Chemistry. Session Electrochemistry - 2 Session Objectives Electrolysis Faradays Laws of...
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Transcript of Chemistry. Session Electrochemistry - 2 Session Objectives Electrolysis Faradays Laws of...
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
+NaCl ( ) Na ( ) Cl ( )
Na e Na Reduction At cathode
At anode21
Cl Cl (g) e Oxidation2
For example,
Electrolysis of molten sodium chloride
Electrolysis of aqueous sodium chloride
+NaCl (aq.) Na (aq.) + Cl (aq.)
2H O ( ) H (aq.) + OH (aq.)
H+ ions are discharged at cathode because discharge potential of H+ ion is much lower than Na+ ion
21
H + e H (gas)2
At anode, Cl– is discharge as Cl2 (gas) because discharge potential of Cl– is much lower than that of OH– ion.
21
Cl Cl g e2
Electrolysis of aqueous copper sulphate
Electrolysis of aqueous copper sulphate solution using inert electrodes
2+ 24 4CuSO (aq.) Cu (aq.) + SO (aq.)
+2H O ( ) 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.
2Cu 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.
2 24OH 2H O ( ) O (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 i t
or W = Z it
I = current in amperes
t = time in seconds.
Z = constant of proportionality (electrochemical equivalent.)
Faraday’s LawE
By definition Z96500
I . t . EW
96500
E=Equivalent mass of the substance
1 Faraday=96500 coulomb
Na e Na E 23g1F 23g
3Al 3e Al E 9g27g3F
2Cu 2e Cu E 31.75g2F 63.5g
Illustrative Example
Find the total charge in coulombson 1 g ion of N3– .
Electronic charge on one mole ion N3–
= 3 ×1.602 ×10–19 × 6.023 × 1023 coulombs
= 2.89 ×105 coulombs
Solution :
No. of moles in 1g N3- ion = 1/14 =0.0714286.
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)
27w 0.1 0.9 g
3
Solution:
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?
Number of Faraday of electricity passed
325 10 6096500
325 10 6096500 2
32325 10 60
6.023 1096500 2
= 4.68 × 1018 atoms of calcium.
Solution:
moles of Ca atoms
atoms of Ca
Illustrative Example
What current strength in amperes will be required to liberate 10 g of bromine from KBr solution in half an hour?
Bromine is liberated by the reaction at anode as follows
22Br Br 2e
Solution:
E itW
96500
160 i 30 6010
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
1 1 1 1
2 2 2 2
1 1
2 2
W E Z It Eor
W E Z It E
Z EHence,
Z E
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 saltEquivalent ofi ron from ferric salt
M32=
M 23
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)
SolutionWatt = Volt × Current
100 = 110 × Current
100 10current ampere
110 11
But we know, Q = i × t
1010 60 60 coulombs
11
Number of equivalent =10 10 60 6011 96500
= 0.339
Mass of cadmium deposited
0.339 112.42
= 19.06 g
Electrode Potential
The electrode potential depends upon:
• Nature of the metal
• Concentration of the metallic ions in solution
• Temperature of the solution.
Zn (s) Zn++ (aq) + 2e–
Zn++ (aq.) + 2e– Zn (s)
Oxidation electrode potential
Or
Reduction electrode potential
Standard Hydrogen Electrode
21
H g H aq. e2
21
H aq. e H g2
OR
Pt, H2 (g) (1 atm)/ H+ (aq) (c = 1 M)
When elements are arranged in increasing order of standard electrode potential as compared to that of standard hydrogen electrode,
It is called electrochemical series.
Electrochemical series
0E SRP of cationnM / M
0E SRP of anion1X / X22
Higher the RP , more tendency to get reduced (strong oxidising agent) and vice versa.
Note: Oxidation-Reduction-Potential of an element havesame magnitude and different sign
(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:
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.
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?
Illustrative Example
Zinc has lower reduction potential than iron. Therefore, it can reduce iron.
Solution:
Standard electrode potential
When the ions are at unit activityand the temperature is 25°C (298 K),the potential difference is called the standard electrode potential (E°).
Electrochemical cell
Daniel cell
Chemical energy electrical energy.
Galvanic cell:-
–(s) (aq)Zn Zn e 2 2
Oxidation at anode
Reduction at cathode
–(aq) (s)Cu e Cu 2 2
Cell reaction
(s) (aq) (aq) (s)Zn Cu Zn Cu 2 2
Electromotive force(EMF)
E0cell=E0
cathode-E0anode
Reactions involved
Electromotive force(EMF) of a cell
As per IUPAC convention if we consider standard reduction potentialof both the electrodes then
Emf = ER.H.Electrode - EL.H.Electrode
Emf E Ecathode anode
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
2+ 3+Ni/Ni 1.0 M || Au 1.0 M | Au
= –0.25 V for and 1.5 V Au3+/Au.
oE 2Ni | Ni
o o ocell cathode anodeE E E
= 1.5 – (–0.25) = 1.75 V
Solution:
Illustrative Example
The standard oxidation potential Eo for the half reaction are given below.
o
2 o
Zn 2e Zn E –0.76 V
Fe 2e Fe E –0.41 V
Calculate E0 for the following cell reaction Zn + Fe++ Zn++ + Fe
Solution
o o ocell cathode anodeE E E
= –0.41 – (–0.76)
= +0.35 V
Thank you