electrochemistry

1. 뇨 뇨 뇨 뇨al 뇨i ㈠〱 0 i 60 a0nl
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‫׼‬ꤐ 꾨
 “ 6ꤐ  ! 6ꤐ 1‵  ‫׺‬͸ ‫׼‬ 꾨l, 6ꤐ ‫׼‬ꤐ〴 ͸ 〴 ‫׼‬ 꾨l,
6ꤐ 〴꾨 Ͷꤐ ‫׼‬ 〴6 ꤐ ‫׼‬ 꾨l, ꤐ ‫׼‬ 6ꤐ ‫׼‬ ꤐ Ͷ ꤐ Ͷ 6ꤐ
〴l ‫׼‬〴l ͸ ꤐ‵‫׼‬ l꾨 ‫׼‬ ‫׼‬ 꾨l 6ꤐ 꾨Ͷꤐ 꾨〴l ‫׼‬ ” (͸ ꤐ- Ͷꤐ ꤐ, ‵ꤐ -
1- 1)

2. What is electrochemistry?
Electrochemistry is the part of
chemical sciences that deals with
heterogeneous electron transfer
reactions.
Applications
1) Sensor : Glucose senor, pH sensor, arsenic sensor etc
2) Energy: (a) Battery: Dry cell, lead storage cell etc
(b) Fuel cell: Methanol fuel cell
3) Electroplating
4) Wastewater treatment etc

3. In chemistry and manufacturing, electrolysis is a
technique that uses a direct electric current (DC) to drive
an otherwise non-spontaneous chemical reaction.
Faraday's law of electrolysis : The amount of substance produced at each
electrode is directly proportional to the quantity of charge flowing through the
cell.
Electrolysis

4. Problem 1: Find the charge in coulomb on 1 mole-ion of N3-.
Solution:
Charge on one ion of N3-
= 3 × 1.6 × 10-19 coulomb
Thus, charge on one mole-ion of N3-
= 3 × 1.6 10-19 × 6.02 × 1023
= 2.89 × 105 coulomb
Problem 2
How much charge is required to reduce (a) 1 mole of Al3+ to Al
and (b)1 mole of KMnO4 to Mn2+ ?
Solution:
( a ) T h e r e d u c t i o n r e a c t i o n i s
Al3+ + 3e- → Al
Thus, 3 mole of electrons are needed to reduce 1 mole of Al3+
Q = 3 × F = 3 × 96500 = 289500 coulomb
(b) The reduction is
M n 4 - + 8 H + 5 e - → M n 2 + + 4 H 2 O
1 m o l e 5 m o l e
Q = 5 × F = 5 × 96500 = 48500 coulomb
Electrolysis

5. Problem3.
How much electric charge is required to oxidise (a) 1 mole of H2O
to O2 and (b)1 mole of FeO to Fe2O3?
Solution:
(a) The oxidation reaction is
H 2 O → 1 / 2 O 2 + 2 H + + 2 e -
Q = 2 × F = 2 × 96500 =193000 coulomb
(b) The oxidation reaction is
F e O + 1 / 2 H 2 O → 1 / 2 F e 2 O 3 + H + + e -
Q = F = 96500 coulomb
Problem 4.
Exactly 0.4 faraday electric charge is passed through three electrolytic cells in series, first containing AgNO3, second CuSO4 and third
FeCl3 solution. How many gram of rach metal will be deposited assuming only cathodic reaction in each cell?
Solution:
The cathodic reactions in the cells are respectively.
Ag+ + e- → Ag
Cu2+ + 2e- → Cu
and Fe3+ + 3e- → Fe
Hence, Ag deposited = 108 × 0.4 = 43.2 g
Cu deposited = 63.5/2×0.4=12.7 g
and Fe deposited = 56/3 ×0.4=7.47 g
Electrolysis
FeCl3
1F can react with 1 mole charge

6. Problem5.
An electric current of 100 ampere is passed through a molten liquid of
sodium chloride for 5 hours. Calculate the volume of chlorine gas liberated at the
electrode at NTP.
Solution:
The reaction taking place at anode is
Cl- → 1/2Cl2 + e-
Q = I × t = 100 × 5 × 3600 = 180,0000 coulomb = 18.65 F
1F can liberate 0.5 mol Cl2 gas
Hence, 18.65F can liberate 9.32 mol Cl2 gas
Now PV= nRT,
or V = nRT/P = 9.32 mol x 22.4 L (at NTP) = 208.91 L
Electrolysis

7. Ionic mobility

8. I = nuAq

11. Solvation of ions having same charge

12. • Small ions may have larger electric field.
• An ion with stronger electric field carries more solvent (water)
molecules with it when it moves through the solution.
• Small ions may have bigger Stokes radius.
a: hydrodynamic radius
(Stokes radius)
fs
fric 
F
-----2r------
Ffric
Hydration sphere
r
f 
6


13. The mobilities of ions
+ -
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
+
+
+
+
+ +
+
+
+
+
+
+
+ +
+
Electric field E
+ -
s
Mobility:
Drift speed s :
f
ze
s
E
 E
u
s 
r
ze
f
ze
u

6


+
-----2a------
S Ffric=fs
Felec=zeE

14. Relative ionic size and mobility
For different ions with the same charge such as Li+, Na+ and K+ the
electrical forces are equal, so that the drift speed and the mobility are
inversely proportional to the radius. In fact, conductivity measurements
show that ionic mobility increases from Li+ to Cs+.
Ionic mobility
solvation

15. Questions
1) Define drift velocity and Ionic mobility.
2) Show how ionic mobility of a ion is related to physical parameters such as
viscosity of medium, charge of ion and size of the ion.
3) Although K+ is larger than Li+ but ionic mobility of K+ is higher. Explain.
4) At infinite dilution, calculate molar conductance of ions of group-1
elements.
5) What is the order of ionic mobility of ions of alkaline metals?
6) Why KCl and KNO3 are used as supporting electrolytes or used in salt
bridge?
7) Why KCl is better than NaCl in reducing blood pressure?

16. Questions
8) Define hydrodynamic radius and ionic radius.
9) The Cs+ cations with z=1 and hydrodynamic radius (taken as the ion
radius plus the first hydration sphere) r=170 pm move in a solution
with viscosity of 1.0 cP (1.0╳ 10-3kg m-1s-1).
Estimate the mobility of the cation.
Mah-che@sust.edu
mahtazim@yahoo.com

17. End of lecture 2
May Allah SWT be with you and keep you safe from all evil effects and
diseases
Stay fine
Allahhafez

18. Lecture 3
What different ways ions and molecules use to move in solution?
Ions
1) Convection
2) Migration
3) Diffusion
Molecules
1) Convection
2) Diffusion

21. Steps of electrochemical process
(1) mass transfer
(2) adsorption-
(4) desorption
(3) Electron transfer
(5) mass transfer

22. • When we perform electrochemical experiments, we stop covection by keeping
the system ( electrochemical cell) rest such that no external force can cause any
movement of ions.
• Diffusion of ions are natural as near the electrode surface mass vacancy is
created when electrolysis is performed. Hence, often ionic mobility near the
electrode surface is strongly dependent on difussion of ions/molecules.

23. 1) If electrode is highly active then near
the electrode difusssion is dominent
near the electrode surface
Diffision Diffision
(bulk)
2) If electrode is less active then in all
cases ion moves due to migration only.
Important

24. Diffusion
For an efficient electrolysis process, near the electrode
surface concentration of ions is decreased, so ionic mobility
is highly dependent on diffision of ions.
Diffision layer
ET
ET rate mass transfer rate

26. Molar conductivity
Around the diffusion layer J = sc where s= dirift velocity and c = concentration
Hence, J = sc =
RT
DF
dx
dc
c
D
s
or
dx
dc
D
sc
or
sc
dx
dc
D
J








1
,
,
Fick's first law of diffusion

27. We know that uE
s 
Ionic mobility is proportional to ionic charge , hence zs
uE 
E
zs
u 



zD
zsl
l
zs
u 


/
where, D= sl, l is the distance between two electrodes and  is the
potential between two electrodes
Note that potential  = RT/F, thus the above expression can be written as
RT
zFD
u 
Relation between Molar conductivity and diffusion
s= drift velocity ( cm s-1)
u= ionic mobility (cm2 V-1s-1
E= potential applied per unit distance (Vcm-1)
= applied potential (V)
l= distance between electrodes (cm)
D=diffusion coefficient (cm2s-1)
z= charge number

28. Now we know that  = zuF and
Combining these two equations , we get
RT
F
D
z
v
D
z
v
2
2
2
)
( 




 


RT
zFD
u 
This equation is known as Einstein- Nernst equation and explains that
(i) Molar conductivity is proportional to diffision coefficient
(ii) Molar conductivity is a function of square of charge on the ion
This equation can be written to represent molar conductivity of an electrolyte as
RT
DF
z 2
2


Important: This equation represents molar conductivity assuming that electrolytes are fully
dissociated. For weak or sparigly soluble salts, molar conductivity is valid at infinite dilution.

29. Molar conductivity: is the conductivity of one mole ion or electrolyte at infinite dilution.
Hence, one can compare conductivity at infinite dilution only. It is only applicable at infinite dilution to compare two or more
ionic species.
But at a given concentration, the specific conductivity can best explain the relative strength of conductivity.
Specific conductivity is the conductivity of ions/eelctrolytes in present in 1cm3 volume of solution
Units:
Molar conductivity- Scm2 mol-1
Specific conducttivity- Scm-1

30. Consequently, specific conductivity largely dependent on concentration of
electrolyte/ions and charge.
How specific conductivity is related to ionic mobility and concentration?
We know that  = zuF (1)
(2) where is specific conductance and C is
concentration
Combining these two equations we get
or, (3)
so, k depends on charge of ions, ionic mobility and concentration

31. Both salts are sparingly soluble in water and even under saturated condition they are only slightly
dissociated. So molar conductivity calculation will not give perfect result.
Hence, we have to calculate specific conductivity.
in case of satuarated Ba(NO3)2
ksp = 4.64x10-3
Thus, s = 0.1 molL-1 = 1.0x10-4 mol cm-3,
i.e CBa2+ = 1.0x10-4 mol cm-3 CNO3-= = 2.0x10-4 mol cm-3
u (Ba2+) = 6.6x10=4 cm2V-1s-1 and u (NO3-) = 7.41x10-4 cm2V-1s-1. Now, using
the following equation specific conductivity of barium and nitrate ions could
be calculated separately
k(Ba2+) = 1.27x10-5 Scm-1 and k(NO3-) = 1.43x10-5 Scm-1
and solution conductivity is (1.27+1.43)x10-5 Scm-1 =2.70x10-5 Scm-1

32. Comparison of saturated Ba(NO3)2 and BaSO4 solutions
Similarly, in case of satuarated BaSO4
ksp = 1.08x10-10
Thus, s = 1.04x10-5 molL-1 = 1.04x10-8 mol cm-3,
i.e CBa2+ = 1.04x10-8 mol cm-3 CSO4= = 1.04x10-8 mol cm-3
u (Ba2+) = 6.6x10-4 cm2V-1s-1 and u (SO4) = 8.29x10-4 cm2V-1s-1. Now, using the
following equation specific conductivity of barium and sulphate ions could
be calculated separately
k(Ba2+) = 1.32x10-9 Scm-1 and k(SO4) = 1.67x10-9 Scm-1
and solution conductivity is (1.32+1.67)x10-9 Scm-1 =2.98x10-9 Scm-1
This means that kBa(NO3)2/kBaSO4  10,000

33. In two beakers A and B (containing 100ml water each), 0.74g
and 1.48g Ca(OH)2 was added. Which one was more
conductive? What was their pH values?
Add of salt to a fixed volume of solvent/ g
Ca(OH)2 (s) Ca2+(aq) + 2OH- (aq)
concentration
solubility, s
saturation point
saturated solution
u
n
s
a
t
u
r
a
t
e
d
Ca(OH)2 is a sparingly soluble base

34. Ca(OH)2
Add of salt to a fixed volume of solvent/ g
Ca(OH)2 (s) Ca2+(aq) + 2OH- (aq)
concentration
solubility, 0.011 M
saturation point, 0.0814 g/100 ml
saturated solution
u
n
s
a
t
u
r
a
t
e
d
solubility, 0.011 M = 0.814 g/L Ca(OH)2
or, 0.0814 g /100 ml Ca(OH)2 makes the
solution satuartaed. But we added 0.74 and
1.48 g salt.........!
A
ksp = 5.02x10-6
solubility, s= 0.011 molL-1

35. So maximum concentration of dissolved Ca(OH)2 both in A and B beakers were
same and it was 0.011 mol/L = 1.1x10-5 mol cm-3 . Hence, their conductivity and
pH were also same.
CCa2+= 1.1x10-5 mol cm-3 , and COH-= 2.2x10-5 mol cm-3 ,
uCa2+= 6.17x10-4 cm2V-1s-1, uOH- = 20.64x10-4 cm2V-1s-1, uCa2+= 2, uOH-=1
Now applying
kCa2+ =1.31x10-6 Scm-1, and kOH- =4.38x10-6 Scm-1
Total conductivity k= kOH-+kCa2+ = 5.70x10-6 Scm-1

36. pH
We found that s= 0.011 molL-1, [Ca2+] =0.011 molL-1 ,
Hence, [OH-] = 2x0.011 molL-1 = 0.022 molL-1
pOH= -log[OH-]= 1.66
pH= 14-pOH=14-1.66= 12.34 for both solutions

38. Questions:
1) Write down the modes of mass transfer. Define each terms.
2) Draw an electrochemical system illustrating bulk and diffusion layer.
3) Under what condition diffusion coefficient can be used inplace of ionic mobility to
determine molar conductivity?
4) Write the basis of Fick's first law of diffusion.
5)Deduce Einstein-Nernst equation.
6) Define weak electrolyte and sparingly soluble salt.
7) Between 'molar conductivity' and 'specific conductivity', which one will you select to
know the relative conductivity of a solution? Explain.
8) Calculate molar conductivity of NaCl, Na2SO4, Ca(NO3)2 using diffusion coeifficient and
specific conductivity of 0.05 M solutions using ionic mobility. Repeat the calculation for
barium salts.
9) Write down the different steps of an electrochemical process ocuuring at an electrode
surface.
10) When we a reaction to be diffusion controlled?

39. End of lecture 3