The study of chemical applications of electricity is known as Electro-Chemistry . It deals with chemical reaction produced through an electrolyte or production of electricity through chemical reactions .
Conductor : A material that allows electric current to pass through it is known as a conductor .
Types of Conductors :
1) Metallic conductors : In these conductors the flow of electrons is responsible for the passage of electric current and there is no change in physical and chemical nature of the metal .
Eg : Graphite , all metals etc .,
2) Electrolytic conductors : In these conductors current is carried by ions . the electrolyte which is present in either liquid or fused state and the decomposition of electrolyte take place . Eg : H2SO4 , HCl , HNO3 , NaOH , KOH
Ampere : The current which deposits 0.001118 gms of silver per second from a 15% solution of AgNO3 in a voltmeter of definite dimensions .
Coulomb : The quantity of current which passes in 1 second with current strength of 1 ampere is known a a Coulomb .
Ohm : The resistance at 0° c of a column of mercury 106.3 long and weighing 14.4521 gms is known as a Ohm .
Volt : The difference in potential required to send a current of 1 amp strength through 1 ohm resistance is known as a Volt .
Conductance : The conductance of an electrolytic solution is defined a the reciprocal of its own resistance . Conductance is given by L/C = 1/R Its units are ohm-1 (or) mho (or) siemen
Specific conductivity : The conductivity of 1 cm3 of solution is known as specific conductivity .
k = L/a x 1/R = cell constant x Conductance
UNITS : ohm-1 cm-1
Equivalent Conductivity : The equivalent conductivity of an electrolyte is defined as the conductance of all ions produced by dissociation of 1gm equivalent of an electrolyte a certain volume V of a solution at constant temperature .
λ eq or λ V = K x 1000 / C or N
Where k = Specific conductivity
C = conc of solution in gm equivalent per litre
λ eq = KV and V=1000/C
UNITS : ohm-1 cm2 eq-1
Molecular conductivity : It is defined as the conductance of all the ions produced by dissociation of 1 mol wt of an electrolyte dissolved in certain volume V of the solvent at a constant temperature .
λm = k x 1000 /M
UNITS : ohm-1 cm2 m-1
Problems :
1) The specific conductivity of a deci-normal solution of an electrolyte is 3 x 10-4 what is equivalent conductance ???
SOl : λ eq = k x 1000/c
= 3 x 10-4 x 103 / 0.1
λ eq = 3 ohm-1 cm2 eq-1
2) Specific conductance of 0.02 normal solution of an electrolyte is 0.004 ohm-1 cm-1 . The resistance of solution is 250 ohms . find the cell constant ???
SOl : k = L/a x 1/R = cell constant x Conductance
cell constant (L/a) = 0.004 x 250 = 1 cm-1
3) Resistance of an conductivity cell is found to be 180 ohms & 220 ohms when filled with 0.1 M NaCl & 0.01 N of CuSO4 electrolyte . Calculate the equi conductivity of CuSO4 if the specific conductivity of 0.1 N NaCl solution is 0.0125 ohm-1 cm-1 ???
Sol : k = L/a x 1/R of NaCl
0.0125 = L/a x 1/180
L/a = 2.25 cm-1
K of CuSO4 = Cell constant / resistance = 2.25/220
= 0.0102 ohm-1 cm-1
λ eq of cuso4 = k x 1000 / 0.01 = 0.0102 x 1000/0.01
= 1020 ohm-1 cm2 eq-1
Kohlraush's law : The equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the equivalent conductivity of its cation and anion
λ eq (at infinite) = λ+(at infinite) + λ-(at infinite)
APPLICATIONS :
1) Determination of equivalent conductivity at Infinite dilution of weak electrolytes
Eg: λ eq (at infinite) CH3COOH = λ eq (at infinite) HCl , CH3COONa , NaCl
λ eq (at infinite) CH3COOH = λ eq(at infinite) [H+] +λ eq(at infinite) [Cl-] +λ eq(at infinite)[CH3COO-] + λ eq(at infinite) [Na+] - λ eq(at infinite) [Na+] - λ eq(at infinite) [Cl-]
= λ eq(at infinite) [H+] +λ eq(at infinite)[CH3COO-]
= λ eq (at infinite) CH3COOH
2) Determination of degree of dissociation : It is the fraction of no. of molecules dissociated into ions .
α (alpha)= No of mol. dissociated into ions / total No of mol. in solution
α (alpha) = λmc/λmα NO UNITS
PROBLEMS on 2nd application:
a) At 25° c the resistance of cell filled with deci-normal KCl solution is 425 ohm the resistance of same cell when filled with 0.1 normal NH4OH is 2300 ohms . The specific conductivity of a deci-normal KCl solution is found to be 0.0012 & equivalent conducatnce of NH4OH at Infinite dilution is found to be 380 ohm-1 cm2 eq-1 . What is the degree of dissociation of NH4OH ???
Sol: k(kcl) = L/a x 1/R
0.0012 = L/a x 1/425
L/a = 0.51 cm-1
K(NH4OH) = 0.51/2300 = 2.2 x 10-4
λ eq (NH4OH) = k x 1000 / N = 2 x 10-4 x 103 /0.1
= 2.2 ohm-1 cm2 eq-1
Degree of dissociation = 2.2/380 = 0.0057
b) At 20° c equivalent conductivity at infinite dilution of NH4Cl , NaOH , NaCl are 130 , 220 & 110 the equivalent conductivity of 0.01 N NH4OH is 9.20 . calculate the degree of dissociation of NH4OH ???
Sol: λ eq (NH4OH) = λ eq NH4Cl + λ eq NaOH - λ eq NaCl
= 130+220+110 = 240
α (alpha) = λm/λ(infinite)
= 9.20/240 = 0.038
3) Determination of solubility of sparingly soluble salt : The conc of sparingly soluble salt is the solubility of the salt .
S = k x 1000 /λ(infinite) gm.eq/lit or gm mol/lit
S = k x 1000 /λm(infinite) x m gm/lit
S = k x 1000 /λeq(infinite) x e gm/lit
Problems on 3rd application :
1) The specific conductivity of saturated solution of AgCl at 25° c was found to be 3 x 10-6 ohm-1 cm-1 the K of H2) used to make solution 1.6 x 10-6 ohm-1 cm-1 the equivalent conductivity of AgCl at infinite dilution is 140 ohm-1 cm2 eq-1 find the solubility of Agcl in water at 25° c ???
SOl: K(Agcl) = K(saturated sol of Agcl) - K (H2O)
= (3.4-1.6) x 10-6 = 1.8 x10-6 ohm-1 cm-1
S = 1.8 x 10-6 x 1000 / 140 = 1.28 x 10-5 gm.eq/lit
Ionic mobility : It is defined as the total amount of current is carried by any one of the ions .
Ionic mobility α speed of ions
Ionic mobility α speed of cations (u)
Ionic mobility α speed of anions (v)
Total ions α (u+v)
Under the influence of electric field the cation and anion of electrolyte moves at different speeds and hence carry different amounts of electricity .
K = 96,500 .
Ionic mobility = Ionic conductivity / 96,500 cm/sec
since 1 equivalent of an ion under unit potential gradient carrier a charge of 1 Faraday per sec . The proportionality constant k = 96,500 coulombs .
Transport no also directly proportional to the speed of ions
Transport no of cations t+ = U/U+V
Transport no of anions t- = V/U+V
Galvanic cell : an electro chemical cell is a device for converting chemical energy into electrical energy (or) Cell which is used for production of electrical energy from chemical reactions is known as a Galvanic cell . eg: danial cell
CLICK ON THE PICTURE FOR BETTER VIEW :)
The danial cell is a typical example of galvanic cell and consists of copper rod dipping in a solution of CuSO4 which is separated with the help of a semi permeable membrane or a salt bridge from solution of ZnSO4 in which zinc rod is dipped . The Porous wall prevents the diffusion of two liquids but allows the passage of ions through it when the flow of electric current takes place .
OXIDATION : anode: Zn ----->Zn2++2e-
REDUCTION : cathode: Cu2++ 2e-----> Cu
The total cell along with the flow of electrons is described as follows :
E.M.F = 1.09 volts
Concentration Cell : A conc cell is a galvanic cell in which electrical energy is produced by transfer of material from a system of high concentration to low concentration .
TYPES :
1) Electrode conc cell : In these cells the E.M.F arises as a result of two like electrodes at different conc's dipping in the same solution of electrolyte . eg: Two hydrogen electrodes at unequal gas pressures p1&p2 immersed in same solution of H+ ions . The following cell reactions takes place
OXIDATION : H2(p1) ------> 2H+ + 2 electrons
REDUCTION : 2H+ + 2 electrons ------> H2(p2)
TOTAL REACTION: H2(p1) -----> H2(p2)
It is clear that in this process there is no over all chemical change and there is only transfer of H2 gas from electrode with pressure (p1) to the electrode with pressure (p2) . The EMF depends only on pressure and conc of H2 ions solution in which electrodes are dipped .
2) Electrolyte conc cells : These cells consists of two identical electrodes dipped in a two electrolyte solutions of different conc's . A conc cell with 2 ZN electrodes dipped in 2 sloutions of Znso4 with different conc's c1&c2 joined through semi permiable membrane or salt bridge can be represented as follows :
ZN/ZNSO4(c1) // ZNSO4(c2)/Zn
OXIDATION : Zn(c2) -----> Zn(2+) + 2 electrons
REDUCTION : Zn(2+) + 2 electrons -----> Zn(c1)
The EMF of cell can be calculated by following expression based on nernst equation :
Electrodes :
Single electrode potential (S.E.P) : The potential difference between metal and solution (or) electrode and electrolyte is called Single electrode potential .
Reference electrode : It is an electrode of standard potential with which we can compare the potentials of other electrodes .
TYPES :
1) Primary reference electrode
2) Secondary reference electrode
1) Primary reference electrode : The standard hydrogen electrode is the primary reference electrode . The EMF of such all has orbitally fixed at zero . Hydrogen electrode may be employed to find the PH value of unknown solution . The solution whose PH value is to be determined is taken in a vessel and platinum electrode is half dipped in it . Hydrogen gas at 1 atmosphere is bubbled through the solution . The platinum catalyses the reaction
H+(aq) + e- -----> 1/2 H 2 (g)
This reaction develops a definite potential at the electrode depending on the H+ ions conc of the solution under test .
The above half cell so formed is connected to a standard or normal Hydrogen electrode . The two solutions are separated by a salt bridge to eliminate liquid junction potential . The EMF of full cell is then determined by means of potentiometer . The EMF of the reference electrode being zero the observed EMF gives directly the EMF of half cell containing solution under test .
2) Secondary reference electrode :
Click on the picture for a better View :-)
Calomel electrode : It consists of tube at bottom where a small amount of Hg is placed . It is covered with a paste of solid Hg2Cl2 . A solution of Kcl is then placed over the paste . A platinum wire dipping into the mercury layer is used for making electrical contact . The side tube is used for making contact with a salt bridge . The saturated calomel electrode is formulated as mercury , Hg2Cl2 , Kcl salt solution . The electrode reaction taking place in the half cell is
1/2 Hg2Cl2 + e------> Hg + Cl -
The electrode can be coupled with H2 electrode containing solution of unknown PH. The total cell assembly is :
Pt , H2/H+// KCl.Hg2Cl2/Hg
The EMF of the cell is :
Ohm : The resistance at 0° c of a column of mercury 106.3 long and weighing 14.4521 gms is known as a Ohm .
Volt : The difference in potential required to send a current of 1 amp strength through 1 ohm resistance is known as a Volt .
Conductance : The conductance of an electrolytic solution is defined a the reciprocal of its own resistance . Conductance is given by L/C = 1/R Its units are ohm-1 (or) mho (or) siemen
Specific conductivity : The conductivity of 1 cm3 of solution is known as specific conductivity .
k = L/a x 1/R = cell constant x Conductance
UNITS : ohm-1 cm-1
Equivalent Conductivity : The equivalent conductivity of an electrolyte is defined as the conductance of all ions produced by dissociation of 1gm equivalent of an electrolyte a certain volume V of a solution at constant temperature .
λ eq or λ V = K x 1000 / C or N
Where k = Specific conductivity
C = conc of solution in gm equivalent per litre
λ eq = KV and V=1000/C
UNITS : ohm-1 cm2 eq-1
Molecular conductivity : It is defined as the conductance of all the ions produced by dissociation of 1 mol wt of an electrolyte dissolved in certain volume V of the solvent at a constant temperature .
λm = k x 1000 /M
UNITS : ohm-1 cm2 m-1
Problems :
1) The specific conductivity of a deci-normal solution of an electrolyte is 3 x 10-4 what is equivalent conductance ???
SOl : λ eq = k x 1000/c
= 3 x 10-4 x 103 / 0.1
λ eq = 3 ohm-1 cm2 eq-1
2) Specific conductance of 0.02 normal solution of an electrolyte is 0.004 ohm-1 cm-1 . The resistance of solution is 250 ohms . find the cell constant ???
SOl : k = L/a x 1/R = cell constant x Conductance
cell constant (L/a) = 0.004 x 250 = 1 cm-1
3) Resistance of an conductivity cell is found to be 180 ohms & 220 ohms when filled with 0.1 M NaCl & 0.01 N of CuSO4 electrolyte . Calculate the equi conductivity of CuSO4 if the specific conductivity of 0.1 N NaCl solution is 0.0125 ohm-1 cm-1 ???
Sol : k = L/a x 1/R of NaCl
0.0125 = L/a x 1/180
L/a = 2.25 cm-1
K of CuSO4 = Cell constant / resistance = 2.25/220
= 0.0102 ohm-1 cm-1
λ eq of cuso4 = k x 1000 / 0.01 = 0.0102 x 1000/0.01
= 1020 ohm-1 cm2 eq-1
Kohlraush's law : The equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the equivalent conductivity of its cation and anion
λ eq (at infinite) = λ+(at infinite) + λ-(at infinite)
APPLICATIONS :
1) Determination of equivalent conductivity at Infinite dilution of weak electrolytes
Eg: λ eq (at infinite) CH3COOH = λ eq (at infinite) HCl , CH3COONa , NaCl
λ eq (at infinite) CH3COOH = λ eq(at infinite) [H+] +λ eq(at infinite) [Cl-] +λ eq(at infinite)[CH3COO-] + λ eq(at infinite) [Na+] - λ eq(at infinite) [Na+] - λ eq(at infinite) [Cl-]
= λ eq(at infinite) [H+] +λ eq(at infinite)[CH3COO-]
= λ eq (at infinite) CH3COOH
2) Determination of degree of dissociation : It is the fraction of no. of molecules dissociated into ions .
α (alpha)= No of mol. dissociated into ions / total No of mol. in solution
α (alpha) = λmc/λmα NO UNITS
PROBLEMS on 2nd application:
a) At 25° c the resistance of cell filled with deci-normal KCl solution is 425 ohm the resistance of same cell when filled with 0.1 normal NH4OH is 2300 ohms . The specific conductivity of a deci-normal KCl solution is found to be 0.0012 & equivalent conducatnce of NH4OH at Infinite dilution is found to be 380 ohm-1 cm2 eq-1 . What is the degree of dissociation of NH4OH ???
Sol: k(kcl) = L/a x 1/R
0.0012 = L/a x 1/425
L/a = 0.51 cm-1
K(NH4OH) = 0.51/2300 = 2.2 x 10-4
λ eq (NH4OH) = k x 1000 / N = 2 x 10-4 x 103 /0.1
= 2.2 ohm-1 cm2 eq-1
Degree of dissociation = 2.2/380 = 0.0057
b) At 20° c equivalent conductivity at infinite dilution of NH4Cl , NaOH , NaCl are 130 , 220 & 110 the equivalent conductivity of 0.01 N NH4OH is 9.20 . calculate the degree of dissociation of NH4OH ???
Sol: λ eq (NH4OH) = λ eq NH4Cl + λ eq NaOH - λ eq NaCl
= 130+220+110 = 240
α (alpha) = λm/λ(infinite)
= 9.20/240 = 0.038
3) Determination of solubility of sparingly soluble salt : The conc of sparingly soluble salt is the solubility of the salt .
S = k x 1000 /λ(infinite) gm.eq/lit or gm mol/lit
S = k x 1000 /λm(infinite) x m gm/lit
S = k x 1000 /λeq(infinite) x e gm/lit
Problems on 3rd application :
1) The specific conductivity of saturated solution of AgCl at 25° c was found to be 3 x 10-6 ohm-1 cm-1 the K of H2) used to make solution 1.6 x 10-6 ohm-1 cm-1 the equivalent conductivity of AgCl at infinite dilution is 140 ohm-1 cm2 eq-1 find the solubility of Agcl in water at 25° c ???
SOl: K(Agcl) = K(saturated sol of Agcl) - K (H2O)
= (3.4-1.6) x 10-6 = 1.8 x10-6 ohm-1 cm-1
S = 1.8 x 10-6 x 1000 / 140 = 1.28 x 10-5 gm.eq/lit
Ionic mobility : It is defined as the total amount of current is carried by any one of the ions .
Ionic mobility α speed of ions
Ionic mobility α speed of cations (u)
Ionic mobility α speed of anions (v)
Total ions α (u+v)
Under the influence of electric field the cation and anion of electrolyte moves at different speeds and hence carry different amounts of electricity .
K = 96,500 .
Ionic mobility = Ionic conductivity / 96,500 cm/sec
since 1 equivalent of an ion under unit potential gradient carrier a charge of 1 Faraday per sec . The proportionality constant k = 96,500 coulombs .
Transport no also directly proportional to the speed of ions
Transport no of cations t+ = U/U+V
Transport no of anions t- = V/U+V
Galvanic cell : an electro chemical cell is a device for converting chemical energy into electrical energy (or) Cell which is used for production of electrical energy from chemical reactions is known as a Galvanic cell . eg: danial cell
CLICK ON THE PICTURE FOR BETTER VIEW :)
The danial cell is a typical example of galvanic cell and consists of copper rod dipping in a solution of CuSO4 which is separated with the help of a semi permeable membrane or a salt bridge from solution of ZnSO4 in which zinc rod is dipped . The Porous wall prevents the diffusion of two liquids but allows the passage of ions through it when the flow of electric current takes place .
OXIDATION : anode: Zn ----->Zn2++2e-
REDUCTION : cathode: Cu2++ 2e-----> Cu
The total cell along with the flow of electrons is described as follows :
E.M.F = 1.09 volts
Concentration Cell : A conc cell is a galvanic cell in which electrical energy is produced by transfer of material from a system of high concentration to low concentration .
TYPES :
1) Electrode conc cell : In these cells the E.M.F arises as a result of two like electrodes at different conc's dipping in the same solution of electrolyte . eg: Two hydrogen electrodes at unequal gas pressures p1&p2 immersed in same solution of H+ ions . The following cell reactions takes place
OXIDATION : H2(p1) ------> 2H+ + 2 electrons
REDUCTION : 2H+ + 2 electrons ------> H2(p2)
TOTAL REACTION: H2(p1) -----> H2(p2)
It is clear that in this process there is no over all chemical change and there is only transfer of H2 gas from electrode with pressure (p1) to the electrode with pressure (p2) . The EMF depends only on pressure and conc of H2 ions solution in which electrodes are dipped .
2) Electrolyte conc cells : These cells consists of two identical electrodes dipped in a two electrolyte solutions of different conc's . A conc cell with 2 ZN electrodes dipped in 2 sloutions of Znso4 with different conc's c1&c2 joined through semi permiable membrane or salt bridge can be represented as follows :
ZN/ZNSO4(c1) // ZNSO4(c2)/Zn
OXIDATION : Zn(c2) -----> Zn(2+) + 2 electrons
REDUCTION : Zn(2+) + 2 electrons -----> Zn(c1)
The EMF of cell can be calculated by following expression based on nernst equation :
Electrodes :
Single electrode potential (S.E.P) : The potential difference between metal and solution (or) electrode and electrolyte is called Single electrode potential .
Reference electrode : It is an electrode of standard potential with which we can compare the potentials of other electrodes .
TYPES :
1) Primary reference electrode
2) Secondary reference electrode
1) Primary reference electrode : The standard hydrogen electrode is the primary reference electrode . The EMF of such all has orbitally fixed at zero . Hydrogen electrode may be employed to find the PH value of unknown solution . The solution whose PH value is to be determined is taken in a vessel and platinum electrode is half dipped in it . Hydrogen gas at 1 atmosphere is bubbled through the solution . The platinum catalyses the reaction
H+(aq) + e- -----> 1/2 H 2 (g)
This reaction develops a definite potential at the electrode depending on the H+ ions conc of the solution under test .
The above half cell so formed is connected to a standard or normal Hydrogen electrode . The two solutions are separated by a salt bridge to eliminate liquid junction potential . The EMF of full cell is then determined by means of potentiometer . The EMF of the reference electrode being zero the observed EMF gives directly the EMF of half cell containing solution under test .
2) Secondary reference electrode :
Click on the picture for a better View :-)
Calomel electrode : It consists of tube at bottom where a small amount of Hg is placed . It is covered with a paste of solid Hg2Cl2 . A solution of Kcl is then placed over the paste . A platinum wire dipping into the mercury layer is used for making electrical contact . The side tube is used for making contact with a salt bridge . The saturated calomel electrode is formulated as mercury , Hg2Cl2 , Kcl salt solution . The electrode reaction taking place in the half cell is
1/2 Hg2Cl2 + e------> Hg + Cl -
The electrode can be coupled with H2 electrode containing solution of unknown PH. The total cell assembly is :
Pt , H2/H+// KCl.Hg2Cl2/Hg
The EMF of the cell is :
Saturated Kcl Reduction potential is 0.2422 Volts , 0.1N Kcl reduction potential is 0.3335 Volts , 1 N Kcl reduction potential is 0.2810 Volts .
Problems :
1) Saturated Kcl Reduction potential is 0.2422 Volts , 0.1N Kcl reduction potential is 0.3335 Volts , 1 N Kcl reduction potential is 0.2810 Volts and the EMF of the following cell at 25° c is 0.445 volts .
cell : pt,H2 (g)/H+//Kcl(saturated)/Hg2Cl2 ,Hg
Calculate PH . Given that O.P of saturated calomel electrode is 0.2422 .???
Sol: Ecell = ER - EL
0.445 = 0.2422 - EL
EL = 0.2422 - 0.4450
PH = 0.445 - 0.2422
_____________
0.0592
2) What is single electrode potential (S.E.P) of half cell for Zn electrode dipping in a 0.01M Zn solution . Eo= 0.763 Volts . Find Electrode potential of Zn ???
Sol: E= Eo - E(unknown)
E((unknown) = - 0.0591 log[0.01]
--------------
2
= -0.0591
E = 0.763 + 0.0591 = 0.8814
3) Calculate EMF of following reaction at 25° c .
Cu2++Zn -----> Cu + Zn2+ .. EZn=0.763 V , ECu =-0.337
Sol : E= ER - EL = 0.763 = 0.337
= 0.1 Volts
Glass electrode : It is made of a special glass of relatively low M.P and high electrical conductivity . It is a corning glass containing Na2O - 22% , CaO - 6% and SiO2 - 72% . The assembly of glass electrode consists of a thin glass bulb and this electrode is filled with 0.1M HCl and a silver wire coated with AgCl is immersed in it here Ag/AgCl acts as internal reference electrode . The glass electrode is representated as Ag/AgCl/0.1M HCl/Glass .
The measurement of PH the Glass electrode is immersed in the solution whose PH is to be determined . It is then combined with a reference saturated calomel electrode . The cell assembly is represented as
Ag/AgCl/0.1M Hcl/Glass/Unknown solution (H+)//Kcl.Hg2Cl2/Hg
The EMF of the cell at 298k is as
Ecell = ER - EL
= 0.2422-[EG]+(0.0592 PH)
PH = 0.2422 -Ecell -EoG
__________________
0.0592
Glass electrode potential(EoG ) is 1st measured by dipping the electrode in buffer solution of known PH values .
ClicK on the picture for a better view :-)
Quin Hydrone electrode : It can be very easily set up by adding a pinch of quin hydrone powder to the experimental solution is stirred until the solution is saturated and a slight excess of it remain undissolved . Then indicator electrode of brigth platinum is inserted in it for determining PH value this half cell is combined with other reference electrode i.e., saturated calomel electrode and the EMF of cell is determined potentiometrically . The cell is may be represented as
Pt/QH2 QH+//Kcl.Hg2Cl2/Hg+
When a platinum electrode is immersed in a solution containing quin hydrone . A potential is established the value of which is given by
E=EoQ - 2.303 x 8.318 x 107 x298 Log10 QH2
--------------------------------- ----------
2 x 96,500 [Q][H+]2
The quinone & hydro quinone are taken in equi-molar conc's then
Quinone = Hydro quinone
Potentiometric Titration's : A titration in which the equivalent or end point of the reaction is determined with help of the potentials of the reaction mixture is known as potentiometric titration .
TYPES:
1) Acid-Base Titration : The acid solution whose strength has to be determined is taken in a beaker and hydrogen electrode and calomel electrode were dipped in the solution with the help of a salt bridge . The electrodes are connected to the potentiometer and EMF is measured . A known volume of standard alkali solution (NaOH) is added from a burette stirred thoroughly and EMF of cell is recorded . Like this 10-15 readings are recorded by repeating the procedure of addition of standard alkali solution . The volume of alkali added is plotted against EMF observed. The sweetest portion of the curve indicates the equivalent point of the solution .
2) Oxidation & Reduction titration : The procedure adopted for oxidation & reduction titration's is same as in acid-base titration's . The only difference is that the electrode reversible to Hydrogen ions is replaced by a bright platinum electrode. The EMF of the electrode is determined by the activity of ratio of substance being oxidised are reduced .
Eg : The Ferrous iron is taken in a beaker and titrated with potassium chromate and platinum electrode and calomel electrode are dipped. The two electrodes are connected to potentiometer (PM) and EMF of solution after addition of potassium chromate . The step rise is the end point of the titration .
3) Precipitation reaction : In these reactions also an electrode reversible to one of the ions involved is made use of , for example : titration of AgNO3 with NaCl where AgCl is precipitated out . Ag electrode is used along the calomel electrode . The AgNO3 is placed in the micro burette and NaCl is taken in a beaker containing electrodes . The EMF of cell is measured and plotted against volume of AgNO3 added . The step rise in the curve shows the end point of the titration .
4) Determination of PH by EMF method : The EMF of a solution depends on the conc of hydrogen ions or PH of the solution . A hydrogen electrode containing solution of unknown PH is paired with a standard calomel electrode . The complete cell may be represented as
Pt,H2/H+//KCl.Hg2Cl2/Hg+
The EMF of cell is measured by potentiometer and PH of unknown solution can be calculated as follows :
Batteries :
Definition : When two or more electro-chemical cells are electrically inter connected each of which consists of 2 electrodes and an electrolyte is called a battery .
Types : There are two types of batteries depending on recharging capacities , they are :
1) Primary batteries /cells (non - rechargeable)
2) Secondary batteries / cells (rechargeable)
1) Primary batteries : These are non-rechargeable and meant for a single use and meant to discharge after the use . Lithium cells belong to this cells . The cells having Lithium can be classified into 2 categories :
a) Lithium cells with solid cathode : In these cells lithium is anode and MnO2 is cathode . The cathode MnO2 should be heated to more than 300o c to remove water by incorporating it in the cathode . This is very important to improve the efficiency of the battery . The electrolyte is a mixture of propylene carbonate and 1-2 di methoxy methane .
Anodic reaction : Li ------> Li+ + e-
Cathodic reaction : MnO2 + e- + Li
Net reaction : Li + MnO2 ------> LiMnO2
Applications :
1) The cylindrical cells are used in fully automatic cameras .
2) Coil cells are widely used in electronic devices such as calculators , watches .
b) Li cells with liquid cathode : In this cell the cathode constitute 50% - SO2 with solvents such as acrylonytryl or propylene carbonate or mixture of both . The cell reaction can be represented as
Li+ + 2SO2 ----> LiS2O4
Thyonil chloride also can be used as a liquid cathode . The cell reaction can be represented as :
4Li + 2SOCl2 ------> 4LiCl + S + SO2
Applications :
1) Used on electronic circuit board for supplying fixed voltage for memory protection and other stand by functions .
2) Used for military and space applications .
3) Used in medical devices such as neuro-stimulator's , drug delivery systems .
2) Secondary Batteries : These are rechargeable and have very large capacitance and long period of low current discharge . eg : Lead acid storage cells & Nickel - cadmium cells .
Lead acid storage cells : This cell has a great advantage of working both as an electrolytic cell and as a voltaic cell . It consists of lead anode and lead dioxide as cathode . Lead dioxide is used in form of a paste . A no of lead plates are connected in parallel and a no of lead di oxide plates are also connected in parallel . The lead plates fit in between lead dioxide plates . The plates are separated by insulators like strips of wood rubber or glass fibre . The entire combination is immersed in 20-21% dil H2SO4 . The discharging of storage cell is operating as a voltaic cell wher the oxidation of Pb takes place .
Anodic reaction : Pb ----> Pb2+ + 2e-
The Pb2+ combines with SO42- (sulphate ions) to produce PbSO4
Pb2+ + SO42- -----> PbSO4
The 2 electrons released at anode close to the lead dioxide electrode and causes reduction of lead dioxide to produce Pb2+ which finally combine with SO42- to produce PbSO4 .
Cathodic reaction : PbO2 + 4H+ + 2e- ------> Pb2+ + 2H2O
Pb2+ + SO42- -------> PbSO4
The lead accumulator for car consists of lead and lead storage cells in series and is capable of delivering 12 Volts . The net reaction is
Pb + PbO2 + 4H+ + SO42- -------> 2PbSO4 + 2H2O + Energy .
Charging : During the charging or re charging an external EMF> 2 volts is passed so that the cell reactions are reversed as
anode : 2PbSO4 + 2e- ------> Pb + SO42-
Cathode : PbSO4 + 2H2O + 2e- ------> PbO2 + 4H+ + SO42- .
Net reaction : 2PbSO4 + 2H2O + Energy ------> Pb + PbO2 + 4H+ + SO42- .
The positive pole of the generator is attached to positive pole of battery and negative pole is connected to the negative pole of the battery of generator . During the discharging operation the conc of H2SO4 decreases while the conc of the acid is restored during charging . Battery containers are made up of hard rubber and palstic .
Applications : Used to supply current for electric vehicles , gas engine ignition in telephone exchanges , electric trains , laboratories , hospitals , broad castings stations , auto mobiles , power stations .
Nickel - Cadmium cells : This battery consists of a Cadmium anode and a cathode composed of a paste of NiO(OH) . The cell reactions are
Cd(s) + 2OH- -----> Cd(OH)2 + 2e-
NiO(OH) + 2e- + 2H2O ----> 2Ni(OH)2 + 2OH-(aq) .
The reaction can be easily reversed because the reaction products nickel hydroxide & cadmium hydroxide adhere to the electrode surface . Nickel-Cadmium cell give a voltage1.4 V and can be connected in series to give nickel-cadmium storage batteries . There are two types :
1) Pocket - plate Nickel-cadmium battery : The material used for cathode and anode are filled into pocket plates (two perforated nickel coated steel strips) and connected in series . The cells have no longer life i.e., more than 20 yrs with capacities between 10 - 1000 amp and maintains the voltage of 1.4-1.45 Volts .
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2) Sintered plate Nickel-cadmium cells : The cathode materials are finely nickel- sintered in a mould . A Nickel screen impregnated with Nickel nitrate and process to produce Ni(OH)2 in the course for anode the moulds are impregnated with Cadmium-salt processed to get Cd(OH)2 . The electrodes are separated by separators such as porous polymer membrane the electrolytic solution is KOH . The sealed cells are made in 2 different designed plates :
1)Horizantal disc (button cells)
2)spirally wound cylindrical cells
Applications :
1) Used for aircrafts & diesel engine starting .
2) lighting of trains .
3) Emergency power supply .
4) many military applications .
Fuel cell : It is an electro chemical cell which converts chemical energy contained in a easily available fuel oxidant system into electrical energy . The basic principle of fuel cells are identical to those of the well known identical cell . The only difference is that in fuel cell the chemical energy is provided by a fuel and an oxidant stored outside the cell .
H2O2 Fuel cell : The cell consists of two inert porous electrodes made of graphite impregnated with finely divided platinum or 75%of alloy of lead with 25% of silver or nickel and an electrolytic solution 2.5% of KOH . Through the anode Hydrogen gas is bubbled and through the cathode Oxygen gas is bubbled . The following reactions takes place :
2H2 -----> 4H+ + 4e-
O2 + 4H+ + 4e- -----> 2H2O
The product discharged as water and the standard EMF of cell is 1.23 Volts . A no of such fuel cells are stacked together in series to make a battery .
Applications :
1) Used as axillary energy source in space vehicles , sub marines or other military vehicles .
2) Because of light weight these fuel cells are preferred for space craft and product water is valuable space water for astronauts .
Advantages :
1) energy conservation is high 70-82% .
2) The product water is drinking water source for astronauts .
3) noise and thermal pollution are low .
4) fuel cell offer an excellent method for use of fossil fuels .
5) maintenance cause is low, fuel transportation costs , etc., .
6) This cell is an energy storage systems for space applications .
Limitations :
1) the main limitation of fuel cell lie in high initial costs associated with electrode material and design costs . The later could be brought down with mass production and suitable R&D in this direction . The low maintenance and longetivity of cells could compensate some of these negative factors .
2) large weight and volume of Hydrogen gas-fuel storage system .
3) lack of infrastructure for distribution and marketing of hydrogen gas . The developing natural gas distribution net-work could pave way for this in future .
4) High cost of Hydrogen gas . Technological advances should bring the costs down .
5) Most alkaline fuel cells suffer from carbon dioxide leakages and should be properly filtered and prevented from entering the cell and reacting with electrolyte .
