Transcript Document

Chemistry
Session
Electrochemistry - 2
Session Objectives
•
Electrolysis
•
Faradays Laws of electrolysis
•
Electrode Potential
•
Electromotive force
•
Electrochemical cells
Electrolysis
The process of decomposition of an electrolyte by the
passage of electricity is called electrolysis. In
electrolysis electrical energy is used to cause a
chemical reaction.
Electrolysis
For example,
Electrolysis of molten sodium chloride
NaCl ( ) 
 Na+ ( )  Cl ( )
Na  e 
 Na
Cl 

Reduction
1
Cl2 (g)  e
2
Oxidation
At cathode
At anode
Electrolysis of aqueous sodium chloride
NaCl (aq.)  Na+ (aq.) + Cl (aq.)
H (aq.) + OH (aq.)
H2O ( )
H+ ions are discharged at cathode because discharge
potential of H+ ion is much lower than Na+ ion
H + e 
1
H2 (gas)
2
At anode, Cl– is discharge as Cl2 (gas) because discharge
potential of Cl– is much lower than that of OH– ion.
Cl 
1
Cl2  g  e
2
Electrolysis of aqueous copper sulphate
Electrolysis of aqueous copper sulphate solution using inert electrodes
CuSO4 (aq.)  Cu2+ (aq.) + SO42 (aq.)
H2O ( )
H+ (aq.) + OH (aq.)
At cathode, Cu2+ ions are discharged in preference to H+ ions
because discharge potential of Cu2+ is much lower than H+ ions.
Cu2 aq.  2e  Cu aq.
At anode, OH– ions are discharged in preference to SO42– ions because
discharge potential of OH– is much lower than SO42– ions.
4OH  2H2O ( )  O2 (g)  4e
Faraday’s law
If W grams of the substance is deposited by
Q coulombs of electricity, then
WQ
But Q = it,
Hence W
 it
or W = Z it
I = current in amperes
t = time in seconds.
Z = constant of proportionality
(electrochemical equivalent.)
Faraday’s Law
By definition Z 
W
E
96500
I.t.E
96500
E=Equivalent mass of the substance
1 Faraday=96500 coulomb
Na  e 
 Na
1F
23g
E  23g
Al3  3e 
 Al
27g
3F
E  9g
Cu2  2e 
 Cu
E  31.75g
2F
63.5g
Illustrative Example
Find the total charge in coulombs
on 1 g ion of N3– .
Solution :
No. of moles in 1g N3- ion = 1/14
=0.0714286.
Electronic charge on one mole ion N3–
= 3 ×1.602 ×10–19 × 6.023 × 1023 coulombs
= 2.89 ×105 coulombs
Therefore, charge on 1g N3– ion
=0.0714286 x 2.89 ×105 coulombs
=2.06 x 104 Coulombs
Illustrative Example
On passing 0.1 Faraday of electricity through
aluminium chloride what will be the amount of
aluminium metal deposited on cathode (Al = 27)
Solution:
w  0.1
27
 0.9 g
3
Illustrative Example
How many atoms of calcium will be deposited
from a solution of CaCl2 by a current of 25
milliamperes flowing for 60 s?
Solution:
Number of Faraday of electricity passed
25  103  60

96500
25  103  60

96500  2
moles of Ca atoms
25  103  60

 6.023  1023
96500  2
atoms of Ca
= 4.68 × 1018 atoms of calcium.
Illustrative Example
What current strength in amperes will be
required to liberate 10 g of bromine from
KBr solution in half an hour?
Solution:
Bromine is liberated by the reaction
at anode as follows
2Br   Br2  2e
W
10 
E it
96500
160  i  30  60
2  96500
i = 6.701 ampere
Faraday’s second law of electrolysis
When the same quantity of electricity is passed through different
electrolytes the masses of different ions liberated at the electrodes are
directly proportional to their chemical equivalence
W1 E1
Z It E

or 1  1
W2 E2
Z2It E2
Z
E
Hence, 1  1
Z2 E2
Illustrative Example
The electrolytic cells, one containing acidified
ferrous chloride and another acidified ferric chloride
are connected in series. What will be the ratio of
iron deposited at cathodes in the two cells when
electricity is passed through the cells ?
Solution
Ratio of iron deposited at cathode will be in
their ratio of equivalents

Equivalent of iron from ferrous salt
Equivalent of iron from ferric salt
M
3
=2 
M 2
3
Illustrative Example
A 100 W, 110 V incandescent lamp is connected in
series with an electrolyte cell containing cadmium
sulphate solution. What mass of cadmium will be
deposited by the current flowing for 10 hr?
(Gram atomic mass of Cd = 112.4 g)
Solution
Watt = Volt × Current
100 = 110 × Current
current 
100 10

ampere
110 11
But we know, Q = i × t

10
 10  60  60 coulombs
11
10 10  60  60

Number of equivalent =
= 0.339
11
96500
Mass of cadmium deposited

0.339  112.4
= 19.06 g
2
Electrode Potential
Zn (s)
Zn++ (aq) + 2e–
Oxidation electrode potential
Or
Zn++ (aq.) + 2e–
Zn (s)
Reduction electrode potential
The electrode potential depends upon:
• Nature of the metal
• Concentration of the metallic ions in solution
• Temperature of the solution.
Standard Hydrogen Electrode
1
H2  g
2
H  aq.  e
OR
H  aq.  e
1
H2  g
2
Pt, H2 (g) (1 atm)/ H+ (aq) (c = 1 M)
Electrochemical series
When elements are arranged in increasing order
of standard electrode potential as compared to
that of standard hydrogen electrode,
It is called electrochemical series.
E0 n
 SRP of cation
M /M
E01
 SRP of anion

X2 / X
2
Higher the RP , more tendency to get reduced
(strong oxidising agent) and vice versa.
Note: Oxidation-Reduction-Potential of an element have
same magnitude and different sign
Applications of electrochemical series:
(a) To compare the relative oxidizing
and reducing powers.
(b) To compare the relative activities of metals.
(c) To calculate the standard EMF of any
electrochemical cell.
(d) To predict whether a redox reaction is
spontaneous.
Applications of Electrochemical Series
Higher the standard reduction potential, the lesser will
be its reducing strength.
Li is strongest reducing agent in aqueous solution.
The lesser the standard reduction potential of an
element, greater will be its activity.
A more active metal displaces a less active one from
its salt solution.
Those metals which have positive oxidation potential
will displace hydrogen from acids.
The metals above hydrogen are easily rusted and
those situated below are not rusted.
Illustrative Example
E0 values of Mg+2/Mg is –2.37V, Zn+2/Zn is –0.76V,
Fe+2/Fe is –0.44V.Using this information predict
whether Zn will reduce iron or not?
Solution:
Zinc has lower reduction potential than
iron. Therefore, it can reduce iron.
Standard electrode potential
When the ions are at unit activity
and the temperature is 25°C (298 K),
the potential difference is called the
standard electrode potential (E°).
Electrochemical cell
Galvanic cell:Chemical energy
electrical energy.
Daniel cell
Reactions involved
Oxidation at anode
Zn(s)  Zn2(aq)  2e–
Reduction at cathode
Cu2(aq)  2e–  Cu(s)
Cell reaction
Zn(s)  Cu2(aq)  Zn2(aq)  Cu(s)
Electromotive force(EMF)
E0cell=E0cathode-E0anode
Electromotive force(EMF) of a cell
As per IUPAC convention if we
consider standard reduction potential
of both the electrodes then
Emf = ER.H.Electrode - EL.H.Electrode
Emf  Ecathode  Eanode
Salt Bridge
Salt bridge Agar—agar mixed with
KCl, KNO3, NH4NO3) etc.
• Eliminates liquid - liquid junction potential
• Maintains the electrical neutrality of solutions.
• Completes the circuit.
Illustrative Example
Calculate the emf of the cell
Ni/Ni2+ 1.0 M || Au3+ 1.0 M | Au
2
Eo = –0.25 V for Ni | Ni
1.5 V Au3+/Au.
Solution:
o
o
Eo

E

E
cell
cathode
anode
= 1.5 – (–0.25) = 1.75 V
and
Illustrative Example
The standard oxidation potential Eo
for the half reaction are given below.
Zn  2e 
 Zn
Eo  –0.76 V
Fe2   2e 
 Fe
Eo  –0.41 V
Calculate E0 for the following cell reaction
Zn + Fe++  Zn++ + Fe
Solution
o
o
Eo

E

E
cell
cathode
anode
= –0.41 – (–0.76)
= +0.35 V
Thank you