PHYSICAL CHEMISTRY
ELECTROCHEMISTRY
4. ELECTROLYTIC CONDUCTANCE :
4.1 Factors Affecting Conductance & Resistance :
(i) Solute – Solute interactions (Inter – Ionic force of attraction) Greater the force of
attraction, greater will be the resistance.
Force Charge
(ii) Solute – Solvent Interaction (Hydration/Solvation of Ions)
Greater the solvation
1
Solvation Charge greater will be resistance
size
Li+ (Hydrated largest) Cs+ (Hydrated smallest)
resistance of LiCl > resistance of CsCl
(iii) Solvent-solvent interaction (Viscosity) : greater the viscosity greater will be resistance
(iv) Temperature
T R
(v) Nature of electrolyte
Weak electrolyte – High resistance Strong electrolyte – Low resistance
4.2 Some definations :
Resistance (R) :
V
R= (Ohm's law ())
R=
A
– resistivity/specific resistance
– resistance of unit length wire of unit area of cross section = constant = ( m)
RA
=
Resistivity of a solution is defined as the resistance of the solution between two electrodes of 1
cm2 area of cross section and 1 cm apart.
or
Resistance of 1 cm3 of solution will be it's resistivity.
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Conductance () :
1
C= = mho = –1
R
= S (Siemens)
Conductivity/specific conductance
1
= = =
RA A
unit –1 cm–1
= conductivity of 1 cm3 of solution
concentration of ions
1 1
= C=
R
( no. of ions) no. of charge carriers
Since conductivity or resistivity of the solution is dependent on it's concentration, so
two more type of conductivities are defined for the solution.
Molar conductivity/molar conductance (m) :
Conductance of a solution containing 1 mole of an electrolyte between 2 electrodes which are
unit length apart.
Let the molarity of the solution 'C'
C moles of electrolyte are present in 1 Lt. of solution.
so molar conductance = m
κ × 1000 κ ×1000
m = V m = m =
C molarity
–1 2 –1
Its units are Ohm cm mol
Equivalent conductance ( eq): Conductivity of a solution containing 1 g equivalent of the
electrolyte.
κ ×1000
eq =
Normality
Its units are Ohm–1 cm2 eq–1
Ionic Mobility
Ionic Mobility = speed of the ion per unit electrical field
speed speed
= =
electrical field potential gradient
Its units are V–1 cm2 sec–1
Λ 0M Λ0
Ionic mobility = u = = M
96500 F
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Transport Number
Transport Number of any ion is fraction of total current carried by that ion.
Λ0M
Transport Number of cation =
Λ0Melectrolyte
Ex.17 If resistivity of 0.8 M KCl solution is 2.5 x 10 3 cm calculate m of the solution.
Sol. = 2.5 x 10–3 cm
103
K= = 4 x 102
2.5
4 102 1000 10
m = = 5 x 10 5 –1 cm2 mole–1
0.8
4.3 Variation of conductivity and molar conductivity with concentration
Conductivity always decreases with the decrease in concentration both for weak and
strong electrolytes.
The number of ions per unit volume that carry the current in a solution decreases on
dilution.
Molar conductivity increases with decreases in concentration. This is because the total
volume, V of solution containing one mole of electrolyte also increases.
Molar conductivity is the conductance of solution.
When concentration approaches zero, the molar conductivity is known as limiting molar
conductivity and is represented by the symbol °.
4.3.1 Strong Electrolytes :
For strong electrolytes. increases slowly with dilution and can be represented by the
equation
= – A C1/2
The value of the constant 'A' for a given solvent and temperature depends on the type of
electrolyte i.e. the charges on the cations and anion produced on the dissociation of the
electrolyte in the solution.
Example : Thus NaCl, CaCl2, MgSO4 are known as 1-1 , 2-1 and 2-2 electrolyte respectively.
All electrolytes of a particular type have the same value for 'A'.
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4.3.2 Weak electrolytes
Weak electrolytes like acetic acid have lower degree of dissociation at higher concentration and
hence for such electrolytes, the change in with dilution is due to increases in the number of
ions in total volume of solution that contains 1 mol of electrolyte.
At infinite dilution (i.e. concentration c zero) electrolyte
dissociates completely ( = 1), but at such low concentration the
conductivity of
the solution is so low that it cannot be measured accurately.
Molar conductivity versus c1/2 for acetic acid
(weak electrolyte) and potassium chloride
(strong electrolyte in aqueous solutions.
Kohlarausch's Law :
"At infinite dilution, when dissociation is complete, each ion makes a definite contribution
towards equivalent conductance of the electrolyte irrespective of the nature of the ion with
which it is associated and the value of equivalent conductance at infinite dilution for any
electrolyte is the sum of contribution of its constituent ions."
i.e., = + + –
At infinite dilution or near zero concentration when dissociation is 100%, each ion makes a
definite contribution towards molar conductivity of electrolyte irrespective of the nature of the
other ion. (because interionic forces of attraction are zero)
0
m electrolyte = + + Λ 0m + – Λ m-
0
+ = no. of cation in one formula unit of electrolyte
– = no. of anions in one formula unit of electrolyte
For NaCl, + = 1 – = 1
For Al 2(SO4)3, + = 2 – = 3
0eq electrolyte = eq + eq
Λ0m Λ 0m Al3+
0eq = Λ 0eq .Al3+ =
charge on the cation 3
Λ 0m Λ 0m electrolyte
0
Λ eq = Λ 0eq , electrolyte =
charge on the anion total + ve charge on cation
or
total - ve charge on anion
0 3+
Λ Al Λ SO 24 –
0
Λ0eq Al2(SO4)3 = Λ 0eq Al 3+ + Λ 0eq SO 2–
4 =
m
+ m
3 2
0
2Λ 0mAl3+ + 3Λ 0mSO 2-
Λ eq Al2(SO4)3 = 4
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Ex.18 Λ 0m Na+ = 150–1 cm2 mole–1 ; Λ eq Ba = 100 –1 cm2 eq–1
0 2+
Λ 0eqSO 24 – = 125 -1 cm2 eq–1 ; 3+ –1 2
Λ 0m Al = 300 cm mole
–1
Λ 0m NH +4 = 200 –1 cm2 mole–1 ; Λ 0m , Cl– = 150 –1 cm2 mole –1
Then calculate
(a) Λ0eq , Al3+ (b) Λ0eq Al2(SO4)3 (c) Λ 0m (NH4)2SO4
(d) Λ0m NaCl, BaCl2. 6H2O (e) Λ0m , (NH4)2 SO4 Al 2(SO4)3 . 24H2O (f) Λ0eq NaCl
300
Sol. (a) Λ0eq Al 3+ = = 100 (b) Λ0eq Al2(SO4)3 = 100 + 125 = 225
3
(c) Λ0m (NH4)2SO4 = 2 x 200 + 2 x 125 = 650
(d) Λ0m NaCl.BaCl2.6H2O = 150 + 200 + 3 x 150 = 800 r–1
(e) Λ0m (NH4)2 SO4 Al2(SO4)3.24H2O = 400 + 600 + 4 x 250 = 2000
(f) Λ 0eq NaCl = 300 –1 cm2 eq–1
Ex.19 To calculate Λ0m or Λ0eq of weak electrolyte
Sol. Λ 0mCH3COOH = Λ 0mCH COO- + Λ 0m H+
3
0
= (Λ mCH3 COO –
+ Λ 0mNa ) – Λ 0mNa + + Λ 0mH + + Λ 0mCl+ – Λ 0mCl–
+
Λ 0CH3COOH = Λ 0mCH3COONa + Λ 0mHCl – Λ 0mNaCl
Ex.20 Calculate Λ 0m of oxalic acid, given that
Λ0eq Na2C2O4 = 400 –1 cm2 eq –1
Λ 0m H2SO4 = 700 –1 cm2 mole –1
Λ0eq Na2SO4 = 450 –1 cm2 eq–1
Sol. Λ 0m H2C2O4 = 700 + 800 – 900 = 600–1 cm2 mole
700
Λ0eq = 400 + – 450
2
Λm
= 350 – 50 = 300
2
m = 600
Applications of Kohlaraushch's law
Calculate ° for any electrolyte from the ° of individual ions.
Determine the value of its dissociation constant once we known the ° and ° at a given
concentration c.
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Degree of dissociation : At greater dilution the ionization become 100%, therefore called
infinite dilution.
At lower dilution the ionization (dissociation into ions) is less than 100% and equivalent
conductance become lower,
i.e., eq < °eq
degree of dissociation
Λeq equivalent conductance at a given concentration
= =
Λ0eq equivalent conductance at at infinite dilution
Dissociation constant of weak electrolyte :
C 2
KC = ; = degree of dissociation
1
C = concentration
The degree of dissociation then it can be approximated to the ratio of molar conductivity c at
the concentration c to limiting molar conductivity, º, Thus we have :
= /º
But we known that for a weak electrolyte like acetic acid.
C 2 c 2 c 2
Ka = = =
(1 ) 1 /
Solubility(s) and KSP of any sparingly soluble salt.
Sparingly soluble salt = Very small solubility
Solubility = molarity = 0
so, solution can be considered to be of zero conc or infinite dilution.
K ×1000 K ×1000
m, saturated = M = S = 0
KSP = S2 (for AB type salt)
Solubility ΛM
Example-21 If conductivity of water used to make saturated solution of AgCl is found to be 3.1 x 10–
5–1 cm–1 and conductance of the solution of AgCl = 4.5 x 10–5 –1 cm–1
If Λ 0M AgNO3 = 200 –1 cm2 mole–1, Λ 0M NaNO3 = 310 –1 cm2 mole–1
calculate KSP of AgCl
Solution Λ 0M AgCl = 140
Total conductance = 10 –5
140× 4×10-5 ×1000 1.4×10-4
S= =
140 14
–4
S = 5.4 x 10
S2 = 1 x 10–8
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Example-22 To calculate KW of water
H2O() + H2O() H2O+(aq) + OH–(aq)
m = Λ 0M,H 2 O = Λ 0M H+ + Λ 0M OH–
K ×1000
= - Concentration of water molecules 100% dissociated Ask
molarity
K 1000
molarity = [H+] = [OH–] =
M
2
K 1000 [H ][OH ]
KW = [H+][OH–] = 0 K a or Kb =
M H2 O
Variation of , m &eq of solutions with Dilution
conc. of ions in the solution. In case of both strong and weak electrolytes on dilution the
concentration of ions will decrease hence will decrease.
m or eq ( C) strong electrolyte
1000 × κ
m = ( K a C ) weak electrolyte.
molarity
1000 × κ
eq =
normality
κ C
For strong electrolyte m = constant
C C
κ Ka C 1
For weak electrolyte m
C C C
5. TYPE OF BATTRIES :
5.1 Primary cells : These cells can not be recharge i.e., dry cell (lechlanche cells)
mercury cells (miniature cell used in the electronic devices)
Ecell = constant
as all substances used are either pure solids or pure liquids.
5.1 DRY CELLS and alkaline batteries :
• Cell potential = 1.5 V
• Anode : Zn (s) Zn+2(aq) + 2e–
• Cathode : MnO2 + NH4+ + e– MnO(OH) + NH3
Zn2+ + 4NH3 [Zn(NH3)4]2+
• Alkaline batteries contain basic material inside it.
• NaOH / KOH is used instead of the acidic salt NH4Cl
• Cathode : 2MnO2(s) + H2O() + 2e– Mn2O3(s) + 2OH–(aq)
Anode : Zn(s) + 2OH–(aq) ZnO(s) + H2O() + 2e–
• Voltage produced by these cells = 1.54 V
• The cell potential does not decline under high current loads because no gases are formed.
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