The electric cells and Faraday s Laws of electrolysis

Electric cells

The electric cells: The electric energy from external source is converted into chemical energy through a non-spontaneous oxidation – reduction .

Composition of electric cell:

it consists of:

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)

Electrolysis:

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 :

CuCl_{2 (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)⇒   Cl⁰_2(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

Faradays Laws of electrolysis

The  quantity of electricity in (in coulombs) depends on:

1)current intensity 

2)The time of the passage of the current

Quantity of electricity (in coulombs (c))= Current strength ( Amperes(A)X Time(S)

Faraday :

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)

Faraday s First 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

• Compare between the mass of silver that is formed at the cathode and the passed quantity of electricity every time

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

 Faraday ‘s Second Law

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)

Gram equivalent mass of element:

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

Faraday s First law:

– The quantity of electricity  α  liberated mass of substance

Faraday s second law:

-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

Faraday

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

The Coulomb and general law of electrolysis 

Quantity of electricity

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

The coulomb

It is the quantity of electricity required to deposit 1.118 mg of silver

General Law of electrolyte

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-   >>>>>>> H_{2 (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) >>>>  Cl_{2 (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

Example4:

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

Answer

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

Example5:

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