The electric cells: The electric energy from external source is converted into chemical energy through a non-spontaneous oxidation – reduction .
1) a container contains an electrolyte solution
–Solution of one of (salts ,acids or bases)
OR
-A molten of a salt or a metal oxide
2)2 electrodes are partially dipped in the electrolyte
a)The anode :it is an electrode that is connected to the positive pole of the battery (which an oxidation reaction occurs)
b)the cathode :it is the electrode that is connected to the negative pole of the battery (which reduction occurs)
On connection the two electrodes with a source of electric current (has slightly high potential than the reversible cell ) the electric current flows in the electrolytic cell :
-the electrons transfer in the outer wire from the anode to the cathode
-the ions transfer in electrolyte as the following:
1)the anions (electron rich material particles) they move in the solution towards the anode (positive electrode) where they neutralize their charge by losing electrons(oxidation)
2)the cations (electron poor material particles)they are moving in the solution towards the cathode (negative electrode)where they neutralize their charge by accepting electrons (reduction)
The process of chemical decomposition of the electrolytes due to the effect of passing electric current in it.
Example :Electrolytic cell of copper(II)chloride
1)the electrolyte copper(II)chloride decomposes as the following :
CuCl2 (aq) ⇒ Cu2+ (aq) +2Cl– (aq)
2)the chloride anions(Cl–) at the anode(positive electrode) are oxidized and converted into chlorine gas (Cl2)
oxidation :
2Cl– (aq)⇒ Cl02(g) +2e– E=-1.36 V
3)Copper(II) Cations (Cu2+) at the cathode (negative electrode)are reduced and converted into copper atoms (Cu(s)) deposit on the cathode
Reduction :
Cu2+ (aq) + 2e– ⇒Cu0 (s) E=+0.34 V
The Total equation:
emf cell = E oxid + E red
emf cell =(-1.36 )+ (+0.34) = -1.02 V
Note : The negative sign of the cell potential means that the reaction is non-spontaneous
GR: The type of cells known as electrolytic cells ?
-Because by passing a direct electric current through it the electrolyte material decomposes to simpler substances
The quantity of electricity in (in coulombs) depends on:
Quantity of electricity (in coulombs (c))= Current strength ( Amperes(A)X Time(S)
He established the relation between the quantity of electricity which flow in a solution and the quantity of material liberated at the electrodes in 2 Laws (First and second Law)
The quantity of material (gas or solid) formed or consumed at any electrode is directly proportional to the quantity of electricity that passes in the electrolyte (solution or molten)
Verifying Faraday s first law:
Pass different quantity of electricity in the silver nitrate solution
Observation:
The silver at the cathode increases by increasing the quantity of electricity
Conclusion :
The mass of the deposited silver is directly proportional to the quantity of electricity that passed in silver nitrate
The mass of the different material deposited or consumed by the same amount of electricity is directly proportional to their equivalent masses
The gram equivalent mass = The element gram atomic mass (g/atom)/ Number of charges of elements ion(oxidation ion state)
The mass of the element that has ability to gain or lose one mole of electron in chemical reaction
Example:
1)calculate the gram equivalent mass of Cl=35.5 ,Cu=63.5 , Al=27)
Solution:
The gram equivalent mass of Cl=atomic mass of chlorine / number of charge of chlorine ion
= 2×35.5/2= 35.5 g
The gram equivalent mass of (Cu+2) = 63.5/2= 31.75 g
The gram equivalent mass of (Al+3)Al=27/3=9 g
Verifying Faraday s second Law:
1)By passing the same amount of electricity in each of 3 electrolytes
2)Compare between masses of the elements deposited at the cathode of each cell and their gram equivalent masses
Conclusion :
The masses of the deposited elements by passing the same amount of electricity in the electrolytes of cells connected in series are directly proportional to The gram equivalent mass of each of the elements
Example1:
A quantity of electricity was passed in 2 electrolytes connected in series the first electrolyte contained gold(III) ions Au3+ and the second contained copper(II) ions Cu2+ , so the mass of gold deposited at the first cell cathode equaled 9.38g.
Calculate the mass of copper deposited on the cathode of the second cell [Au=197,Cu=63.5]
Answer:
The gram equivalent mass of Au=atomic mass / numbers of charges
The gram equivalent mass of Au=197/3=65.67 g
The gram equivalent mass of Cu =atomic mass /number of charges
The gram equivalent mass of Cu=63.5/2=31.5
Copper liberated mass / Cu Gram equivalent mass =
gold liberated mass / Au gram equivalent mass
Mass of liberated Cu =(gold liberated mass /gold gram equivalent mass ) x copper gram equivalent
Mass of liberated Cu = 9.38/65.4 x 31.75 = 4.535 g
Example 2
A cell has an electrode B and electrode D ,if 12.7 g of copper deposited on the electrode B and 14 of cerium (Ce) were deposited on the electrode D after a definite time
Calculate cerium oxidation sate if the oxidation state of Cu is +2 [Cu=63.5 ,Ce=140]
Answer:
gram equivalent of Cu= atomic mass of Cu/charges of Cu=64.5/2=31.5 g
liberated Cu / gram equivalent of Cu=liberated Ce / gram equivalent
12.7/31.5 = 14/gram equivalent of Ce
Gram equivalent mass of Ce = 31.5 x 14 / 12.7 =35 g
The oxidation state of Ce = gram atomic mass of Ce /gram equivalent mass
The oxidation state of Ce =140/35 =+4
– The quantity of electricity α liberated mass of substance
-The gram equivalent mass of substance α liberated mass of this substance
quantity of electricity X gram equivalent mass α liberated mass
quantity of electricity X gram equivalent mass = constant X liberated mass
The liberated mass (g)= quantity of electricity(C) X gram equivalent mass(g) / Constant
It is the quantity of electricity required to liberate (deposit ,dissolve or evolve) a gram equivalent of a substance
Example
Calculate the quantity of electricity (in faradays) required to deposit 21.6 g of silver [Ag=108] from silver solution if the reaction at the cathode:
Ag+(aq) + e ⇒ Ag0(s)
Answer:
The quantity of electricity needed to deposited the gram equivalent is = 1 F
Gram equivalent >>>> 1F
108/1 >>>>1 F
21.6 >>>>?? F
The quantity of electricity = 21.6/108
The quantity of electricity =0.2F
Example :
On passing a quantity of electricity of 0.F in an electrolyte solution 4.5 g of M element has been deposited
Calculate the gram equivalent mass of M element .
Answer:
The gram equivalent >>> needs 1 F
4.5 g >>> needs 0.5 F
The gram equivalent mass = 4.5 / 0.5
The gram equivalent mass = 9 g
Quantity of electricity(coulomb)=current intensity(Ampere) X time (second)
1C = 1A X 1S
On passing a quantity of electricity 1C in a solution of silver ions a 1.118 milligram of sliver are precipitated
On passing 1C in AGNO3 solution >>>> 0.001118g of Ag precipitated
On passing 1F in AgNO3 solution >>>> 107.88 g of Ag gram equivalent mass precipitated
1F = 1 x 107.88 / 0.001118 =96500 C
It is the quantity of electricity required to deposit 1.118 mg of silver
When 1F(96500C) passes in an electrolyte this will lead to liberation of a gram equivalent mass of the substance at any of the electrodes
Gram equivalent mass = the atomic mass(g/atm)/number of charges
The quantity of electricity to liberate(g/atm)=1F(96500C) x number of charges
Examples:
Calcualte the quantity of electricity required to deposit (g/atom) of silver
Answer: Ag+(aq) + 1e- ⇒ Ag0 (s)
The quantity =1F x no.of charges
=1 x1 1F=96500 C
2-Calculate the quantity of electricity required to deposit (g/atm) of copper
Cu2+(aq) +2e- >>>>> Cu0 (s)
The quantity = 1 F X no. of charges =
1x 2=2F = 2x 96500 C
3)calculate the quantity of electricity required to evolve 1 mole of hydrogen gas
2H+(aq) +2e- >>>>>>> H2 (g)
The quantity of electricity =1F x 2 = 2 x 96500 C
4)calculate the quantity of electricity to evolve 1 mole of oxygen gas
2O2+(L) >>>>>> O2 + 4e–
The quantity of electricity = 1F x no.of charges
= 1F x 4 = 4F=4 x 96500 C
Liberated mass (g) = quantity of electricity(F) x gram equivalent / 1(F)
Liberated mass (g) = quantity of electricity (C) x gram equivalent/96500
The gram equivalent = atomic mass (g/atom)/ No.of charges(oxidation state-valance)
The liberated mass of the first substance/Gram equivalent of the first substance
=The liberated mass of the second substance / Gram equivalent of the second substance
Examples1
1)calculate the mass of each of gold and chlorine produced from passing a quantity of electricity of 10000 C in an aqueous solution of gold (III) chloride knowing that the reaction at the two electrode are:
2Cl– (aq) >>>> Cl2 (g) + 2 e– ,
Au3+ + 3e– >>>> Au0 (s)
[Au=196.98 , Cl=35.5]
Answer
Gram equivalent mass of Cl=35.5/1=35.5 g
Liberated mass = quantity of elect. X gram equivalent mass /96500
= 10000 x 35.5/96500=3.68g
Gram equivalent of Au=196.98/3=65.66 g
Liberated mass of Au= quantity of elect. X gram equivalent mass /96500
=65.5 x 10000/96500= 6.8 g
Another answer:
Au3+ + 3e– >>>>> Au0(s)
3 mol >>>>> 1 mol
3 x96500C >>>>> 196.5 g
10000 C >>>>> ? g
Mass of liberated gold = 196.5 x 10000 / 3 x 96500
Mass of liberated gold= 6.8 g
Example2
Calculate the quantity of electricity (C) required to liberates 5.6 g of iron from iron (III) chloride solution knowing that the reaction at the cathode is:
Fe3+(aq) + 3e– ⇒ Fe0 (s) [Fe=55.86]
Answer:
Gram equivalent mass of Fe= atomic mass / charges
Gram equivalent mass of Fe=55.86/3=18.62g
The liberated mass =The quantity of elect. X mass equivalent /96500
The quantity of electricity = the liberated mass x96500/mass equivalent
The quantity of electricity= 5.6 x 96500 /18.62=29022.56 C
Example3
Calculate the time (s) required to deposited 2.16 g of silver by passing an electric current with strength 32A in an aqueous solution of silver nitrate
[Ag=108]
Answer:
Gram equivalent mass of Ag = 108/1=108 g
The liberated mass =The quantity of elect. X mass equivalent /96500
The quantity of electricity = the liberated mass x96500/mass equivalent
The quantity of electricity =2.16 x 96500 /108=1930 C
The quantity of elect.=current intensity x time
1930 = 32 x time
Time=1930/32 =60.31seconds
on passing an electric current with strength =7A in an aqueous solution of nitrate for 4 min the cathode mass increased by 1.88 g calculate the gram equivalent mass of the cation which is present in the solution
Quantity of elect.= current intensity x time
Quantity of elect.= 7 x 4 x 60 = 1680 C
The liberated mass =The quantity of elect. X mass equivalent /96500
Gram equivalent= Liberated mass x 96500 /The quantity of electr.
Gram equivalent= 1.88 x 96500 /1680=10.7.8 g
Calculate aluminum moles [Al=27] produced from the electrolysis of the molten of aluminum oxide (Al2O3) on passing an electric current of strength =9.65 A for 5min knowing that the reaction at the cathode is:
Al3+ +3e– ⇒ Al0
Answer:
Quantity of elect.= current intensity (A) x Time(s)
Quantity of elect.=9.56 x 5 x 60=2895 C
Gram equivalent of Al= 27/3 = 9 g
The liberated mass =The quantity of elect. X mass equivalent /96500
The liberated mass= 2895 x 9 / 96500 = 0.27 g
Number of moles of Al = mass of Al /atomic mass of Al
Number of moles of Al = 0.27/27=0.01mol
Another answer:
Al3+ +3e– ⇒ Al0
3 moles >>>> 1 mole
3 x96500 >>>>> produces 1mol
2895 C >>>>> ??? moles
??? moles of Al=2896 / 3 x 96500
??? moles of Al=0.01 mol
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