chemistry

EAMCET PROBLEMS WITH SOLUTIONS FROM ELECTROCHEMISTRY

 PART-3

Example 21:  Calculate the electrode potential at a copper electrode dipped in a 0.1 M solution of copper sulphate at 25o C. The standard electrode potential of Cu2+/Cu system is 0.34 volt at 298 K.

Solution:  We know that Ered = Eored + 0.0591/n  log10[ion]

           Putting the values of Eored =0.34 V, n = 2 and [Cu2+]= 0.1 M

                                Eored = 0.34+0.0591/2  log10[0.1]

                                = 0.34 + 0.02955 × (-1)

                                = 0.34 - 0.02955 = 0.31045 volt 

  

Example 22:  What is the single electrode potential of a half-cell foe zinc electrode dipping in 0.01 M ZnSO4 solution at 25o C? The standard electrode potential of Zn/Zn2+ system is 0.763 volt at 25o C.

Solution: We know that Eox = Eored - 0.0591/n log10[ion]

           Putting the value of Eoox=0.763 V,n=2  and

[Zn2+]=0.01 M

Eoox = 0.763-0.0591/2 log_10 [0.01]

        = 0.763 - 0.02955 × (-2)

        = (0.763 + 0.0591) volt = 0.8221 volt

Example 23:  The standard oxidation potential of zinc is 0.76 volt and of silver is -0.80 volt. Calculate the emf of the cell:

    Zn|Zn(NO3)2||AgNO3|Ag

     0.25 M          0.1 M

     at 250C.

Solution: The cell reaction is

     Zn + 2 Ag+ --> 2Ag + Zn2+

    Eoox of Zn = 0.76 volt

    Eoox  of Ag = 0.80 volt

    Eocell = Eoox   of Zn +  of Ag = 0.76 + 0.80

            = 1.56 volt

            = 1.56 - 0.0591/2×1.3979

            = (1.56-0.0413) volt

            = 1.5187 volt

Example 24:  The emf(E°) of the following cells are:

Ag|Ag|(1 M)||Cu2+(1 M)|Cu;  E° = -0.46 volt

Zn|Zn2(1 M)||Cu2|(1 M)|Cu; E° = +1.10 volt

Calculate the emf of the cell:

Zn|Zn2+(1 M)||Ag+(1 M)|Ag

Solution:   Zn|Zn2+(1 M)||Ag+(1 M)|Ag

Ecell = Eox(Zn/Zn2+) + Ered (Ag+/Ag)

With the help of the following two cells, the above equation can be obtained.

Ag|Ag+(1 M)||Cu2+(1 M)|Cu,  E° = -0.46 volt

or  Cu|Cu2+(1 M)||Ag+(1 M)|Ag, E° will be +0.46 volt

or   +0.46 = Eox(cwcu2+) + Ered (Ag+/Ag)          .... (i)

Zn|Zn2+(1 M)||Cu2+1|Cu,  E° = +1.10 volt

 + 1.10 = Eox(Zn/Zn+) + Ered(Cu2+/Cu)                             ....... (ii)

 Adding Eqs. (i) and (ii),

+ 1.56 = Eox(Cu/Cu2+) + Ered(Ag+/Ag) + Eox(Zn/Zn2+) + Ered(Cu2+/Cu)

 Since    Eox(Cu/Cu2+) - Ered(Cu2+/Cu)

 So        +1.56 = Em(Zn/Zn+) + Ered(Ag+/Ag)

Thus, the emf of the following cell is

Zn|Zn2+(1 M)||Ag+(1 M)|Ag is +1.56 volt.

Example 25: Calculate the e.m.f of the cell.

Mg(s)|Mg2+(0.2M)||Ag+(1×10-3)|Ag

 EoAg+/Ag = +0.8 volt,     EoMg2+/Mg  = -2.37 volt

What will be the effect on e.m.f. if concentration

of Mg2+ ion is decreased to 0.1 M?

Solution:   Eocell = EoCathode - Eoanode

                        = 0.80-(-2.37) = 3.17 volt

              Cell reaction

             Mg + 2Ag+  --> 2Ag + Mg2+

Ecell = Ecello - 0.0591/n log(Mg2+)/[Ag+]2

       = 3.17 -0.0591/2 log 0.2/[1× 10-3 ]2

       = 3.17 - 0.1566 = 3.0134 volt

      when  Mg2+ = 0.1 M

      Ecell = Eocell - 0.0591/n log(0.1)/[1 x 10-3]2

             = (3.17 - 0.1477) volt

             = 3.0223 volt

Example 26:  To find the standard potential of M3+/M electrode, the following cell is constituted:

Pt|M|M3+(0.0018 mol-1L)||Ag+(0.01 mol-1L)|Ag

The emf of this cell is found to be 0.42 volt. Calculate the standard potential of the half reaction M3+ + 3e-  M3+. = 0.80 volt.

Solution:  The cell reaction is

      M + 3Ag+ ---> 3Ag + M3+

Applying Nernst equation,

   Ecell = Ecello - 0.0591/n log(Mg2+)/[Ag+]3

   0.42 =  Ecello - 0.0591/n log (0.0018)/(0.01)3 =  Ecello - 0.064

   Ecello =(0.042+0.064)= 0.484 volt

   Eocell = Eocathode - Eoanode

 or Eoanode  = Eocathode  - Eocell

           = (0.80-0.484) = 0.32 volt

Example 27.     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 0.118volt 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

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 

Example 28.    The emf of the cell Ag|Agl in 0.05 MK\Sol. NH4NO3|10.05 M AgNO3\Ag is 0.788 volt at 25°C. The activity coefficient of KI and silver nitrate in the above solution is 0.90 each. Calculate (i) the solubility product of Agl, and (ii) the solubility of Agl in pure water at 25°C.

Solution:    Ag+ ion concentration on AgN03 side

        = 0.9 × 0.5 = 0.045 M

Similarly I- ion concentration in 0.05 M KI solution

            = 0.05 × 0.9 - 0.045 M

Ecell = 0.0591/1 log[Ag+ ](R.H.S.)/[Ag+ ](L.H.S.) = 0.0591 log 0.045/[Ag+ ](L.H.S.)

or     log 0.045/[Ag+ ](L.H.S.) = 0.788/0.0591 = 13.33

[Ag+]L.H.S. = 0.045/(2.138× 1013 )

                = 2.105 × 10-15 M

Solubility product of Agl = [Ag+][I-]

= 2.105 × 10-15 × 0.045

= 9.427 × 10-17

Solubility of Agl = √(Solubility product of Agl)

                      = √(9.472×〖10〗^(-17) )

                        = 9.732 × 10-9 g mol L-1

                        = 9.732 × 10-9 × 143.5 g L-1

                        = 1.396 × 10-6 g L-1

Example 29:  The observed emf of the cell

Pt|H2(1 atm)|H+(3×10-4 M)||H+(M1)|H2(1 atm)|Pt

is 0.154 V. Calculate the value of M1 and pH of cathodic solution.

Solution:    Ecell = 0.0591 log M1/(3×10-4 )

                                or log M1/(3×10-4 ) = 0.0154/0.0591 = 2.6058

                                 M1/(3×10-4) = 4.034 × 102

                                M1 = 4.034 × 102 × 3 × 10-4 M

                                        = 0.121 M

                                pH = -log [H+]=-log 0.121 = 0.917

Example 30: Calculate the emf of the following cell at 25oC.

                                Pt H2|HCl|H2 Pt

                                2 atm        10 atm

Solution:  Ecell =0.0591/2 log P1/P2

                               = 0.0591/2 log 2/10

                              = -0.0206 volt