Go + RT lnQ where
or at 25°C
which is called Nernst Equation.
• E.g. Determine potential of Daniell cell at
25C if [Zn2+] = 0.100 M and [Cu2+] =
0.00100 M.
b
a
n
m
]
B
[
]
A
[
]
N
[
]
M
[
Q
Q
ln
nF
RT
E
E o
Q
log
n
0592
.
0
E
E o
126
122. Example
Calculate the electrode potential at a copper electrode dipped in a
0.1M solution of copper sulphate at 250C . The standard potential
of Cu2+/Cu system is 0.34 volt at 298 K.
Solution:
0 2
red
Putting the values of E 0.34 V,n 2 and [Cu ] 0.1 M
red 10
0.0591
E 0.34 log [0.1]
2
0.34 0.02955 ( 1)
0.31045 volt
Cu2+ + 2e- Cu
2 2
0 2
10
Cu /Cu Cu /Cu
0.0591
We know that E E log [Cu ]
n
127
123. Calculate the EMF of the cell
Zn(s) | Zn2+ (0.024 M) || Zn2+ (2.4 M) | Zn(s)
Solution
Zn2+ (2.4 M) + 2 e = Zn
Reduction Zn = Zn2+ (0.024 M) + 2 e
Oxidation----Zn2+ (2.4 M) = Zn2+ (0.024 M), DE° = 0.00 - - Net
reaction
Using the Nernst equation:
0.0592 (0.024) DE = 0.00 - --- log -- 2 (2.4)
= (-0.296)(-2.0) = 0.0592 V
128
124. Nernst Equation Applications
The Nernst equation can be used to calculate:
Single electrode reduction or oxidation potential at any
conditions
Standard electrode potentials
Comparing the relative ability as a reductive or oxidative
agent.
Finding the feasibility of the combination of such single
electrodes to produce electric potential.
Emf of an electrochemical cell
Unknown ionic concentrations
The pH of solutions and solubility of sparingly soluble salts
can be measured with the help of the Nernst equation.
129
125. The Relationship Between Ksp And the Solubility of a Salt
• Ksp is called the solubility product because it is literally the product of the
solubility of the ions in moles per litre. The solubility product of a salt can
therefore be calculated from its solubility, or vice versa.
• The higher the solubility product constant, the more soluble the
compound. The Ksp expression for a salt is the product of the
concentrations of the ions, with each concentration raised to a power
equal to the coefficient of that ion in the balanced equation for the
solubility equilibrium
• Example: Let's calculate the solubility of AgBr in water in grams per liter,
to see whether AgBr can be removed by simply washing the film.
• We start with the balanced equation for the equilibrium:
• AgBr(s) <----> Ag+(aq)+ Br-(aq)
• We then write the solubility product expression for this reaction.
• Ksp = [Ag+][Br-] = 5.0 x 10-13
130
126. Solution
The dissociation reaction of AgCl in water is
AgCl (s) ↔ Ag+ (aq) + Cl- (aq)
For this reaction, each mole of AgCl that dissolves produces 1 mole of both
Ag+ and Cl-.
The solubility would then equal the concentration of either the Ag or Cl ions.
solubility = [Ag+] = [Cl-]
Ksp = [Ag+][Cl-]
since [Ag+] = [Cl-]
Ksp = [Ag+]2 = 1.6 x 10-10
[Ag+] = (1.6 x 10-10)½
[Ag+] = 1.26 x 10-5 M
solubility of AgCl = [Ag+]
solubility of AgCl = 1.26 x 10-5 M
131
Que: The solubility product of AgCl is 1.6 x 10-10 at 25 °C.
127. Limitations of Nernst Equation
• The activity of an ion in a very dilute solution is close to
infinity and can, therefore, be expressed in terms of the ion
concentration. However, for solutions having very high
concentrations, the ion concentration is not equal to the ion
activity. In order to use the Nernst equation in such cases,
experimental measurements must be conducted to obtain the
true activity of the ion.
• Another shortcoming of this equation is that it cannot be used
to measure cell potential when there is a current flowing
through the electrode. This is because the flow of current
affects the activity of the ions on the surface of the electrode.
Also, additional factors such as resistive loss and over
potential must be considered when there is a current flowing
through the electrode.
132
128. Problem: The solubility of barium sulphate at 298 K is 1.05 x 10-5 mol dm-3.
Calculate the solubility product.
The equilibrium is:
Notice that each mole of barium sulphate dissolves to give 1 mole of barium
ions and 1 mole of sulphate ions in solution.
That means that:
• [Ba2+] = 1.05 x 10-5 mol dm-3
• [SO42-] = 1.05 x 10-5 mol dm-3
All you need to do now is to put these values into the solubility product
expression, and do the simple sum.
133
129. Ksp for Ca3(PO4)2 is 2.07 × 10−33 at 25°C. Calculate the aqueous solubility of
Ca3(PO4)2 in terms of the following:
• A) the molarity of ions produced in solution
• B) the mass of salt that dissolves in 100 mL of water at 25°C
A)
Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO4
3−(aq)
Ksp=[Ca2+]3[PO3
−4]2 =(3x)3(2x)2
2.07×10−33 =108 x5
1.92×10−35 =x5
1.14×10−7 M= x
This is the molar solubility of calcium phosphate at 25°C. However, the molarity
of the ions is 2x and 3x, which means that [PO4
3−] = 2.28 × 10−7 and [Ca2+] = 3.42 ×
10−7
B)To find the mass of solute in 100 mL of solution, we assume that the density of this dilute solution is
the same as the density of water because of the low solubility of the salt, so that 100 mL of water
gives 100 mL of solution. We can then determine the amount of salt that dissolves in 100 mL of water:
(1.14×10−7 mol/1 L)100 mL(1 L/1000 mL)(310.18 g Ca3(PO4)2/1 mol)
=3.54×10−6 g Ca3(PO4)2
134
