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What is Electrochemistry

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                                     Chapter 1

                             Electrochemistry

1.1 Introduction

Electrochemistry is an area of chemistry concerned with the interconversion of chemical
and electrical energy. It is important in modern science and technology not only because
of batteries but also because it makes possible the manufacture of essential industrial
chemicals and materials. For example, sodium hydroxide which is used in the
manufacture of paper, textiles, soaps and detergents, is produced by passing electric
current through an aqueous solution of sodium chloride. Chlorine which is essential for
the manufacture of Polyvinyl chloride (PVC) is obtained in the same process. Aluminum
metal is also produced in an electrochemical process, as is pure copper for use in
electrical wiring.
Electrochemistry is a branch of chemistry dealing with chemical reactions that
involve electrical currents and potentials. Some chemical reactions that proceed
spontaneously can generate electrical current, which can be used to do useful work; while
other chemical reaction can be forced to proceed by using electrical current. While all
this may sound rather esoteric, many practical devices based on these reactions, and
many products made by these reactions are well-known, everyday household items.
In this chapter, we look at the principles involved in the design and operation of
electrochemical cells. In addition we will explore the principle of electrochemistry in
various fields.

1.2 Electrochemical cells
Electrochemical cell is the device in which an electrochemical change takes place in the
system. Whenever a redox reaction is allowed to take place in a beaker, it is found that
the solutions becomes hot due to the production of chemical energy in the form of heat.
For instance, when a Zn rod is placed in CuSO 4 solution (Figure 1.1), the solution
becomes hot as the reaction proceeds according to the following equation


Zn(s) + CuSO 4 ( aq)                      ZnSO 4 (aq)+ Cu(s)




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    (a)                                (b)
Figure (1.1) The redox reaction of Zinc metal with aqueous Cu2+ ions
              (a)A strip of Zinc metal is immersed in an aqueous copper sulfate solution.
              (b)The redox reaction takes place at the metal solution interface and directly
          transfers two electrons from Zn to Cu2+ ions to produce copper metal.
It is a device used to convert the chemical energy produced in a redox reaction into
electrical energy. Electrochemical cells can be classified into two types, viz electrolytic
cells and galvanic cells (also known as voltaic cells and commonly known as batteries).
The names galvanic and voltaic honor the Italian Scientist (Luigi Galvani (1737-1798)
and Alexandro Volta (1745-1827) who conducted he pioneering work in the field of
electrochemistry
1.3 a) Electrolytic cells (Fig.1.2 )
It is a device in which the electrical energy from an external source can be used to bring
about chemical reactions. In this case, the reaction is being driven in the non-spontaneous
direction by external electrical force and the free energy change is positive. i.e, - nFE= G
Example of electrolytic cells are charging of lead storage battery, and electrolytic purification of
metals


1.3 Galvanic cells (Voltaic cells)

It is device in which the free energy change of a spontaneous chemical reaction is
converted into electrical energy by keeping the oxidation and reduction parts of a reaction
physically separated.
Examples of galvanic cells are: dry cell or lead storage battery



The differences between the galvanic cell and the electrolytic cells are described in table
1.1

Table 1.1 Differences between electrolytic cell and       galvanic cell
Sl.No     Galvanic cell                                   Electrolytic cell
1         It is a device to convert chemical              It is a device to convert electrical
          energy into electrical energy                   energy into chemical energy
2         The redox reactions occurring at the            The redox reactions occurring at the
          electrodes are spontaneous                      electrodes are non- spontaneous and


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                                                       these reactions takes place only when
                                                       electrical energy is supplied
3           The electrodes are of dissimilar types     The electrodes used may be
                                                       dissimilar or of the same metal
4           Each electrode is dipped in its own        Both the electrodes are inserted at the
            ions and both have separate                same electrolyte solution
            compartments
5           The electrolyte solutions are              No salt bridge is used
            connected by a salt bridge
6           The electrode on which oxidation           The electrode, which is connected to
            takes place is called the anode            the negative terminal of the battery, is
            (Negative pole ) and the electrode on      called cathode, the cations migrate to
            which reduction takes place is called      it, which gain electrons, and hence
            cathode (or positive pole )                reduction takes place here. The other
                                                       electrode is called anode.

1.3.1 A Daniel cell

A Daniel cell is the best example of a galvanic cell (Figure 1.3) It consists of two
electrodes of dissimilar metals, Zn and Cu, each electrode is in contact with a solution of
its own ions ; ZnSO 4 and CuSO 4 respectively. Each electrode of a galvanic cell may be
regarded as a „half cell‟. The two solutions are externally connected through a voltmeter/
potentiometer and internally connected by a salt bridge. The redox reactions of the Daniel
cell can be represented as two half-cells reactions.

They are

At anode,
                Zn                              Zn2+ + 2e- (oxidation half cell reaction)

At the cathode,
        Cu 2+ + 2e-                       Cu ( reduction half cell reaction)


The overall redox reaction is

                Zn + Cu2+                         Zn 2+ + Cu
In this case, Zn is oxidized to Zn2+ and Cu 2+ is reduced to Cu. Recall that an oxidation is
a loss of electrons (an increase in oxidation number) a reduction is a gain of electrons (a
decrease in oxidation number)




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Daniel cell was first
constructed
in 1836 by John
Frederick Daniel
an English Chemist)




Figure 1.3 ii) A galvanic cell uses the oxidation of zinc metal to Zn2+ ions and the
reduction of Cu2+ ions to copper metal.
In a Daniel cell, when Zn electrode is coupled with the Cu electrode, the tendency of Zn
(lower reduction potential) to dissolve and form Zn2+ ions is greater and therefore Zn
half-cell acquires negative charge. The electrons released during the oxidation process
reached at the Cu electrode and are available for Cu 2+ ions to get reduced at the cathode.
Here the tendency of the Cu electrode (high reduction potential) to go into solution, as
Cu2+ is less than the tendency to Cu2+ ions to get deposited as Cu as the Cu half cell. Thus
the Cu half- cell becomes positively charged (Fig.1.3 b)




Fig 1.3 iii) Zinc- Cu galvanic cell

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The free energy change of the above reaction is converted into electrical energy. Here a
salt bridge separates both the electrolytic solution.
1.3.2 Salt bridge

A salt bridge is a U-shaped tube containing concentrated solutions of an electrolyte like
KCl, KNO 3 or K 2 SO4 or NH4 NO3 etc. or solidified solutions of such an electrolyte in
agar agar and gelatin. The specialty of these electrolytes is that the cation and anion of
the electrolyte migrate at the same rate. ie they carry the same fraction of electric charge
( K + and Cl-, or K+ and NO 3 -, ions have similar current carrying mobilities).
The fraction of electric charge carried by an ion is called its transport number ( t+ for
cation and t- for anion) The t+ of the cation and the t- of the anion of the salt used in
the salt bridge should be same so that the overall current will be carried in either d irection
equally.

1.3.3.The functions of the salt bridge are

       1. To maintain the electrical neutrality of the circuit

       In Daniel cell, the transference of electrons from Zn anode to Cu cathode leads to the
       accumulation of positive charge around the anode due to the formation of Zn2+ ions
       and the accumulation of negative charge around the cathode due to the deposition of
       Cu2+ions as Cu on the cathode. The positive charge so accumulated around the anode
       will prevent the flow of electrons from it. Similarly the accumulation of negative
       charge around cathode will prevent the acceptance of electrons from the anode. As
       the transference of electrons stops, the current in the electrical circuit also stops.
       Because of the charge imbalance, the electrode reactions wo uld quickly come to a
       halt; electron flow through the wire would cease.
       At this stage salt bridge comes to the aid and retain the electrical neutrality of the
       solution in the two half cells. When the concentration of Zn2+ ions around the anode
       increases, sufficient number of Cl- ions migrates from the salt bridge to the anode half
       cell. Similarly sufficient number of K+ ions migrates from the salt bridge to the
       cathode half-cell, for neutralizing excess negative charge due to the addition of SO 4 2-
       ions in the cathode half cell. Thus the salt bridge provides cations and anions to

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       replace the ions lost or produced in the two half cells. It maintains electrical neutrality
       of the two solutions in the two half cells. The anode and cathode get their names from
       the direction of ion flow between the two compartments. Anions move towards the
       anode, cations move toward the cathode.
       2. It completes the electrical circuit internally by allowing the flow of ions, while the
          electrons are allowed to flow externally through the wire
       3. It prevents intermixing of solution as well as ions in the two half cells, which may
          lead to precipitation reaction
       4. It eliminates the liquid junction potential. It is the potential developed at the
       junction due to the accumulation of positive charge at the anodic part and negative
       charge at the cathodic part.
1.3.4. Designing a galvanic cell
Design a galvanic cell that uses the redox reaction
Fe (s) + 2 Fe3+ ( aq)                 3Fe2+(aq)
Identify the anode and cathode half reactions, and sketch the experimental set up. Label
the anode and cathode, indicate the direction of electron flow, and identify the sign of
each electrode.


Strategy
First, separate the overall reaction into anode ( oxidation) and cathode ( reduction) half
reactions, Then set up the two half –cells that use these half reactions, and connect the
half –cells with a conducting wire and a salt bridge.


Solution
In the overall cell reaction, iron metal is oxidized to iron(II)ions, iron (III) ions are
reduced to iron(II)ions. Therefore, the half-cells reactions are


Anode (oxidation)              Fe(S)                      Fe2+(aq) +2e-
Cathode (reduction) 2 X [Fe3+(aq)+e-                  Fe2+(aq)]
Overall reaction: Fe(s) +2Fe3+                    3Fe2+(aq)



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The cathode half-cell reaction is multiplied by a factor of 2 so that the two half-reactions
will add to give the overall cell reaction. Whenever half- reaction are added, the electrons
must cancel. No electrons can appear in the overall reaction because all of the electrons
lost by the reducing agent are gained by the oxidizing agent.
Experimental set up is as follows: the anode compartment consists of an iron metal
electrode dipping into an aqueous solution of Fe(NO 3 )2 . Since Fe3+ is a reactant in the
cathode half reaction, Fe(NO 3 )2 would be good electrolyte for the cathode compartment.
The cathode can be any electrical conductor that doesn‟t react with iron in the solution.
A platinum wire is a common inert electrode but iron metal can be used because it would
react directly with Fe3+, thus short-circuiting the cell. The salt bridge contains NaNO 3 .
Electrons flow through the wire from the iron anode (-) to the platinum cathode (+).




Fig 1.3.4 A galvanic cell based on Fe2+/Fe3+ system

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1.3.5 Representation of a galvanic cell.
It is convenient to use a shorthand notation for representing the galvanic cell. The
following points have to be noted to represent an electrochemical cell.
       a) Anode is always written on the left hand side
                    M/Mn+ (c1 ) ----- oxidation (metal/ metal ion), single vertical line represent
                    the phase boundary which separate between the electrode and the aqueous
                    solution.
       b) Cathode of the cell is written on the right hand side of the anode
                    Mn+( c2 )/M------- ( Reduction) (Metal ion/metal)
       c) Two vertical lines ( || ) represent the ( salt bridge) between the two salt solutions
            therefore a complete cell can be represented as
                 M/Mn+ ( c1 ) (||) Mn+( c2 )/M
            The reaction involved in the cell is
            Zn + Cu2+                            Zn 2+ + Cu
We can write the following
                                       Salt Bridge
                         Anode half cell |Cathode half cell
             Example: Zn (s)/ Zn2+ (C1 ) || Cu2+(C2 )/Cu
            Phase boundary Electrons flow this way             Phase boundary
.
1.3.6 Worked out examples
 Q1.Write the cell representation of the following reactions
               SnCl2                      FeCl2 + Sn
       a) Fe +
       Solution Step 1- Find the anode (electrode at which oxidation takes place and cathode
       (electrode at which reduction takes place)
       Here Fe is getting oxidized as Fe2+ and Sn2+ is getting reduced as Sn.
                                                       2+    2+
       Therefore, the cell can be represented as Fe/ Fe 11 Sn /Sn

            2 Cr (s)+3 Pb2+(aq)             3Pb(s)+2Cr3+(aq)
       b)
       Solution



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       Here Chromium is getting oxidized as Cr 3+ and Pb2+ is getting reduced as Pb. So Cr
       will act as anode and Pb will act as cathode. Therefore the cell can be represented as

            Cr/Cr3+ (aq)   Pb/Pb 2+ (aq)
       c)

d) The representation of a cell is given as H2(g)|HCl(aq)|AgCl(s)|Ag|PtR . Write the
anode , cathode and overall reactions of the given cell.
Anode: H2(g)=2H++2e-
Cathode: [AgCl(s)+e=Ag+Cl- ] ×2
Overall: 2AgCl(s)+ H2(g)=2Ag+ 2H ++2Cl-
e) Cd(s)|CdCl2(0.1M)|AgCl(s)|Ag(s)
Anode: Cd=Cd2++2e-
Cathode: [AgCl + e- =Ag++Cl- ] ×2
Overall: Cd+2AgCl=2Ag+Cd+2+2Cl-


Q2 Construct the cells in which the following reactions are taking place. Which of
the electrodes shall act as anode (negative electrode) and which one as cathode
(positive electrode)?

(a) Zn + CuSO4 = ZnSO4 + Cu

(b) Cu + 2AgNO3 = Cu(NO3 )2 + 2 Ag

(c)Zn + H2 SO4 = ZnSO4 + H2




Solution: It should always be kept in mind that the metal which goes into solution in
the form of its ions undergoes oxidation and thus acts as negative electrode (anode)
and the element which comes into the free state undergoes reduction and acts as
positive electrode (cathode):

(a) In this case Zn is oxidized to Zn2 + and thus acts as anode (negative electrode)
while Cl2 + is reduced to copper and thus acts as cathode (positive electrode). The cell
can be represented

As      Zn|ZnSO4||CuSO4|Cu

or Zn|Zn2 +||Cu2 +|Cu

Anode (-) Cathode (+)




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 (b) In this case Cu is oxidized to Cu2 + and Ag+ is reduced to Ag. The cell can be
represented as

Cu|Cu(NO3 )2||AgNO 3|Ag

or Cu|Cu2 +||Ag+|Ag

Anode (-) Cathode (+)

 (c)In this case Zn is oxidized to Zn2 + and H+ is reduced to H2. The cell can be
represented as

Zn|ZnSO4||H2 SO4|Cu

Or Zn|Zn2 +||2H+|H2 (Pt)

Anode (-) Cathode (+)

1.4. Electromotoive force (E.M.F). or cell potential of a cell

Electricity cannot flow from one point to another unless there is a potential difference
between the two points. Flow of electricity from one electrode to another electrode in a
galvanic cell indicates that the two electrodes have different potentials. The difference in
potential, which causes the flow of current from one electrode of higher potential to
another electrode of lower potential, is called electromotive force or E.M.F. or cell
potential. The E.M.F. of a cell can be calculated from the values of electrode potentials of the
two half-cells constituting the cell.
       E.M.F of cell = reduction potential of the cathode (ERight) – reduction potential of
       anode (ELeft )
       Ecell = E cathode- E anode or
           = Eright – Eleft.
       The cell reaction is feasible only when Cell EMF is positive. Negative value of cell
       emf indicates that the cell reaction is not feasible. If all reactants and products are
       present in their standard states at 25 0 C, the EMF is called standard EMF.

ie., E0cell =E0 (cathode) - E0 (anode).
The value of E0 is not changed when a half reaction is multiplied by an integer.

1.4.1 Worked out example
Find out the emf of the following cell.
Cu+2Fe+ →2Fe2++Cu+

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2Fe++2e- →2Fe2+      E0(cathode = 0.77 V
Cu →Cu2++2e-          E0 (anode) =0.34 V
E0cell=E0(cathode)-E0 (anode)

E0cell=0.77-0.43=0.43V
1.4.2 Measure ment of the emf of galvanic cells.
a) Voltmeter Measure ment
The simplest method to determine the emf of a cell is to join the two electrodes to a
voltmeter and take the reading directly. But the emf of a cell cannot be determined
accurately by connecting a voltmeter between two electrodes. During the measurement
the voltmeter draws some current from the cell, thereby causing a change in the emf due
to the formation of reaction products at the electrodes and a change in the concentration
of electrolyte around the electrodes. Consequently such a measurement would indicate
potential difference less than the actual emf of the cell. Due to the appreciable flow of
current, a part of the emf. is used up in over coming the internal resistance of the cell
b) Potentiometric or Poggendorff’s compensation method
A potentiometeric method, based on Poggendorff‟s compensation principle, is used to
out the emf of galvanic cell.
In the measurement of emf a galvanic cell by compensation method (cell), a standard cell
with known e.m.f. is used. In this method, the unknown e.m.f. is opposed by another
known e.m.f. until the two balance each other. An adjustable external potential is applied
and the cell is connected opposite to this potential. The value of this potential is
determined, when it prevents the current from flowing in the given cell. The imposed
potential is consequently equal in magnitude to the potential of the cell. A schematic
representation of the circuit diagram is shown in fig 1.5. It consists of a uniform wire AB
of high resistance, a storage battery (C) of constant, but large emf is connected to the
ends of A and B of the wire, through a rheostat, R (an adjustable resistance). The
standard cell S is included in circuit and joined to the pole A and other end of the cell is
connected to a sliding contact J through a galvanometer.
The second circuit is completed through cell X of unknown emf and one of its end is
connected to pole A and other end is connected to the sliding contact through

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galvanometer. A two way key is used to introduce either of the cells in the circuit as
shown in the scheme.
The sliding contact is fixed at point . Now Key is put in K 1 and thus the standard cell, S is
introduced in the circuit. It sends an emf in a direction that is opposite to that of cell C.
The variable resistance R is adjusted till zero deflection is obtained in galvanometer
while the sliding contact is slowly moved over AB to find a point D' of null deflection.
The distance AD' is measured. Therefore, if Es is the emf of the unknown cell,
                          Es  AD'-----------------------------------------(1)

Similarly Key is put in K 2 so that the unknown cell, X is in the circuit. The jockey
moved over wire AB to get a null deflection in the galvanometer. The distance AD is
measured. If EX is the e.m.f. of the unknown cell, Ex is proportional to AD

Thus             Ex  length AD------------------------------------------(2)

                                        Ex AD
Therefore (2)/(1) gives                                                  (3)
                                        Es AD'

       AD
Ex        xEs                                (4)
       AD'




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Fig 1.5. A schematic sketch showing the poggendorff‟s method for the determination of
emf of a cell

Since length AD, AD‟ and Es are known, Ex can be found out. If we know the value of
Es, we can also calibrate the wire AB first and then we can directly get the reading in
volts. By adjusting the rheostat, we can repeat the experiments for a different potential in
the circuit.

To get accurate results, the following conditions must be satisfied

1.Es > Ex

2. AB should be uniform

3. Galvanometer should be highly sensitive




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1.5. Nernst’s Equation

Nernst equation is an equation that can be used to determine the equilibrium reduction potential
of a half-cell in an electrochemical cell. It c an also be used to determine the total voltage
(electromotive force) for a full electroc hemical cell.

It is named after the German physical chemist Walther Nernst ,   who first formulated it,

The relation between the electrode potential and the concentration of the electrolyte
solution is given by Nernst Equation.


Nernst equation is given as


                     2.303RT     M
Ecell  E 0 cell            log n
                        nF      M
Where.


         Ecell = Electrode potential at a particular concentration
         E0 cell = Standard electrode potential.
         R = Universal Gas constant, R = 8.314  472(15) J  K−1  mol−1
         T = Temperature in degree K.
         F = One Faraday ie, 96500 coulombs.
         [M] = Concentration of electrode =1.
         [Mn+] = Concentration of electrolyte.
Derivation:
         Consider the reversible redox reaction of the type
        Mn+ (aq) ne-
              +                       M (s)
                                                  (1)
According to Van‟t Hoff‟s reaction isotherm, the free energy change is for a reversible
reaction is given as
                      Pr oducts
 G  G 0  RT ln                ( 2)
                     Re ac tan ts
where G is the free energy change. And G0 is the standard free energy change ( ie is
the change in free energy, when the concentration of the reactants and the products are
unity each)

While applying ( 1) in ( 2), (2)becomes



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                         M
G  G 0  RT ln                           (3)
                        M n

In a reversible reaction, the electrical energy produced is at the decrease of free energy

       Ie, G  nFE and G 0  nFE 0 --------(4)
       where E is the electrode potential; E0 is the standard electrode potential; F is the
       Faraday constant ( 96, 500 coulombs)


       Substitute (4) in (3), and (3) becomes

                                    M
        nFE  nFE 0  RT ln           ----- (5)
                                   M n
       Divide equation ( 5) by – nF and then ( 5) becomes

                RT    M
       E  E0     ln n (6) since [M] =1, ( 6) becomes
                nF M
                RT    1
       E  E0     ln n ---------- (7)
                nF M

       E  E0 
                   RT
                   nF
                              
                      ln M n --------(8)


       E  E0 
                   2.303RT
                      nF
                                    
                           ln M n ------(9)

       This expression is known as Nernst equation for electrode potential.
                   2.303RT 0.0591
       At 250 C,          
                      nF      n
       2
       Therefore (9) becomes
                0.0591              
       E  E0         log[ M n  ]  (10)
                   n                



       EMF of cell


        Case 1
       Suppose the reaction occurring in a reversible cell represented by the following equation

       aA+bB                       cC+dD –(11)


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       By applying 11 in Nernst equation, we gets
                            2.303RT     [c ] c [ D ] d
       Ecell  E 0 cell            log                (12)
                               nF       [ A] a [ B]b


       Or in general
                            2.303RT      Pr oducts
       Ecell  E 0 cell            log
                               nF       Re ac tan ts
       Case 2.

       Let us consider a cell

       M1 + M2 n+                   M1n+ + M2

       The EMF of the cell is

                                                  n
                            2.303RT     [M 1 ]
       Ecell  E 0 cell            log     n
                               nF       [M 2 ]

       E0 cell = E0 M2 n+/M2 – E0     n+
                                    M1 /M1   = E0 cathode - E0 anode
       Where,
       M1 n+ = Oxidised state of metal M1 ,
       M2 n+ = Oxidized state of metal M2


1.5.1 Woked out examples based on Nernst Equation



Q       Calculate the electrode potential of a copper electrode placed in 0.015 M
        CuSO4 solutions EoCu =- 0.34v.
                E = Eo + 0.059 log [Mn+]
                           n
        The reduction reaction is
                   Cu2+ + 2e _       Cu (n = 2)
                   E = 0.34 + 0.059 / log [0.015]
                                2
                   = 0.286 V

Q.      What is the potential of Ca2+/Ca electrode in which the concentration of Ca2+ is
        0.01 M

                   Eo Ca2+/ Ca = -2.87V
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Ans:           Ca2+ + 2e _ Ca
               E = Ca2+ /Ca = E0 Ca2+ / Ca + 0.059 log [Ca2+ ]
                                                n       [Ca]
               Eo Ca2+/ Ca  = -2.87V
                                n = 2
                               [Ca] = 1
                         [Ca2 +]     = 0.01M = 10 -2
                     E Ca2+/ Ca = -2.87 + 0.059 log [10 -2 ]
                                             2
                                = -2.929 V

Q.     The standard calculation potential of zinc is -0.76 V and silver is 0.80 V.
       Calculate the emf of the cell.

               Zn | Zn (NO3)2 || AgNO3               | Ag
                        0.1M     (0.01M)
Ans: The cell reaction is
               Zn + 2Ag+  Zn2+ + 2 Ag
               Eo cell               =     EC - EA
                                     =     0.80 - (-0.76)
                                     =     1.56 V
                                 0.0591      [ Zn 2 ]
                Ecell  E cell 
                           0
                                        log
                                    n       [ Ag 2 ] 2


                                   0.0591     [ Zn 2 ]
                Ecell  E cell 
                           0
                                          log
                                      n       [ Ag 2 ]
                                   0.0591         [0.1]
                Ecell  E 0 cell         log
                                      2       [0.01x0.01]


                Ecell  E 0 cell 
                                     0.0591
                                        2
                                                
                                            log 0.1x10 4   
                                   0.0591
                Ecell  1.56             log 10 3
                                      2
                      = 1.4715 V

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Q.     What ratio of Pb2+ to Zn2+ concentration is needed to reverse the following cell
       reaction.

                Zn + Pb2+ (aq)                 Zn2+ (aq) + Pb
                EO Zn2+ / Zn         =    -0.76 V       EO Pb2+ | Pb = -0.126 V

Ans:
                Zn + Zn2+ || Pb+2 | Pb
                                     0.0591     [ Zn 2 ]
                Ecell  E 0 cell           log
                                        2       [ Pb 2 ]

                                 0.0591     [ Zn 2 ]
                Ecell  E cell 
                           0
                                        log
                                    2       [ Pb 2 ]
                                           0.0591     [ Pb 2 ]
                Ecell  0.126  (0.76)         log
                                              2       [ Zn 2 ]
                                0.0591     [ Pb 2 ]
                Ecell  0.634         log
                                   2       [ Zn 2 ]

                 At Equilibrium Ecell = 0

                               0.0591     [ Pb 2 ]
                 0  0.634           log
                                  2       [ Zn 2 ]

                                      [ Pb 2 ]
                                         2
                                                = 3.22 X 10 -22
                                      [ Zn ]
1.5.2 Difference between Emf a nd potentia l diffe rence

The potential difference is the difference between the electrode potentials of the two
electrodes of the cell under any condition w hile emf is the potential generated by a
cell when there is zero electron flow, i.e., it draws no current. The points of
difference are given below:

1.6,.Electrode potential
The natural tendency of en electrode to undergo oxidation or reduction reaction when it is in
contact with an aqueous solution of its own ions is expressed in terms of electrode potential.
Thus there are two types of electrode potential, i) Oxidation potential:- Tendency of an electrode
in a half cell to lose electrons and ii) Reduction potential:- Tendency of an electrode to gain




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electrons. When a metal is in contact with a solution of its own ions it may undergo

                M        Mn+ + ne-
oxidation,
                        Mn+ + ne-        M
or reduction reaction                        .




       (Fig. 1.6a)                                         (Fig 1.6 b)
Because of this, negative or positive charge is developed on the metal, which attracts
positively or negatively charged free ions in the solution. This leads to the formation of a
layer of negative ions ( Fig a).) or positive ions ( Fig b, ) around metal electrode. Thus a
dynamic equilibrium is established between the +ve /-ve charges at the electrode.
Therefore due to the interaction of opposite charges at the respective electrodes, +ve (a)
or –ive (b) ions remain close to the metal interface. This charged layer around an
electrode is called Helmholtz electrical double layer (Fig.1.6).. Because of the formation
of this electrical double layer a potential difference exists between the metal electrode
(M) and its ionic solution (Mn+). This potential difference will persist as long as the
charge is allowed to remain on the metal. This potential difference becomes constant at
equibillium and is known as electrode potential of a metal.




19
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Thus the single electrode potential can be defined as the equilibrium potential difference
exits at the interface between the metal electrode (M) and its ionic solution (Mn+), when
it is in contact with a solution of its own ions.
The single electrode potential depends on factors such as

             The nature of the metal, M,
             The concentration of metal ions, Mn+ and.
             The temperature
1.6.1 Standard Electrode potential
It is defined, as the equilibrium potential difference exists at the interface between the metal
electrode and its surrounding ions of unit concentration at 25o C. Standard reduction potential is
denoted as E0 Mn+/M and Standard oxidation potential is denoted as E0 M/ Mn+.


1.6.2 Electrochemical series.
The electrodes are arranged in the increasing or decreasing order of standard electrode potential. If the
electrodes are arranged in the increasing order of standard reduction potential or decreasing order of
standard oxidation potential on hydrogen scale, it is known as electrochemical series . Standard electrode
potentials in aqueous solution at 298 K is given in table 1.2.
Table 1.2. Standard Electrode reduction potential at 298 K.

           Element                      Electrode Reaction            Standard Electrode
                                                                      Reduction potenti al
                                            (Reducti on)
                                                                             Eo , volt

              Li                            Li+ + e- = Li                     -3.05

               K                            K+ + e- = K                      -2.925

              Ca                          Ca2+ + 2e- = Ca                     -2.87

              Na                           Na+ + e- = Na                     -2.714

              Mg                          Mg 2+ + 2e- = Mg                    -2.37

              Al                           Al3+ + 3e- = A l                   -1.66

              Zn                          Zn 2+ + 2e- = Zn                   -0.7628

              Cr                           Cr3+ + 3e- = Cr                    -0.74




20
````


              Fe         Fe2+ + 2e- = Fe    -0.44

              Cd         Cd 2+ + 2e- = Cd   -0.403

              Ni         Ni2+ + 2e- = Ni    -0.25

              Sn         Sn 2+ + 2e- = Sn   -0.14

              H2         2H+ + 2e- = H2      0.00

              Cu         Cu 2+ + 2e- = Cu   +0.337

              I2          I2 + 2e- = 2I-    +0.535

              Ag         Ag + + e- = Ag     +0.799

              Hg         Hg 2+ + 2e- = Hg   +0.885

             Br2         Br2 + 2e- = 2Br-   +1.08

             Cl2         Cl2 + 2e- = 2Cl-   +1.36

              Au         Au 3+ + 3e- = Au   +1.50

              F2          F2 + 2e- = 2F-    +2.87




Figure                                                   1.6.b
Varation                                                    of
properties                                                  of
elements                                             based on


electrochemical series



21
````

Metals at the top of the series are good at giving away electrons. They are good reducing agents. The
reducing ability of the metal increases as you go up the series.

Metal ions at the bottom of the series are good at picking up electrons. They are good o xidising agents. The
oxidising ability of the metal ions increases as you go down the series. The more negative the E° value, the
more the position of equilib riu m lies to the left - the more readily the metal loses electrons. The more
negative the value, the stronger reducing agent the metal is.

The more positive the E° value, the more the position of equilibriu m lies to the right - the less readily the
metal loses electrons, and the more readily its ions pick them up again. The more positive the value, the
stronger oxidizing agent the metal ion is.

1.7 Types of electrodes
There are different types of electrodes, which can act as anode or cathode in an electrochemical
cell.
        a) Metal-metal ion electrode. This electrode consists of a metal rod placed in a solution
           containing its own ions. The half cell is represented as
           Mn+ (aq)/M(s)


           The redox reaction at the electrodes is


            Mn+ (aq) +ne-       M


           Eg. Zn2+/Zn or Cu2+/Cu
         The electrode potential is

                            RT      [M ]
         Ecell  E 0 cell      ln
                            nF [ M n  ]
         Since [M]=1, equation becomes
         Ecell  E 0 cell 
                             RT
                             nF
                                      
                                 ln M n  

        b) Gas electrode
A gas electrode consists of a particular gas passing through an inert electrode ( Pt), which
is dipped in a solution containing ions to which it is reversible. At the metal interface,
there is equilibrium between the gas and the ions.
Consider hydrogen electrode. The half cell is represented as
                H+(aq)/H2 (g)/Pt and the electrode reaction is
              H+ (aq) + e-                    ½ H2 (g)


22
````


        Chlorine electrode is also obtained by bubbling chlorine gas through Pt electrode,
        which is dipped in a solution of Chloride ions.
        Chloride electrode is represented as Cl- /Cl2 (g), Pt
        And the electrode reaction is 1/2Cl2 +e-                   Cl-.

c) Metal – Metal salt ion electrode.
        It consists of a metal and a sparingly soluble salt of the same metal dipp ed in a
        solution of salt having the same anion. In this electrode, a metal is in contact with
        a sparingly soluble salt (MX) of the same metal dipped in a solution containing
        anion of the salt. It is represented as
        X-(aq)MXM(s)
        The electrode reaction for this type of electrode is written as
        MX(s)+ne-               M (s) +x-(aq)
        and the electrode potential is
        EX-/MX/M= E0 X-/MX/M - RT ln [X-]
                                         F
        Examples, a) Calomel electrode- Cl-/Hg2 Cl2 /Hg
                        b) Silver chloride electrode Cl-/AgCl/Ag
        d) Oxidation reduction electrode.
It consist of an inert metal (Pt) dipped in a solution containing ions in two oxidation states of the
substance
It is represented as Pt/Mn1+ (c1 )/ Mn2+(c2 )
Electrode reaction is
        Mn2+ (c2 ) +ne-             Mn1+ (c1 )
                                                 RT [C1 ]
The electrode potential is given as E  E 0       ln
                                                 nF [C 2 ]
For example, an oxidation –reduction electrode is formed when a Pt wire is placed in a solution of
iron sulphate. Here Fe 2+ is getting oxidized to Fe 3+ by releasing an electron. This electron is
picked up by the platinum wire and that goes to the circuit.
Fe2+               Fe3+ +e-
The developed potential could be assigned due to the tendency of one oxidizing state to change
into a more stable oxidized state. The potential is sensed by a Pt electrode.


23
````

Pt,Fe2+/ Fe3+, Pt.Sn2+/Sn4+
1.8 Reference electrodes
Reference electrode is an electrode of standard electrode potentia l, with which we can compare
the potential of any other electrode, by coupled with it. The reference electrode provides a stable,
reliable and reproducible potential against which the unknown electrode is compared. Therefore
they can be used repeatedly with accuracy. Reference electrodes can be broadly classified into
two types.
1.8.1Primary reference electrodes
Standard hydrogen electrode is the primary reference electrode which is used to measure the
potential of another electrode. The electrode potential is taken as zero at all temperatures
ie E0 = 0.
The important features of hydrogen electrode are as follows
       1.
            Primary standard hydrogen electrode is represented as. Pt,H2 (g)/H+
       2. A schematic sketch is shown in figure.1.7
       3.
            It consists of small platinum strips coated with platinum black to absorb hydrogen gas.
       4.
            A Pt wire is welded to the Pt strip and sealed in a glass tube so as to make contact with
            outer circuit through Hg.
       5.
            The Pt strip and glass tube is surrounded by an outer glass tube, which has an inlet for H 2
            gas at the top and a number of holes at the base for escape of excess hydrogen gas.
       6.
            The Pt is placed in acid solution, which has H + ion.
       7.
            Pure hydrogen gas is circulated at one atmospheric pressure.
       8.
            A part of the gas is absorbed and the remaining escape through the holes.




24
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       9.
            This establishes equilibrium between the absorbed hydrogen and the hydrogen ions in




            the solution.


Figure 1.7: A schematic sketch of standard Hydrogen electrode

The half cell reaction is 1/2H2 + +e -                      H

The electrode is reversible with respect to hydrogen ion. Since hydrogen gas is non
conducting, Pt or some other metal which is not attacked by the medium and which can
easily arrive equilibrium with H2 is commonly used.


1.8.2 Measurement of standard electrode potential
The single electrode potential of a half-cell is determined by coupling it with a reference
electrode.     Standard hydrogen electrode.[S.H.E.] is usually taken as reference electrode for
finding out the single electrode potential.. The electrode potential of reversible hydrogen
electrode is taken as zero.

25
````

1.8.2.a Measurement of electrode potential. of     Zn| Zn2+ electrode.

Zn rod dipped in 1M ZnSO4 solution. This half cell is combined with standard hydrogen electrode
through a salt bridge both electrodes are connected with a voltmeter (fig 1.8). Oxidation takes
place at the zinc electrode. Therefore E0 (Zn2+, Zn) electrode is negative. The cell representation
is given as
             +
Zn/Zn2+(c1) H (c =1M) H2(g)(p=1atm} ,Pt

The electrode reactions of the cell are
                           Zn2+ + 2e-
At anode , Zn
                       -
At cathode,    2H+ + 2e               H2
                                           . The current flows from hydrogen electrode to metal
electrode. The e m f of the cell is found to be 0.76volt from the voltmeter.
Ecell = E0 cathode – E0anode
 Ie, 0.76 = 0- E0 anode
Therefore E0 anode = -0.76volt




Figure 1.8 .a. A cell to determine the potential of anode ( Zn electrode) using standard hydrogen
electrode


1.8.2.b Measurement electrode potential of Cu2+/Cu electrode
        Cu rod dipped in 1M CuSO4 solution. This half cell is combined with standard hydrogen
electrode through a salt bridge both electrodes are connected with a voltmeter (fig 1.8 b).
Reduction takes place at the Cu electrode. Therefore Potential of Cu electrode is positive . ie,
E(Cu2+, Cu) is positive.
The cell representation is given as

26
````


             +
Zn/Zn2+(c1) H (c =1M) H2(g)(p=1atm} ,Pt

The electrode reactions of the cell are
                             Zn2+ + 2e-
At anode , Zn
                         -
At cathode,      2H+ + 2e             H2
                                           . The current flows from copper electrode to hydrogen
electrode. The e m f of the cell is found to be 0.34 volt from the voltmeter.
E cell = E0 cathode – E0 anode
 Ie, 0.34 = E0 cathode - 0
Therefore E0 cathode = 0.34 volt




Fig 1.8.b. A cell to determine the potential of cathode( Copper electrode)
All other single electrode potentials are referred to as potentials on the hydrogen scale.
Positive electrode potential indicate that, the metal which is connected to SHE undergoes
reduction process and negative electrode potential indicate that the metal which is
connected to SHE undergoes oxidation.

1.8.2.c Limitations of primary reference electrode ( Standard hydrogen electrode)

          It is difficult to maintain H+ ion concentration as 1M
          It is also difficult to maintain a flow of 1atm of hydrogen at the interface of pt
           electrodes. It cannot be used in solutions containing redox systems.
          It readily affected by compounds of Hg, As, S and oxidisng agents like Fe3+,
           Mn4+, Cr2 O 72- etc and therefore, this electrode cannot be used in solutions
           containing these ions.
          Impurities present in hydrogen flow may poison the Pt electrodes which alters the
           potential value.

1.8.3 Determination of PH using standard hydrogen electrode

27
````



SHE can be used to find the PH of the unknown solutions

Procedure

The unknown solution is taken in a vessel and a platinum electrode is half dipped in it.
Hydrogen gas at 1 atmosphere is bubbled through the solution. The platinum catalyses
the following electrochemical reaction at the electrode.

H+ (aq) + e-                   ½ H2

This generates a potential which depends on the H+ ion concentration of the unknown
solution.

On applying Nernst equation on above equation we gets,


                   RT    [M ]
        E  E0       ln        ,   since [M]=1, E0 = 0 , and n= 1, then the equation becomes
                   nF [ M n  ]

                            [1]
       E  0  0.0591log
                           [H  ]

                            
       E  0.0591x  log H 

       E  0.0591P H

The half cell formed is connected to a standard or normal hydrogen electrode. The two solutions
are connected by a salt bridge. The e.m.f of the cell is determined by using a potentiometer.
Since the emf of the SHE is zero, The observed emf gives directly the emf of the half cells
containing the solutions under test. A schematic diagram showing the determination of P H using
standard hydrogen electrode is given in figure 1.8.c.




28
````



Figure 1.8.c. A schematic diagram showing the determination of P H            using standard
hydrogen electrode.

The cell is represented as

The emf of the cell is

Ecell= Ecathode- Eanode

       E anode = E hydrogen electrode with solution of unknown P H

               = -0.0591PH


E cathode= EH+/H = 0

Therefore E cell = 0-(-0.0591PH)
          E cell= 0.0591PH
          PH= 0.0591/Ecell

1.8.4. Calomel electrode

Calomel electrode is used as a reference electrode. It is represented as : Hg, Hg2 Cl2 (s);
KCl(solution). A sketch of calomel electrode is shown in figure 1.8.d

It consists of Hg of high degree of purity, placed at the bottom of a glass tube having a
side tube an each side. Hg is covered by a paste of Hg2 Cl2 (calomel). A solution of KCl is
introduced above the paste through the side tube. The concentration of the solution is
either decinormal, normal or else the solution is fully saturated. A Pt wire sealed into a
glass tube serves to make electrical contact of the electrode with the circuit.




29
````



Fig 1.8.d : A sketch of calomel electrode.

If the electrode reaction involves oxidation,

                2Hg (l)                          Hg22+ + 2e-

               Hg22+ + 2Cl-                       Hg2Cl2 (s)

               2Hg(l) + 2Cl-                  Hg2 Cl2 (s) + 2Cl-


If the electrode reaction involves reduction thus

          Hg2 Cl2 (s)               Hg22+ + 2Cl-

               Hg22+ 2e-                  2Hg (l)-

       Hg2 Cl2 (s) + 2e-                    2Hg (l) + 2Cl-


Hence the electrode reaction is reversible with respect to the chloride ion
concentration. The potential of the electrode depends upon the con. of KCl solution
taken.
       The electrode potential is given by
                                  2.303RT     [ Hg ] 2 [Cl  ] 2
       Ecalomel  E 0 calomel            log
                                     nF         [ Hg 2 Cl 2 ]
       Here [Hg]=[Hg2Cl2 ]=1, n=2

        Ecalomel  E 0 calomel 
                                 2.303RT
                                    2F
                                                    
                                         2 log [Cl  ]     
                                 2.303RT
        Ecalomel  E 0 calomel          log[ Cl  ]
                                    F
        Ecalomel  E calomel  0.0591log[ Cl  ] , if T = 298 O K
                     0




Thus the potential of the calomel electrode depends on the concentration of the chloride
[Cl-] ions and decreases as the concentration of the chloride ions decreases. The potential
of the calomel electrode is found to depend on the concentration of KCl solution used.


30
````
Concentraion of KCl               E0 calomel (volts)
0.1N                              0.3334
1.0N                              0.2810
Saturated                         0.2422V


Advantages of calomel electrode (Secondary reference electrodes over primary reference
electrode).
     It is simple to construct
     Cell potential measurements are reproducible and stable over a large period and
         does not vary with temperature
     It is calibrated with respect to SHE
     It is having least potential gradient with temperature
     Potential values are reliable and measurements can be made with great accuracy

Disadvantages of calomel electrode

It should not be used above 500 C because the mecurous chloride brakes down, yielding
an unstable readings. Recently calomel electrode has fallen into disfavor because of its
toxicity.
Determination of PH using calomel electrode
A calomel electrode can be coupled with hydrogen electrode containing unknown
solution of PH for its determination. A sketch of determination of PH is shown in figure
1.8.e.




The cell is represented as




31
````


               Pt,H2(g)(p=1atm}   H+ (c = ?)   Hg2Cl2 (s) Hg+

               The e.m.f of the cell is

               E cell = E cathode - E anode

               = Ecalomel + E hydrogen electrode with solution of unknown P H

               E cell = Ecalomel- (-0.0591PH)

               E cell= E calomel +0.0591PH

                               Ecalomel
               P H  Ecell 
                                0.0591

Ion-selective electrode (ISE)
A ion selective electrode have the ability to respond to certain specific ions and develop a
potential. The potential developed is a measure of the concentration of the species of
interest. These electrons use a membrane which is sensitive to particular chemical
species. When the membrane of the electrode is in contact with solution of containing the
specific ion, a voltage dependent on the level of that ion in solution develops at the
membrane. All ISE‟s measure the specific ion concentration directly. Samples need to be
aqueous to avoid contaminating or dissolving the membrane.

TYPES OF ION SELECTIVE ELECTRODES

1. Glass Membrane Electrodes Glass membrane electrodes are formed by the doping of
the silicon dioxide glass matrix with various chemicals. The most common glass
membrane electrodes is the pH electrode. Glass membrane electrodes are also available
for the measurement of sodium ions. Glass membrane electrodes exchange H ions from
an acid solution.

2. Polymer Membrane Electrodes Polymer membrane electrodes consist of various ion-
exchange materials in an inert matrix such as PVC, polythene or silicone rubber. After
the membrane is formed, it is sealed to the end of a PVC tube. The potential developed at
the membrane surface is related to the concentration of the species of interest. Potassium,
calcium and nitrate ion sensitive electrodes are the examples.

3. Solid State Electrodes      Solid state electrodes utilize relatively insoluble inorganic
salts in a membrane. Solid state electrodes exist in homogeneous or heterogeneous forms.
In both types, potentials are developed at the membrane surface due to the ion-exchange
process. Examples include silver/ sulphide, chloride and fluoride.

4. Gas Sensing Electrodes Gas sensing electrodes are available for the measurement of
ammonia, carbon dioxide, nitrogen oxide and sulfur dioxide. These electrodes have a gas
permeable membrane and internal buffer so lution. The pH of the buffer solution changes

32
````


as the gas reacts with it. The change is detected by a combination pH sensors with the
housing. Due to the construction, gas sensing electrodes do not require an external
reference electrode.

1.8.2 Glass electrode
A glass electrode consists of a thin glass bulb filled with 0.1 N HCl and a silver wire
coated with AgCl is immersed in it. Here Ag/AgCl acts as the internal reference
electrode. Glass electrode made of a special glass of relatively low melting point and high
electrical conductivity. It is a ac corning glass containing Na2 O ( 22%), CaO (6%) and
SiO2 (72 %)The glass electrode is represented as

Ag AgCl(s) 0.1HCl glass




Fig 1.8.f. A sketch shows the representation of glass ele ctrode
When the glass electrode is immersed in an unknown solution whose PH is known, there
develops a potential between the two surfaces of the membrane. The potential difference
developed is proportional to the difference in PH value.
Theory
The glass membrane of the glass electrode undergoes an ion exchange reaction, The Na +
ions on the glass are exchanged for H+ ions . The potential difference or the boundary
potential (Eb) at the interface is the result of the difference in potential ( E1-E2)
developed across the gel layer of the glass membrane between the two liquids. This

33
````


boundary potential is related to the concentration of acid solution inside the glass bulb
(c1) and the concentration of the acid solution into which the glass bulb is dipped (c2) by


                             RT C 2
           E B  E1  E2      ln
                             nF C1
As C1 = 0.1M (constant)


Boundary potential (EB ) = RT/nF lnC2 -L where L is a constant and it depends on the PH
of the solution taken in the bulb of the glass electrode
when C1 -C2- EB = 0
But it was observed that even when C1 =C2 a small potential is developed. This is called
asymmetric potential (Eassy )
The potential of the glass electrode (EG) is given by
EG =EB + EAg/AgCl+Eassy
EG = RT/nF lnC2 -L+ EAg/AgCl+Eassy
     = RT/nF lnC2+ (-L+ EAg/AgCl+Eassy)
     = RT/nF lnC 2 + E0 G where E0 G = -L+ EAg/AgCl+Eassy


EG = - 2.303 RT/nF (- log H+)+ E0 G
At 298 K
       EG = E0 G -0.0591 pH
Advantages of glass electrode
       o It has not salt or protein error
       o It can be used in turbid, colored and colloidal solution
       o It can used in both alkaline and oxidizing solutions
       o PH Measurements are possible even with few millmeters of solutions
       o Most convenient and simple to use
       o It is not easily poisoned
       o Equilibrium is rapidly achieved
       o The results are accurate
Limitations of glass electrode

34
````


       o For the measurement of emf of glass electrode, we cannot use ordinary
          potentiometers because of high resistance of glass. Electronic potentiometers are
          used in such cases


       o The glass electrode can be used upto a P H of 13 but becomes sensitive to Na+ ions
          above PH = 9 resulting in an alkaline error.

       o It is not suitable for in presence of ethyl alcohol, acetic acid, gelatin and sodium
          salt solutions with PH more than9.0


       o Due to absorptive nature, a lot of washing is required. Before use, it is kept
          immersed first in 0.1 M HCl and then in water for 24 hours.


Measurement of PH of a solution using glass electrode
The glass electrode is immersed in the solution whose is to be determined. It is combined
with a reference electrode. Saturated calomel electrode is used as the reference electrode.
The cell is represented as


                                             H
Ag AgCl(s) 0.1HCl glass Solution of unknown P KCl (sat) Hg2Cl2 (s) Hg

The emf of the cell at 298 k is given as


Ecell= E right- Eleft
        =Ecalomel - EG
       =Ecalomel- E0 G-0.0591PH
    ( Ecalomel  E 0 G)  Ecell
P 
  H

             0.0591
The cell emf is a function of the P H of solution in which the glass electrode is dipped.
 For PH measurements, E0 g is fist measured by dipping the glass electrode in buffer
solutions of known PH values. Once the E0 G is determined, then the electrode is placed
in the solution of unknown P H and, the emf of the cell is measured and using above
equation, its PH can be calculated.

PH meter

35
````



It is an instrument used for the direct measurement of PH of unknown solution. It makes
use of a combination of glass electrode and a calomel electrode sealed in single tube.

1.8.6 Worked out examples
Q. Determine the pH of the solution at 25 0 C from the following data. The given cell is
pt/H2 (1 atm)/H+ (c=?)/saturated calomel electrode. The EMF of this cell is 0.6346 V and
the electrode potential of normal calomel Electrode is + 0.24 V.

Ans: . He re , the EMF of the cell,
Ecell= E cathode- E anode
E cell =E calomel – E (2H+ / H2 )
         = Ecalomel + 0.0591PH
        H
       P = Ecell- E calomel/0.0591

Therefore, substituting respective values
Or 0.6346 = 0.24 + 0.0591pH

Or pH = (0.6346-0.24)/0.0591 = 6.67



Q. The Cell, Ag / AgCI (s) / HCI ( 0.IN) / glass / unknown solution / saturated KCI
solution / Hg2 CI2 (s) / Hg gave an EMF of 0.1120 V at 25 0 C, when the pH of the buffer
used was 6.0. When a buffer of unknown pH was used, the EMF of the cell was found to
be 0.3865 V at 250C. What is the pH of the unknown solution ? Ecalomel ( Saturated=
0.2415 V at 25o C)
 Solution. The expression for pH of the solution obtained by using glass electrode is
given by
                               PH = E cell + E0 G- E calomel
                                        0.0591

where E0 G is the standard electrode potential of glass electrode
            Here E cell = 0.1120 V, E calomel ( saturated)= 0.2415 V
            PH of the buffer used for standardizing glass electrode =6.0
            Putting these values in the above expression,
                                     6.0 =0.1120+ E0 G- 0.2415
                                              0.00591

36
````



or                                       E0 G =6.0 x 0.0591 – 0.1120 + 0.2415 =0.4841 V


In the case of solution of unknown pH, the EMF of the cell = 0.3865V
Hence, again using the same expression, we find

                                        PH = 0.3865 + 0.4841 – 0.2415
                                                   0.0591

                                                    PH = 10.64

1.9 Reversi ble and irreversible cells.

When a cell work in a thermodynamically reversible manner it is called reversible cell. A cell is said to
work reversibility when it is sending out infinitesimally small current so that the reaction of the cell
remains virtually in state of equilib riu m.

A reversible cell will have to satisfy the following conditions
             In an emf exactly equal to the cell emf is supplied fro m an external source, the cell reaction
               will stop and so no current will be g iven out by the cell. On application of E.M.F. equal to that
               of the cell no reaction should occur in the cell.


              When an opposing E.M.F. infinitesimally greater than the cell is applied, the cell reaction
               should be reversed.
              If the external emf is infinitesimally less than that of the cell emf, an extremely small current
               will flo w through the cell which is proportional to the chemical reaction taking place in the
               cell.
Eg. Reversib le cell – Daniel cell.


Daniel cell has the emf value 1.09 volt. If an opposing emf exactly equal to 1.09 volt is applied to the cell,
the cell reaction stops.




           Zn + Cu 2+ --> Cu + Zn 2+

If the external emf is increased infinitesimally beyond 1.09 volt, the cell reaction is reversed.

                                Cu + Zn 2+ --> Zn + Cu 2+

If the applied emf is slightly less than 1.09 volt , the normal cell reaction taking place.

                       Zn + Cu 2+ --> Cu + Zn 2+

The reaction is not a reverse of the first equation and the products are different . Therefore

37
````

Daniel cell is termed as reversible cell.

Any other cell wh ich does not obey the above conditions is termed as irreversib le. A cell consisting of zinc
and copper electrodes dipped into the solution of sulphuric acid

[ Zn/H2SO4 /Cu] is irreversible.

When the external emf is just smaller than the emf of the cell, the following reactions taking place

At anode Zn → Zn 2+ + 2e-

At cathode 2H+ + 2e- →H2

The complete cell react ion is

Zn + 2H+ → Zn 2+ + H2 (1)

When the external emf is just greater than the cell emf , the following reactions taking place.

At anode, Cu →Cu 2+ + 2e-

At cathode, 2H+ + 2e- → H2

The overall reaction is

Cu + 2H+ → Cu 2+ + H2 (2)

The reaction 2 is not exactly reverse of the equation (1) and therefore Zinc acid cell is not irreversible .

Similarly, the cell

          Zn|H2 S04 (aq)|Ag

is also irreversible because when the external emf is greater than the emf of the cell, the cell reaction,

          Zn + 2H+ --> Zn 2+ + H2

becomes

          2Ag + 2H+ --> 2Ag + + H2

There it is an irreversible cell.

1.10 CONCENTRATION CELLS

Concentration cells are electrochemical cells in which electrical energy is produced not
due to any chemical reaction but due to the transfer of matter from one half cell to the
other because of the difference in the concentration of the species involved. Theses
Concentration cells are of two types


38
````


1. Electrode concentration cells and electrolye concentration cells.

(i) Electrode concentration cells:

In these cells, the potential difference is developed between two like electrodes at
different concentrations dipped in the same solution of the electrolyte. For example, two
hydrogen electrodes at different gas pressure in the same solution of hydrogen ions
constitute a cell of this type.

         (Pt,H2 (Pressure p1 ))/Anode |H+ | (H2 (Pressure p2 )Pt)/Cathode

If p1 , p2 oxidation occurs at L.H.S. electrode and reduction occurs at R.H.S. electrode.

       Ecell = 0.0591/2 log(p1 /p2 ) at 25o C

In the amalgam cells, two amalgams of the same metal at two different concentrations are
interested in the same electrolyte solution.

(ii) Electrolyte concentration cells:

In these cells, electrodes are identical but these are immersed in solutions of the same
electrolyte of different concentrations. The source of electrical energy in the cell is the
tendency of the electrolyte to diffuse from a solution of higher concentration to that of
lower concentration. With the expiry of time, the two concentrations tend to become
equal. Thus, at the start the emf of the cell is maximum and it gradually falls to zero.
Electrolyte concentration cells are of two types

i) Electrolyte concentration cell without transfer of substance and ii) electrolyte
concentration cell with transfer of substance.

i) Electrolyte concentration cell without transfer. In this type of cell, the two electrolytic
solutions are not in direct contact with each other and the transference of ions from one
solution to the other does not take place directly. The two solutions are separated by a salt
bridge or by a third electrode.

Example; (Ag|AgNO 3 (C1 ) || (AgNO 3 (C2 )|Ag ( Figure 1.9 )




39
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Fig 1.9 ; Electrolyte concentration cell without transference using salt bridge.

Theory:

When a metal electrode (M) is placed in a solution containing its own ions[ M n+], then a
potential ( E) is developed at the electrode, and the potential varies with the
concentration ( C) of the ions in accordance with the Nernst‟s eq uation:

           2.303RT
E  E0            log C
              nF

Let us consider a general concentration cell represented as

          M | Mn+ (C1 ) || Mn+ (C2 ) | M where c1 and c2 are concentrations of active metal
ions (Mn+) in contact with the two electrodes respectively and C2 > C1.

The cell reactions are

At anode

M → Mn+ (C1 ) + ne-

At cathode

Mn+ (C2 ) + ne- → M

The overall reaction is


40
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Mn+ (C2 )→ Mn+ (C1 )

EMF of cell = E cathode – Eanode

                                 0.0591     1             0.0591    1
               Ecell  [ E 0           log    ]  [E 0         log ]
                                    n       C2               n      C1

                  0.0591                    0.0591
Ecell  [ E 0           log C2 ]  [ E 0         log C1 ]
                     n                         n

            0.0591    C
Ecell  [          log 2 ]
               n      C1

And the general equation for the emf of concentration cell is given by

           2.303RT    C
Ecell             log 2
              nF      C1

Thus, the emf developed is due to the transference of metal ions from the solution of
higher concentration ( C 2 ) to the solution of lower concentration ( C 1 ).

       c) Concentration cell without transfer using a third electrode

          Consider an electrochemical cell such as

          Pt, H2 (g), HCl (a1)│AgCl(s), Ag

          Let the activity of H+ ions in the solution be (a+ )1 and that of Cl- ions be (a-)1 . since
          reduction takes place on the right hand electrode and oxidation on the left hand
          electrode, the two half cell reactions will be as follows:

          Oxidation ( at anode)

          ½ H2 (g)→ H+ (a+)1 + e- ----- (1)

          AgCl(s) +e- → Cl- (a-)1 +Ag (s) ----- (2)

          The net reaction is

           ½ H2 (g)+ AgCl(s) + → H+ (a+)1 + Cl- (a-)1 +Ag (s) ----- (3)

          Consider the same cell with the difference that the activity of HCl solution is now
          a2



41
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        Pt, H2 (g), HCl (a2 )│AgCl(s), Ag

The net reaction for one Faraday of electricity will now be as follows:

                    ½ H2 (g) + AgCl(s) → H+ (a+)2 + Cl- (a-)2 +Ag (s) ------- (4)

Consider the situation in which cells are connected to each other in such a way that t hey
send current in opposite direction. Thus,

Pt, H2 (g) (1atm), HCl (a1 )│AgCl(s), Ag│ Ag, AgCl(s), HCl (a2 )│ H2 (g) )1atm), Pt

         The overall reaction of the combined cell for the passage of one Faraday of
        electricity will be obtained by subtracting equation (4) from (3)

        i.e., H+ (a+ )2 + Cl- (a-)2   →    H+ (a+)1 + Cl- ----- (5)

Thus for the flow of one Faraday of electricity, the overall reaction is the transfer of one
mole of each of H+ and Cl- ions or one mole of HCl, from a solution of activity a2 to that
of activity a1

The emf of such a cell is given by

                  RT ([(a )1 (a )1 ] 0
E w.o.t  E 0      ln                  , E =0 (6)
                  F    [(a ) 2 (a )2]

            RT ([(a  ) 2 (a  ) 2 ]
E w.o.t      ln                     (7)
            F    [(a  )1 (a  )1 ]

         RT ([(a  ) 2 ]
E w.o.t     ln             (8) where (a±)1 and (a±)2 are the mean ionic activities of the
          F      [(a  )1 ]
electrolytes in the two solutions and the subscript stands for without transference.

         RT ([a 2 ]
E w.o.t   ln       where a1 and a2 are the activities of hydrochloric acid in the two
         F    [a1 ]
solutions.

Here the cell reaction does not involve transfer of electrolyte from one solution to the
other directly. It takes place indirectly.

The middle electrode, viz Ag, AgCl(s) is withdrawn; the two solutions of HCl will be in
direct contact with each other. The cell will then become a concentration cell with
transference

Applications of Concentration cells

42
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The concentration cells are used to determine the solubility of sparingly soluble salts and
valency of the cation of the electrolyte.

Relation between Equ ilibriu m constant, Gibbs free energy and EMF o f the cell

Concept of equili brium in electrochemical cell

In an electrochemical cell a reversib le redo x process takes place, e.g., in Daniel cell:

Zn(s)+Cu 2+(aq) <==> Zn 2+(aq)+Cu(s)

(1) At equilibriu m mass action ratio becomes equal to equilibriu m constant,

i.e., Q = Ke

(2)    Oxidation potential of anode = -React ion potential of cathode

i.e., emf = o xidation potential of anode + Reduction potential of cathode

          =0

Cell is fu lly d ischarged

cording to Nernst equation:

E = Eo - 0.0591/n log 10 Q at 25o C

At equilibriu m, E = 0, Q = K

0 = Eo 0.0591/n log 10 K

K = Antilog [(n Eo )/0.0591

1.9.3 Worked out example

Q. Consider an electrolyte concentration cell of silver electrodes without transfer is
represented as

Ag/Ag+ (0.00475 M) || Ag+ (0.043 M)/Ag. Find out the emf of the cell.

An electrode of higher concentration of Ag+ (0.043 M) acts as cathode, while the other
silver electrode with Ag+ ion concentration (0.00475 M) acts as anode.
Solution
          At anode                    Ag(s)              Ag +    + e
                                                   (0.00475 M)

          At cathode            Ag+      + e  Ag(s)
                                   (0.043 M)

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        The net reaction of the concentration cell is
                               Ag+         Ag+
                             (0.043 M)        (0.00475 M)

The EMF of the concentration cell is
                           E cell =             0.0591 log10 [0.043 ]
                                                   1         [0.00475]
                                         =      0.056 V

Q. A concentration cell is constructed by dipping two copper electrodes in 0.001 M
and 0.1 M CuSO4 and the two solutions are connected that is in contact with higher
Cu2+ ion concentration acts as cathode. Represent the cell, write the cell reactions and
find out the emf of the cell.
Solution
(i)    Representation of concentration cell
                    Cu/Cu2+ (0.001M) || Cu2+ (0.1M)/Cu
(ii)    The two electrode reactions are as follows:
               At anode
                                         Cu       Cu2+
                                                    (0.001M )
               At cathode
                                     Cu2+ 2e  Cu
                                     (0.1M)
(iii)   The net reaction of the concentration cell is
                                     Cu2+  Cu2+
                                     (0. 1 M)      (0.001 M)

        The EMF of the concentration cell is
                           E cell = 0.0591 log10                0.1
                                            2                   0.001

                                         = 0.0591V

Q A cell contains two hydrogen electrodes. The negative electrode is in contact with a
solution of 10-6 M hydrogen ions. The emf of the cell is 0.118 volt at 25° C. Calculate the
concentration of hydrogen ions at the positive electrode.

Solution:       The cell may be represented as

Pt|H2 (1 atm)|H+||H+|H2 (1 atm)|Pt

10-6 M CM


44
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Anode                          Cathode

(-ve)                         (+ve)

H2 ---> 2H+ + 2e-         2H+ + 2 ---> H2

Ecell = 0.0591/2 log([H+ ]cathode2 )/[10-6 ]2

0.081 = (0.0591) log ([H+ ])/10-6

log[H+ ]cathode/10-6 =0.118/0.0591=2

[H+ ]cathode/10-6 = 102

[H+]cathode = 10-6 = 10-4 M


1.9.4 Polarization
        It is a phenomenon in which there is a variation of electrode potential, owing to
the inadequate diffusion of species from the bulk of the electrolyte solution to the vicinity
of the electrode [or owing to back emf brought about by the product of electrolysis is termed
polarization].
        When a voltaic cell consisting of Zinc and copper electrodes in sulphuric acid is set up
and a current is taken from it, the emf of the cell readily decreases. This is because the copper
electrode becomes covered with the bubbles of the gas which make a gas electrode with emf
opposite to that of the cell. If the bubbles of the gas are removed mechanically or chemically the
emf of the cell remains constant.
                 For a given concentration of electrolyte, the polarization emf for any given type
of cell is a constant.. The new electrode formed as a result of accumulation of product of
electrolysis has a emf opposite to that of the cell is called cell back emf. Thus unless the applie d
emf is greater than the polarization emf, electrolysis almost stops.
Reasons for polarizations
            1)   The gases collected at the electrodes produce local cells and offer resistance to
                 normal flow of current through the cell (gas polarization).
            2)   The product of electrolysis may convert the inert platinum electrodes in to active
                 electrodes which can exercise a back emf.
            3)   The deposited metal may also form a cell, function in the opposite direction
                 similar to the gas electrode.


45
````

           4)    The concentration of electrolyte becomes different due to movement of ions.
                 Thus the difference in concentration of electrolyte forms a concentration.cell
                 working in opposite direction. –concentration polarization.
The extent of polarization depends on
                       Size of electrode: Large surface area decreases the polarization. More
                          reaction takes place.
                      Nature of the electrode surface: Smooth surface is more polarized than
                       rough surface. [Polarization effects on Pt. Black electrodes is much less
                       than that on smooth pt, electrode]
                      Nature of ions deposited: Extent of polarization increases when the
                       liberated ions, form adherent and non – porous film on them.
                      Concentration of electrolyte: Low con of electrolyte decreases the
                       polarization effects.
                      Temperature: Polarization effects are minimized by increasing the
                       temperature, since rate of diffusion of ions increases.
                      Conductivity of electrolytic solution: By adding certain substances which
                       reduce conductivity decreases polarization.
                      Use of Depolarizer: Depolarizer is a substance which is employed to
                       reduce the extent of polarization: Use of oxidizing agents like HNO3 ,
                       MnO2 and chromic acid as depolarizer helps in converting evolved
                       hydrogen into H2 O. Thus hydrogen is not permitted to be absorbed on the
                       electrode. There by polarization effects are reduced.
                      Bruising out the gases present on electrodes.

1.9.5 Decomposition potential
        It is the minimum voltage that is required to bring about electrolysis of an electrolyte
without any hindrance or it is defined as minimum potential which must be applied between the
two electrodes immersed in given electrolytic solution in order to bring about continuous
electrolytic decomposition. When electrolysis carried out, the product of electrolysis accumulate
around the electrode(s). This causes, change in concentration around the electrodes and an
opposing emf (back emf) is produced.
When two pt electrodes are dipped in H 2 SO4 , as the electrolysis of H 2 O starts by the evolution of
hydrogen and oxygen, it stops very soon because the back emf (produced by the absorption of



46
````

oxygen and hydrogen on the two electrodes is greater than the optimized voltage. Electrolysis
proceeds smoothly because the applied voltage just exceeds the back emf.




Figure 1.10 Variation of applied voltage with current. From the figure it is clear that with
increase in applied voltage the current increases and after applying a particular emf, the
current sharply increases. The emf corresponds to the sharp increase of current is called
the decomposition potential.

The decomposition potential of dil. acids and bases using bright Pt electrodes is 1.7 volts in each
case, because the back emf exerted due to accumulation of hydrogen .
Variation of decomposition potential for different metals.
       I. The decomposition potential for different metals like Zinc, iron etc are vary because they
           possess hugh solutions pressures, ie, they have greater tendencies to go into solutions as
           cat ions and they send back the deposited ions into solution. Thus, deposition of these
           metals at the cathode is difficult.
       II. Decomposition Potentials of metals like Cu, Ni etc. are low, because they possess low
           solution pressures, ie, they possess greater tendency to be deposited.
         Significance
                  Separation of metal ions mixture by electrolysis.
                   A solution containing Cu and Zinc ions can be separated by electrolysis. Since
                   the decomposition potential of Cu is low (1.2V). So at this voltage, only Cu is

47
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               deposited. When all Cu is deposited, then zinc ions get deposited later at a higher
               decomposition potential (2.55 V).
             It is possible to predict the order of the discharge of ions from a solution
             containing several ions. For example, from an electrolyte containing Cu , Zn
             and Cd the order of deposition is copper, aluminium and then zinc. This is
             due to the fact that the E of Cu, Cd and Zn are are +0.34 V – 0.44 V, -
             0.76V respectively.

              In the refining of metals.
              In the electroplating processes.
1.9.6 Over voltage or Over potential
When an electrochemical cell is polarized, its actual potential is different than that
expected according to Nernst equation. The extent of polarization is measured as the
over-potential. In some cathodic reactions, the reduction of the electro-active species may
involve several steps and if any one of the steps is slow and irreversible it may result in
polarization. Polarization of the cell may take place at once or both the electrodes.
Polarization may be caused by both the effects. But concentration polarization may be
eliminated by stirring of the electrolyte and then polarization is mainly due to
overvoltage.

Over-voltage can be defined as the excess voltage to be applied over and above the
theoretical decomposition potential of an electrolyte, to start the electrolysis of the
electrolyte. Or It is the difference between the potential of the electrodes at which the
electrolysis actually proceeds continuously and the theoretical decomposition potential
for the same solution is called over voltage.


 Ie Over-voltage, = Experimental decomposition potential- theoretical decomposition
potential

For evolution of hydrogen gas , this potential difference is called hydrogen over voltage.
For example, theoretical decomposition potential value for the electrolysis of dil H2 SO 4
solution with Pt electrodes is 1.23V, but actual decomposition takes place at a potential
value of 1.70 V.
Overvoltage = 1.70 – 1.23 V= 0.47v
  This overvoltage is referred to as the bubble overvoltage as it is observed just at the
point at which gas bubbles begin to appear. The values for hydrogen overvoltage, using
different metals as the cathode at low current density are given in table 3.
Table 3. Hydrogen overvoltage


48
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Electrode                       Hydrogen          overvoltage   Oxygen overvoltage
                                (volt)
Platnised platinum              0.00                            0.25
Iron                            0.08                            0.24
Smooth platinum                 0.09                            0.43
Nickel                          0.22                            0.06
Cadmium                         0.48                            0.43
Lead                            0.64                            0.32
Zinc                            0.70
Mercury                         0.80




Factors affecting over voltage value
        Nature and Physical state of electrode metal:
         eg. Rough electrodes (Platinized Pt) possess lower over voltage than smooth
         electrodes (smooth pt). This is because Platinized Pt possesses much larger
         surface area than the smooth Pt.
        Nature of the substance deposited.
        Metals have low over voltage than hydrogen.
         Eg. In general, metals have low over voltage than hydrogen. Evolution of
         hydrogen takes place in three stages, whereas that of metal in one stage only.
         H3 O+H+ + H2 O
         H++e-H                   Three stages
         H+HH2 (g)
         Mn+ + ne-M           One stage
        Current density. Over voltage increases with current density in a logarithmic
         manner.
         D = a + b log I where „D‟ is the overvoltage, a and b are the constants, I is the
         current density. This equation is known as Tafel‟ equation.

        PH In the absence of strongly absorbed ionsthe overvoltage at most cathodes are
         indepenedent of hydrogen ion concentration over a large range of Ph values. In

49
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          strongly acd or alkaline solutions devaiosn may occur due to the large
          concentration of hydrogen or hydroxyl ions.


         Temperature
          The increase of temperature will decrease overvoltage. The change is generally 2
          millivolts per degree.
Applications

       1. The electrode deposition of metal having negative potential can be achieved to
          existence of hydrogen over voltages.
          The hydrogen over voltage on Cd metal is about 1.2 volts at current density of
          0.001 amp cm-2. But no H2 is produced, although both Cd and H2 are expected to
          be formed together at pH equal to 7 due to their nearly equal electrode potentials.
          ( H+/H=-0.41 V and Cd2+ / Cd =-0.403V). Thus, metal ions may be reduced one
          at a time by controlling the electrolysis potential. The metal that possesses the
          most positive electrode potential is electrodeposited first. Hence, electrode
          position technique can be used in quantitative process.
          Consider the electrolysis of acidic solution of Zn2+ ions. The electrode potential of
          hydrogen and Zinc is 0.00 V and –0.76v. So, only hydrogen should be liberated at
          cathode. The Zn2+ ions should be deposited on cathode only after all H+ ions have
          been reduced to hydrogen gas. However, zinc starts depositing along with
          liberation of hydrogen. This is due to the high over voltage ( 0.7 V) of hydrogen
          at Zn cathode. As this value of over voltages is very close to electrode potential of
          zinc, so zinc starts getting deposited simultaneously.
       2. Corrosion of metals. Consider the dissolution of lead (Pb) or Zinc in HCI Due to
          negative standard electrode potential of Pb(-0.12 V) and Zn (-0.76 V), they
          should get dissolved in HCI with liberation of hydrogen. But, no hydrogen is
          liberated due to high over voltage of Pb ( 0.64 V) and Zn (0.70 V) So Pb and Zn
          are not corroded by HCI. However, Zn and Pb will get corroded if it contains a
          metal with low over voltages as impurity.



1.10. Batteries

A Battery: A system which converts chemical energy into electrical energyMore
correctly, a battery is an electrochemical cell:








50
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1.10.2 Terminology
Galvanic Cells convert the energy from spontaneous chemical reactions into electricity
Electrolytic Cells use electricity to drive non-spontaneous chemical reactions

All galvanic cells produce electricity from reactions which involve the transfer of
electrons from one species to another
There are two components to each cell – the species donating the electrons, and the
species accepting them
We write “half- reactions” to represent these two components, and to explicitly show the
transfer of electrons
The oxidation half-reaction shows the species which is donating electrons
The reduction half-reaction show the species which is receiving electrons
We can also write the net reaction (or overall reaction) for the cell, the balanced sum of
the two half-reactions
LEO the lion says GER:
        Loss of Electrons is Oxidation; Gain of Electrons is Reduction
Electrodes are electrical conductors in the cell where chemical reactions take place
        The anode is the electrode where oxidation takes place
        The cathode is the electrode where reduction takes place
        The cathode receives the electrons given off at the anode and passes them along
The voltage of the whole cell is the electrical energy that it gives off, measured in volts
(V)
The current is the rate at which electrons pass through the cell, measured in amperes
(A)Open –circuit voltage- This is the voltage measured across the terminals of the cell
or battery when no external current is flowing.
Closed –circuit voltage- This is the voltage measured across the terminals of the cell or
battery when current is flowing into the external circuit.
Separator-It is the barrier between the positive and negative electrodes to prevent direct
shorting of the electrodes. They must be permeable to ions, must not conduct electrons.
They must be inert in the total environment
Amps/Ampere-Hour. This is the rate at which electrons flow in a wire. Unit- coulombs
per second.
Powe r density- The ratio of the power delivered by a cell or battery to its weight. During
discharge of a battery, the power density decreases ( W/kg)
Energy density- The ratio of the energy out put of a cell to its weight Wh/kg
Efficiency- The ratio of the out put of battery on the discharge to the input required to
restore it to the initial state of charge under specified conditions



51
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Cycle life- A battery does have a limited life. This life is expressed of number of cycles
that the battery has gone through. One cycle means one discharge followed by a charge.
Each cycle the battery is loosing a bit of its initial capacity. This loss of capacity is
initially a linear process, but the closer the battery gets to its end of life the bigger the loss
per cycle will be. At one stage the loss of capacity is getting so big that the battery
becomes technically dead. This depends very much on the technology used.
Shelf life-The duration of storage at the end of which a cell or battery still retains the
ability to give a specified performance. Or, it is the length of time a battery can remain in
storage without losing its energy capacity. The metal plates eventually leak and react with
each other, even though not in use.
Capacity- The total quantity of electricity involved in the electrochemical reaction
( coulombs per ampere –hour- Ah) or the total number of ampere hour or watt hour that
can be withdrawn from a fully charged cell or a battery
ie , Q = xnf, x- no of moles of reaction, n- no of e-, F- Faraday constant
Batteries can be classified into primary batteries and secondary batteries.
     The primary battery converts chemical energy to electrical energy directly, using
           the chemical materials within the cell to start the action.
          The secondary battery must first be charged with electrical energy before it can
           convert chemical energy to electrical energy.
           The secondary battery is frequently called a storage battery, since it stores the
           energy that is supplied to it. The difference between primary battery and
           secondary battery are described in Table


           .Table. 4 Difference between primary battery and secondary batteyr
           Primary cells(P)                         Secondary cells (2o )
           1. A cell which act as a source of 1. A cell in which electrical energy is
           electrical    energy     without    being stored in the form of chemical energy
           previously charged by an electrical by previous charging of an electrical
           current from on external source,            current from an external source
           2. It only act as a voltaic cell.           2. It act both as a voltaic cell and
                                                       electrical cell.
           3. One term action only                     3. It can be used over and over again
           4. Once exhausted cannot be recharged       4. Once exhausted it can be recharged


52
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                                                  for a no. of times.
       5. Eg. Dry cell, Daniel cell               5. Pb acid cell, Ni-Cd battery(alkaline
                                                  battery)

1.10.4 Secondary cells
               Lead- acid cell (storage cell) – that can operate both as voltaic cell and as
an electrical cell. These cells can be recharged by passing a current through them in a
reverse direction. The chemical reaction becomes reversed [from those during discharge]
and the cell regains the original state. Since they accumulate electric energy they are also
called accumulator or storage cell.




Figure 1.11 Pb-Acid cell
Lead–sulphuric acid cell consists of one of its electrode made of lead. The other electrode
is made of (PbO 2 ). These are alternatively kept and are insulated from each other by
porous separators made of wood, rubber, plastics or glass fibers. The entire combination
is then immersely 20% H2 SO4 . A representation of Pb-acid cell is shown in figure

53
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Chemical reactions during discharging
          When the storage cell is operating for supplying electrical energy the lead
electrode loses electrons which flow through the wire.
                                          Pb Pb2+ +2e-
The Pb2+ ion then react with SO 2-4 ions forming PbSO 4 at the anode
                                       Pb2+ +SO42-  PbSO4
The e- released from the anode flows to the cathode where lead in PbO 2 gains electron to
form Pb2+ ions. The Pb2+ ions then combine with SO 4 2- ions.
                         PbO 2 +4H+ +2e-  Pb2+ +2H2O (at cathode)

                                       Pb2+ + SO42-  PbSO4

The net reaction is Pb+PbO 2 +4H+ + 2SO4 2-  2 PbSO 4 +2 H2 O+energy. It may be noted
that lead sulphate is ppted at both the electrodes. The voltage of each cell is about 2 Volt.
Chemical reaction at charging
When both anode and cathode becomes covered with PbSO 4 , the cell eases to function
as a voltaic cell. To recharge a lead storage cell, the reaction taking place during
discharging are reversed by passing as external emf greater than 2V from a generator.
The positive pole of the generator is attached to the positive pole of the cell.
PbSO 4 + 2e- Pb + SO 42- [cathode]
PbSO 4 + 2H2O+ 2e-  PbO2 +4H+ + SO 42- [anode].
Net reaction
          2 PbSO4 +2H2O+energyPb+ PbO 2 +4H++2 SO 42-
During charging the electrolyte density rises to 1.28 which is maximum. Over charging
will concentrate electrolyte which eat up the lead plates. Both Pb and PbO 2 plates are
regenerated. Water produced during discharging is consumed during the harging process.
Merits
       1. Emf of each cell 2V
       2. Internal resistance is low
       3. A battery has six such cells in series to make a car battery ( 12 V).




54
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Note : Lead-acid batteries are referred to as “storage batteries”, because this charge-
discharge cycle is so reliable. These batteries were used in every automobile until quite
recently. The battery is discharged in order to start the engine. Once the engine is running
and burning gasoline, it turns an alternator which recharges the battery. This process can
continue for up to 5 years of normal driving After that time, enough of the lead sulfate
product has been shaken off the plates that it can no longer recharge. Lead-acid batteries
are also used in environments where vehicles cannot emit combustion products: Indoor
forklifts, golf carts, handicapped carts in airports, wheelchairs
However, lead is an environmental concern! How do we dispose of the millions and
millions of batteries which die each year?
There is a very successful recycling program in the U.S. – 97% of spent batteries are
recycled But environmentally healthier options are under investigation. A leading
contender is the magnesium- acid batteryNote on Fuel cells
Fuel cells are galvanic cells in which chemical energy of fuel is directly converted to
electrical energy. Since the cell is operated at a temperature above 100 o C, this can be
condensed and used.
Characteristics.
They do not store chemical energy.
React arils are supplied constantly, while products are removed constantly.
Advantages of Fuel cells
       1. Very efficient and convert 75% of available chemical energy to electrical energy.
       2. Electrode metals are………………
These are considered as volatile cells in which electrical materials usually in the form of
gases are supplied continuously and consumed to produce electricity.
Difference between conventional, electrochemical cell and fuel cell.
In a fuel cell, the materials undergoing oxidation at anode or reduction at the cathode are
stored out side the cell. These are not the part of the cell as in dry cell. It does not require
any recharging. The efficiency of a Carnot cycle varies from 20 to 40, where as a fuel cell
convert 75% of the available chemical energy into electrical energy.
Hydrogen oxygen fuel cell
Based on the combustion of H2 to form water where hydrogen act as the fuel and O2 act
as the oxidation in the presence of an aqueous electrolyte.


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                                    2H2 (g) +O2 (g)  2H2 O(g)
IT consists of two electrodes made of porous graphite impregnated with a catalyst (Pt,Hg
metal oxide). The electrodes are placed in an aqueous solution of KOH and NaOH. O 2
and H2 are continuously fed in to the cell under a pressure of about 50 atm. The gases
diffuse into the electrode porous and so does the electrolyte solution.
Oxidation half reaction (Anode)
As the H2 diffuse through the anode it is absorbed on the electrode surface in the form of
hydrogen atom which react with hydrogen ion of the electrolyte to form water.
                                              H2 2H
                                      2H + 2OH-  2H2 O +2e


                                Net reaction H2 + OH- 2H2 O + 2e-
The e- flows through the external circuit to the cathode. The O 2 diffused through the
cathode on the electrode surface, where it is reduced to OH- ions.
                                     O2 +2H2 O +4 e-  4 OH-
The OH- ions thus formed migrate through the electrolyte from the oxygen electrode
(cathode) to the H2 electrode( anode).
The electrolyte thus remains invariant. The electrical out put of the cell results from the
flow of e- through the external circuit. The overall cell reaction is the combination of H 2
and O 2 to form H2 O.
                                    2 H2 +4 OH-  4 H2 O+4e-
                                     O2 +2H2 O +4 e-  4 OH-


                                         2H2 + O2  2H2 O
Merits
       1. Efficiency is very high and does not depend on working temperature.
       2. 75% of chemical energy is converted into electrical energy
       3. They are very light
       4. It takes little time to go into operation.
       5. The land requirement is considerably less compared to conventional power plants.



56
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       6. The individual cells can be stacked and connected in series to generate higher
           voltages.
       7. It produce electric current from direct reaction with fuel and oxidizer without
           going through wasteful intermediate.
Demerits
       1. Cost of materials
       2. Lack of long term reliability
       3. about the fuels.
Merits
       8. Efficiency is very high and does not depend on working temperature.
       9. 75% of chemical energy is converted into electrical energy
       10. They are very light
       11. It takes little time to go into operation.
       12. The land requirement is considerably less compared to conventional power plants.
       13. The individual cells can be stacked and connected in series to generate higher
           voltages.
       14. It produce electric current from direct reaction with fuel and oxidizer without
           going through wasteful intermediate.
Demerits
       1Cost of materials
       2 Lack of long term reliability



1.10.5 The Nickel-Cadmium Battery
The Ni-Cd battery is represented as
Cd (s), Cd(OH)2(s)|KOH (6H)| NiO(OH(s), Ni(OH)2 (s)

Ni-Cd batteries have a metal case with a sealing plate equipped with self- sealing safety
valve. The positive and negative electrode plates, isolated from each other by the
separatorand. Are rolled in a spiral shape inside the case. This is known as the jelly-roll
design and allows a NiCd cell to deliver much higger maximum current.
The cell contains a paste of KOH.This provides the OH- ions needed for the reaction,
while also providing a medium to pass charge (electrolyte). The anode consists of solid
metal which is transformed into cadmium hydroxide. The cathode consists of Ni3+ ions


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Figure 1.12 A schematic sketch of Ni-CD battery


in a NiO(OH) paste which are transformed into nickel hydroxide. A sketch of Ni-Cd
battery is given in figure 1.12 a) and b)

In a nickel-cadmium battery, the reactions actually look like this:

            At Anode, Cd(s) + 2 OH- (aq) → Cd(OH)2(s) + 2 e-

           At cathode, 2NiO(OH)(s) + 2 H2O(l) + 2 e- → 2Ni(OH)2(s) + 2 OH- (aq)

The net reaction is Cd(s) + 2NiO(OH)(s) + 2 H2O(l) → Cd(OH)2 (s) + 2Ni(OH)2(s)

 During oxidation Cd is oxidized as Cd2+ and combine with OH ions forming Cd(OH)2.
The electrons produced during oxidation moves into cathode a Ni3+ accepts the electrons
and reduced to Ni(OH)2. The emf of the cell produced is 1.4 volt.
The reactions are to be reversed beacsed the products Cd(OH)2(s) and 2Ni(OH)2 (s) are
solid products which adhere to the electrode surfaces.

During charging the overall reaction is
Cd(OH)2(s) + 2Ni(OH)2(s) → Cd(s) + 2NiO(OH)(s) + 2 H2O(l)Ni-cd batteries can be
recharged hundreds of times because the solid products of the electrode reaction are
adhere to the surface of the electrodes. The solid hydroxides are sticky, cling to the
innards of the battery, and remain in place.
If current is applied, the reaction can be driven backwards

Features

          Voltage of Ni-Cd cell- 1.4 volt

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          Compact, light weight, good cycle life, capacity and long life
          Good high discharge rate (for power tools)
          Relatively low energy density
          Over 1000 cycles (if properly maintained)
          Fast, simple charge (even after long storage
          vercome by 60% discharges to 1.1V
          NiCd batteries may suffer from a "memory effect" if they are discharged and
           recharged to the same state of charge hundreds of times.
          The apparent symptom is that the battery "remembers" the point in its charge
           cycle where recharging began and during subsequent use suffers a sudden drop in
           voltage at that point, as if the battery had been discharged.
          The capacity of the battery is not actually reduced substantially. Some electronics
           designed to be powered by NiCds are able to withstand this reduced voltage long
           enough for the voltage to return to normal.
Advantages
          Deliver high current (10-1000Ah)
          The batteries are more difficult to damage than other batteries, tolerating deep
           discharge for long periods.
          Longer shelf life
          In fact, NiCd batteries in long-term storage are typically stored fully discharged.
           This is in contrast, for example, to lithium ion batteries, which are less stable and
           will be permanently damaged if discharged below a minimum voltage.
          NiCd batteries typically last longer, in terms of number of charge/discharge
           cycles, than other rechargeable batteries.
          Compared to lead-acid batteries, NiCd batteries have a much higher energy
           density. A NiCd battery is smaller and lighter than a comparable lead-acid
           battery. In cases where size and weight are important considerations (for example,
           aircraft), NiCd batteries are preferred over the cheaper lead-acid batteries
Disadvanatges




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          They are more costly than lead-acid batteries because nickel and cadmium are
           more costly
          The battery exhibits a very marked negative temperature coefficient. This means
           that as the cell temperature rises, the internal resistance falls. This can pose
           considerable charging problems, particularly with the relatively simple charging
           systems employed for lead-acid type batteries.
          Cadmium is an environmental toxin. Hence its use is not encouraging.


          Applications- Ni-Cd battery is widely used in photography, phones, computers,
           transmitters, hearing aids, emergency light etcBecause of their good performance,
           Ni-Cd battery is widely used in military operations even at low temperatures
           stems, In railroad track and signal operations, electrical switch gear, train lighting
           , security and alarm systems and in emergency power suppliers.


1.10.6. Lithium Batte ries
.Lithium metal ( Li) belongs to the family of alkali metal. Atomic weight is 3 and it is
highly reactive and possess high energy density.
1.10.6 a) Li-MnO2 battery
- MnO2 as Cathode. Lithium salts such as LiCl, LiBr, LiClO 4 dissolved in a mixed
organic solvent (Propylene carbonate and 1, 2 dimethoxy-ethane) is used as the
electrolyte. The anode and cathode are separated by anon-woven polypropylene
separator.




                                                                           Figure 1`.13
Reactions during discharging are

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At anode-
xLi(s)→ xLi+ + xe-
Cathode
Mn(IV)O 2 (s) + xLi+ + xe-               Mn(iii)O2 ( Li+)x(s)
Overall reaction
xLi + Mn(IV)O 2 (s ) → Mn(iii)O 2 ( Li+)
FeaturesMn4+ - reduced to Mn3+ by the intestinally occupied Li ions in the MnO 2
           intercalation compound.
          LiMnO 2 signifies the intestinal Li ions in the host MnO 2 lattice
          High energy density ( 200 Wh/kg)
          Long shelf life (20 years at 70°C)
          Capable of high rate discharge
          Wide operating temperature range
          Exhibits' both electronic and ionic conductivity
          Voltage- 3.0V
          Applicaions
          Uesd for long term back up, safety and security devices, cameras,                    lighting
           equipment
In the 1970‟s, Lithium metal was used for making batteries but its instability rendered it
unsafe and impractical. Lit hium- coba lt o xide and grap hite are now used as the lithium-
ion- moving electrodes. The Lithium-Ion battery has a slightly lower energy density than
Lithium metal, but it is much safer. It is             introduced by Sony in 1991. A lithium-ion
battery ( Li-ion battery or LIB) is a family of rechargeable battery types in which lithiu m ions
move fro m the negative electrode to the positive electrode during discharge, and back when charging.
Unlike lithiu m primary batteries (wh ich are disposable), lithiu m-ion cells use an intercalated lithiu m
compound as the electrode material instead of metallic lithiu m.




Figure 1.14: A schematic sketch of Li- ion battery is shown in figure

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Construction
Anode is made up of graphite with Li atoms inserted between its layers of carbon ato ms
(Litigated graphite (LixC6 ). It contains no Li metal as such, that is why Li- ion battery
isn‟t called lithium battery. Cathode consist of – CoO2 (Li is incorporated into its
structure. The electrolyte- is a solution of Li salt in an organic solvent or a solid state or a
polymer electrolyte that can transport Li ions. Both the anode and cathode are materials
into which, and from which, lithium can migrate. During insertion (or intercalation )
lithium moves into the electrode. During the reverse
process, extraction (or deintercalation) lithium moves back out. When a lithium-based
cell is discharging, the lithium is extracted from the anode and inserted into the cathode.
When the cell is charging, the reverse occurs.

The electrode reactions during charging can be written as

The anode half- reaction is:


               The cathode half-reaction is:



and during discharging the electrode reactions were occour in the revrese directions.

In a lithium- ion battery the lithium ions are transported to and from the cathode or anode,
with the transition metal, cobalt (Co), in LixCoO 2 being oxidized from Co3+ to Co4+ during
charging, and reduced from Co4+ to Co3+ during discharge.

Advantages
   o PO WER – High energy density means greater power in a smaller package.1
          60% greater than NiMH battery
          220% greater than NiCd battery
       o HIGHER VOLTAGE – a strong current allows it to power complex mechanical
          devices.
       o LO NG SH ELF- LIF E – only 5% discharge loss per month.
       o 10% for NiMH, 20% for NiCd

       o Voltage- 3.6-3.7 (Li CoO 2), LiMn2O 4 (4)


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       o Wide variety of shapes and sizes efficiently fitting the devices they power.
       o Much lighter than other energy-equivalent secondary batteries
       o High open circuit voltage in comparison to aqueous batteries (such as lead
         acid, nickel- metal hydride and nickel-cadmium). This is beneficial because it
         increases the amount of power that can be transferred at a lower current.
   o No memory effect.
Disadvantages of Li-Ion

       o EXP ENSIV E -- 40% more than NiCd.DELICATE -- battery temperature must be
         monitored from within (which raises the price), and sealed particularly well.

Applications
          1. They are good in           portable electronics, with one of the best energy-to-
                  weight ratios, no memory effect, and a slow loss of charge when not in
                  use. Beyond consumer electronics, LIBs are growing in popularity for
                  military, electric   vehicle,   and aerospace applications   due   to   their
                  high energy density. They are sued in cell phones, laptop computers,
                  digital cameras and , power tools etc.,




1.11.1 Model question-Ans wers

Q1. Why does an electrochemical cell stop giving current after sometimes?
The flow of electric current is due to the potential difference between two electrodes.
With time, concentration of metal ions increases at anode due to oxidation and
concentration of metal ions decreases at cathode due to reduction. This results in
increases in reduction potential of anode and decreases in deduction potentional of
cathode. After some time, the electrode potent ional of both electrode become equal and
flow of current is stopped,.
Q2. What are the functions of salt Bridge?




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Salt bridge allows the flow of current by completing the electrical circuit. It maintains
electrical neutrality of solutions of both the half cells. It eliminates the liquid junction
potential.

Q3. What happens when a copper rod is dipped in a solution of (a) FeSO 4 (b) FeCI3 ?
Ans. (a) As reduction potential of copper E(-)Cu2+ / Cu = 0.34 V is more than that of iron
(E (-)Fe 2+ / Fe = -0.44V), so copper cannot displace Fe    2+
                                                                 ions. No reaction takes place.‟
(b) As reduction potential of copper is less than that of iron (E(-)Fe3+ / Fe    2+
                                                                                      = 0.77V, so
copper will displace Fe 2+ ions from FeCI3 and following reaction will take place.
Cu (s) + 2 FeCI3 (aq) → CuCI2 (aq) + 2 Fe Cl2 (aq)
Q4.Why the EMF of a concentration cell gradually decrease?
In concentration cell, electrical energy is produced due to transfer of ions from the
solution of high concentration (C2) to a solution of low concentration (C1) The
difference in concentrations decrease with time which results in decrease in EMF of cell.


         0.0591     C2
EMF            log
            n       C1


As the ratio of C 2 / C1 decrease with time, so the EMF of cell gradually decreases.


Q5. CuSO 4 can be stored in a silver container but not in zinc container. Why ?


                                                     2+
The standard electrode potential of zinc ((E(-)Zn         / Zn =-- 0.76V) is less than that of
                 2+
copper (E(-)Cu        / Cu = 0.34V, so zinc can displace copper from CuSO4 solution by
following reaction
        Zn + CuSO 4 -→-- ZnSO 4 + Cu
On the other hand, the electrode potential of silver (E(-) Ag+ / Ag = 0.80) is more than
that a of copper, so silver cannot displace copper from CuSO4 solution.
Q6. Why is KCI used in preparing salt bridge?
KCI is used in salt bridge because the speed of diffusions of K + ion and CI- is equal. So,
there is no net transfer of these ions and liquid junction potential is minimized.


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Q7. .Can free iodine react with silver or not?
     E (-)Ag+ /Ag = 0.80 V, E(-) I2 = 0.54V.
Ans. The reaction between free iodine and silver is
Ag I2 →Ag+ + AI-
EMF of this redox reaction is
E (-) cell = E (-) cathode – E (-) Anode = (E(-) I2 / I- – E (-) Ag+ / Ag = 0.54 – 0.80 =-
0.26 V
As EMF of cell is negative, so iodine cannot react with silver.


Q.8 Define the term : activity” of a solution.
The effective concentration of ions in solution is called activity and is denoted by symbol
„a”. For ideal solutions, activity a = concentration
For non- ideal solutions, activity is obtained by multiplying concentration with a
correction factor called activity coefficient (y).
Activity a= y x c
Q9. What is a reference electrode? Name two reference electrodes.
The electrode which is used measuring the electrode potential of another electrode by
combining with it to from a cell is called a reference e lectrode. Standard hydrogen
electrode (SHE) and Calomel electrode are commonly used reference electrodes.
Q10- What is the cell reaction for the following cell?
Zn/Zn2+ (a= 1) III Pb2+ (a=1) / Pb
Ans. At Anode Zn → Zn2+ = 2 e- (Oxidation)
At Cathode Pb2 + 2e- → Pb ( Reduction)
Overall reaction Zn + Pb2 → Zn2 + + Pb
Q11. Write down the cell for which the overall reaction is
Cd + Cu 2+ ( a=1)→ – Cd 2 + (a=1) + Cu
The cell representation is Cd / Cd 2+ ( a=1) // cu 2+ (a-1)/ Cu
Q12. Can we measure the absolute value of electrode potential ?
Electrode potential is a measure a flow of electric charge. An electrode has to be coupled
with a reference electrode to set up an electromechanical cell. Hence, the electrode



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potential is relative potential with reference to a standard electrode and absolute value of
electrode potential can not be measured.
Q13. Write the cell reactions for following cell Pt, Ti 3+/ Ti 4+, Fe 3+/ Fe 2+, Pt.
Ans. The cell reactions are]
At Anode Ti 3+ → Ti 4+ + e- ( Oxidation)
At Cathode Fe 3 + e- → Fe 2+ ( Reduction)
Overall reaction Ti 3+ Fe 3+ → Ti 4 + Fe 2+
Q14 Can a nickel rod be used to stir a CuSO 4 solution if
E Ni / Ni = 0.025 V and E Cu 2+ / cu= ).0.14 V


Ans. As electrode potential of nickel is less than that of copper, so nickel will displace
copper from CuSO 4 solution by following reaction.
Bi + Cuso4 → NiSO4 + Cu
Hence, nickel rod cannot be used to stir CuSO 4 solution
Q15. Predict whether the following reaction is feasible or not
2AI (s) + 3Sn4 + (aq)- 2AI 3+ + 3Sn 2+ (aq)
E(-)AL 3+ /AI =- 1.66 V, 3Sn 4+ /Sn 2+ = 0.14V
Ans. As reduction potential of aluminum is less than that of tin, aluminum can displace
Sn2+ ions from Sn 4+ (aq) solution. Hence, the reaction is feasible.
Q15. Why does blue colour of CuSO4 is discharged when iron rod is dipped in it?
As the standard electrode potential of iron (E → Fc 2+ /Fe = -0.044 V) is less than that of
copper ( E-Cu 2+ /Cu-=0.34V) so iron can displace copper from CuSO4 solution resulting
in discharge of blue colour.
Fe + CuSO 4 → FeSO4 + Cu
(Blue) (Colorless)
Q16 Why the electrode potential of Zn is assigned a negative value whereas that of CU is
assigned a positive value.
On comparison with standard hydrogen electrode, in zinc electrode the election flow is
from zinc electrode to hydrogen electrode. So electrode potential of Zn is assigned a
negative value. In copper electrode, the electron flow from hydrogen electrode to copper
electrode, so copper is assigned a positive value.

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Q17. What are the units of E- cell and G ?
The units of E cell is volts ( Joule/ Coulomb) and units of G is joules.
Q18 . Which out of Na, Zn, Cu will be strongest reducing agent?
Given:
E(-)Na+ /Na=-2.17 E Zn 2+ /Zn=-0.76V, E Cu 2+ /Cu=0.34V
A reducing agent helps in reductions but itself undergoes oxidation. Na, having least
electrode potential, will undergo oxidation easily/ hence, out of Na, Zn and Cu; Na will
act as strongest reducing agent.
Q19. Which out of F2, Cla2and L2 will be strongest oxidation agent?
Given: E(-)F2/F= 2.87V, Eci2/C1-=1.36, EI2/I-=0.54V
An Oxidizing agent oxidizes other species but itself undergoes reduc tion . F2 having
highest electrode potential, will undergo reduction easily. Hence , out of F 2 CI2 and I2 ; F2
will act as strongest oxidizing agent.


1.11.2 Short Ans wer type (Part A) Questions
1. What is an electrochemical cell?
2. What is the working principle of an electrochemical cell?
3. What is a galvanic cell? Give an example.
4. Define the following : (i) Galvanic cell (ii) Electrolyte cell
5. Describe the construction and working principle of a galvanic cell. Give their
     electrode and net cell reactions?
6. Mention the electrochemical conventions of an electrochemical cell.
7. Represent any galvanic cell. Write the net cell reaction and give an expression for
     the EMF of the cell.
8. Why is the anode of galvanic cell - ve and its cathode +ve? Write the different
     electrode reactions which occur at the electrodes.                      .
9.    Explain the origin of potentials at the electrodes of a galvanic cell.
10. What is the criterion that decides anode and cathode of a galvanic cell consisting of
     two dissimilar metals that are immersed in their own salt solutions?
11. Discuss the origin of electrode potential? Derive Nernst equation for single electrode
     potential.
12. Define single electrode potential? How is it denoted conventionally for both
     oxidation and reduction processes.
13. Define the terms
     (i) Standard electrode potential
     (ii) EMF of a cell
           16. Define standard electrode potential?
14. Mention the factors that affect the electrode potential.



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15. How is the standard electrode potential of a cathode of a cell measured? Mention its
     sign.
16. How is the standard electrode potential of an anode measured? Mention its sign?
17 Derive Nernst equation for a single electrode potential of a cell for a reduction
     electrode process.
18. The net cell reaction is M1 + M2 n+           1
                                                    n+ + M
                                                           2
     Give an expression for the EMF of a cell, in terms of the ionic concentrations M 1 n+
     and M2 n+ respectively.
19 Mention various types of electrodes of half-cells.
20 Write a note on ion-selective electrode.
21. What is a metal- metal salt- ion electrode? Give an example
22. What is a gas electrode? Mention an example.
23. What is an oxidation-reduction electrode? Give an example.
24. What is a ion selective electrode? Mention an example.
25 What do you mean by a reference electrode? Mention the different types of reference
     electrode?
26. Describe standard hydrogen electrode (SHE)?
27 Define a concentration cell.
28. Distinguish a galvanic cell and a concentration cell. Illustration with the help of
     example.
30. Write a note on calomel electrode.
31. What are ion selective electrodes? Explain the measurement of pH of a solution
           using a glass electrode?
32. Explain the origin of single electrode potential.
33. Differentiate between primary cells and secondary cells
34. Give the reactions of Pd-.acid cell
35. What are the merits and demerits of Pb-acid cells?
36 Mention the advantages of Li- ion battery over Ni-Cd battery

1.11.3Analytical/proble m type (Part b) type questions
1.    The EMF of the following cell is found to be 0.20 V at 298 K,
               Cd(s)/Cd2+ (aq)(?) // Ni2+ (aq) (2.0M)/Ni(s)
      What is the molar concentration of Cd2+ ions in solution?
2     Calculate the EMF of the the cell in which the reaction is
               Mg (s) + / 2Ag+             2+ (aq) + 2Ag (s)
      When [Mg2+ (aq)] = 0.130 M and [Ag+ (aq)] = 1 x 10-4 M
3.     Calculate the equilibrium constant for the reaction
                Zn (s) + Cd2+            2+ (aq) + Cd (s)
       Given : EOcell = 0.36 V
4.     Calculate the standard free energy change and maximum work obtained for the
       reaction occurring in the Daniel cell.
                Zn + Cu2+        2+ + Cu


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       Given : EO Zn2+ /Zn = -0.76 V; EO Cu2+ /Cu = 0.34 V and F = 96,500 C mol-1
       How is it related to the equilibrium constant for the reaction?

5.     Calculate the equilibrium constant for the reaction
               Cu (s) + 2 Ag+        2+ (aq) + 2Ag (s)
                 0
       Given : E cell = 0.46 V
6.     Consider the cell, Mg (s)/Mg 2+ (aq) // Ag+ (aq)/Ag (s). The standard reduction
       potential of Mg and Ag are -2.38 and 0.80 V respectively. Calculate the equilibrium
       constant at 298 K and also the maximum work that can be obtained using the cell,
7      Consider an electrochemical cell,
       Al (s)/Al3+ (0.01 M) // Fe2+ (0.20)/Fe (s)
       The standard reduction potentials of Al and Mg are -1.66 and -0.44 V respectively.
       Calculate the W max that can be obtained by the cell.(Ans: EOcell = 1.22 V; W max = 705.8 kJ)
8.     Compute the potential of the Ag+ /Ag couple with respect to Cu2+/Cu if the
       concentration of Ag+ and Cn2+ are 402 x 10-6 M and 1.3x10-3 M respectively. EO
       of Ag & Cu are 0.8 V & 0.34V
9.     Calculate the voltage of the cell Mg/Mg2+ (aq soln)//Cd2+ (aq soln)/Cd at 25%C.
       When [Cd2+] = 0.1 M, [Mg2+ ] = 1.0 M and EO cell = 1.97 V

10. Hydrogen electrode an normal calomel electrode, when immersed in a solution at 25
o
  C showed a potential of 0.0664 V. Calculate the pH of the solution.

11. Consider a cell of hydrogen electrode ( H+ ion concentration = X molar ) and calomel
electrode ( reduction potential = 0.281 V at 25 0 C). This cell has the EMF at 250 C equal
to 0.50 V. What is the pH value of hydrogen electrode solution?

12. A solution has a pH of 4.0 when measured with a standard hydrogen electrode and a
standard calomel electrode. What is the potential of the cell, if the electrode potential of
saturated calomel electrode is 0.2415 V at 250 C ?

13. Find the pH of an electrolyte used with hydrogen electrode in the following cell, Pt/
H2 (g) / H+ (aq) (aH+=x) / NCE; NCE is the normal calomel electrode (atm); E cell at 298
K is 0.830 V, E NCE at 298 K is 0.280 volt.

14.. The cell, Ag / Ag CI (s) / HCI ( 0.IN) glass/ unknown solution / saturated KCL
solution / Hg2 CI2(s) / Hg, gave n e.m f. of 0.11118 V at 298 K, when the pH of buffer
used was 4.0. When a buffer of unknown pH was used, the e.m.f. of the cell was fund to
be 0.3996 V at 298 K. What is the pH of the unknown solution ? E cal ( Saturated)=
0.2422 V at 298 K
                                                                        ( Ans. 8.878)

15.. Calculate the EMF of the cell Fe/Fe2+(0.01M)//Ag+ at 298 K, if the standard
          electrode potential of Fe and Ag electrodes are -0.42 V and 0.8 V respectively.

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16.     The standard cell potential Eo for the reaction Fe + Zn2+   2+ + Zn is -0.353 V.
        If a piece of iron is placed in a 1M Zn2+ solution, what will be the equilibrium
        concentration of Fe2+
17. .Calculate the EMF for the cell Sn (s)/Sn2+ (0.15M) // Ag+ (0.30M)/ Ag (s) at 25oC.
     The standard reduction potential of Sn and Ag are -0.14 and +0.80 V respectively.

18.     Consider the given cell, Al (s)/ Al3+ (aq) // Mg2+ (aq.) /Mg (s). Write the
        electrode and net cell reaction and calculate the Eo for the cell. The standard
        reduction potentials of Al and Mg are -1.66 and -2.38 V respectively.
19..    A Copper wire is dipped in AgNO3 solution kept in a beaker A. A silver wire is
        dipped in a solution of copper sulphate kept in a beaker B. If standard reduction
        potential for copper and silver are 0.34 and 0.80 V respectively, predict in which
        beaker the ions present will get reduced? (Ans: Ag+ will get reduced to Ag in
        beaker A)
20.    Using standard electrode potential, predict the reaction, if any reaction that occurs
        between Fe3+ (aq.) and I_ .
        Given : EO Fe3+ / Fe 2+ = 0.77 V; EO I2/I- = 0.54 V
21.     Write the half cell and net cell reactions for the cell, Cd (s)/Cd 2+ (0.01 M)//Cu2+
        (0.5M)/Cu (s) the standard reduction potentials of cadmium and copper are -0.4V
        and 0.34 V respectively. Calculate the EMF of the cell.
22.     An electrochemical cell consists of an iron electrode dipped in 0.1 M FeSO 4 and
        silver electrode dipped in 0.05 M AgNO3. Write the cell representation, cell
        reaction and calculate EMF of the cell in 298 K. Given the standard reduction
        potentials of iron and silver electrodes are -0.44 and +0.80 respectively.

1.11.4 Descriptive (Part C type) questions
1. Define single electrode potential and standard electrode potential? Derive Nernst
equation for electrode potential.
2.Derive Nernst equation for single electrode. Explain the method of determining the
electrode potential using an hydrogen electrode?
3. Explain the construction and working of a calomel electrode? Mention its
     advantages.
4.. How is potential of an electrode measured using a calomel electrode?
5. Write brief notes on:
     (i) Calomel electrode
     (ii) Glass electrode
.6. Describe the construction of a calomel electrode? Why is it called a secondary
     reference electrode?
7.. Construct and describe a calomel electrode?
8. What is a glass electrode? Describe the construction and working principle of a glass
     electrode?
9. Give the principle of a glass electrode? How is it constructed.


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10. What are ion selective electrode? Give the construction of glass electrode and
     explain the experimental method of determining pH using glass electrode.
11.. What is a glass electrode? How is it constructed? Describe the experimental
     determination of a pH of a solution using a glass electrode.
           (From Univ. QP)
12.. Describe a method by which the pH of an unknown solution is determined using a
     glass electrode.
13.. What are concentration cells? Derive an expression for the EMF of a concentratio n
     cell.
14.. What are concentration cells? Explain with an example. Calculate the EMF of the
     given cell at 298 K. Ag (s)/AgNO3 (0.018M)//AgNO3 (1.2M)/Ag(s).                   (From Univ. QP)
15. Derive the Nernst equation for the potential of a single electrode from
     thermodynamic principle. From this, deduce an expression for the EMF of a copper
     concentration cell in which the copper ions ratio is 10. Calculate the EMF of this
     cell at 25oC.                                                                     (From Univ. QP)
16. What is a single electrode? How is it constructed? Describe the experimental
     determination of pH of a solution using glass electrode.
17.. What are concentration cells? Derive an expression for the EMF of a concentration
     cell.
18.. What are concentration cells? Explain with an example. Calculate the EMF of the
     given cell at 298 K. Ag (s)/AgNO3 (0.018M)//AgNO3 (1.2M)/Ag(s).                   (From Univ. QP)
19.. Derive the Nernst equation for the potential of a single electrode from
     thermodynamic principle. From this, deduce an expression for the EMF of a copper
     concentration cell in which the copper ions ratio is 10. Calculate the EMF of this
     cell at 25oC.                                                                     (From Univ. QP)
20.. What is a single electrode? How is it constructed? Describe the experimental
     determination of pH of a solution using glass electrode.
21.. (i) Explain the construction of calomel electrode. Explain how this electrode is
           used to determine the potential of unknown electrode.
     (ii) The spontaneous galvanic cell Tin/tin ion (0.024 m) 11 tin ion (0.064 m) Tin
           develops an EMF of 1.126 25oC. Calculate the valency of tin.
     (iii) Explane the principles of a membrane electrode Mention the different types of
           membranes available.
22. (i) Derive the Nernst equation for a single electrode. Explain the determination of
           a single electrode potential using a standard hydrogen electrode.
     (ii) Explain the construction and working of a calomel electrode.
     (iii) A concentration cell was constructed by immersing two silver electrodes in
           0.05 M and 0.2 M AgNO 3 solution. Write the cell representation, cell reaction
           and calculate the EMF of the concentration cell.
23. (i)          What are reference electrodes? Explain the construction and working of an
      calomel electrode
     (ii) Write the half- cell and net-cell reaction and also calculate the voltage generated
           in the following cell Mn/Mn2+//Fe2+/Fe when an ion rod is immersed in 6.9 X
           10-4m FeSO 4 solution and an Mn rod is immersed in 2.6 X 10-6 m MnSO 4
           solution. Give Eo for Fe2+ /Fe is -0.4V and Mn/Mn2+ is -1.18 V.

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            (iii)   What are ion selective electrode? Explain the construction and
                    application of glass electrodes in pH determination.

24 (a)     Define (i) Single electrode potential and (ii) Standard electrode potential.
     (b) An electrochemical galvanic cell is obtained by coupling silver (EoAg+ /Ag =
           0.08 V) with a standard hydrogen electrode (Eo SHE = 0) at 298 K. How would
           you determine the potential of silver electrode? Represent the electrochemical
           cell and write the cell reaction.
     (c) Consider an electrochemical as given, Zn/Zn2+ (0.005M)//Ag+ (0.1M)/Ag.
           The cell reaction is spontaneous at 298 K. The standard reduction potential of
           zinc and silver are -0.76 and 0.80 V respectively. Write the redox electrode
           reaction with their respective electrode potential, net cell reaction calculate the
           EMF of the cell.
  (d) What are reference electrodes? Explain the construction and working of a calomel
electrode.
25. Describe on the construction and working of Pb-acid cells. Mention its merits and
    demerits too.
26. Discuss the functioning of Li-MnO2 battery
27. Give an account of Ni-Cd battery. Mention its uses.
28. What is Li- ion battery? How does it work ? Explain with a neat diagram and
equations?




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