electro chemistry type 1

32 Pages·2011·0.64 MB·English

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Subject : CHEMISTRY
Topic : ELECTRO CHEMISTRY
R
Index
1. Key Concepts
2. Exercise I
3. Exercise II
4. Exercise III
5. Exercise IV
6. Answer Key
7. 34 Yrs. Que. from IIT-JEE
8. 10 Yrs. Que. from AIEEE
Student’s Name :______________________
Class :______________________
Roll No. :______________________
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KEY CONCEPTS
M.
E
H
ELECTROCHEMICAL CELLS C
O
An electrochemical cell consists of two electrodes (metallic conductors) in contact with an electrolyte R
T
(an ionic conductor). C
E
An electrode and its electrolyte comprise an Electrode Compartment. L
E
2
3
Electrochemical Cells can be classified as: of
2
(i) Electrolytic Cells in which a non−spontaneous reaction is driven by an external source of current. e
g
a
P
(ii) Galvanic Cells which produce electricity as a result of a spontaneous cell reaction
Note: In a galvanic cell, cathode is positive with respect to anode.
L
In a electrolytic cell, anode is made positive with respect to cathode. A
m P
O
o H
cELECTROLYSIS B
.
s
asse Tmheeta dlse coor mlibpeorsaittiioonn o off g ealesecstr aotl yetleec storoludteios nis b kyn poawsnsa agse e olefc etlreocltyrsiics .current, resulting into deposition of 81 ,
l 8
c 8
o 5
k 0
eELECTROLYTIC CELL 3
w.t This cell converts electrical energy into chemical energy. 89
w 9
w The entire assembly except that of the external battery is 0
known as the electrolytic cell 0,
e: 0
t 0
bsiELECTRODES 00
we The metal strip at which positive current enters is called anode; anode is positively charged in electrolytic 32
m cell. On the other hand, the electrode at which current leaves is called cathode. Cathodes are negatively )-
5
o 5
r charged. 7
f 0
ge H: (
ka Anode Positive Loss of electron Positive P
Pac or oxidation current Sir)
y takes place enters K.
ud R.
St Cathode Negative Gain of electron Current S.
d or reduction leaves A (
a Y
o takes place I
l R
n
A
w
K
DoELECTROLYSIS OF MOLTEN SODIUM CHLORIDE R.
E NaCl(molten) → Na+ + Cl– G
E A
RReactions at anode (oxidation) : cathode (reduction) UH
F 2Cl– → Cl (g) + 2e–: 2Na+ + 2e– → 2Na(l) S
There are two types of electro2des used in the electroytic cell, namely attackable and non - attackable. r :
o
t
The attackable elecrodes participitate in the electrode reaction. They are made up of reactive metals like c
e
r
Zn, Cu, Ag etc. In such electrodes, atom of the metal gets oxidised into the corresponding cation, which Di
is passed into the solution. Thus, such anodes get dissolved and their mass decreases. On the other S,
E
hand, non-attackable electrodes do not participiate in the electrode reaction as they made up of unreactive S
S
elements like Pt, graphite etc. Such electrodes do not dissolve and their mass remain same. A
L
C
O
K
E
T
FARADAY’S LAWS OF ELECTROLYSIS:
M.
E
(i) First law of electrolysis : H
C
Amount of substance deposited or liberated at an electrode is directly proportional to amount of charge O
R
passed (utilized) through the solution. T
C
E
w∝Q L
E
W = weight liberated, Q = charge in coulomb 2
3
w = ZQ of
3
Z = electrochemical equivalent ge
a
P
when Q = 1 coulomb, then w = Z
Thus, weight deposited by 1 coulomb charge is called electrochemical equivalent.
Let 1 ampere current is passed till ‘t’ seconds .
Then, Q = It ∴ w = ZIt
L
A
m 1 Faraday = 96500 coulomb = Charge of one mole electrons P
O
o H
c One faraday is the charge required to liberate or deposit one gm equivalent of a substance at B
.
s
asse cLoert r‘eEs’p ios neqduinivga leelnetc wtreoidgeh.t then ‘E’ gm will be liberated by 96500 coulomb. 81 ,
l 8
c 8
o E E 5
k ∴ 1 Coulomb will liberate gm ; By definition, Z= 0
e 96500 96500 3
w.t ItE 89
ww W∴h eWn =a 9g6a5s0 is0 evolved at an electrode, then above formula changes as, 0, 0 9
te: ItV 00
si V= e 0
b 96500 0
we where V = volume of liberated gas, V = equivalent volume of gas. 32
m Equivalent volume may be defined as:e 5)-
o 5
r The volume of gas liberated by 96500 coulomb at STP. 7
f 0
ge (ii) Second law of electrolysis : H: (
a
k P
Pac Wwehigenh t soafm seu basmtaonucnets o dfe cphoasrigteed is o pra dsissesdo ltvherodu agt ha ndoifdfeer oenr tc ealtehcotdroel yartee sino lruattiioon osf c tohnenire ecqteudiv ianl seenrties then Sir)
y weights. i.e. w /w = E /E K.
ud 1 2 1 2 R.
St QUALITATIVE ASPECTS OF ELECTROLYSIS S.
d A (
a In the electrolysis process we have discussed above, we have taken molten salt as electrolyte, which Y
o I
l R
n contains only one cation and anion. Now, if the electrolyte taken contains more than one cation and A
w
K
Do anion (for example, aqueous solution of the ionic electrolyte), then the cation and anion that will get R.
E discharged depends on the ability of cation to get reduced and the ability of anion to get oxidised. G
E A
The ability of an ion to get oxidised or reduced depends upon the size, mass, positive charge, negative H
R U
F charge etc. Thus, it is not possible to predict qualitatively that which ion would be discharged first, as one S
factor might enhance the ability to discharge while the other factor may hamper it. This can only be r :
o
predicted on the basis of quantitative value assigned based on the cumulative effect of all the factors reponsible ct
e
for an ion's ability to discharge. The value is referred as standard potential, which is determined by keeping Dir
the concentration of ion as 1 M, pressure of gas at 1 atm, and the measurement done at 25°C. For a cation, S,
E
the standard reduction potential (SRP) values are compared. The cation having higher standard reduction S
S
potential value is discharged in preference to cation with lower SRP value provided the ions are at 1 M A
L
concentration. For an anion, the standard oxidation potential (SOP) values are compared and anion having C
O
higher SOP is preferentially discharged, if the concentration is 1 M for each of the ion. The SRP values at K
E
25°C for some of the reduction half reactions are given in the table below. T
S.NO. Reduction half cell reaction E° in volts at 25°C
M.
E
1. F + 2e– 2F– + 2.65 H
2 C
2. S O2−+ 2e– 2SO2− + 2.01 O
2 8 4 R
3. Co3+ + e– Co2+ + 1.82 T
C
E
4. PbO + 4H+ + SO2−+ 2e– PbSO + 2H O + 1.65 L
5. MnO2−+ 8H+ + 5e4– Mn2+ + 4H O4 2 + 1.52 32 E
4 2 of
6. Au3+ + 3e– Au + 1.50 4
e
g
7. Cl + 2e– 2Cl– + 1.36 a
2 P
8. Cr O2−+ 14H+ + 6e– 2Cr3+ + 7H O + 1.33
2 7 2
9. O + 4H+ + 4e– 2H O + 1.229
2 2
10. Br + 2e– 2Br– + 1.07
11. NO2 −+ 4H+ +3e NO + 2H O + 0.96 AL
m 3 2 P
O
o12. 2Hg2+ + 2e– Hg2+ + 0.92 H
c 2 B
.
asses1143.. ACug2++ ++ eI–– + e– A g CuI ++ 00..78969 81 ,
l 8
oc15. Hg2++ 2e– 2 Hg + 0.79 58
k 2 0
e16. Fe3+ + e– Fe2+ + 0.77 3
w.t17. I + 2e– 2I– + 0.535 89
w 2 9
w18. Cu+ + e– Cu + 0.53 0
19. Cu2+ + 2e– Cu + 0.34 0,
te:20. Hg Cl + 2e– 2Hg + 2Cl– + 0.27 00
bsi21. AgC2 l +2 e– Ag + Cl– + 0.222 00
we22. Cu2+ + e– Cu+ + 0.15 32
m 23. Sn4+ + 2e– Sn2+ + 0.13 )-
5
o 5
r24. 2H+ + 2e– H 0.00 7
f 2 0
ge 25. Fe3+ + 3e– Fe – 0.036 H: (
ka26. Pb2+ + 2e– Pb – 0.126 P
Pac27. Sn2+ + 2e– Sn – 0.14 Sir)
y 28. AgI + e– Ag + I– – 0.151 K.
ud29. Ni2+ + 2e– Ni – 0.25 R.
St30. Co2+ + 2e– Co – 0.28 S.
d 31. Cd2+ + 2e– Cd – 0.403 A (
a Y
o32. Cr3+ + e– Cr2+ – 0.41 I
l R
n
A
w33. Fe2+ + 2e– Fe – 0.44
K
Do34. Cr3+ + 3e– Cr – 0.74 R.
E 35. Zn2+ + 2e– Zn – 0.762 G
E A
R36. 2H2O + 2e– H2 + 2OH– – 0.828 UH
F37. Mn2+ + 2e– Mn – 1.18 S
38. Al3+ + 3e– Al – 1.66 r :
o
t
39. H + 2e– 2H– – 2.25 c
40. M2g2+ + 2e– Mg – 2.37 Dire
41. Na+ + e– Na – 2.71 S,
E
42. Ca2+ + e– Ca – 2.87 S
S
43. Ba2+ + 2e– Ba – 2.90 A
L
44. Cs+ + e– Cs – 2.92 C
O
45. K+ + e– Κ – 2.93 K
E
46. Li+ + e– Li – 3.03 T
When solution of an electroyte contains more than one type of cations and anions at concentrations
M.
different than 1 M, the discharge of an ion does not depend solely on standard potentials but also E
H
depends on the concentration of ion in the solution. This value is refered as potential, called as C
O
reduction potential for cation and oxidation potential for anion. The relation between reduction R
T
potential and standard reduction potential is given by Nernst equation, as C
E
L
RT [concentrationof product] E
ERP = E°RP – nF ln [concentrationof reactant] of 32
5
where E = Reduction potential of cation and E° = Standard reduction potential of cation. e
g
RP RP a
Thus, it is possible that a cation (A+) with lower standard reduction potential getting discharged in P
preference to cation (B+) having higher standard reduction potential because their concentration
might be such that the reduction potential of A+ is higher than that of B+.
When two metal ions in the solution have identical values of their reduction potentials, the simultaneous
L
deposition of both the metals will occur in the form of an alloy. A
m P
O
o H
cGALVANIC CELL B
.
s
asse This cell converts chemical energy into electrical energy. 81 ,
l 8
c 8
o 5
k 0
e 3
w.t 89
w 9
w 0
0,
e: 0
t 0
si Galvanic cell is made up of two half cells i.e., anodic and cathodic. The cell reaction is of redox kind. 0
b 0
we Oxidation takes place at anode and reduction at cathode. It is also known as voltaic cell. It may be 32
m represented as shown in Fig. Zinc rod immersed in ZnSO4 behaves as anode and copper rod immersed )-
5
o in CuSO behaves as cathode. 5
r 4 7
ge f OZnx idat iZonn2 +t a +k e2se p– l a(lcoes sa to af neloedcetr:on : oxidation) H: (0
a
k P
Pac RCued2+u +ct 2ioen– tak e Cs pu l(agcaei na to cf aetlhecotdroen: ; reduction) Sir)
y Over all process: K.
Stud Zn(s) + Cu2+ Cu(s) + Zn2+ S. R.
d A (
a In galvanic cell like Daniell cell; electrons flow from anode (zinc rod) to the cathode (copper rod) Y
o I
l R
n through external circuit; zinc dissolves as Zn2+ ; Cu2+ ion in the cathode cell picks up two electron and A
w
K
Do become deposited at cathode. R.
E G
ESALT BRIDGE A
H
R U
Two electrolyte solutions in galvanic cells are seperated using salt bridge as represented in the Fig. salt
F S
bridge is a device to minimize or eliminate the liquid junction potential. Saturated solution of salt like r :
o
KCI, KNO , NH Cl and NH NO etc. in agar-agar gel is used in salt bridge. Salt bridge contains high t
3 4 4 3 c
concentration of ions viz. K+ and NO − at the junction with electrolyte solution. Thus, salt bridge carries re
3 Di
whole of the current across the boundary ; more over the K+and NO3− ions have same speed. Hence, S,
salt bridge with uniform and same mobility of cations and anions minimize the liquid junction potential & E
S
completes the electrical circuit & permits the ions to migrate. S
A
L
C
O
K
E
T
Representation of a cell (IUPAC conventions ): Let us illustrate the convention taking the example of Daniel
M.
cell. E
H
C
(i) Anodic half cell is written on left and cathodic half cell on right hand side. O
R
Zn(s) | ZnSO (sol) || CuSO (sol) | Cu(s)
4 4 CT
(ii) Two half cells are separated by double vertical lines: Double vertical lines indicate salt bridge or any type E
L
of porous partition. E
2
(iii) EMF (electromotive force) may be written on the right hand side of the cell. 3
of
(iv) Single vertical lines indicate the phase separation between electrode and electrolyte solution. 6
e
Zn | Zn2+ || Cu2+ | Cu ag
P
(Illustration of Phase boundary)
(v) Inert electrodes are reprsented in the bracket
Zn | ZnSO || H+ | H , Pt
4 2
L
A
mCONCEPT OF ELECTROMOTIVE FORCE (EMF) OF A CELL P
O
o Electron flows from anode to cathode in external circuit due to a pushing effect called or electromotic H
c B
s. force (e.m.f.). E.m.f. is some times called as cell potential. Unit of e.m.f. of cell is volt.
asse EMF of cell may be calculated as : 81 ,
cl E = reduction potential of cathode − Reduction potential of anode 88
o cell 5
k Similarly, standard e.m.f. of the cell (E°) may be calculated as 0
e 3
w.t E°cell = Standard reduction potential of cathode − Standard reduction potential of anode 89
w 9
w 0
e: SIGN C EOMNFV oEf NceTll IsOhoNul dO bFe EpoMsitFive other wise it will not be feasible in the given direction . 00,
t 0
bsi Zn | ZnSO4 || CuSO4 | Cu E = +1.10 volt (Feasible) 00
we Cu | CuSO4 || ZnSO4 | Zn E = −1.10 volt (Not Feasible) 32
m NERNST EQUATION )-
o Walter Nernst derived a relation between cell potential and concentration or Reaction quotient. 55
r 7
f 0
ge ∆G = ∆G° + RT ln Q ..(1) H: (
a
Pack wLehte nre, F∆aGra adnady c∆hGar°g aer eis f rtaeke eenn eorugty f raonmd sat acneldl aorfd e f.rmee.f .e (nEer)g tyh ecnh aenlegcetr; i‘cQal’ wiso rreka cdtoionne qbuyo thtiee ncte.ll Sir) P
y may be calculated as, K.
ud Work done = Charge x Potential = nFE R.
St S.
d From thermodynamics we know that decrease in Gibbs free energy of a system is a measure of A (
a reversible or maximum obtainable work by the system if there is no work due to volume expansion Y
o I
nl ∴ −∆G = nFE and −∆G° = nFE° AR
w
K
Do Thus from Eq. (i),we get −nFE = -nFE° + RT lnQ R.
E 0 0.0591 G
E At 25°C, above equation may be written as E = E − logQ A
n H
R U
F Where ‘n’ represents number of moles of electrons involved in process. S
E, E° are e.m.f. and standard e.m.f. of the cell respectively. r :
o
t
In general , for a redox cell reaction involving the transference of n electrons c
e
aA + bB → cC + dD, the EMF can be calculated as: Dir
S,
E
S
S
A
L
C
O
K
E
T
0.0591 [C]c[D]d M.
E = E° – log E
Cell Cell n [A]a[B]b CH
O
Prediction and feasibility of spontaniety of a cell reaction. R
T
Work done by the cell = nFE; C
E
It is equivalent to decrease in free energy ∆G = –nFE L
E
Under standard state ∆ G0 = –nFE0 (i) 2
3
(i) From thermodynamics we know, ∆ G = negative for spontaneous process. Thus from eq.(i) it is 7 of
e
clear that the EMF should be +ve for a cell process to be feasible or spontaneous. g
a
P
(ii) When ∆G = positive, E = negative and the cell process will be non spontaneous.
(iii) When G = 0 , E = 0 and the cell will attain the equilibrium.
Reactions ∆∆∆∆G E
Spontaneous (–) (+)
L
Non- spontaneous (+) (–) A
m P
o Equilibrium 0 0 HO
.c Standard free energy change of a cell may be calculated by electrode potential data. B
s
lasse aSnuobdsteit)u itnin egq t.h (ei) v waleu em oafy E g0e t( i∆.eG., s0t.andard reduction potentialof cathode- standard reduction potential of 881 ,
c 8
o Let us see whether the cell (Daniell) is feasible or not: i.e. whether Zinc will displace copper or not. 5
k 0
w.te Zn | (s) | ZnSO4 (sol) || CuSO4 (sol) | Cu(s) 893
te: ww EE00Zcne2l+l=/ZEn0C=u−2+0/.C7u6−voElt0zn ;2 +E/ZC0nu2+/Cu=+0.34volt 000, 0 9
si 0
b =0.34 –(–0.76) = +1.10 volt 0
we Since E0 = +ve , hence the cell will be feasible and zinc will displace copper from its salt solution. In the 32
m other words zinc will reduce copper. 5)-
o 5
r 7
f 0
ge THERMODYNAMIC TREATMENT OF NERNST EQUATION H: (
a
k P
Pac Determination of equilibrium constant : We know, that Sir)
udy E = E0 − 0.0591logQ ..(i) R. K.
St n S.
d At equilibrium, the cell potential is zero because cell reactions are balanced, i.e. E = 0 A (
oa ∴ From Eq. (i), we have IY
l R
n
A
E Dow 0 = E0 − 0.0n591logKeq or Keq=antilog0.n0E5091 G R. K
E A
H
R U
FHeat of Reaction inside the cell: Let n Faraday charge flows out of a cell of e.m.f. E, then S
−∆G = nFE (i) r :
o
t
Gibbs Helmholtz equation (from thermodynamics ) may be given as, c
e
r
∂∆G Di
∆G = ∆H + T   (ii) S,
 ∂T  E
p S
S
From Eqs. (i) and (ii), we have A
L
C
∂(−nFE) ∂E O
−nFE=∆H+T  =∆H−nFT  K
 ∂T  ∂T E
p p T
∂E M.
∴ ∆H=−nFE+nFT∂T HE
p C
O
Entropy change inside the cell : We know that G = H - TS or ∆G = ∆H − T∆S ...(i) R
T
where ∆G = Free energy change ; ∆H = Enthalpy change and ∆S = entropy change. C
E
According to Gibbs Helmoholtz equation, L
E
2
∆G=∆H+T∂∆G 8 of 3
 ∂T  e
p g
a
P
∂∆G
∆G=∆H=T
 
 ∂T 
p
From Eqs. (i) and (ii), we have
L
A
m ∂∆G ∂∆G P
O
o −T∆S=T  or ∆S=−  H
c  ∂T   ∂T  B
s. p p
asse ∂E 81 ,
cl or ∆S=nF  88
o ∂T 5
k p 0
e 3
w.t ∂E 89
ww where ∂T is called temperature coefficient of cell e.m.f. 0 9
p 0,
e: 0
t 0
siDIFFERENT TYPES OF HALF-CELLS AND THEIR REDUCTION POTENTIAL 0
b 0
we 32
m (1) Gas-Ion Half Cell: )-
o In such a half cell, an inert collector of electrons, platinum or graphite is in contact with gas and a solution 55
r 7
f containing a specified ion. One of the most important gas-ion half cell is the hydrogen-gas-hydrogen ion 0
ge half cell. In this half cell, purified H gas at a constant pressure is passed over a platinum electrode which H: (
Packa is in conHta+c(t awqi)t h+ a en– a cid s1o/l2u tHion. 2 Sir) P
y 2 K.
ud 0.0591 (pH )1/2 R.
St E =E0 − log 2 S.
ad H+ /H2 H+ /H2 1 H+ YA (
o I
l R
n
A
w
K
Do(2) Metal-Metal Ion Half Cell: R.
E This type of cell consist of a metal M in contact with a solution containing Mn+ ions. G
E Mn+(aq) + ne– M(s) HA
R U
F S
E =E0 −0.0591log 1 r :
Mn+/M Mn+ /M n Mn+ to
  c
e
r
Di
(3) Metal-Insoluble Salt - Anion Half Cell: S,
In this half cell, a metal coated with its insoluble salt is in contact with a solution containing the anion of the E
S
insoluble salt. eg. Silver-Silver Chloride Half Cell: S
A
This half cell is represented as Cl–/AgCl/Ag. The equilibrium reaction that occurs at the electrode is CL
AgCl(s) + e– Ag(s) + Cl–(aq) O
K
E
T
E − =E0 − −0.01591logCl− HEM.
Cl /AgCl/Ag Cl /AgCl/Ag C
O
potential of such cells depends upon the concentration of anions. Such cells can be used as Reference Electrode. R
T
(4) Oxidation-reduction Half Cell: C
E
This type of half cell is made by using an inert metal collector, usually platinum, immersed in a solution L
E
which contains two ions of the same element in different states of oxidation. eg. Fe2+ - Fe3+ half cell. 2
3
Fe3+(aq) + e– Fe2+(aq) of
9
e
g
  a
Fe2+ P
0.0591  
E =E0 − log
Fe3+ /Fe2+ Fe3+ /Fe2+ 1 Fe3+
 
CONCENTRATION CELL L
A
m The cells in which electrical current is produced due to transport of a substance from higher to lower P
O
o H
c concentration. Concentration gradient may arise either in electrode material or in electrolyte. Thus there B
.
s
asse(i) Earlee cttwrood tey pcoesn coefn ctorantcioenn tcrealtlion cell . 81 ,
l 8
c(ii) Electrolyte concentration cell 8
o 5
k 0
e 3
w.tElectrode Gas concentration cell : 89
w 9
w HPte, rHe,2 h(Pyd1)r o| Hge+n( Cga) s| His 2b(uPb2b),l ePdt at two different partial pressures at electrode dipped in the solution of 0, 0
e: same electrolyte. 0
t 0
si 0
web Cell process : 1/2H2(p1)→H+(c)+e− (Anode process) 32 0
m + − )-
o H (c)+e →1/2H2(p2) 55
r 7
ge f 1/2H2(p1)⇔1/2H2(p2) H: (0
a
y Pack ∴ E= −2.30F3RTlogpp121/2 K. Sir) P
ud R.
d St or E = 2.303RTlogp2 , At 250C , E = 0.059logp1 A (S.
loa  2F  p1 2F p2 RIY
n
A
w For spontanity of such cell reaction, p >p K
DoElectrolyte concentration cells: 1 2 R.
E Zn(s) | ZnSO (C ) || ZnSO (C ) | Zn(s) G
E 4 1 4 2 A
R In such cells, concentration gradient arise in electrolyte solutions. Cell process may be givewn as, UH
F S
Zn(s)→Zn2+(C1)+2e (Anodic process) r :
o
t
Zn2+(C )+2e→Zn(s) ec
Zn2+(2C )(cid:2)Zn2+(C ) (Over all process) Dir
2 1 S,
E
S
S
∴ From Nernst equation, we have A
L
C
O
K
E
T
2.303RT C  2.303RT C  M.
E = 0 − log 1 or E= log 2 HE
2F C  2F C  C
2 1 O
For spontanity of such cell reaction, C > C R
2 1 T
C
CONDUCTANCE E
L
Introduction: E
2
Both metallic and electrolytic conductors obey Ohm's law of 3
i.e. V = IR 0
1
e
where V = Potential difference in volt; I = Current in ampere ; R = resistance in Ohm g
a
P
We know, resistance is directly proportional to length of conductor and inversely proportional to
cross sectional area of the conductor.
l l
R∝ or R=ρ ( ρ= Specific resistance ) L
A A A
m P
Specific resistance is the resistance of a conductor having lengths of 1 cm and corss sectional O
o H
.c area of 1 cm2. B
s
asse Unit of R is ohm and unit of specific resistance is ohm cm 81 ,
clReciprocal of resistance is called as conductance and reciprocal of specific resistance is called as 88
o 5
kspecific conductance. 0
e 3
w.t 1 1A A 89
ww R=ρ l or C=K l 0 9
where C = conductance ohm−1 ; K = specific conductance ohm−1cm−1 . 0,
e: 0
t Mho and siemens are other units of conductance 0
si 0
b 0
we K= lC 32
m A )-
5
ro Specific conductance= Cell constant x Conductance 75
f 0
age SPECIFIC CONDUCTANCE IS CONDUCTANCE OF 1 CM3 OF AN ELECTROLYTE SOLUTION. H: (
k P
y Pac dIne fciansitee o cfo enleccetnrtoralyttioicn s eonluctloiosne,d t hine as pceeclli fhiacv cionngd tuwcota enlceec tirso ddeefsin seodf ausn tith aer ceoan sdeupcatraantecde obfy a1 s comls uatpioarnt .of K. Sir)
ud1. Equivalent Conductance R.
St Equivalent conductance is the conductance of an electrolyte solution containing 1 gm equivalent S.
d of electrolyte. It is densoted by ∧. A (
a Y
o ∧= K x V I
l R
n
w (∧ = ohm−1 cm−1 x cm3 = ohm−1 cm2) KA
Do Usually concen ration of electrolyte solution is expressed as C gm equivalent per litre. R.
E G
E 1000 A
R Thus, V= C UH
F S
{Volumehaving1gmequivalentelectrolyteinthesolution}Thus, ∧=K×1000 or :
t
C c
e
2. Molar Conductance Dir
Molar conductance may be defined as conductance of an electrolyte solution having 1 gm mole S,
electrolyte in a litre. It is denoted by ∧ . SE
m S
∧ = K × V A
m L
Usually concentration of electrolyte solution is expressed as ‘M’ gm mole elecrtrolyte per litre. C
O
K
E
T