Ch 07 Phys Chem

8/8/2019 Ch 07 Phys Chem

1/40

1

Electrochemical Equilibrium

CHAPTER 7

Electrochemical reactions involve the transfer of electrons

We recognize the importance of electrochemical equilibria in

galvanic cells for example; which can produce electriccurrent.

Battery technology relies on this knowledge where it ispossible to convert Gibbs energy change into useful form of

work

On the other hand, we also have electrolytic cells whichconsume electric power to induce a chemical reaction

This is used in the industries such as in the processelectro latin and electro-refinin Aluminum roduction


2/40

2

The measurement ofelectromotive force

of an electrochemical cell at varioustemperature makes it possible to obtain thethermodynamic properties

i.e. E = f(temp)

The chemical potential (Q) of theelectrolytes, for instance, can be calculated

based on these measurements


3/40

3

Equilibria Involving Potential Differences

When a small charge dQ moves through an electric potentialJ, the work done on the charge is given by

dw (joule) = J (volt or Joule/coulomb).dQ (coulomb)

If ions (that carry electric charge) with charge number ofziare transferred, the differential charge dQ can be related to thedifferential amount of mol dni by :

dQ(coul.) = zi .F(coul./mol).dni (mol)

Where F is the Faraday Constant which is equal to theavogadros number x charge of proton (or electron) :

F = Na.e = (6.0221367x1023 / mol) x (1.60217733x10-19 C )

F = 96485.309 C/mol


4/40

4

So when we consider a multiphase system with phases atdifferent potential the electrical work term is of the form:

w = 7 zi F Ji ni where Ji is the potential of the phase containing species i

When we want to relate this to the thermodynamic

equation, however, it should be noted that Ji is not anatural variable of Gibbs Energy, Hence it is necessary tomake the necessary transformation :

G = G - 7 zi F Ji ni

Substituting G = 7 Qi ni gives G = 7 (Qi - ziFJi) nior G = 7 Qi.ni where Qi = Qi - zi F J i


5/40

5

Based on this, the differential of the transformed Gibbs

Energy :

dG = dG - 7 ziFJi dni - 7 zi F ni dJiSubstituting dG = - SdT + VdP + 7Qidniwe then have :

dG = - SdT + VdP + 7Qidni - 7 ziFJi dni - 7 zi F ni dJiOR

dG = - SdT + VdP + (7Qi - 7 ziFJi ) dni - 7 zi F ni dJi

and as before we have : Qi = Qi - zi F Ji thendG = - SdT + VdP + Qidni - 7 zi F ni dJI

and consistently for chemical potential :i

injPTidn

Gd

QJ d!

d

,,,


6/40

6

Equation for an electrochemical cell

Electrochemical cell can be classified as a galvanic cell in

which chemical reaction occur spontaneously. The Gibbs(free) energy of this reaction is harnessed to generate usefulwork (recall: it produces potential differences in Volt)

On the other hand we can also have electrochemical cell that

work in reverse, in which, chemical reaction is caused/inducedby an externally applied potential differences

(i) Cathode is the electrode in which reduction occurs; in agalvanic cell this electrode is positive since electrons are

flowing into it and reacting to form a reduced species

(ii) Anode is the electrode in which an oxidation reactionoccurs; (it is negative in a galvanic cell)


7/40

7

Galvanic Cell :

At cathode : A + e- p A- (reduction, electron is supplied)At Anode : A p A+ + e- (oxidation, electron is released)


8/40

T Nugraha / LS / bachelor / PhysicalChemistry

8

Swiss German

University As the potential difference is

increased, a point will be

reached that no current is

flowing

This is the equilibrium

potential that is of primaryconcerned in this chapter

This is a thermodynamic

measurement as the direction of

the current through the cell can

be changed by an infinitesimal

change in the applied voltage


9/40

9

A wide variety of galvanic cell can be constructed :

Cells without liquidjunctions :

Pt(s)H2(g)HCl(m)AgCl(s)Ag(s)

The vertical lines represent the phaseboundary

In the figure on the side:H2(g) a

non-metal is passed over a platinum

electrode (oxidized to H+), on theother side Ag+ is reduced to Ag(s)

while the Cl- is released to the solution


10/40

10

Cells with liquidjunctions :

Zn(s)Zn2+(m1)1 Cu2+(m2)Cu(s)

Zn(s)Zn2+(m1)1 Zn2+(m2)Zn(s)In the above example Zn is oxidizedinto Zn2+ while Cu2+ is reduced toCu(s)

1 represent a junction betweentwo liquids

1 1 represent a salt bridgetypically made up ofpotassiumchloride orammonium chloride inwhich the anion and cations havenearly equal mobility

salt bridge prevents mixing/reaction

Note: for a galvanic

cell there is aconvention to draw

oxidation on the left

while reduction on

right


11/40

11

To obtain the relationship between electromotive force (E)

and the chemical potential or activities of reactants &

products consider the following cell without liquid junction :

PtLH2(g)HCl(m)AgCl(s)Ag(s)PtRNote: m indicates molality of the species and we assume

international convention in which the left side is oxidationand the right side is reduction

So the reactions are :

2AgCl(s) + 2e-(PtR) p 2Ag(s) + 2Cl-(aq) (reduction)H2(g) p 2H+(aq) + 2e-(PtL) (oxidation)


12/40

12

The sum of these two electrode reactions is :

H2(g) + 2AgCl(s) + 2e-(PtR) p 2H+(aq) + 2Ag(s) + 2Cl-(aq) + 2e-(PtL)

alternatively this can also be written (shorter version) :

H2(g) + 2AgCl(s) + 2e-(PtR) p 2HCl(aq) + 2Ag(s) + 2e-(PtL)

Consider the cell in the figurebeside; in which the electrodes are

not connected electrically (open

circuit)

Under this condition, when thesystem is at equilibrium the

relationship

7Ri Qi = 0 must be satisfied


13/40

13

Such that for :

H2(g) + 2AgCl(s) + 2e-(PtR) p 2HCl(aq) + 2Ag(s) + 2e-(PtL)

We have ( for7RiQ

i= 0) :

2Q(HCl,aq) + 2Q(Ag,s) + 2Q(e-,PtL)

- Q(H2,g) - 2Q(AgCl,s) - 2Q(e-,PtR) = 0From before we have : Qi = Q+ ziFJiand for electrically neutral species zi = 0 such that Qi = Qonly for electrons: Q(e-,PtL) = Q(e-,PtL) FJL and

Q(e-,PtR) = Q(e-,PtR) FJR

But for electrons at equilibrium we have : Q(e-,PtL) = Q(e-,PtR)So the Q term can cancel out ; Therefore we have in summary :


14/40

14

2Q(HCl,aq) + 2Q(Ag,s) 2FJL- Q(H2,g) - 2Q(AgCl,s) + 2FJR= 0

But we also know that change in chemical potential :

2Q(HCl,aq) + 2Q(Ag,s) - Q(H2,g) - 2Q(AgCl,s) = (rG

Therefore : (rG = 2F(JL-JR) = -2F (JR-JL) =-2FE

Hence : (rG = =-2FE note that the 2 is coming from Ri

where E = JR- JL which is the electromotive force which canbe measured using a potentiometer


15/40

15

The relationship (rG = -2FE can actually be generalizedto any electrochemical reaction in which it can be written as

FEG er Y!(

Where Re is the absolute value of the stoichiometric

number of the electron in the electrochemical reaction (orsimply the number of electron being transferred during theredox reaction)

Since we are dealing with Gibbs energy change, then we

know ifthe right hand electrode has more potential thanthe left hand electrode, the value of the electromotive force(E) of the cell is positive

This will give a negative value of(rG which means that the

electrochemical reaction is spontaneous


16/40

16

If the schematic diagram of the cell is reversed(i.e. the electrochemical reaction is reversed),

the sign of E and therefore the sign of(rG

would also be reversedThe reversed reaction then should have (rGwhich is positive and this means that the reverse

reaction is not a spontaneous reaction


17/40

17

Relation between E and the equilibrium constant (K)

Since the value ofE can be obtained under equilibrium

condition, then we can somehow relate it to the value ofK

Recall from the previous chapter we have (rGo = -RT ln K

but before we also have (rG= -|Re|FE or equivalently forstandard state we have (rGo = -|Re|FEo (note we will seehow Eo is determined in later discussion)

Hence we have : |Re|FEo = RT ln K such that :

!!

RT

FEKorK

F

RTE

o

e

e

oY

Y

expln F = 96485.309 C/mol


18/40

18

Example : Three different galvanic cells have standard

electromotive forces (Eo) of0.01 V, 0.1 V and 1.0 V at

25C. Calculate Kfor each reaction assuming |Ye| = 1 forthese reactions

Ans. :

!

RT

FEK

o

eY

exp So by entering the values, we get :

476.1)15.298)(3145.8(

)01.0)(96485(exp

.

!

!

K

VK

KmolJ

molCoul

(Note thatequivalentunitforVoltisJoule/coulomb)similarly we have K= 49.0 for Eo = 0.1 V,

K= 8.02x1016 for Eo = 1 V


19/40

19

Calculate enthalpy, Entropy and Gibbs Energy. Also, if

spontaneous calculate how much energy is available for work.

Zn + Cu2+ p Zn2+ + Cu

(fHo (kJ) 0 64.77 -153.89 0 (Ho = -218.66

So (J/K) 41.63 -99.6 -112.1 33.15 (So = -21.0Therefore: (G = (H - T(S = -218.66x103 - 298x(-21.0)

= - 212.402 kJ/mol < 0 ; spontaneous

While -T(S is equal to qsurr: heat transferred to surrounding(G if negative signifies the amount of energy that is availablefor work(our purposes) hence also called Gibbs Free Energy

RECALL FROM CHAPTER FOR (USE TABLE C2 or C3


20/40

20

Furthermore, knowing that (rGo = -|Re|FEo if we have dataon the variation ofGibbs energy as a function of Temp

than we can also obtain the expressions for(S and (H

recall that (S = -(HG/HT) then we have (So = |Re|F(H Eo/ HT)

moreover, (G =(H - T(S or(H =(G + T(S, thus

(Ho = -|Re|FEo + |Re|FT(H Eo/ HT)

So, if we know the variation ofEo as a function of

temperature then we can calculate the above entities.


21/40

21

Example : the electrochemical reaction

Pt|H2(g)|HCl(aq)|AgCl(s)|Ag(s) has Eo (Volt) as a

function of temperature (in celcius) as follow (0 to90Celcius) :

Eo= 0.23659 - 4.8564x10-4 T - 3.4205x10-6 T2 + 5.869x10-9 T3

calculate (Go

, (So

, (Ho

for this reaction at 25C(i) for(Go we simply need to calculate Eoby substitutingthe value of T, and then calculate (rGo = -|Re|FEo

so Eo =

0.22240 V (at 298K), hence :(rGo = - (1) x (96485 Coul/mol) x (0.22240 Volt)

= - 21.458 kJ/mol


22/40

22

(ii) (So = |Re|F(H Eo/ HT) =

1 x (96485 Coul/mol) x [-4.8564x10-4 - 2(3.4205x10-6)T +

3(5.869 x 10-9) T2] Volt = -62.297 J/mol.K(at 298K)

(iii) (H = (G + T(S

= -21.458 kJ/mol+ (298.15K)(-62.297x10-3 kJ/mol.K)

= -40.032 kj/mol (at 298K)


23/40

23

StandardElectrode Potential(standardreduction

potential, Eo)

For both research and industrial purposes, the value of

standard electrode potential (Eo) for many substances

are tabulated

For this purpose, we need to define the referenceelectrode against which all of the other

substances/electrodes are measured

In this case, the standard

measurement is measuredagainst a hydrogen electrode in

whichH2(g) is oxidized to

H+(aq).


24/40

24

Hence, the measurement is done against the oxidation of

H2(gas) at the anode i.e. H2(g) p H+(aq) + e-

To facilitate the measurement, a convention was adopted

such that the standard electrode potential of the hydrogen

(H2(g) at 1 atm on Pt electrode) is arbitrarily assigned

the value zero:

H+(aq) + e- p H2(g) Eo = 0.0000 V

(Note that it is written in terms ofreduction reaction)

for example: for the measurement of Eo

for the reductionof Ag+(aq) + e- p Ag(s) we have :

AgCl(s) + e- p Cl-(aq) + Ag(s) Eo = 0.2224 V

Hence Eo

is called the standard reduction potential


25/40

25

Hence, we can also see the Eo orstandard electrode

potential is the measured tendency of an electrode reaction to

occur in the direction of reduction

A brief listing ofstandard electrode potentials (25C) can be

seen in table 7.2 (p.240)

Note that fluorine, for instance, electrode has the most positiveelectrode potential F2(g) + e

- p F- - Eo = +2.87 V

if we connect this to the expression : (rGo = -|Re|FEo andsince we expect, if G < 0, then we know this reaction is

spontaneous

Furthermore, as the value is also the highest among other

listed electrode potential for other substances


26/40

26

The fluorine electrode has the highest tendency to pull

electrons from the other electrode such that it will undergo

reduction

On the other hand, Li is at the bottom of the list i.e. it has the

lowest Eo, in fact it has negative value of Eo

For such electrodes, the highest tendency is not to undergo

reduction (or to pull electrons from the other electrode), butrather they have a higher tendency to give up/donate electron

i.e. to undergo oxidation reaction

Similarly, we can see it from the point of view (Go

= -|Re|FEo

such that Gibbs energy change is positive for the reduction of

Li, therefore, we know that it is the reverse

reaction/oxidation that should occur spontaneously (Note:

against H2(g) on Pt electrode)


27/40

27

Earlier we discussed the electromotive force of a cell is given

by E = JR- JL

Based on this we can express the standard electromotive forcefor a cell as : Eo = ERo - EL

o

Hence basedon this we can :

(i) Calculate the standard electrode potential of any cellinvolving any substances that are available on the list (table

7.2)

(ii) Moreover, we can also determine which electrode will act

as anode and cathode when the cell delivers current

(iii) We can also evaluate equilibrium constant K based on the

relation ln K = |Ye| FEo/(RT) ; [also (rG]

Eo = Ereduction Eoxidation


28/40

28

Example: Calculate Eo for : Pt|H2(g)|HCl(aq)|Cl2(g)|Pt

Ans.: Write the relevant half reaction (reduction) i.e.

(1) Cl2 + 2e-

m 2Cl-

(aq) Eo =

1.3604 V (Eo

R)(2) H2(g) + 2e

- m 2H+(aq) Eo = 0.0000 V (EoL)

Note: The two reactions must have the same number ofelectron (donated = accepted), if not multiplied one of them

to have the same electron. The value of Eo is not affectedbythis multiplication

Then Eocell = EoR - E

oL = 1.3604 - 0.0000 = 1.3604 Volt

Hence the full reaction can be written as :

H2(g) + Cl2(aq) m 2H+(aq) + 2Cl-(aq) (i.e. HCl(ai)) Eo = 1.3604 V

NOTE: Eo = Ereduction Eoxidation


29/40

T Nugraha / LS / bachelor / PhysicalChemistry 29

If necessary we can calculate (Go from (Go = -

|Re|FEo

(Go= -(2)x(96458 coul)(1.3604 J/coul) = -262.44kJ/mol

we can also calculate equilibrium constant K :

451078.9

15.298.3145.8(

)3604.1()96485(2expexp x

KKmolJ

coulJxcoulx

RT

FEK

o

e !

-

!

!

i.e. volt


30/40


Swiss German

University

KF

RTEEfromhence

HClaK

e

o

oCl

oH

ln)(

22

2

Y

!

!

!

oCl

oHe

o

P

P

P

P

HCla

F

RTEE

22

2)(

lnY

If we need to evaluate E not at standard state (eg. not at

P=1 bar)


31/40


Activity of Electrolytes

When we deal with activity coefficient, it is customary to useconcentration in molal unit (rather than mole fraction ormolarity)

molality (m) is the amount in mole of the electrolyte in 1 kgof solvent such that the unit would be mol/kg

One of the advantage with working with molality is that theunit is not affected by temperature (unlike molarities forexample)

the activity of a solute on the m scale is defined by :

o

ii

im

ma

K!

Where mo is the standard value of

the molality ( 1 mol/kg solvent)


32/40

32

For this relationship, for very low concentration (m p 0)

1lim0

!p imi

K

Such that for very dilute solution the activity aio

ii

im

ma

K!

Would simplify to ai = mi/mo (note that mo is simply 1 molal)

At very low concentration the distribution of ions in the

solution can be considered to be completely random and the

ions are too far apart to exert any interaction among

themselves, thus the Ki is unity

At higher concentration, the attraction and repulsion become

more important. Hence the K is reduced from unity (


33/40


34/40


Swiss German

University

Fuel Cell

A fuel cell is a type of battery which is continuously supplied

with oxidant and reductant so that the redox reactions that

generate electricity can occur indefinitely

Some of the currently designed fuel cell can work at the lowerrange temperature (25-100C), mid range temperature (100-

500C), high temperature (500 to 1000C) and very high

temperature (above 1000C)

at very low temperature, the use of generally expensive catalyst

is necessary. While at high temperature, the challenge is in the

choice of material that can withstand the temperature as well as

the energy required to maintain the high temperature


35/40


Swiss German

University

Determination of pH

In many cases in practice, chemists needs to deal with

the concentration of hydrogen ions (H+) which could range

from 10-1 M to 10-14 M

Back in 1909, a chemist named Sorenson adopted

exponential notation for [H+]. He defined pH as thenegative exponent of 10 that gives the H+ concentration

i.e. [H+] = 10-pH

Today, pH is defined to be as close as possible to the

negative base 10 logarithm ofhydrogen ion (H+

) activity :pH = -log (aH+)_


36/40


Swiss German

University The pH may be measured with a hydrogen electrode

conn

ected with a calomel electrode through a salt bridge :Pt | H2(PH2) | H+(aH+) 1 1 Cl- | Hg2Cl2 | Hg

The electromotive force of this cell may be considered tobe made up of three contributions :

junctionliquidP

P

HEaE

oH

-!

21

2

log0591.02802.0

The contribution of the calomel electrode is 0.2802 V,

while the Eliquid junction can be assumed to be 0.00 V

the number 0.0591 is coming from RT/nF but still

multiplied by conversion factor from ln to log (base 10),

recall originally the middle term is :

-

21

2

lno

H

P

P

Ha

nF

RT

**


37/40


Swiss German

University

Furthermore, to make the situation more simple we can

use PH2 = 1 bar such that the partial pressure contributio

nis reduced to 1

So after the simplification the equation (**) is :

E 0.2802 = -0.0591 log aH+

and since pH is defined as log (aH+) then

E 0.2802 = 0.0591 log aH+ or0591.0

2802.0!E

pH

Currently, the use of H2 in the measurement of pH isavoided. The use of glass electrode is typically used inpractice in most laboratories


38/40


Swiss German

University

The basic process is shown

in the figure beside which isa hydrogen-oxygen fuel cell

with solid electrolyte

The membrane is

impermeable to reactant

gases but not to H+ ions

The (ion exchange)

membrane is also acting as

the electrolyte

For low temp operation, the

membrane is covered with

finely divided Pt catalyst


39/40


Swiss German

University

The half cell reactions are :

O2(g) + 2H+ + 2e- p H2O(l) Eo = 1.2288 V

2H+ + 2e- p H2(g) Eo = 0 V (reversed)

H2(g) + O2(g) p H2O(l) Eo

= 1.2288 V


40/40


Swiss German

University

Date post:	10-Apr-2018
Category:	Documents
Upload:	laviniakartika
View:	226 times
Download:	0 times

Ch 07 Phys Chem

Documents