ELECTRO CHEMISTRY. For this Unit    12 Hours is prescribed p

in the  Syllabus.It carries Questions of 19 marks in It carries Questions of 19 marks in 

annual Exam.   Out of these  usually O k i  i   t A    F k  Onemark is in part A,   Fourmarks are in part B,   11 marks are in part C 

d h k fand Three marks are in D1 part of the Question Paper. It is the very 

3 February 2012

Q p yimportant chapter in syllabus

ELECTROCHEMISTRYELECTROCHEMISTRY

It deals with the inter convertion of electrical energy into chemical energy.electrical energy into chemical energy.Electrolytes – Conducts electricity in 

lt      St t   E      N Cl KClmolten or aq. State.  Ex :   NaCl,KClNon–Electrolyte – Not to conduct yelectricity either in molten or aq.  State    Ex:Glass,Benzene Ex:Glass,Benzene 

3 February 2012

Two types of electrolytes:Two types of electrolytes:Strong electrolyte: Ionize completely in a moderate concentration Ex: KNO NaOHmoderate concentration. Ex: KNO3, NaOH

i i ll i dWeak electrolyte: Ionize partially in a moderate concentration. Ex: CH3COOH, NH4OH3 4

Degree of ionization =eg ee o o at o

3 February 2012

Factors  influencing electrical conduction.1) Nature of electrolyte 2)Nature of solvent1) Nature of electrolyte. 2)Nature of solvent.3)High salvation or water of hydration,4)Dil ti 5)T t4)Dilution 5)Temperature.Faradays laws of electrolysis.i  l   f  dFirst law of Faraday:

m  α Q   or   m = z . I.t.  Where  Q=I. t  m= mass depositedZ =  E/96500 = Electrochemical equivalenceZ    E/96500   Electrochemical equivalence

I = strength of current in amperest =  time in seconds t =  time in seconds 

3 February 2012

Faraday’s second law of electrolysis : = Faraday s second law of electrolysis :mass deposited  α 1/ equivalent massm /E = m /Em1/E1 = m2/E2

=

3 February 2012 Dr.N.N.HOSUR

Arrhenius theory of ionization.1 Splitting up of molecules electrolyte into ions1.Splitting up of molecules electrolyte into ions 

is called ionization .2 Ionization is reversible2.Ionization  is reversible.3.Ions  migrate towords oppositely charged 

electrode4.Total  ‐ve charge is equal to total  +ve charge 

and solution is electrically neutral5.Charge possesed by  ions is equal to its 5.Charge possesed by ions is equal to its

valency.6 properties of electrolyte is equal to sum of6.properties of electrolyte is equal to sum of

the properties of its ions3 February 2012

7.Equilibrium is establishedMerits of Arrhenius theoryMerits of Arrhenius theory.1.  X – ray spectrum   of electrolyte indicate 

the  presence of ionsthe  presence of ions.2. Constance of enthalpy of Neutralisation of 

d bstrong acid & strong base.3. Abnormal colligative properties can be 

explained.4. Common colour of the salt containing 4. Common colour of the salt containing 

common ion.5 Qualitative analysis of  inorganic salts 5.Qualitative analysis of  inorganic salts 

indicate  the exhistance of ions3 February 2012

Demerits of Arrhenius theory.1 Fail to explain conduction in molten state1.Fail to explain conduction in molten state.2.Fail to explain mechanism of ionisation.3 Fail to explain behavior of St electrolyte3.Fail to explain behavior of St. electrolyte

S ifi   d   (k)  Specific conductance  (k): The conductance produced by the solution of 1m3 placed between two electrode  is called specific conductance. specific conductance. Unit of Sp. Conductance  Sm‐1. 

3 February 2012

Molar conductance.. The conductance of a solution placedThe conductance of a solution placed

between two electrodes containing one mole of electrolyte is called Molarmole of electrolyte is called Molar conductance.

f 2 1Unit of Molar conductance Sm2 mol-1.Theories of Acid base concept.

Arrhenius theory of acid base concept:Acid donate  H+ ion in aq. Medum.Acid donate  H ion in aq. Medum.

HCl     → H+ + Cl‐Base donate  OH‐ ion in aq  MedumBase donate  OH ion in aq. Medum.

NaOH     → Na+ + Cl‐3 February 2012

Limitations: 1) cannot explain the acidic nature of1) cannot explain the  acidic nature ofCO2,   SO2  and  the basic nature of CaO , MgO.2) Not capable to explain acidic and basic2) Not capable to explain acidic and basic nature of compounds other than aq. medium) h f3) Non exhistance of  H3O+

L   B t d  id b   tLowry –Bronsted acid‐base concept.Acid donate proton ( H+ ),  Base accept proton.

HCl    +   H2O           H2O+   + Cl‐2 2Acid‐1       Base‐2 Acid‐2       Base‐1

3 February 2012

Acid ‐base  conjugative pair.A pair of acid and base with a differenceA pair of acid and base with a difference

of proton is called acid – base conjugative pairr

HCl +   NH3 NH+4 +  Cl‐

Weaker is the acid, then stronger is the , gconjugate base and vice –versa.

Limitation: Limitation: 1)  Fail to explain acidic nature of Al Cl3 ,   BF3 2)  Fail to explain the basic nature of CaO & MgO2)  Fail to explain the basic nature of CaO & MgOSubstance which act as both acid and  base    ll d  hi tiare  called amphiprotic

3 February 2012

Lewis concept of acid – base  Acid is a substance which accept a pair ofAcid is a substance which accept a pair of 

electron and base is a substance which donate lone pair of electronslone pair of electrons.Al Cl3,   BF3 are acids.       NH

+3 , Cl

‐ , OH‐ are baseLimitations:‐1) Fail to explain acidic nature H Cl, HNO3etc.3 2) Cannot explain rel.strength of acid , base naIonic product  of water:Ionic product  of water:

Water ionise to a small extent as fallows.H2O (l)   +  H2O (l)    H3O+

(aq) +  OH‐ ( aq)H2O (l)   +  H2O (l)    H3O (aq) +  OH ( aq)K[H2O]2 =   [H3O+][OH‐]

3 February 2012

KW = [H+][OH‐]=   1x10‐14   mol2  dm‐6  at298KNuetral  solution  Nuetral  solution  

[H+]=[OH‐]=  1.x10‐7 mol2  dm‐3

Solution is acidic if Solution is acidic if = [H+] >[OH‐],…[H+] >  10‐7 M

S l i  i  b i   if   Solution is basic  if   [OH‐] >[H+],…[OH‐] >  10‐7 M

3 February 2012

pH of a solutionpH = ‐log [H+] and    pOH = ‐log  [OH‐]pH    log10[H ] and    pOH    log10 [OH ]Relation between pH and pOHFrom the ionic product of water From the ionic product of water 

Kw = [H+][OH‐]= 10‐14T ki     l     b h  idTaking   ‐log  as both side,‐log (H+)   ‐log (OH‐) = +14pH   +  pOH =  14

Neutral solution   pH  = 7  pAcid solution    pH   <   7Basic solution pH > 7

3 February 2012

Basic  solution   pH   >   7

Ostwalds dilution law:Ostwalds dilution law:Weak electrolyte :dissociate partially into ions  with degree ofdissociate partially into ions, with degree ofionization  α at molar concentration. “C” 

HA( )   H       AHA(aq)   H+   +   A‐At equi. C(1‐α)         Cα     Cα.Applying law of mass action K = 

3 February 2012 Dr.N.N.HOSUR

Ka=[H+] =

= 

Ka= Cα2 (α<<1)

]

[H+]  =Ka    Cα (α<<1) [H ]  =

Similarly for a weak base   [OH ] [OH‐]  =

3 February 2012

Common   ion effect:Supression of the degree of dissociation ofSupression of the  degree of dissociation of

weak electrolyte  by adding common ion.NH OH          NH+ + OH‐ (weak base)NH4OH          NH+

4 + OH (weak base)NH4Cl      → NH+

4 +  Clff l iBuffer solution :Resist  to the change of pH  value by the 

addition of small quantity and of acid   or  small quantity of base.small quantity of base.Types of of buffer  i) Acidic buffer  

ii) Basic buffer      iii) Neutral bufferii) Basic buffer      iii) Neutral buffer3 February 2012

Mechanism of acidic buffer: k l h bweak acid and its salt with strong base 

CH3COOH  CH3COO‐ + H+        (Weak Acid)3 3

CH3COONa     →      CH3COO‐ +   Na+ 

By the addition of acidBy the addition of acid H++ CH3COO- CH3COOH+ H2O

By  the addition of base   By  the addition of base   OH‐+ CH3COOH CH3COO‐+ H2O.H  th   H  l  th   l ti   i    Hence the pH value the solution remain un change by the addition small quantity of acid or base.

3 February 2012

Mechanism of basic Buffer:NH OH  NH + + OH‐NH4OH  NH4 + OHNH4Cl→ NH4

+ + Cl‐

addition of acid,H  NH OH  NH OH   H OH++ NH4OH  NH4OH + H2O

NH4+ +OH‐ NH4OH

Hence the pH of the solution remain punchange by the addition small quantity of acid or baseacid or base.

3 February 2012

For a weak acid pKa = ‐log10kaFor a weak base pKb = ‐log10kbFor a weak base pKb =  log10kbSmaller is pka  stronger is acid acidSmaller is pkb stronger is baseSmaller is pkb stronger is base.

Handerson equation:For acidic buffer      pH=pKa+log [Salt]/[Acid]For basic bffer pOH=Kb+log [Salt]/[Base]For basic bffer pOH Kb+log [Salt]/[Base]

3 February 2012

Solubility product:K =[Ag+] [C1‐]Ks=[Ag ] [C1 ]   

Salt                Ksp Salt                 KspAgCl 1 7x10‐10 PbCl 1 7x10‐5AgCl 1.7x10 10 PbCl2 1.7x10 5

CdS 3.6x10‐29 PbS 3.4x10‐28C S SCuS 8.5x10‐45 ZnS 2.5x10‐22MnS 1.4x10‐15 Al(OH)3 8.5x10‐233Fe(OH)3 1.1x10‐26Precipitation occurs, when Ksp< Ionic Precipitation occurs, when Ksp< Ionic product  then the solution is super saturatedNo Precipitation when Ksp> Ionic product  No Precipitation when Ksp> Ionic product  then the solution is unsaturated

3 February 2012

Application of solubility product and common ion affect in qualitative analysis common ion affect in qualitative analysis 

of in organic analysisII Group: Reagents    HCl and H SII Group: Reagents,   HCl and H2SHCl    → H+ +  Cl‐H S  H SH2S  2H+ +S‐

Role of HCl: To suppress the dissociation of H2S. d  [S2 ]      i i   l   h  to decrease [S2‐],  to precipitate only the 

second group  basic  radical, as Ksp < Ionic products and to prevent the precipitation of higher group basic radicalsg g p

3 February 2012

III  GroupNH OH  NH + + OH‐NH4OH  NH4 + OHNH4Cl→ NH4

+ + Cl‐

Role of   NH4 Cl  :T     h  di i i   f  NH OH      To suppress the dissociation of  NH4OH , to  decrease the [OH], to precipitate only the III group basic radicals ,as Ksp < Ionic products   and to prevent the higher group basic radicals p g g pfrom precipitation

3 February 2012

IV   Group: G t NH Cl NH OH H SGroup reagent.  NH4 Cl+ NH4OH+H 2SNH4OH  NH+

4+ OHNH4Cl →        NH4

+ CTH2S          2H+ + S‐2H+ + OH‐ H 2O

Role of NH Cl and NH OHRole of  NH4 Cl   and   NH4OH

To increase the [S‐2] of ion and to precipitateTo increase the [S 2] of ion and to precipitatethe      IV group basic radicals

3 February 2012

ELECTRODICSSingle electrode potential (E)Single electrode potential (E)

Zn Zn2+ + 2eZn Zn + 2e

C 2+ 2 CCu2+ + 2 Cue3 February 2012

• Zinc ionise by leaving two electrons on itssurface the negatively charged electrode issurface, the negatively charged electrode isattracted by positively charged Zn2+ ions.As a result a electrical double layer isAs a result a electrical double layer isdeveloped. Formation of an electricald bl l i i t t ti ldouble layer gives rise to potentialdifference across the layer. This potentialdifference is called electrode potential.Oxidation occures at zinc electrode, thisO da o o u a od ,electrode act as anode.

3 February 2012

Opposite process occures on copperelectrode Cu2+ ions get two electrons fromelectrode. Cu ions get two electrons fromCu electrode to form Cu, electrical doublelayer is developed Reduction occures onlayer is developed. Reduction occures onits surface. This electrode act as cathode.St d d l t d t ti l (E0) It iStandard electrode potential (E0): It isthe electrode potential developed when theelectrode is in contact with a solution ofunit concentration of its ion at 298K, if theu o a o o o a 98 ,gases are involved then the gas must be atone atm of pressureone atm of pressure.

3 February 2012

NERNST EQUATION For areversible electrode reactionreversible electrode reaction,

Mn+ + ne M.

E = E0 + [Mn+]logn

0591.0

E = E0 + [Mn+] Q [M] =1log303.2xRT

n

E = E0 + [M ], Q [M] =1lognF

Problem: 1 [MSO4] = 0.5Mdegree of dissociation = 0 95 at 298Kdegree of dissociation = 0.95 at 298K.E0 = – 0.76 V E = ?

3 February 2012

Solution: M2+ + 2 M.e

∴[M2+] = 0.95x 0.05 = 0.045M.

E = E0 + log [M2+]0591.0

0 76 l 0 04505910

E E + log [M ]n

0591.0

= - 0.76 + log 0.0452

0591.0

0 76 0 0397 V= - 0.76 – 0.0397 V

= 0 7997 v= ‐0.7997 v3 February 2012

2. For a cell reactiono a a oCd + Cu2+ Cd2+ + Cu.

[Cu2+] = 0.02M, [Cd2+] = 0.002 M,EO

Cu = + 0.34V E0Cd = - 0.40VE Cu + 0.34V E Cd 0.40V

E0cell=E0

Cu–E0Cdcell Cu Cd

= + 0.34 – (- 0.40)= 0.74 v0.74 v

3 February 2012

E = E0 – log.0591.0

][][][][

2

2

CdCuCuCd

+

+

][][

CdCu

n ][][ CdCu

= + 0 74 – log0591.0 002.0][Cd

= + 0.74 – log.2 02.0

0 74 0 0295 l 0 1= 0.74 – 0.0295 log 0.1= 0.74 – 0.0295 log 0.1 = 0.7695V.

3 For a cell reaction3. For a cell reactionFe + 2Ag+ Fe2+ + 2Ag

3 February 2012

Given that [Fe2+] = 0.01M, [Ag+] =0.1 M.E0

Ag = 0.8 V E0Fe = - 0.44 V, E0

cell= E0

Ag – E0FeAg Fe

=+0.80–(-0.44)= 1 24 V1.24 V

From the Nernst equation.22 ]Ag][Fe[ +

Ecell = E0cell - log

n0591.0

2]Ag[]Fe[]Ag][Fe[

+

= 1.24 - log0591.0

2

2

]A[

]Fe[+

+

gn 2]Ag[ +

3 February 2012

1. Galvanic cell or Voltaic cell:These cells are spontaneous∆G = -ve. These cells are used in∆G ve. These cells are used inwatches, hearing aids, flash lights, RadioSets electric clock photo flash vehicleSets, electric clock, photo flash, vehiclebatteries, etc.

3 February 2012

2. Electrolytic cell: These cellsare not spontaneous and ∆G = +ve

Eg: Nelson cell, Down’s celltetc.

3 February 2012

Comparison of Electrochemical Cells

galvanic electrolyticg y

needpower

produces electrical power

sourcetwoelectrodes

electrical current

anode ( ) anode (+)d i anode (-)cathode (+)

anode (+)cathode (-)

salt bridge vessel

conductive medium

salt bridge vessel

∆G < 0∆G > 0

3 February 2012

Galvanic cell, Daniell Cell

LUIGI GALVANI

ALESSANDROVOLTA

3 February 2012

Zn acts as anode or negatively charged,Cu acts as cathode or +vely charged.Each of these electrodes are called halfcells.

EMF of a cell (E): It is a driving forceEMF of a cell (E): It is a driving forceto drive the electrons from anode tocathodecathode.

3 February 2012

Standard EMF (E0): It is the EMF of thell h th t t d d t i thcell when the reactants and products in the

cell reaction are at unit concentration, at298 K and at one atm. of pressure.

∆G = - nFE,For a cell E must be positive as n andFor a cell E must be positive as n andF are positiveE = ER – EL = Ered – E0x.

3 February 2012

Hydrogen Electrode

Equation

H2 2H+ + 2Platinised Blacke2

Limitations:

3 February 2012

Determination of electrode potential

3 February 2012

Electrochemical series:Electrode System E0 in v.Li+ + Li - 3 05eLi + Li 3.05 Mg2+ +2 Mg - 2.37Al3+ + 3 Al - 1.66e

ee

3 66Zn2+ + 2 Zn - 0.76Fe2+ + 2 Fe - 0.44

e

e2H+ + 2 H2 0.00Cu2+ + 2 Cu + 0.34

ee

Ag+ + Ag + 0.80Cl2 + 2 2C + 1.36A 3+ 3 A 1 50

ee

Au3+ + 3 Au + 1.50F2 + 2 2 + 2.87F

ee3 February 2012

Application of electrochemical series:1 Predicting the oxidising and1. Predicting the oxidising and

reducing ability 2. To Predict the displacement of metal

from other metalic salt soln.o ot e eta c sa t so3. To predict the cell reaction.

4. To predict the displacement ofhydrogen from acids.y g

5. To predict the EMF of cell

f llEMF of cell = ERed - EOxi3 February 2012

Corrosion: It is a electrochemicalphenomenonphenomenon

3 February 2012

At anodic regionF F 2+ 2At Cathodic regionFe → Fe2+ + 2 e

In acidic medium4H+ + O2 + 4 → 2H2O.e2 2In neutral and alkaline medium

e

H

2H2O + O2 + 4 → 4Oe

3 February 2012

2Fe2+ + 4O → 2Fe (OH)2.H

4 Fe (OH)2 + O2 + 2H2O → 2 Fe2 O3. 3H2O

F t i fl i th i

(Rusted Iron)

Factors influencing the corrosion1. Nature of metal

2. Difference in potential between2. Difference in potential between anodic and cathodic region

3 February 2012

3. Nature of corrosion product

4. Ratio of anodic and cathodic area

5. Hydrogen over voltage

6. Temperature

7 pH7. pH

8 Presence of Oxidising agents8. Presence of Oxidising agents

3 February 2012

Corrosion control:

1. Protective coating i. Metal Coating ii. Inorganic coatingiii. Organic coatingg g

2 Corrosion inhibitors i Anodic inhibitors2. Corrosion inhibitors i. Anodic inhibitorsii. Cathodic inhibitors

3. Cathodic protection i. Sacrificial methodsii I d th dii.Impressed methods

3 February 2012