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An Introduction to Electrochemistry



A Tutorial on The Basics of Electrochemistry: Notes, Figures and Problems with Solutions


Target Audience: These notes on Atomic Structure are meant for college freshmen, or high school students in Grades 11 or 12.

They might be of use to Indian students preparing for the ISC or CBSE Class 11 and 12 Examinations, IIT JEE (main and advanced), AIEEE; students from across the world preparing for their A Level Examinations, IB (International Baccalaureate) or AP Chemistry.


This compilation of notes has been prepared by Anirud Thyagarajan of IIT Kharagpur

Faraday’s Laws of Electrolysis

The relationship between the quantity of electric charge passed through an electrolyte and the amount of the substance deposited at the electrodes was presented by Faraday in 1834, in the form of laws of electrolysis.


Faraday's First Law

When an electric current is passed through an electrolyte, the amount of substance deposited is proportional to the quantity of electric charge passed through the electrolyte.

If W be the mass of the substance deposited by passing Q coulomb of charge, then according to the law, we have the relation:

W ∝Q

A coulomb is the quantity of charge when a current of one ampere is passed for one second. Thus, amount of charge in coulombs,

       Q = current in amperes × time in seconds

               = 1 × t

So       W ∝ 1 × t

or       W = z × 1 × t

where z is a constant, known as electro-chemical equivalent, and is characteristic of the substance deposited.

When a current of one ampere is passed for one second, i.e., one coulomb (Q = 1), then

W = Z

Thus, electrochemical equivalent can be defined as the mass of the substance deposited by one coulomb of charge or by one ampere of current passed for one second. For example, when a charge of one coulomb is passed through silver nitrate solution, the amount of silver deposited is 0.001118 g. this is the value of electrochemical equivalent of silver.


Faraday's Second Law

When the same quantity of charge is passed through different electrolytes, then the masses of different substances deposited at the respective electrodes will be in the ratio of their equivalent masses.

The law can be illustrated by passing same quantity of electric current through three voltametres containing solutions of H2SO4, CuSO4 and AgNO3 respectively as shown in Fig.12.1. In the first voltameter, hydrogen and oxygen will be liberated; in the second, copper will be deposited and in the third, silver will be deposited.

(Mass of hydrogen)/(Mass of copper) = (Equivalent mass of hydrogen)/

                                                                 (Equivalent mass of copper)

or    (Mass of copper)/(Mass of silver) = (Equivalent mass of copper)/

                                                                 (Equivalent mass of silver)

or    (Mass of silver)/(Mass of hydrogen) = (Equivalent mass of silver)/

                                                                (Equivalent mass of hydrogen)

       It is observed that by passing one coulomb of electric charge.

       Hydrogen evolved = 0.00001036 g.

       Copper deposited = 0.0003292 g.

and     Silver deposited = 0.001118 g

       These masses are in the ratio of their equivalent masses. From these masses, the amount of electric charge required to deposit one equivalent of hydrogen or copper or silver can be calculated.

For hydrogen = 1/0.0001036= 96500 coulomb

For copper = 31.78/0.0003292= 96500 coulomb

For silver = 107.88/0.001118 = 96500 coulomb

This follows that 96500 coulomb at electric charge will deposit one g equivalent of any substance. 96500 coulomb us termed as one Faraday and is denoted by F.

Again according to first law,

                               W = Z×Q

Then Q = 96500 coulomb, W becomes gram equivalent mass (E).

               Thus, E = Z×96500

               or     Z = E/96500

                      z1/z2 = E1/E2

Fundamental unit of charge:

As one g-equivalent of an ion is liberated by 96500 coulomb, it follows that charge carried by one g-equivalent of an ion is 96500 coulomb. If the valency of an ion is 'n', then one mole of these ions will carry a charge of nF coulomb. One g-mole of an ion contains 6.02 × 1023 ions.

Then,

               The charge carried by an ion = nF/(6.02 × 1023 ) coulomb

For n = 1,

               The fundamental unit of charge = F/(6.02 × 1023 )

i.e.,   96500/(6.02 × 1023 ) = 1.6 × 10-19  coulomb

or     1 coulomb*=6.24 × 1018 electrons

The rate of following of electric charge through a conductor is called the electric current.

Coulomb:

It is the unit of electric charge. It is the amount of charge that moves past may given point in a circuit when a current of 1 ampere is supplied for one second.

1 coulomb = 1 ampere - second

       It is also defined as the amount of charge which is required to deposit by electrolysis 0.001118 g of silver from a solution of silver nitrate.

An electron has 1.6×10-19 coulomb of negative charge. Hence, one coulomb of charge is carried by 6.24 × 1018 coulombs. 1 mole of electrons carry a charge of 96500 coulomb. This quantity of charge is called Faraday.

Charge carried by 1 mole of electrons

                                      = (6.023×1023)(1.6×10-19)

                                      = 96368 coulomb

                                      = 96500 coulomb

Electric current = (Electric charge)/Time

1 ampere = (1 coulomb)/(1 second)

Volt is a unit of electrical potential difference. It is defined as potential energy per unit charge.

               1 volt = (1 joule)/(1 coulomb) = (1 newton×1 metre)/(1 ampere×1 second)

Electrical energy = Potential difference × Quantity of charge

                       = V × Q

                      = V × 1 × 1                       (1 = ampere; 1 = second)

                      = Watt-second

Faraday's law of gaseous electrolytic product

       We know            W = ZQ

                                   = Zlt

                               W = (It E)/96500                     ....... (i)

where                Z = E/96500

Equation, V = Volume of gas evolved at S.T.P. at an electrode

Ve = Equivalent volume

= Volume of gas evolved at an electrode at S.T.P. by 1 Faraday charge


CONDUCTANCE AND CONDUCTORS

ARRHENIUS THEORY OF ELECTOLYTIC DISSOCIATION

In order to explain the properties of electrolytic solutions, Arrhenius put forth, in 1884, a comprehensive theory which is known as theory of electrolytic dissociation or ionic theory. The main points of the theory are:

(i)     An electrolyte, when dissolved in water, breaks up into two types of charged particles, one carrying a positive charge and the other a negative charge. These charged particles are called ions. Positively charged ions are termed cations and negatively charged as anions.

AB --> A+  + B-

NaCl -->  Na+ + CL-

K2SO4 --> 2K++ SO42-

Electrolyte             Ions

In its modern form, the theory assumes that solid electrolytes are composed of ions which are held together by electrostatic forces of attraction. When an electrolyte is issolved in a solvent, these forces are weakened and the electrolyte undergoes dissociation into ions. The ions are solvated.

A+B- --> A+  + B-

or               A+B-+ aq -->  A+(aq)+B- (aq)

(ii)    The process of splitting of the molecules into ions of an electrolyte is called ionization. The fraction of the total number of molecules present in solution as ions is known as degree of ionization or degree of dissociation. It is denoted by

α= (Number of molecules dissociated into ions)/(Total number of molecules)

It has been observed that all electrolytes do not ionize to the same extent. Some are almost completely ionized while others are feebly ionized. The degree of ionization depends on a number of factors (see 12.6).

(iii)    Ions present in solution constantly re-unite to form neutral molecules and, thus, there is a state of dynamic equilibrium between the ionized the ionized and non-ionised molecules, i.e.,

                       AB <-->  A+ + B-

Applying the law of mass action to above equilibrium

[A+ ][B- ] /[AB] =K

K is known as ionization constant. The electrolytes having high  value of K are termed strong electrolytes and those having low value of K as weak electrolytes.

(iv)   When an electric current is passed through the electrolytic solution, the positive ions (cations) move towards cathode and the negative ions (anions) move towards anode and get discharged, i.e., electrolysis occurs.

The ions are discharged always in equivalent amounts, no matter what their relative speeds are.

(v)    The electrolytic solutions is always neutral in nature as the total charge on one set of ions is always equal to the total charge on the other set of ions. However, it is not necessary that the number of two sets of ions must be equal always.

AB <-->  A+ + B-                              (Both ions are equal)

NaCl <-->  Na+ + Cl-                         (Both ions are equal)

AB2 <-->  A2+ + 2B-                  (Anions are double that of cations)

BaCl2 <-->  Ba2+ + 2Cl-              (Anions are double that of cations)

A2B <--> 2a+ + B2-                   (Anions are double that of cations)

Na2SO4 <--> 2Na+ +           (Anions are double that of cations)

(vi)   The properties of electrolytes in solution are the properties of ions present in solution. For example, acidic solution always contains H+ ions while basic solution contains OH- ions and characteristic properties of solutions are those of H- ions and OH- ions respectively.

(vii)   The ions act like molecules towards depressing the freezing point, elevating the boiling point, lowering the vapour pressure and establishing the osmotic pressure.

(viii)  The conductively of the electrolytic solution depends on the nature and number of ions as the current is carried through solution by the movement of ions.


Electrolytic Conductance

The conductance is the property of the conductor (metallic as well as electrolytic) which facilitates the flow of electricity through it. It is equal to the reciprocal of resistance i.e.,

               Conductance = 1/Resistance = 1/R                            ..... (i)

It is expressed on the unit called reciprocal ohm (ohm-1 or mho) or siemens.

Specific conductance or conductivity:

The resistance of any conductor varies directly as its length (l) and inversely as its cross-sectional area (a), i.e.,

R ∝ 1/α or R ρ 1/α                                         ..... (ii)

where  is called the specific resistance.

If l = 1 cm and a = 1 cm2, then

R = ρ                                                         ..... (iii)

The specific resistance is, thus, defines as the resistance of one centimeter cube of a conductor.

The reciprocal of specific resistance is termed the specific conductance or it is the conductance of one centimeter cube of a conductor.

It is denoted by the symbol . Thus,

Κ= 1/ρ, Κ = kappa - the specific conductance              ...... (iv)

       Specific conductance is also called conductivity.

       From Eq. (ii), we have

       ρ = a/l.R or 1/ρ = 1/a.1/R

       K = 1/a×C     (1/z = cell constant)

      or Specific conductance = Conductance × cell constant


KOHLRAUSCH'S LAW

"At infinite dilution, when dissociation is complete, each ion makes a definite contribution towards equivalent conductance of the electrolyte irrespective of the nature of the ion with which it Is associated and the value of equivalent conductance at infinite dilution for any electrolyte is the sum of contribution of its constituent ions", i.e., anions and cations. Thus,

/\ = λa + λc

The  and  are called the ionic conductance of cation and anion at infinite dilution respectively. The ionic conductances are proportional to their ionic mobilities. Thus, at infinite dilution,

λc = kuc

and  λa = kua

where uc and ua are ionic mobilities of cation and anion respectively at infinite dilution. The value of k is equal to 96500 c, i.e., one Faraday.

Thus, assuming that increase in equivalent conductance with dilution is due to increase in the degree of dissociation of the electrolyte; it is evident that the electrolyte achieves the degree of dissociation as unity when it is completely ionized at infinite dilution. Therefore, at any other dilution, the equivalent conductance is proportional to the degree of dissociation. Thus,

Degree of dissociation

α = /\/(/\ )

  =(Equivalent conductance at a given concentration)/(Equivalent

                                                                conductance at infinite dilution)


Calculation of absolute ionic mobilities:

It has been experimentally found that ionic conductance is directly proportional to ionic mobilities.

         λ+ ∝ u+

        λ- ∝ u-

where u+ and u- are ionic mobilities of cations and anions.

λ+ = Fu+                where F = Faraday

λ- = Fu-                         = 96500 coulomb

Ionic mobility= (Ionic velocity)/(Potential gradient)

              =(Ionic velocity (cm/sec))/(Potential gradient (volt)/elctrode seperation)


Nernst Equation

ELECTRODE AND CELL POTENTIALS /NERNST EQUATION

       The electrode potential and the emf of the cell depend upon the nature of the electrode, temperature and the activities (concentrations) of the ions in solution. The variation of electrode and cell potentials with concentration of ions in solution can be obtained from thermodynamic considerations. For a general reaction such as

               M1A + m2B .....  n1X + n2Y + ....   .......(i)

occurring in the cell, the Gibbs free energy change is given by the equation

    G = ∆Go + 2.303RT log10 (axn1 × ayn2)/(aAm1 × aBm2) ....... (ii)

where 'a' represents the activities of reactants and products under a given set of conditions and ∆Go refers to free energy change for the reaction when the various reactants and products are present at standard conditions. The free energy change of a cell reaction is related to the electrical work that can be obtained from the cell, i.e., ∆Go = -nFEcell and ∆Go = -nFEo. On substituting these values in Eq. (ii) we get

-nFEcell = -nFEo + 2.30eRT log10 (axn1 × ayn2)/(aAm1 × aBm2)   ....... (iii)

or   Ecell = Ecello - 2.303RT/nF log10  (axn1 × ayn2)/(aAm1 × aBm2) ....... (iv)

This equation is known as Nearnst equation.

Putting the values of R=8.314 JK-1 mol-1, T = 298 K and F=96500  C, Eq. (iv) reduces to

       E = Eo - 0.0591/n log10 (axn1 × ayn2)/(aAm1 × aBm2) ....... (v)

         = Eo  - 0.0591/n log10 ([Products])/([Reactants])   ....... (vi)




(Solved) Problems On Electrochemistry:



Example 1.        Find the charge in coulomb on 1 g-ion of

Solution:             Charge on one ion of N3-

                                = 3 × 1.6 × 10-19 coulomb

                        Thus, charge on one g-ion of N3-

                                = 3 × 1.6 10-19 × 6.02 × 1023

                                = 2.89 × 105 coulomb



Example 2.        How much charge is required to reduce (a) 1 mole of Al3+ to Al and (b)1 mole of  to Mn2+?

Solution:             (a) The reduction reaction is

                        Al3+       + 3e- --> Al

                      1 mole       3 mole

Thus, 3 mole of electrons are needed to reduce 1 mole of Al3+.

                        Q = 3 × F

                       = 3 × 96500 = 289500 coulomb

                  (b) The reduction is

                       Mn4-  + 8H+ 5e- --> MN2+ + 4H2O

                       1 mole       5 mole

                                Q = 5 × F

                                  = 5 × 96500 = 48500 coulomb




Example 3.        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

                        H2O --> 1/2 O2 + 2H+ + 2e-

                       1 mole                       2 mole

                                Q = 2 × F

                                   = 2 × 96500=193000 coulomb

                        (b) The oxidation reaction is

                                FeO + 1/2 H2O --> 1/2 Fe2O3 + H++e-

                                Q = F = 96500 coulomb



Example 4.      An ammeter and a copper voltameter are connected in series through which a constant current flows. The ammeter shows 0.52 ampere. If 0.635 g of copper is deposited in one hour, what is the percentage error of the ammeter? (At. mass of copper = 63.5)

Solution :          The electrode reaction is:

                   Cu2+    + 2e-  -->   Cu

                       1 mole    2 × 96500 C

                        63.5 g of copper deposited by passing charge = 2 × 96500 Coulomb

0.635 g of copper deposited by passing charge                    

                       =(2×96500)/63.5×0.653 coulomb

                        = 2 × 965 coulomb

                        = 1930 coulomb

We know that

                        Q = l × t    

                       1930 = I × 60 × 60

                       I= 1930/3600=0.536 ampere

                       Percentage error = ((0.536-0.52))/0.536×100=2.985    



Example 5.      A current of 3.7 ampere is passed for 6 hours between platinum electrodes in 0.5 litre of a 2 M solution of Ni(NO3)2. What will be the molarity of the solution at the end of electrolysis? What will be the molarity of solution if nickel electrodes are used? (1 F = 96500 coulomb; Ni = 58.7)

Solution:           The electrode reaction is

                        Ni2+   + 2e‑  --> Ni                    

1 mole 2 × 96500 C

Quantity of electric charge passed

                        = 3.7 × 6 × 60 × 60 coulomb = 79920 coulomb

Number of moles of Ni(NO3)2 decomposed or nickel deposited = (1.0 - 0.4140) = 0.586

Since 0.586 moles are present in 0.5 litre,

Molarity of the solution = 2 × 0.586 = 1.72 M

When nickel electrodes are used, anodic nickel will dissolve and get deposited at the cathode. The molarity of the solution will, thus, remain unaffected.


Example 6.        In an electrolysis experiment, a current was passed for 5 hours through two cells connected in series. The first cell contains a solution gold salt and the second cell contains copper sulphate solution. 9.85 g of gold was deposited in the first cell. If the oxidation number of gold is +3, find the amount of copper deposited on the cathode in the second cell. Also calculate the magnitude of the current in ampere.

Solution:             We know that

(Mass of Au deposited)/(Mass f Cu deposited)=(Eq.mass of Au)/(Eq.Mass of Cu)                       

                        Eq. mass of Au =  197/3; Eq. mass of Cu 63.5/2

                       Mass of copper deposited

                        = 9.85 × 63.5/2 x 3/197 g = 4.7625 g

                        Let Z be the electrochemical equivalent of Cu.

                        E = Z × 96500

                or     Z =E/96500=63.5/(2×96500)

                Applying W = Z × I × t

                        T = 5 hour = 5 × 3600 second

                4.7625 =  63.5/(2×96500) × I × 5 × 3600

        or     I = (4.7625 × 2 × 96500)/(63.5 × 5 × 3600)=0.0804 ampere



Example 7.        How long has a current of 3 ampere to be applied through a solution of silver nitrate to coat a metal surface of 80 cm2 with 0.005 cm thick layer? Density of silver is 10.5 g/cm3.

Solution:             Mass of silver to be deposited

                                 = Volume × density

                                = Area ×thickness × density

Given: Area = 80 cm2, thickness = 0.0005 cm and density = 10.5 g/cm3           

                       Mass of silver to be deposited = 80 × 0.0005 × 10.5

                                                                = 0.42 g

                        Applying to silver E = Z × 96500

                                                Z = 108/96500 g

                        Let the current be passed for r seconds.

                        We know that

                                W = Z × I × t

                        So, 0.42 = 108/96500 x 3 x t

                        or     t = (0.42 × 96500)/(108×3)=125.09  second



Example 8.        What current strength in ampere will be required to liberate 10 g of chlorine from sodium chloride solution in one hour?

Solution:             Applying E = Z × 96500 (E for chlorine = 35.5)

                        35.5 = Z × 96500

                        or Z = 35.5/96500  g

                        Now, applying the formula

                        W = Z × I × t

                        Where W = 10 g,        Z= 35.5/96500   t=60×60=3600 second

                                        I = 10x96500/35.5x96500 = 7.55   ampere


Example 9.      0.2964 g of copper was deposited on passage of a current of 0.5 ampere for 30 minutes through a solution of copper sulphate. Calculate the atomic mass of copper. (1 faraday = 96500 coulomb)

Solution:             Quantity of charge passed

                        0.5 × 30 × 60 = 900 coulomb

                        900 coulomb deposit copper = 0.2964 g

                       96500 coulomb deposit copper = 0.2964/900×96500=31.78 g

                        Thus, 31.78 is the equivalent mass of copper.

                        At. mass = Eq. mass × Valency

                                        = 31.78 × 2 = 63.56



Example 10.      19 g of molten SnCI2 is electrolysed for some time using inert electrodes until 0.119 g of Sn is deposited at the cathode. No substance is lost during electrolysis. Find the ratio of the masses of SnCI2 : SnCI4 after electrolysis.

Solution:             The chemical reaction occurring during electrolysis is

                        2SnCl2 -->  SnCl4 + Sn

                        2×190 g     261 g     119 g

                        119 g of Sn is deposited by the decomposition of 380 g of SnCl2

                                    So, 0.119 g of SnCl2 of Sn is deposited by the decomposition of

                              380/119×0.119=0.380  g of SnCl2

Remaining amount of SnCl2 = (19-0.380) = 18.62 g

        380 g of SnCl2 produce = 261 g of SnCl4

So 0.380 g of SnCl2 produce = 261/380×0.380=0.261 g of SnCl

Thus, the ratio SnCl2 : SnCl4 =18.2/0.261 ,    i.e., 71.34 : 1



Example 11.      A current of 2.68 ampere is passed for one hour through an aqueous solution of copper sulphate using copper electrodes. Calculate the change in mass of cathode and that of the anode. (At. mass of copper = 63.5).

Solution:           The electrode reactions are:

                        Cu2+ + 2e-  -->  Cu (Cathode)

                        1 mole       2 × 96500 C

                       Cu --> Cu2+ + 2e- (Anode)

Thus, cathode increases in mass as copper is deposited on it and the anode decreases in mass as copper from it dissolves.

Charge passed through cell = 2.68 × 60 × 60 coulomb

Copper deposited or dissolved = 63,5/(2×96500)×2.68×60×60

=3.174 g

Increase in mass of cathode = Decrease in mass of anode = 3.174 g


Example 12.        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

                        1 mole    1 mole

                        108 g        1 F

                        Cu2+    + 2e-  -->Cu

                        1 mole       2 mole

                        63.5 g       2 F

        and           Fe3+   + 3e- --> Fe

                        1 mole       3 mole

                        56 g          3 F

        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



Example 13.        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

                        2Cl-  -->   Cl2      + 2e-

                        71.0 g    71.0 g     2 × 96500 coulomb

                                    1 mole

                        Q = I × t = 100 × 5 × 600 coulomb

The amount of chlorine liberated by passing 100 × 5 × 60 × 60 coulomb of electric charge.

       =1/(2×96500)×100×5×60×60=9.3264 mole

Volume of Cl2 liberated at NTP = 9.3264 × 22.4 = 208.91 L



Example 14.        A 100 watt, 100 volt incandescent lamp is connected in series with an electrolytic cell containing cadmium sulphate solution. What mass of cadmium will be deposited by the current flowing for 10 hours?

Solution:             We know that

                                        Watt = ampere × volt

                                        100 = ampere × 110

                                Ampere = 100/110

                                Quantity of charge = ampere × second

                                                           = 100/110×10×60×60 coulomb

                        The cathodic reaction is

                                Cd2+       +     2e-  -->   Cd

                                112.4 g      2 × 96500 C

Mass of cadmium deposited by passing  100/110×10×60×60                    

               Coulomb charge = 112.4/(2×96500)×100/110×10×60×60=19.0598 g