Lecture 4- Equilibrium Chemistry (4h)

8/11/2019 Lecture 4- Equilibrium Chemistry (4h)

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USTH

Analytical chemistryLecture’s notes

(For 2 nd year BIO and EVNStudents)

Sept. to Dec., 2013

Assoc. Prof. Dr. Ta Thi Thao

Dept. of Anal. Chem.;VNU Universit of Science



Lecture 4: Equilibria in solutions

and titration

- Equilibrium

and Le Satelie principle- Equilibrium Constants for chemical

reactions and titration

+ Acids and bases+ Complexon reaction

+ Precipitation

+ Redox reaction

Lecture outlines:



Equilibrium

* Definition:



* REVERSE REACTION (backward

reaction) reciprocal K ( K’ or K-1)

4.1. Equilibrium



• ADD REACTIONS

Multiply Ks

4.1. Equilibrium



Problem example 1:



• LE CHATELIER’S PRINCIPLE

2.1. Equilibrium

CO2 + H2 H2O(g) + CO a drying agent is added to absorb H2O

H2(g) + I2(g) 2HI(g) Some nitrogen gas is added

Q.: How do the following equilibria shift?

NaCl(g) + H2SO4 (l) Na2SO4(s) + HCl(g) the reaction is carried out in

an open container

Problem example 2: What factors affecting to the equilibrium?



REACTION QUOTIENT, Q



REACTION QUOTIENT, Q

• K is thus the special value that Q has

when the reaction is at equilibrium



Eg.: HABER-BOSCH:

N2 + 3 H2 2 NH3 + E

K IS DIMENSIONLESS!

•Concentrations in mol/liter (M)

•pressures in atmospheres (atm)

•ignore solids

•ignore solvents



4.2. Equilibrium Constants for

chemical reactions-Acid- Base Reaction (Ka, Kb)

- Complexon reaction (Kf )

- Precipitation reaction (Ksp)- Redox Reaction (E0)



4.1. Acid- Base Reactions

* Brønsted-Lowry Law• Acids donate H+ ions to other ions or

molecules, which act as a base.

• In an operational sense:

• an acid is any substance that increases the

concentration of the H+ ion when it dissolves in

water.

• a base is any substance that increases theconcentration of the OH- ion when it dissolves in

water.



* Con jugated Acids and Bases

CH3COOH + H2O CH3COO- + H3O+

CH3COO-: …………………………………………..

H3O+: …………………………………………………







Weak acid and weak base

H l l ti



pH calculation

• For a strong acid and strong base:

pH=- log [H+] with [H+]= C A if C A>10-6MpOH= -log [OH-] with [OH-] = CB if CB>10-6M

• For a mono weak acid and mono weak base:

if C A>10-6

M and K A>10-8

B BC K OH ][

A AC K H

][

if C A>10-6 M and K A>10-8

•For a poly weak acid and poly weak base that K1>>K2>>K3:consider as mono weak acid or weak base has first H+

For a ampholytic salt: 21][ K K H



Buffers solution

• When H+ is added, it reacts essentially to

completion with the weak base present

H+ + A- HA or H+ + B BH+

• When OH- is added, it reacts essentially to

completion with the weak acid present

OH- + HA H2O + A-

OH- + BH+ H2O + B

If a solution containing HA and A- or mixture of B and BH+



Buffers

• pH = pKa

+ log [base]/[acid]

• The maximum buffer capicity: pH=pKa

• Useful range: pH pKa 1

• pH determined byKa of acid and ratio of acid/conjugate base

or

Kb of base and ratio base/conjugate acid



Buffer Table

Formic Acid Ka

1.8 X 10-4

pKa

3.74

Barbituric Acid 9.8 X 10-5 4.01

Butanoic Acid 1.52 X 10-5 4.82

• Choose a pKa near the desired pH

• pH = pKa + log (base/acid)• 4.0 = 3.74 + log (base/acid)

• 0.26 = log (base/acid)

• 10.26 = 1.8 = (Na formate / formic acid)

How to make a solution with pH 4?



Buffer Capacity

• As long as ratio remains virtually constant,

the pH will be virtually constant

• This is true as long as concentrations of

buffering materials (HA/A-) or (B/BH+) are

large compared with H+ or OH- added.



Problem Examples• 4- methylaniline is considered as a weak

base with pKa= 5.084 . Calculate the pH of10-2 M 4- methylaniline solution (2pts)

•

• •

•

• How does pH change if 50 mL of 0.05 MHCl is added to 100 mL of 10-2 M 4-

methylaniline solution (2pts)



Ladder diagram



Ladder diagram of complex

systems• E.g. H3PO4

and it’s salts



4.2.2. Complex-Ion Equilibria

• Many metal ions, especially transition

metals, form coordinate covalent bonds

with molecules or anions having a lone

pair of electrons.

For example, the silver ion, Ag+, can react with

ammonia to form the Ag(NH3)2+ ion.

)NH:Ag:NH()NH(:2Ag 333



Complex-Ion Equilibria

• A complex ion is an ion formed from a metal

ion with a Lewis base attached to it by acoordinate covalent bond.

A complex is defined as a compound containing

complex ions.

A ligand is a Lewis base (an electron pair donor) that

bonds to a metal ion to form a complex ion.

•The number of covalent bonds that a cation tends to

form with electron donors is its coordination number.Typical values for coordination numbers are two, four,

and six.

•The species formed as a result of coordination can be

electricall ositive, neutral, or ne ative.



Complex-Ion FormationThe aqueous silver ion forms a complex ion with

ammonia in steps.

)aq()NH(Ag (aq)NH)aq(Ag 33

)aq()NH(Ag (aq)NH)aq()NH(Ag 2333

When you add these equations, you get the overall

equation for the formation of Ag(NH3)2+.

)aq()NH(Ag (aq)NH2)aq(Ag 233



Complex-Ion Formation

The formation constant, Kf , is the equilibrium

constant for the formation of a complex ion from

the aqueous metal ion and the ligands.

The formation constant for Ag(NH3)2+ is:

23

23

f ]NH][Ag[

])NH(Ag[K

The value of K f for Ag(NH3)2+ is 1.7 x 107.



Complex-Ion FormationThe formation constant, K

f

, is the equilibrium

constant for the formation of a complex ion from

the aqueous metal ion and the ligands.

The large value means that the complex ion is

quite stable.

When a large amount of NH3 is added to asolution of Ag+, you expect most of the Ag+ ion to

react to form the complex ion.

The higher the Kf of complex, the better the selectivity of the ligand

The dissociation constant, Kd , is the reciprocal,

or inverse, value of Kf.



Equilibrium Calculations with Kf

What is the concentration of Ag+(aq) ion in

0.010 M AgNO3 that is also 1.00 M NH3?

The Kf for Ag(NH3)2+ is 1.7 x 107.

In 1.0 L of solution, you initially have 0.010

mol Ag+(aq) from AgNO3.

This reacts to give 0.010 mol Ag(NH3)2+,

leaving (1.00- (2 x 0.010)) = 0.98 mol NH3.You now look at the dissociation of Ag(NH3)2

+.



Equilibrium Calculations with

Kf




The following table summarizes.

Starting 0.010 0 0.98Change -x +x +2x

Equilibrium 0.010-x x 0.98+2x

(aq)NH2)aq(Ag )aq()NH(Ag 323




Kf




The dissociation constant equation is:

f d

23

23

K

1

K ])NH(Ag[

]NH][Ag[




Kf




Substituting into this equation gives:

7

2

107.1

1

)x010.0(

)x298.0)(x(




Kf




If we assume x is small compared with 0.010

and 0.98, then

8

2

109.5)010.0(

)98.0)(x(




Kf




and

10

)98.0()010.0(8

101.6109.5x 2

The silver ion concentration is 6.1 x 10-10 M.

Producing Soluble Complexes



Producing Soluble Complexes Complexation reactions involve a metal ion Mreacting with a ligand L to form a complex ML.

M + L MLComplexation reactions occur in a stepwise fashion,and the reaction above is often followed by additionalreactions:

ML + L ML2

ML2 + L ML3

MLn-1 + L MLn

Unidentate ligands invariably add in a series ofsteps. With multidentate ligands, the maximumcoordination number of the cation may be satisfied

with only one or a few added ligands.

continued



…continued…

The equilibrium constants for complex formation

reactions are generally written as formation

constants.

M + 2L ML2

M + 3L ML3

M + nL MLn

The overall formation constants are products of

the stepwise formation constants for the

individual steps leading to the product.

2

2

1 2

ML

M L K K

33

1 2 3

ML

M L K K K

n

n

n

ML

M L K K K

1 2....

2

3

n



Forming Insoluble Species

The addition of ligands to a metal ion mayresult in insoluble species, such as the familiarnickel-dimethylglyoxime precipitate. In many

cases, the intermediate uncharged complexesin the stepwise formation scheme may besparingly soluble, whereas the addition ofmore ligand molecules may result in soluble

species. AgCl is insoluble, but addition oflarge excess of Cl- produces soluble AgCl2

-, AgCl3

2-, and AgCl43-.

continued



…continued…

In contrast to complexation equilibria, which aremost often treated as formation reactions, solubility

equilibria are usually treated as dissociationreactions

Mx Ay(s) xMy+(aq) + yAx-(aq)

Ksp = [My+]x[Ax-]y

where, Ksp = solubility product. Hence, for BiI3, thesolubility product is written Ksp = [Bi3+][I-]3.

The formation of soluble complexes can be used tocontrol the concentration of free metal ions in

solution and thus control their reactivity.



ORGANIC COMPLEXING AGENTS

Many different organic complexing agents

have become important in analytical

chemistry because of their inherent

sensitivity and potential selectivity in reacting

with metal ions. Such reagents are

particularly useful in precipitating metal and

in extracting metal from one solvent toanother. The most useful organic reagents

form chelate complexes with metal ions.



Reagents for Precipitating Metals

One important type of reaction involving an organic

complexing agent is that in which an insoluble, uncharged

complex is formed. Usually, it is necessary to considerstepwise formation of soluble species in addition to the

formation of the insoluble species. Thus, a metal ion Mn+

reacts with a complexing agent X- to form MX(n-1)+, MX2(n-

2)+, MXn-1+, and MXn(soln).

Mn+ + nX- MXn(soln)

MXn(solid) MXn(soln) Keq = [MXn]

Solubility product expression is:

Ksp = [Mn+][X-]n = Keq / n

n

n

n

MX

M X K K K

1 2....n+ - n



Forming Soluble Complexes for

Extractions

Many organic reagents are useful inconverting metal ions into form that can be

readily extracted from water into an immiscible

organic phase. Extraction are widely used to

separate metals of interest from potential

interfering ions and for achieving a

concentrating effect by extracting into a phase

of smaller volume. Extractions are applicableto much smaller amounts of metals than

precipitations, and they avoid problems

associated with coprecipitation.



Ethylenediamenetetraacetic



Ethylenediamenetetraacetic

acid (EDTA)

EDTA forms 1:1 complexes with metal ions by with 6 ligands: 4 O &

2N. EDTA is the most used chelating agent in analytical chemistry,

e.g. water hardness.



Acid/Base Properties of EDTA



Acid/Base Properties of EDTA• EDTA is a hexaprotic system (H6Y

2+) with 4

carboxylic acids and 2 ammoniums:

• We usually express the equilibrium for the formationof complex ion in terms of the Y4- form (all six

protons dissociated). You should not take this to

mean that only the Y4- form reacts

24.10 p

16.6 p

66.2 p

0.2 p

5.1 p

0.0 p

6

5

4

3

2

1

K

K

K

K

K

K

NH+

NH

+

O OH

O

OH

O

OH

O OH



EDTA Complexes



EDTA Complexes• The equilibrium constant for a reaction of metal

with EDTA is called the formation constant,K f , or the stability constant:

• Again, K f could have been defined for any form

of EDTA, it should not be understood that only

the Y4-

reacts to form complex ion.

4

4

f

44

YM

MY MYYM

n

nnn

K

H D d f



pH Dependence of αY4-

Formation Constants for M-



Formation Constants for MEDTA Complexes

Conditional Formation Constant



Conditional Formation Constant• We saw from the fraction plot that most of the EDTA

is not in the form of Y4- below a pH ~10.

• We can derive a more useful equilibrium equationby rearranging the fraction relationship:

• If we fix the pH of the titration with a buffer,

then αY4- is a constant that can be combinedwith K f

EDTAY

EDTA

Y44 Y

-4-4

Y

EDTAM

MY

YM

MY

-4Y

4

4

4

f

n

n

n

n

K

EDTAM

MY4

f Yf -4

n

n'

K K f Yf

44 MYEDTAM K K ' nn

Example



Example• Calculate the concentration of free Ca2+ in

a solution of 0.10 M CaY2- at pH 10 and pH

6. K f for CaY2- is 4.9x1010

• At low pH, the metal-complex is less stable

22

2

f

1.0

EDTACa

CaY

x

x K

'

6105

f Yf

1010

f Yf

101.1)109.4)(103.2(6.00, pHat

108.1)109.4)(36.0(10.00, pHat

4

4

K K

K K

'

'

f Yf

224 CaYEDTACa K K '

x x x -0.1 Conc

0.1 0 0 Conc

CaYEDTACa

f

i

22

[email protected]

[email protected]

4-

62

x



Problem sets

1)Chapter 6 (Harvey): 18; 22

2) Calculate the conditional formation constants for

AlY- at pH 2 and pH 5 to explain that at pH 2, AlY-

is not favorable but Al3+ could form complexonate

at pH5.• 3) Calculate the free pAl in 100 ml of solution of

0.100 M Al3+ at pH 5 after adding 100 ml of

0.100 M EDTA;

• log Kf of AlY = 16.3; pCa = - log [Ca2+]. EDTA has

K1 = 1.02·10-2, K2 = 2.14x10-3, K3 = 6.92x10-

7 and K4 = 5.50x10-11.

4.2.3. Precipitation reaction



Solubility Product (K sp) = [products]x/[reactants]y but.....

reactants are in solid form, so K sp=[products]x

i.e. A2B3(s) 2A3+ + 3B2 – K sp=[A3+]2 [B2 – ]3

Given: AgBr(s) Ag+ + Br –

In a saturated solution @25oC, the [Ag+] = [Br – ]= 5.7 x 10 – 7 M.

Determine the K sp value.

-132-7

sp 10x3.310x7.5BrAgK

4.2.3. Precipitation reaction



Problem: A saturated solution of silver chromate was to found

contain 0.022 g/L of Ag2CrO4. Find K sp

Eq. Expression: Ag2CrO4 (s) 2Ag+

+ CrO42 –

K sp = [Ag+]2[CrO42 – ]

Problems working from K values



Problems working from K sp values.

Given: K sp for MgF2 is 6.4 x 10 – 9 @ 25 oC

Find: solubility in mol/L and in g/L

MgF2(s) Mg2+ + 2F – K sp = [Mg2+][F – ]2

I.

C.

E.

N/A 0 0

N/A +x +2x

N/A +x +2xK sp= [x][2x]2 = 4x3

6.4 x 10 – 9 = 4x3

2

23-3

-9

MgFMg10x1.24

10x4.6 x

now for g/L:L

MgFmol10x2.1 2

-3

L

MgFg 2

mol

g62.37.3 x 10 – 2

The common ion effect “Le Chatelier”overhead fig 17.16



The common ion effect Le Chatelier

What is the effect of

adding NaF?

CaF2(s) Ca2+ + 2F-

Solubility and pH



Solubility and pH

CaF2(s) Ca2+ + 2F –

Add H+ (i.e. HCl)

2F – + H+ HF

Th i ff t “L Ch t li ”



The common ion effect “Le Chatelier”

Why is AgCl less soluble in sea water than in fresh water?

AgCl(s) Ag+ + Cl –

Seawater contains

NaCl

Problem: The solubility of AgCl in pure water is 1.3 x 10 – 5 M.



Problem: The solubility of AgCl in pure water is 1.3 x 10 5 M.

What is its solubility in seawater where the [Cl – ] = 0.55 M?

(K sp of AgCl = 1.8 x 10 – 10)

AgCl(s)

Ag+ + Cl –

I.

C.

E.

N/A 0 0.55N/A +x +x

N/A +x 0.55 + x

K sp= [Ag+][Cl – ]

K sp= [x][0.55 + x]try dropping this x

K sp = 0.55x

1.8 x 10

– 10

= 0.55xx = 3.3 x 10 – 10 = [Ag+]=[AgCl]

“AgCl is much less soluble in seawater”

more Common ion effect:



more Common ion effect:

a. What is the solubility of CaF2 in 0.010 M Ca(NO3)2?

K sp(CaF2) = 3.9 x 10 – 11

CaF2(s)

Ca2+ + 2F –

[Ca2+] [F – ]

I.

C.E.

0.010 0

+x +2x0.010 + x 2x

K sp= [0.010 + x][2x]2 [0.010][2x]2 = 0.010(4x2)

3.9 x 10 – 11

= 0.010(4x2

)

x = 3.1 x 10 – 5 M Ca2+ from CaF2 so = M of CaF2

Now YOU determine the solubility of CaF2 in 0.010 M NaF.

K sp=[Ca2+][F-]2

7 2+



Answer: 3.9 x 10 – 7 M Ca2+

CaF2(s) Ca2+ + 2F –

0 0.010+x 2x

x 0.010 + 2x

K sp = [x][0.010 + 2x]2

3.9 x 10-11 =x(0.010)2

x(0.010)2

x = 3.9 x 10-7

What does x tell us

Reaction Quotient (Q): will a ppt. occur?



pp

Use Q (also called ion product) and compare to K sp

Q < K sp reaction goes

Q = K sp Equilibrium

Q > K sp reaction goes

No ppt.

Ppt. is dissolved

Problem: A solution is 1 5 x 10–6 M in Ni2+ Na CO is added to



Problem: A solution is 1.5 x 10 6 M in Ni2+. Na2CO3 is added to

make the solution 6.0 x 10 – 4 M in CO32 – .

K sp(NiCO3) = 6.6 x 10 – 9.

Will NiCO3 ppt?

NiCO3 Ni2+ + CO32 –

K sp = [Ni2+][CO32 – ]

Q = [Ni2+][CO32 – ]

Q = [1.5 x 10 – 6][6.0 x 10 – 4] = 9.0 x 10 – 10

Q < K sp no ppt.

We must compare Q to K sp.



Effect of pH on Solubility

• Sometimes it is necessary to account forother reactions aqueous ions might

undergo.

For example, if the anion is the conjugate baseof a weak acid, it will react with H3O+.

if the cation is the conjugate acid

of a weak base, it will react with OH-

You should expect the solubility to be affected

by pH.




• Sometimes it is necessary to account forother reactions aqueous ions might

undergo.

Consider the following equilibrium.

(aq)OC(aq)Ca )s(OCaC2

422

42

H2O

Because the oxalate ion is conjugate to a weak

acid (HC2O4-), it will react with H3O+.

O(l)H(aq)OHC (aq)OH)aq(OC 2423

2

42

H2O




ect o p o So ub ty

• Sometimes it is necessary to account for

other reactions aqueous ions might

undergo.

According to Le Chatelier’s principle, as C2O4

2-

ion is removed by the reaction with H3O+, more

calcium oxalate dissolves.

Therefore, you expect calcium oxalate to be

more soluble in acidic solution (low pH) thanin pure water.

Problem: 0.50 L of 1.0 x 10 – 5 M Pb(OAc)2 is combined with 0.50 L



of 1.0 x 10 – 3 M K 2CrO4.

a. Will a ppt occur? K sp(PbCrO4) = 1.8 x 10 – 14

Pb(OAc)2(aq) + K 2CrO4(aq) PbCrO4(s) + 2KOAc(aq)

then: PbCrO4(s) Pb2+ + CrO42 – K sp= [Pb2+][CrO4

2 – ]

[Pb2+]:

L

Pbmol

2

0.50 L

L

Pb(OAc)mol10x0.12

-5

2

2

Pb(OAc)1

Pb1

1 L5.0 x 10 – 6

[CrO42-]:

L

CrOmol

-2

4

0.50 L

L

CrOK mol10x1.0 42

-3

42

-2

4

CrOK 1

CrO15.0 x 10-4

Q = [5.0 x 10-6][5.0 x 10-4] = 2.5 x 10-9

compare to K sp: Q > K sp so a ppt. will occur

1 L

b find the Eq conc of Pb2+ remaining in solution after the



b. find the Eq. conc. of Pb2+ remaining in solution after the

PbCrO4 precipitates.

Since [Pb2+] = 5.0 x 10-6 and [CrO42-] = 5.0 x 10-4 and there is a

1:1 stoichiometry, Pb2+ is the limiting reactant.

PbCrO4(s) Pb2+ + CrO42 –

I. (after ppt.)

C.

E.

0 5.0 x 10-4 - 5.0 x 10-6 = 5.0 x 10-4

+x +x

x 5.0 x 10 – 4 + x

K sp = [x][5.0 x 10 – 4 + x] Try dropping the “+ x” term.

K sp(PbCrO4) = 1.8 x 10 – 14

1.8 x 10 – 14 = x(5.0 x 10-4) x = [Pb2+] = 3.6 x 10 – 11

This is the concentration of Pb2+ remaining in solution.

Complex ion formation affects to ppt. formation



Ag+(aq) + NH3(aq) Ag(NH3)+(aq)

AgCl(s)

Ag+ + Cl –

Ag(NH3)+(aq) + NH3(aq) Ag(NH3)2+(aq)

Ag+(aq) + 2NH3(aq) {Ag(NH3)2}+(aq)

formation or stability constant: 7

2

3

23f 10x1.7

NHAg)NH(AgK

K sp= 1.8 x 10 – 10

For Cu2+: Cu2+ + 4NH3 [Cu(NH

3)4]2+(aq)

K 1 x K 2 x K 3 x K 4 = K f = 6.8 x 1012

Solubility and complex ions:



Problem: How many moles of NH3 must be added to dissolve 0.050 mol of

AgCl in 1.0 L of H2O? (K spAgCl = 1.8 x 10 – 10 ; K f [Ag(NH3)2]+ = 1.6 x 107)

AgCl(s) Ag+ + Cl –

Ag+

(aq) + 2NH3(aq) Ag(NH3)2+

(aq)

sum of RXNS: AgCl(s) + 2NH3 Ag(NH3)2+(aq) + Cl –

f sp2

3

23overall K xK

NH

Cl)NH(AgK

= 2.9 x 10 – 3

Now use K overall to solve the problem:

K overall= 2.9 x 10 – 3 =

NH

Cl)NH(Ag2

3

23

2

3NH

0.0500.050

[NH3]eq = 0.93 but ..... How much NH3 must we add?

[NH3]total= 0.93 + (2 x 0.050) = 1.03 M

2 ammonia’s for each Ag+

Fractional Precipitation: “ppting” one ion at a time. Compounds must have



different K sp values (i.e. different solubilities)

Example: K sp CdS = 3.9 x 10 – 29 and K spNiS = 3.0 x 10 – 21

? Which will ppt. first? least soluble

Problem: A solution is 0.020 M in both Cd2+ and Ni2+. Just before NiS begins

to ppt., what conc. of Cd2+ will be left in solution?

Approach: Find conc. of S2 –

ion when Ni2+

just begins to ppt. since Cd2+

will already be ppting. Then use this S2 – conc. to find Cd2+.

NiS(s)

Ni2+ + S2 – K sp= 3.0 x 10 – 21 = [0.020][S2 – ]

[S2 – ] = 1.5 x 10 – 19 M when Ni2+ just begins to ppt.

So: CdS(s) Cd2+ + S2 – K sp= 3.9 x 10 – 29 = [Cd2+][1.5 x 10 – 19]

[Cd2+] = 2.6 x 10 – 10 M when NiS starts to ppt.

Activity effect



Activity effect

• Ionic strength:

• where ci and zi are the concentration and

charge of the ith ion.• Activity: True thermodynamic constants

use a species activity in place of its molar

concentration (a).

• : activity coefficient



Debye- Huckel theory









4.2.4.OXIDATION-REDUCTION

Redox reation

DEFINITIONS



• Oxidation: Loss of electrons.

• Reduction: Gain of electrons.• Reductant: Species that loses electrons.

• Oxidant: Species that gains electrons.

• Oxidation number: the oxidation number of acentral atom in a coordination compound is the

charge that it would have if all the ligands were

removed along with the electron pairs that

were shared with the central atom.

Eg,: Cu + HNO3 Cu(NO3)2 + NO + H2O

STRENGTH OF REDUCING AND



STRENGTH OF REDUCING AND

OXIDIZING AGENTS

Zn + Fe2+ Zn2+ + Fe

Which way will the reaction go?

Gr = -16.3 kcal/mole

Zn is a stronger reducing agent than Fe.

Fe + Cu2+ Fe2+ + Cu

Gr = -34.51 kcal/moleFe is a stronger reductant than Cu.

ELECTROMOTIVE SERIES



ELECTROMOTIVE SERIES

Weakest oxidant Strongest

reductant

Zn Zn2+ + 2e- Fe Fe2+ + 2e-

Cu Cu2+ + 2e-

Ag Ag+ + e-

Strongest oxidant Weakest

reductant

Schematic diagram of a Zn-Cu electro-chemical cell. The two

metal electrodes are immersed in solutions of ZnSO and CuSO



metal electrodes are immersed in solutions of ZnSO4 and CuSO4,

which are prevented from mixing by a porous partition.

ELECTROMOTIVE FORCE



ELECTROMOTIVE FORCE

Electromotive force (EMF): Theelectrical potential generated by the

half reactions of an electrochemical

cell.Consider: Zn + Cu2+ Zn2+ + Cu

At equilibrium the electrochemical

cell has to obey24.37

2

2

10][

][

Cu

Zn K



kcal K RT Go

r 8.50log3025.2

oor nFE G

F: (Faraday Constant) = 96,489 C mol-1

= 23.06 kcal/V-1

mol-1

Eo = Gr o/(2F)

= -50.8 kcal/(2 mol·23.06 kcal V-1 mol-1)

= -1.1 V

STANDARD HYDROGEN



STANDARD HYDROGEN

ELECTRODE (SHE)

• In order to assign a ranking of half-cell reactions,we arbitrarily set E = 0.00 V for the reaction:

H2(g) 2H+ + 2e-

when pH2 = 1 bar and pH = 0. In other words, Eo =0.00 V. This is equivalent to the convention: Gf

o

(H+) = Gf o (e-) = 0.00 kcal/mole.

• We then connect this SHE to any other electrode

representing a half-cell reaction and we can

obtain Eo for all half-cell reactions. This is called

the standard electrode potential.



If we always write half-cell

reactions with the electrons on

the right hand side, E0 tells us

the position of the reaction in the

electromotive series relative to

SHE.

THE NERNST EQUATION



THE NERNST EQUATION

• The value E0 refers to the EMF of a half-cellreaction when all reactants are in the

standard state, e.g., for

Zn Zn2+ + 2e-

Eo is the EMF when aZn2+ = 1.0. What if this is

not the case?

)][Re

][log(3025.2d

OxnF

RT E E o

EMF = ElectroMotive Force - a source of energy that can cause

a current to flow in an electrical circuit.



]log[

2

3025.2 2 Zn

F

RT E E o

At 25°C we have in this particular case:

]log[0296.0

]log[2

0592.0

20

2

Zn E

Zn E E o

Or in the most general case at 25°C:

)][Re

][log(

0592.0

d

Ox

n E E o

DEFINITION OF Eh



O OWe define Eh to be the EMF between a half-cell

reaction in any state, and the SHE. Forexample, Eh for the zinc reaction above is

given defined by the overall reaction: Zn +

2H+ Zn2+ + H2(g) for which:

2

2

][][log

20592.0 2

H

p Zn E Eh H o

However, for the SHE by definition, pH2 = 1 bar

and [H+] = 1 mol L-1, so:

]log[0296.0 2 Zn E Eh

o

STABILITY LIMITS OF WATER INEh H SPACE



Eh-pH SPACEUpper limit:

H2O(l) 2H+ + 1/2O2(g) + 2e-

Eh = E0 + 0.0295 log (pO2

1/2[H+]2)

E0 = Gr o/(nF) = -56.687/(2·23.06) = 1.23 V

Eh = 1.23 + 0.0148 log pO2

- 0.0592 pH

At the Earth’s surface, pO2 can be no greater than 1 bar so

Eh = 1.23 - 0.0592 pH

Lower limit:

1/2H2(g) H+ + e-

Eh = 0 + 0.059 log ([H+]/pH2

1/2)

Again, let pH2 = 1 bar.

Eh = -0.059 pH

o



pH

0 2 4 6 8 10 12 14

Eh(

volts)

-1

0

1

T = 25oC

pH

2

= 1 bar

pO2

= 1 bar O 2

H 2 O

H 2

H 2 O

Eh-pH diagram

depicting thelimits of stability

of liquid water.



Eh-pH diagram

showing theredox conditions

of various

natural

environments.

Factors affecting to the standard electrode potential-



Conditional standard electrode potential

1.Effect of pH

2.Effect of complex reagents

3.Effect of precipitation reactions

Electrode potential of a solution.



Electrode potential of a solution.

The potential buffer solution.

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Lecture 4- Equilibrium Chemistry (4h)