CHEMICAL EQUILIBRIUM Chapter 7. Chemical equilibrium Chemical reactions tend to move towards a...

CHEMICAL CHEMICAL EQUILIBRIUMEQUILIBRIUM

Chapter 7Chapter 7

Chemical equilibriumChemical equilibrium Chemical reactions tend to move Chemical reactions tend to move

towards a dynamic equilibrium in towards a dynamic equilibrium in which both reactants and products which both reactants and products are present but have no further are present but have no further tendency to undergo net change.tendency to undergo net change.

Spontaneous Spontaneous chemical reactionschemical reactions

The direction of spontaneous change at constant The direction of spontaneous change at constant temperature and pressure is towards lower values temperature and pressure is towards lower values of Gibbs energy. of Gibbs energy.

This idea also applies to chemical reactions.This idea also applies to chemical reactions. If we can calculate the minimum value of the If we can calculate the minimum value of the

Gibbs energy for a particular reaction mixture, Gibbs energy for a particular reaction mixture, this corresponds to the location of the equilibrium this corresponds to the location of the equilibrium composition.composition.


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A ↔ BIf dξ is an infinitesimal amount of A that turns to BdnA = −dξdnB = +dξ

The quantity The quantity (xi) is called the (xi) is called the extent of the reaction and has units extent of the reaction and has units of moles.of moles.


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nA ,0 → nA ,0 − ΔξnB ,0 → nB ,0 + Δξ

The reaction Gibbs energy The reaction Gibbs energy rrG is G is defined as the slope of the Gibbs defined as the slope of the Gibbs energy plotted against the extent of energy plotted against the extent of reactionreaction

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rG = ∂G∂ξ ⎛ ⎝ ⎜

⎞ ⎠ ⎟p,T


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rG = ∂G∂ξ ⎛ ⎝ ⎜

⎞ ⎠ ⎟p,T

dG = μAdnA + μBdnB = −μAdξ + μBdξdG = (μB −μA )dξ

∂G∂ξ ⎛ ⎝ ⎜

⎞ ⎠ ⎟p,T

= μB −μA = ΔrG


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rG = 0μB = μA

Exergonic and Exergonic and endergonic reactionsendergonic reactions

We can express the spontaneity of a reaction at We can express the spontaneity of a reaction at constant temperature and pressure in terms of constant temperature and pressure in terms of the reaction Gibbs energy.the reaction Gibbs energy.

If If rrG < 0, the forward reaction is spontaneousG < 0, the forward reaction is spontaneous If If rrG > 0, the reverse reaction is spontaneousG > 0, the reverse reaction is spontaneous If If rrG = 0, the reaction is at equilibriumG = 0, the reaction is at equilibrium


If a reaction for which If a reaction for which rrG < 0 is G < 0 is called exergonic.called exergonic.

Because the process is spontaneous Because the process is spontaneous it can be used to drive another it can be used to drive another process, such as another reaction, or process, such as another reaction, or used to do non-expansion work.used to do non-expansion work.


In biological cells, the oxidation of In biological cells, the oxidation of carbohydrates acts as the heavy carbohydrates acts as the heavy weight that drives other reactions weight that drives other reactions such as formation of proteins from such as formation of proteins from amino acids, muscle contractions amino acids, muscle contractions and brain activity.and brain activity.


If a reaction for which If a reaction for which rrG > 0 is G > 0 is called endergonic.called endergonic.

The reaction is not spontaneous and The reaction is not spontaneous and can only proceed by doing work on it, can only proceed by doing work on it, such as electrolyzing water to reverse such as electrolyzing water to reverse its spontaneous formation reaction.its spontaneous formation reaction.

EquilibriumEquilibrium

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rG = μB −μA = (μBθ + RT ln pB ) − (μA

θ + RT ln pA )

ΔrG = ΔrGθ + RT ln pB

pApBpA

=Q (reactant quotient)

ΔrG = ΔrGθ + RT lnQ


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rG = ΔrGθ + RT ln pB

pA= ΔrG

θ + RT lnQ

pB = 0 Q = 0pA = 0 Q = ∞

ΔrGθ =GB ,m

θ −GA ,mθ = μB

θ −μAθ

ΔrGθ = Δ fG

θ (B) − Δ fGθ (A)


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0 = ΔrGθ + RT lnK

RT lnK = −ΔrGθ

K = pBpA

⎛ ⎝ ⎜

⎞ ⎠ ⎟equilibrium


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2A + B → 3C + D0 = 3C + D − 2A − B

0 = ν JJJ

∑ν A = −2 ν B = −1ν C = +3 ν D = +1

A stoichiometric number is positive A stoichiometric number is positive for products and negative for for products and negative for reactantsreactants


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2A + B → 3C + D0 = 3C + D − 2A − B

0 = ν JJJ

∑ν A = −2 ν B = −1ν C = +3 ν D = +1

If If changes by changes by , then the change , then the change in the amount of of any species J is in the amount of of any species J is JJ


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N2 g( ) + 3H2 g( ) → 2NH3 g( )0 = 2NH3 g( ) − N2 g( ) − 3H2 g( )ν NH3 = +2 ν N2 = −1 ν H2 = −3

If initially there is 10 mol of NIf initially there is 10 mol of N22 present, when the extent of reaction present, when the extent of reaction changes from x = 0 to x = 1 (so changes from x = 0 to x = 1 (so = = +1 mol), the amount of N+1 mol), the amount of N22 changes changes from 10 mol to 9 mol.from 10 mol to 9 mol.


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When When = +1 mol, the amount of = +1 mol, the amount of NHNH33 changes by +2 mol and the changes by +2 mol and the amount of Hamount of H22 changes by -3 mol. changes by -3 mol.


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When When = +10 mol, all the N = +10 mol, all the N22 is is consumed.consumed.


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rG = ΔrGθ + RT lnQ

ΔrGθ = νΔ fG

θ −Products∑ νΔ fG

θ

Reactants∑

ΔrGθ = ν JΔ fG

θ

J∑

Q = activites of productsactivities of reactants


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Q = activites of productsactivities of reactants

Q = aJν J

J∏

2A + 3B → C + 2D

Q = aA-2aB

-3aC1 aD

2 = aC1 aD

2

aA2aB

3


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K = aJν J

J∏ ⎛

⎝ ⎜

⎞

⎠ ⎟equilibrium

An equilibrium constant K expressed An equilibrium constant K expressed in terms of activities is called a in terms of activities is called a thermodynamic equilibrium thermodynamic equilibrium constant.constant.

Activities are dimensionless Activities are dimensionless numbers so the thermodynamic numbers so the thermodynamic equilibrium constant is also equilibrium constant is also dimensionless.dimensionless.


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K = aJν J

J∏ ⎛

⎝ ⎜

⎞

⎠ ⎟equilibrium

In elementary applications, activities In elementary applications, activities can be replaced by numerical values can be replaced by numerical values of molalities, molarities or partial of molalities, molarities or partial pressures.pressures.

The resulting expressions are only The resulting expressions are only approximations.approximations.


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K = aCaDaAaB

= γCγDγ AγB

× bCbDbAbB

=KγKb

In elementary applications, KIn elementary applications, K = 1 so = 1 so K ≈ KK ≈ Kbb

How equilibria How equilibria respond to pressurerespond to pressure

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rG = ΔrGθ + RT lnK

RT lnK = −ΔrGθ

∂K∂p ⎛ ⎝ ⎜

⎞ ⎠ ⎟T

= 0

The equilibrium constant depends on The equilibrium constant depends on the value of the value of rrGGθθ, which is defined at , which is defined at a single, standard pressure. Hence K a single, standard pressure. Hence K is independent of the pressure.is independent of the pressure.


The conclusion that K is independent The conclusion that K is independent of pressure does not necessarily of pressure does not necessarily mean that the equilibrium mean that the equilibrium composition is independent of the composition is independent of the pressure.pressure.

It depends on how pressure is It depends on how pressure is applied.applied.


Consider a reaction vessel in which Consider a reaction vessel in which the pressure in the vessel is the pressure in the vessel is increased by injecting an inert gas.increased by injecting an inert gas.

The presence of another gas does The presence of another gas does not alter the equilibrium not alter the equilibrium composition because the partial composition because the partial pressure of each reacting gas pressure of each reacting gas molecules does not changed upon molecules does not changed upon addition of the inert gas.addition of the inert gas.


If however, the pressure is increased If however, the pressure is increased by confining the gases to a smaller by confining the gases to a smaller volume.volume.

Consider the reaction A Consider the reaction A 2B. 2B.

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K = pB2

pA pθ



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K = pB2

pA pθ

For the right hand side of the For the right hand side of the equation to remain constant, pequation to remain constant, pAA must must increase sufficiently to cancel out increase sufficiently to cancel out the increase in the square of pthe increase in the square of pBB..



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K = pB2

pA pθ

In order for pIn order for pAA to increase to increase sufficiently, the equilibrium sufficiently, the equilibrium composition must shift in favor of A composition must shift in favor of A at the expense of B. The number of A at the expense of B. The number of A molecules will increase as the molecules will increase as the volume is decreased.volume is decreased.


The increase in the number of A The increase in the number of A molecules and the corresponding molecules and the corresponding number B molecules in the number B molecules in the equilibrium A equilibrium A 2B is a special 2B is a special case of a Le Chatelier’s principle case of a Le Chatelier’s principle that states:that states:““A system at equilibrium, when A system at equilibrium, when subject to a disturbance, responds in subject to a disturbance, responds in a way that tends to minimize the a way that tends to minimize the effect of the disturbance”.effect of the disturbance”.


The principle implies that, if a The principle implies that, if a system at equilibrium is compressed, system at equilibrium is compressed, then the reaction will adjust to then the reaction will adjust to minimize the pressure. It does this minimize the pressure. It does this by reducing the number of particles by reducing the number of particles in the gas phase.in the gas phase.


Suppose that there is an amount n of Suppose that there is an amount n of A present initially (no B). At A present initially (no B). At equilibrium the amount of A is (1-equilibrium the amount of A is (1-)n )n and the amount of B is 2and the amount of B is 2n, where n, where is the extent of dissociation of A into is the extent of dissociation of A into 2B.2B.

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xA =1−α( )n

1−α( )n + 2αn= 1−α

1+α

xB = 2α1+α

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xA =1−α( )n

1−α( )n + 2αn= 1−α

1+α

xB = 2α1+α

K = pB2

pA pθ = xB

2 p2

xA ppθ = 2α

1+α× 2α

1+α× 1+α

1−α× ppθ

K = 4α 2

1−α 2

ppθ

α = 11+ 4 p Kpθ ⎛ ⎝ ⎜

⎞ ⎠ ⎟

1 2


So even though K is independent of So even though K is independent of pressure, the amounts of A and B do pressure, the amounts of A and B do depend on pressure.depend on pressure.

How equilibria How equilibria respond to respond to

temperaturetemperature Le Chatelier’s principle predicts that Le Chatelier’s principle predicts that

a system at equilibrium will tend to a system at equilibrium will tend to shift in the endothermic direction if shift in the endothermic direction if the temperature is raised.the temperature is raised.

Conversely, an equilibrium can be Conversely, an equilibrium can be expected to shift in the exothermic expected to shift in the exothermic direction if the temperature is direction if the temperature is lowered.lowered.


temperaturetemperature Exothermic reactions: increased Exothermic reactions: increased

temperature favors the reactants.temperature favors the reactants. Endothermic reactions increased Endothermic reactions increased

temperature favors the products.temperature favors the products.


temperaturetemperature The van’t Hoff equation (Justification The van’t Hoff equation (Justification

7.2), is an expression for the slope of 7.2), is an expression for the slope of a plot of the equilibrium constant as a plot of the equilibrium constant as a function of temperature.a function of temperature.

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d lnKdT

= ΔrHθ

RT 2

d lnKd 1 T( )

= − ΔrHθ

R

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d lnKd 1 T( )

= − ΔrHθ

R


temperaturetemperature

For an exothermic reaction, dlnK/dT For an exothermic reaction, dlnK/dT < 0. The negative slope means that < 0. The negative slope means that ln K, and therefore K itself, ln K, and therefore K itself, decreases as the temperature rises.decreases as the temperature rises.

If K decreases, then equilibrium If K decreases, then equilibrium shifts away from products.shifts away from products.

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d lnKdT

= ΔrHθ

RT 2

d lnKd 1 T( )

= − ΔrHθ

R



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rGθ = ΔrH

θ −TΔrSθ

ΔrGθ

T= ΔrH

θ

T− ΔrS

θ

For an exothermic reaction, For an exothermic reaction, rrHHθθ/T is /T is negative and corresponds to the negative and corresponds to the increase of entropy in the increase of entropy in the surroundings.surroundings.

Increasing entropy drives a Increasing entropy drives a spontaneous change.spontaneous change.



When the temperature is increased, When the temperature is increased, rrHHθθ/T decreases and so the /T decreases and so the decreasing entropy of the decreasing entropy of the surroundings has less importance, surroundings has less importance, so there is less driving force for the so there is less driving force for the forward reaction and reactants are forward reaction and reactants are favored.favored.

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rGθ = ΔrH

θ −TΔrSθ

ΔrGθ

T= ΔrH

θ

T− ΔrS

θ



For an endothermic reaction, For an endothermic reaction, rrHHθθ/T /T is positive and corresponds to the is positive and corresponds to the decrease of entropy in the decrease of entropy in the surroundings. surroundings.

Driving force is the increase of Driving force is the increase of entropy in the system.entropy in the system.

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rGθ = ΔrH

θ −TΔrSθ

ΔrGθ

T= ΔrH

θ

T− ΔrS

θ



When the temperature is increased, When the temperature is increased, rrHHθθ/T gets smaller. This /T gets smaller. This corresponds to less loss of entropy in corresponds to less loss of entropy in the surroundings.the surroundings.

This favors a shift towards reaction This favors a shift towards reaction products.products.

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rGθ = ΔrH

θ −TΔrSθ

ΔrGθ

T= ΔrH

θ

T− ΔrS

θ



If we assume If we assume rrHHθθ varies little with varies little with temperature over the temperature range temperature over the temperature range of interest, then we can take it outside the of interest, then we can take it outside the integral.integral.

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d lnKd 1 T( )

= − ΔrHθ

R

lnK2 − lnK1 = − 1R

ΔrHθ d(1 T)

1/T1

1/T2∫

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lnK2 − lnK1 = − ΔrHθ

R1T2

− 1T1

⎛ ⎝ ⎜

⎞ ⎠ ⎟

Calculating an Calculating an equilibrium constantequilibrium constant

Calculate the equilibrium constant for the Calculate the equilibrium constant for the reaction Nreaction N22 + 3H + 3H22 2NH 2NH33 at 298 K. at 298 K.



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rGθ = 2ΔFG

θ (NH3,g) −{ΔFGθ (N2,g) + 3ΔFG

θ (H2,g)}

ΔrGθ = 2(−16.45 kJ mol-1)

ΔrGθ = −32.90 kJ mol-1

ΔrG = ΔrGθ + RT lnK

0 = ΔrGθ + RT lnK



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rG = ΔrGθ + RT lnK

0 = ΔrGθ + RT lnK

lnK = − ΔrGθ

RT= − −32.90 ×103 J mol-1

8.314 J K -1 mol-1 × 298 KlnK =13.28

K = 5.84 ×105

Calculating degree of Calculating degree of dissociationdissociation

The standard Gibbs energy of reaction for The standard Gibbs energy of reaction for the decomposition Hthe decomposition H22O(g) O(g) H H22(g)+ ½ (g)+ ½ OO22(g) is +118.08 kJ mol(g) is +118.08 kJ mol-1-1 at 2300 K. What at 2300 K. What is the degree of dissociation of His the degree of dissociation of H22O at O at 2300 K and 1.00 bar?2300 K and 1.00 bar?

Calculating degree of Calculating degree of dissociationdissociation

The standard Gibbs energy of reaction for The standard Gibbs energy of reaction for the decomposition Hthe decomposition H22O(g) O(g) H H22(g)+ ½ (g)+ ½ OO22(g) is +118.08 kJ mol(g) is +118.08 kJ mol-1-1 at 2300 K. What at 2300 K. What is the degree of dissociation of His the degree of dissociation of H22O at O at 2300 K and 1.00 bar?2300 K and 1.00 bar?

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lnK = − ΔrGθ

RT= − +118.08 ×103 J mol-1

8.314 J K -1 mol-1 × 2300 KlnK = −6.175

K = 2.08 ×10−3

H2O H2 O2

Initial amount

n 0 0

Change at equilibrium

-n +n +1/2 n

Amount at equilibrium

(1-)n n ½(n)

Mole fraction

(1-)/(1+½)

/(1+½) ½/(1+½)

Partial Pressure

(1-)p/(1+½)

p/(1+½) ½p/(1+½)

H2O H2 O2

Initial amount

n 0 0

Change at equilibrium

-n +n +1/2 n

Amount at equilibrium

(1-)n n ½(n)

Mole fraction

(1-)/(1+½)

/(1+½) ½/(1+½)

Partial Pressure

(1-)p/(1+½)

p/(1+½)

½p/(1+½)

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K =pH2pO2

1 2

pH2O

= αp(1+ 1

2α )×

12αp

(1+ 12α )

⎛

⎝ ⎜

⎞

⎠ ⎟

12

×(1+ 1

2α )1−α( )p

= αp(1+ 1

2α )× α

12 p

12

(2 +α )12

×(1+ 1

2α )1−α( )p

= α32 p

12

1−α( )(2 +α )12

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K = α32 p

12

1−α( )(2 +α )12

K = α32 p

12

212

α32 = K(2

12 )

p12

α = 2.08 ×10−3(212 )( )

23

α = 0.0205

Equilibrium Equilibrium electrochemistryelectrochemistry

An electrochemical cell consists of two An electrochemical cell consists of two electrodes, in contact with an electrolyte.electrodes, in contact with an electrolyte.

An electrolyte is a material that allows the An electrolyte is a material that allows the transport of charged species or an ionic transport of charged species or an ionic conductor. This may be a solution, a conductor. This may be a solution, a liquid, or a solid.liquid, or a solid.

An electrode and its electrolyte comprise An electrode and its electrolyte comprise an electrode compartment.an electrode compartment.


There are a number of electrode There are a number of electrode configurations.configurations.

A common electrode configuration A common electrode configuration consists of a metal that participates in the consists of a metal that participates in the electrochemical reaction i.e. M(s)|Melectrochemical reaction i.e. M(s)|M++(aq) (aq) – metal/metal ion electrode type.– metal/metal ion electrode type.

An ‘inert’ metal may make up one of the An ‘inert’ metal may make up one of the electrodes but is only present as a source electrodes but is only present as a source or sink of electrons. It takes no other part or sink of electrons. It takes no other part in the reaction other than acting as a in the reaction other than acting as a catalyst for it i.e. Pt(s)|Xcatalyst for it i.e. Pt(s)|X22(g)|X(g)|X++(aq) or (aq) or Pt(s)|XPt(s)|X22(g)|X(g)|X++(aq) – gas electrode.(aq) – gas electrode.


M(s)|MX(s)|XM(s)|MX(s)|X--(aq) – Metal/insoluble salt(aq) – Metal/insoluble salt Pt(s)|MPt(s)|M++(aq)|M(aq)|M2+2+(aq) - redox electrode.(aq) - redox electrode. Redox – Reduction – OxidationRedox – Reduction – Oxidation OIL RIGOIL RIG A redox reaction implies the transfer of A redox reaction implies the transfer of

electrons.electrons. Oxidizing agent (or oxidant) is the Oxidizing agent (or oxidant) is the

electron acceptor.electron acceptor. Reducing agent (or reductant) is the Reducing agent (or reductant) is the

electron donor.electron donor. A redox equation may be expressed in A redox equation may be expressed in

terms of two half reactions. One oxidation terms of two half reactions. One oxidation reaction and one reduction equation.reaction and one reduction equation.

A galvanic cell is an A galvanic cell is an electrochemical cell electrochemical cell that produces that produces electricity as a electricity as a result of a result of a spontaneous spontaneous reaction. reaction.

An electrolytic cell An electrolytic cell is an is an electrochemical cell electrochemical cell in which a non-in which a non-spontaneous spontaneous reaction is driven by reaction is driven by an external source an external source of current.of current.

Potential difference Potential difference

Varieties of cellsVarieties of cells The simplest type of cell has a single The simplest type of cell has a single

electrolyte common to both electrodes. electrolyte common to both electrodes. In some cases it is necessary to immerse In some cases it is necessary to immerse

the electrodes in different electrolytes as the electrodes in different electrolytes as in the Daniell cell in which the redox in the Daniell cell in which the redox couple at one electrode is Cucouple at one electrode is Cu2+2+/Cu and at /Cu and at the other is Znthe other is Zn2+2+/Zn. /Zn.

Zn(s) Zn(s) Zn Zn2+2+(aq) + (aq) + 2e2e--

CuCu2+2+(aq) + 2e(aq) + 2e-- Cu(s)Cu(s)

Copper is the Copper is the cathodecathode

Zinc is the anodeZinc is the anode

Liquid junction Liquid junction potentialspotentials

In a cell with two different electrolyte In a cell with two different electrolyte solutions, as in the Daniell cell, there is an solutions, as in the Daniell cell, there is an additional source of potential difference additional source of potential difference across the interface of the two across the interface of the two electrolytes. electrolytes.

This potential is called the liquid junction This potential is called the liquid junction potential, Epotential, Eljlj..

A way to reduce this potential is to use a A way to reduce this potential is to use a salt bridge to join the electrolyte salt bridge to join the electrolyte compartments.compartments.

Notation Notation Phase boundaries are represented by Phase boundaries are represented by

vertical bars |vertical bars | CuSOCuSO44(aq)|Cu(s)(aq)|Cu(s) A liquid junction is represented by 3 A liquid junction is represented by 3

vertical dotsvertical dots A double vertical line || denotes an A double vertical line || denotes an

interface in which it is assumed the interface in which it is assumed the junction potential has been eliminated.junction potential has been eliminated.

Zn(s)|ZnSOZn(s)|ZnSO44(aq)||CuSO(aq)||CuSO44(aq)|Cu(s)(aq)|Cu(s) Convention is to write the anode half cell Convention is to write the anode half cell

first and the cathode half cell second.first and the cathode half cell second.

The electromotive forceThe electromotive force The electric current produced by a The electric current produced by a

galvanic cell arises from a spontaneous galvanic cell arises from a spontaneous chemical reaction taking place inside it.chemical reaction taking place inside it.

Zn(s)|ZnSOZn(s)|ZnSO44(aq)||CuSO(aq)||CuSO44(aq)|Cu(s)(aq)|Cu(s) The cathode is where reduction takes The cathode is where reduction takes

place and the anode is where oxidation place and the anode is where oxidation takes place:takes place:Right hand electrode: CuRight hand electrode: Cu2+2+(aq) + 2e(aq) + 2e-- Cu(s)Cu(s)Left hand electrode: Zn(s) Left hand electrode: Zn(s) Zn Zn2+2+(aq) + (aq) + 2e2e--

The Nernst equationThe Nernst equation A cell in which the overall cell reaction A cell in which the overall cell reaction

has not reached chemical equilibrium can has not reached chemical equilibrium can do electrical work as the reaction drives do electrical work as the reaction drives electrons through an external circuit.electrons through an external circuit.

The work that a given transfer of The work that a given transfer of electrons can depends on the potential electrons can depends on the potential difference between the two electrodes.difference between the two electrodes.

The cell potential is measured in volts, V.The cell potential is measured in volts, V. --FE = FE = rrG (Justification 7.3)G (Justification 7.3) is the stoichiometric coefficient of the is the stoichiometric coefficient of the

electrons in the half reactions, F is electrons in the half reactions, F is Faraday’s constant, and E is the emf.Faraday’s constant, and E is the emf.

The Nernst equationThe Nernst equation

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FE = −ΔrG

E = −ΔrGνF

ΔrG = ΔrGθ + RT lnQ

E = − ΔrGθ

νF− RTνF

lnQ

E = Eθ − RTνF

lnQ

EEθθ - - standard EMF of a standard EMF of a cell.cell.

The Nernst equationThe Nernst equation

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E = Eθ − RTνF

lnQ

RTF

= 25.7 mV at 25oC

If ν =1 and Q is increased by a factor of 10, then the emf decreases by 59.2 mV

Cells at equilibriumCells at equilibrium

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E = Eθ − RTνF

lnQ

0 = Eθ − RTνF

lnK

lnK = νFEθ

RT

Cells at equilibriumCells at equilibrium For a Daniell cell: CuFor a Daniell cell: Cu2+2+(aq) + Zn(s) (aq) + Zn(s) Cu(s) Cu(s)

+ Zn+ Zn2+2+(aq)(aq) = 2 and the standard emf is +1.10 V= 2 and the standard emf is +1.10 V

Cells at equilibriumCells at equilibrium For a Daniell cell: CuFor a Daniell cell: Cu2+2+(aq) + Zn(s) (aq) + Zn(s) Cu(s) Cu(s)

+ Zn+ Zn2+2+(aq)(aq) = 2 and the standard emf is +1.10 V= 2 and the standard emf is +1.10 V

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lnK = νFEθ

RT= 2 × 96485 ×1.1

8.314 × 298= 85.7

K =1.6 ×1037

Standard PotentialsStandard Potentials A galvanic cell is a combination of two A galvanic cell is a combination of two

electrodes and each one can be considered electrodes and each one can be considered as making a characteristic contribution to as making a characteristic contribution to the overall cell potential.the overall cell potential.

It is not possible to measure the It is not possible to measure the contribution of a single electrode, so we contribution of a single electrode, so we measure the potential of electrodes by measure the potential of electrodes by defining the standard hydrogen electrode defining the standard hydrogen electrode (SHE) to be zero and then measure the (SHE) to be zero and then measure the electrode of interest in combination with electrode of interest in combination with the SHE.the SHE.

Pt(s)|HPt(s)|H22(g)|H(g)|H++(aq)(aq) EEθθ = 0 V = 0 V

Standard PotentialsStandard Potentials For two redox couples OxFor two redox couples Ox11/Red/Red11 and and

OxOx22/Red/Red22

RedRed11|Ox|Ox11||Red||Red22|Ox|Ox22 EEθθ = E = Eθθ22 – E – Eθθ

11

RedRed11 + Ox + Ox11 Ox Ox22 + Red + Red22 is spontaneous if is spontaneous if EEθ θ > 0> 0

If If EEθ θ < 0 then work has to be done on the < 0 then work has to be done on the system for the reaction occur as written.system for the reaction occur as written.

Standard PotentialsStandard Potentials Pt(s)|HPt(s)|H22(g)|H(g)|H++(aq)||Cu(aq)||Cu2+2+(aq)|Cu(s) E(aq)|Cu(s) Eθθ = =

+0.34 V+0.34 V Pt(s)|HPt(s)|H22(g)|H(g)|H++(aq)||Zn(aq)||Zn2+2+(aq)|Zn(s) E(aq)|Zn(s) Eθθ= -= -

0.76 V0.76 V

Determining other Determining other thermodynamic thermodynamic

functionsfunctions

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− rG = νFE ∂ΔrG∂T

⎛ ⎝ ⎜

⎞ ⎠ ⎟p

= −ΔrS

ΔrSdT = d(νFE) = νFdE

dEθ

dT= ΔrS

θ

νF

ΔrHθ = ΔrG

θ + TΔrSθ = −νF Eθ −T dE

θ

dT

⎛ ⎝ ⎜

⎞ ⎠ ⎟

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