EAMCET QUESTIONS FROM CHAPTER D AND F-BLOCK ELEMENTS PART-2

Problem 9: In the melting point curves of transition metals, one observes a dip in the curves at the end i.e. Cu, Ag & Au and Zn, Cd & Hg have lower melting points when compared to other transition metals. Explain.

Solution: In the last two groups of transition elements i.e. Cu, Ag, Au, Zn, Cd and Hg all the electrons

are paired which can not take part in metallic bonding. As a result, metallic bond in these elements is weak resulting in the lower melting points of these metals.

Problem 10: Enthalpies of atomization of transition elements are higher than those of alkali and alkaline earth metals. Explain.

Solution: The number of unpaired electrons in transition elements are more when compared to those in alkali and alkaline earth metals. As a result, the metallic bonds in transition metals are stronger and enthalpies of atomization are higher than those of alkali and alkaline earth metals.

Problem 11:  Explain the following:

(a) Scandium forms no coloured ions, yet it is regarded as a transition element.

(b) Transition elements have many irregularities in electronic configurations.

Solution:

(a) Scandium in the ground state has one d electron. Hence it is regarded as transition element.

(b) In the transition elements, the (n – 1)d subshell and ns subshell have very small difference in energy. The incoming electron may enter into either ns or (n-1)d subshell. Hence they show irregularities in their electronic configurations.

Problem 12: Explain the following

(a) Chromium is a typical metal while mercury is a liquid metal.

(b) Cobalt (II) is stable in aqueous solution but in the presence of strong ligands, it is a easily oxidised to cobalt (III).

Solution:

(a) chromium has 5 unpaired electrons in its d – orbitals which make its metallic bond very stronger. Whereas in mercury there are no unpaired d electrons, so its metallic bond is very weak.

(b) CO(III) has greater tendency to form complex than CO(II) hence in the presence of ligands CO(II) changes to CO(III).

Problem 13: Write down the products of the following reactions.

(a) CuSO4 solution is treated with KI solution.

(b) AgNO3 solution is added to Na2S2O3 solution.

Solution:

(a) Free iodine is liberated along with the formation of a white precipitate of cupric iodide.

            CuSO4 + 2Kl ————→ Cul2 + K2SO4

             2Cul2 ————→ 2Cul + l2 

(b) A white precipitate of Ag2S2O3 is obtained which turns yellow, brown and finally black on keeping.

            2AgNO3 + Na2S2O3 ————→ Ag2S2O3 + 2NaNO3

            Ag2S2O3 + H2O ————→ Ag2S + H2SO4

                                                         black ppt.

Problem 14: Explain the following

(a) Zinc readily liberates H2 form cold dil.H2SO4 but not from cold conc. H2SO4.

(b) Blue colour of the CuSO4 solution is discharged slowly when an iron rod is dipped into it.

Solution:

(a) Conc. H2SO4 is a covalent compound. Hence does not contain H+ ions. Dilute H2SO4 contains H3O+ which reacts with Zn and liberates H2.

         H2SO4 + H2O ————→ 2H3O+ + SO42–

         Zn + 2H3O+ ————→ Zn2+ + 2H2O + H2O ↑

(b) Fe is more electropositive than Cu, hence it displaces copper form CuSO4solution.

         Fe(s) + CuSO(aq) ————→ FeSO4(aq) + Cu(s) 

Problem 15: An aqueous solution containing one mole of HgI2 and two moles of NaI is orange in colour. On addition of excess NaI the solution becomes colourless. The orange colour reappears on subsequent addition of NaOCl. Explain with equations.

Solution:

         Hgl2 + 2Nal ————→           Na2[Hgl4]

         (orange)                           coloured due to residual Hgl2)

          Hgl2 + Nal(excess) ————→                   Na2[Hgl4]

          (orange)                                        (colourless because there is no residual Hgl2)

         2Na2[Hgl4] + 2NaOCl + H2O ————→ 2Hgl2 + NaCl + 4NaOH + 2Nal3

                                                                          (orange)

EAMCET QUESTIONS FROM CHAPTER D AND F-BLOCK ELEMENTS PART-1

Problem 1: The chemical reactivity of lanthanides resemble to which other elements of the periodic table?

Solution: The chemical reactivity of the starting lanthanides resemble calcium due to similar first and second ionization energy. But latter lanthanides resemble Al due to ability of showing +3 oxidation state and similarity in I.E.

Problem 2: Enthalpies of atomization of transition elements are higher than those of alkali and alkaline earth metals. Explain.

Solution: The number of unpaired electrons in transition elements are more when compared to these in alkali and alkaline earth metals. As a result, the metallic bonds in transition metals are stronger and enthalpies of atomization are higher than those of alkali and alkaline earth metals. 

Problem 3: Explain the following:

(a) Chromium is a typical metal while mercury is a liquid metal.

(b) Zinc readily liberates H2 from cold dil. H2SO4 but not form cold conc. H2SO4.

Solution:

(a) Chromium has five unpaired electrons in its d-orbitals which make its metallic bond very strong, whereas in mercury there is no unpaired d-electrons so its metallic bond is very weak, hence it is a liquid.

(b) Since, conc. H2SO4 act as an oxidizing agent hence does not evolve H2 when it reacts with zinc.

          Zn + 2H2SO4 ————→ ZnSO4 + SO2 + H2O

Problem 4: Compare thermal stability of ZnO, CdO and HgO.

Solution: ZnO > CdO > HgO

Problem 5: Cu+ ion has 3d104s0 configuration and colourless but Cu2O is red and Cu2S is black. Explain.

Solution: Cu+ ion has 3d104s0 configuration, i.e. it has no unpaired electron hence there is no d-d transition possible and it is colourless. But Cu2O and Cu2S are coloured due to charge transfer of electrons from O2- or S2- to the vacant orbital of Cu+ ion.

Problem 6: While Cu, Ag and Au are considered as transition elements but Zn, Cd and Hg are not considered as transition elements although all the mentioned elements have complete d-orbitals. Explain.

Solution: Although Cu, Ag and Au have their d – orbitals complete in the elemental state. They do have incomplete d orbitals in their compound state. So they are included in the transition elements.

      Cu+2 = 3d9

      Au+3 = 5d8

Zn, Cd and Ag have their d-orbitals complete in their elemental state as well as compound state. So they are not included in the transition elements.

      Zn+2 = 3d10

      Hg+2 = 5d10 

Problem 7:

(i) CrO3 is an acid anhydride. Explain.

(ii) Between Na+ and Ag+ which is a stronger Lewis acid and why?

Solution:

(i) CrO3 + H2O ————→ H2CrO4, i.e CrO3 is formed by loss of one H2O molecule from chromic acid.

(ii) Between Na+ and Ag+, Ag+ is stronger Lewis acid. Because Ag+ has pseudo noble gas configuration which makes it more polarizing.

Problem 8: It is well known that alkali and alkaline earth metals displace hydrogen from dilute acids. But most of the transition elements do not behave so. Explain.

Solution: Alkali and alkaline earth metals have positive oxidation potential. But most of the transition elements have negative oxidation potentials. So they are not as good oxidizing agents as the alkali and alkaline earth metal are.

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

EAMCET PROBLEMS WITH SOLUTIONS FROM ELECTROCHEMISTRY

 PART-2

Example 11:      The specific conductivity of 0.02 M KCl solution at 250C is 2.768 × 10-3 ohm-1 cm-1. The resistance of this solution at 250C when measured with a particular cell was 250.2 ohms. The resistance of 0.01 M CuSO4 solution at 250C measured with the same cell was 8331 ohms. Calculate the molar conductivity of the copper sulphate solution.

Solution:     Cell constant   = (Sp.cond.of KCl)/(Conductane of KCl)

                                   = (2.768× 10-3)/(I/250.2)

                                   = 2.768 × 10-3 × 250.2

 For 0.01 M CuSO3 solution

             Sp. conductivity = Cell constant × conductance

                                   = 2.768 × 10-3 × 250.2 × 1/8331

         Molar conductance = Sp. cond. × 1000/c

                                 =  (2.768×10-3 × 25.2)/8331 × 1000/(1/100)

Example 12:    The values of Eo of some of the reactions are given below:

                                I2 + 2e- --> 2I-;           Eo = +0.54 volt

                                Cl2 + 2e- --> 2Cl-;         Eo = +1.36 volt

                                Fe3+ + e- --> Fe2+;        Eo = +0.76 volt

                                Ce4+ + e- --> Ce3+;       Eo = +1.60 volt

                                Sn4+ + 2e- --> Sn2+;      Eo = +0.15 volt

   On the basis of the above data, answer the following questions:

(a)  Whether Fe3+ oxidizes Ce3+ or not ?

(b)  Whether I2 displaces chlorine form KCl ?

(c)  Whether the reaction between FeCl3 and SnCl2 occurs or not ?

Solution:      (a) Chemical reaction,

                                Fe3+ + Ce3+ --> Ce4+ + Fe2+

                                Two half reactions,

       Fe3+ + e --> Fe2+          Reduction  Eo    = 0.76 volt

       Ce3+ --> Ce4+ + e-        Oxidation  Eoox = -1.60 volt

                                          ---------------------------

                                            Adding             = -0.84 volt

Since, emf is negative the reaction does not occur, i.e., Fe3+ does not oxidise Ce3+.

(b) Chemical reaction

                I2 + 2KCl = 2Kl + Cl2

Half reactions

I2 + 2e- --> 2I-             Reduction  Eo = 0.54 volt

2Cl- --> Cl2 + 2e-          Oxidation  Eoox = -1.36 volt

                                  ---------------------------

                                      Adding         = -0.82 volt

Since, emf is negative, the reaction does not occur, i.e., I2 does not displace Cl2from KCl.

(c) Chemical reaction

                SnCl2 + 2FeCl3 --> SnCl4 + 2FeCl2

Half reactions

Fe3+ + e  Fe2+      Reduction Eo = 0.76 volt

Ce2+  Sn4+ + 2e-  Oxidation Eo = -0.15 volt

                         -------------------------

                             Adding       = +0.61 volt

Since, emf is positive, the reaction will occur.

      

Example 13:      The equivalent conductivity of N/10 solution of acetic acid at 250C is 14.3 ohm-1 cm2 equiv-1. Calculate the degree of dissociation of CH3COOH if  is 390.71.

Solution:            

/\∞CH3 COOH = 390.7 ohm-1  cm-2 equiv-1

/\∞CH3 COOH = 143.3 ohm-1 cm-2 equiv-1

   Degree of dissociation,  α = /\v//\∞ =14.3/390.71

                                      = 0.0366   i.e., 3.66% dissociated

Example 14:      A decinormal solution of NaCI has specific conductivity equal to 0.0092. If ionic conductances of Na+ and Cl- ions at the same temperature are 43.0 and 65.0 ohm-1 respectively, calculate the degree of dissociation of NaCl solution.

Solution:             Equivalent conductance of N/10 NaCl solution

                         = Sp. conductivity × dilution

                        = 0.0092 × 10,000

                        = 92 ohm-1

                  /\∞ = λNa+  + λCl-

                        = 43.0 + 65.0

                        = 108 ohm-1

                        Degree of dissociation,

Example 15:     At 180C, the conductivities at infinite dilution of NH4Cl, NaOH, NaCL are 129.8, 217.4, 108.9 mho respectively. If the equivalent conductivity of N/100 solution of NH4OH is 9.93 mho, calculate the degree of dissociation of NH4OH at this solution.

Solution:       /\∞NH4 Cl = λNa4+ + λCl- =129.8       ..... (i)

/\∞NaOH = λNa+  +  λOH- = 217.4                      ..... (ii)

/\NaCl = λNa + λCl- = 108.9                     ..... (iii)

Adding Eqs. (i) and (ii) and subtracting (iii),

λNa4+ + λCl- + λNa+ + λOH-  - λNa- λCl-

λNa4+ + λOH- = 238.3 mho

Degree of dissociation, α=/\v//\∞ = 9.93/238.3 = 0.04167

or 4.17% dissociated

Example 16:      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 + H2SO4 = ZnSO4 + H2

(d)        Fe + SnCl2 = FeCl2 + Sn

  

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 (+)

(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||AgNO3|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||H2SO4|Cu

or     Zn|Zn2+||2H+|H2(Pt)

        Anode (-) Cathode (+)

(d)        Here Fe is oxidized to Fe2+ and Sn2+ is reduced to Sn. The cell can be represented as

Fe|FeCl2||SnCl2|Sn

or     Fe|Fe2+||Sn2+|Sn

Anode (-) Cathode (+)

Example 17:      Consider the reaction,

                        2Ag+ + Cd --> 2Ag + Cd2+

                        The standard electrode potentials for Ag+ --> Ag and Cd2+ --> Cd couples are 0.80 volt and -0.40 volt, respectively.

(i) What is the standard potential Eo for this reaction?

(ii) For the electrochemical cell in which this reaction takes place which electrode is negative electrode?

Solution:             (i) The half reactions are:

                        2Ag+  + 2e- -->  2Ag.

                                Reduction

                                (Cathode)

                        EoAg+/Ag =0.80  volt          (Reduction potential)

                        Cd --> Cd2+    + 2e-,

                                Oxidation

                                (Anode)

                        EoCd+/Cd = -0.40 volt               (Reduction potential)

                or     EoCd+/Cd2 = +0.40 volt

                        Eo = EoCd+/Cd2 + EoAg+/Ag = 0.40+0.80 = 1.20  volt

(ii) The negative electrode is always the electrode whose reduction potential has smaller value or the electrode where oxidation occurs. Thus, Cd electrode is the negative electrode.

Example 18:      Consider the cell,

                        Zn|Zn2+(aq)(1.0M)||Cu2+(aq)(1.0M)|Cu

                        The standard electrode potentials are

                        Cu2+ + 2e- --> Cu(aq)                   Eo = 0.350 volt

                        Zn2+ + 2e- --> Zn(aq)                   Eo = -0.763 volt

                        (i) Write down the cell reaction.

                        (ii) Calculate the emf of the cell

  

Solution:  (i) Reduction potential of Zn is less than copper, hence Zn acts as anode and copper as cathode.

                        At anode                    Zn --> Zn2+ + 2e-    (Oxidation)

                        At cathode    Cu2+ + 2e- --> Cu                 (Reduction)

                     --------------------------------------------------------

                        Cell reaction  Zn + Cu2+ --> Zn2+ + Cu

(ii)  EoCell = EoZn/Zn2+ + EoCu2+/Cu

               = Oxi. Potential of zinc + Red. Potential of copper

           EoZn/Zn2+ = -0.763        (Reduction potential)

           EoZn2+/Zn= +0.763      (Oxidation potential)

and        EoCu2+/Cu = 0.350     (Reduction potential)

So   Ecello= 0.763+0.350 = 1.113    volt

Oxidation potential is EoM/Ma+ while reduction potential is represented as  EoMa+/M. The value of EoZn/Zn2+ (oxidation potential of Zn) is +0.76 volt and the value of EoCu2+/Cu (reduction potential of copper) is +0.34 volt. The electrode having lower value of reduction potential acts as an anode while that having higher value of reduction potential acts as cathode.

Example 19:                Write the electrode reactions and the net cell reactions for the following cells. Which electrode would be the positive terminal in each cell?

(a)        Zn|Zn2+||Br-, Br2|Pt

(b)        Cr|Cr3+||I-, I2|Pt

(c)         Pt |H2, H+||Cu2+|Cu

(d)        Cd|Cd2+||Cl-, AgCl|Ag

Solution:   (a) Oxidation half reaction, Zn --> Zn2++2e-

          Reduction half reaction, Br2 + 2e- --> 2Br-

         -------------------------------------------------

             Net cell reaction          Zn + Br2 --> Zn2+ + 2Br-

                                Positive terminal-Cathode Pt

             (b) Oxidation half reaction, [Cr --> Cr3+ + 3e-]× 2

         Reduction half reaction, [I2 + 2e- --> 2Ir-] × 3

        ----------------------------------------------------

           Net cell reaction          2Cr + 3I2 --> 2Cr3+ + 6I-

                                Positive terminal-Cathode Pt

            (c) Oxidation half reaction, H2 --> 2H+ + 2e-

     Reduction half reaction, Cu2+ + 2e--->  Cu

     -------------------------------------------------

       Net cell reaction          H2 + Cu2+ --> Cu + 2H+

                                Positive terminal-Cathode Cu

           (d) Oxidation half reaction, Cd --> Cd2+ + 2e-

     Reduction half reaction,   [AgCl+e- --> Ag+Cl- ]×2

     ------------------------------------------------------

                 Net cell reaction Cd+2AgCl --> Cd2++2Ag+2Cl-

                                Positive terminal-Cathode Ag

Example 20: Will Fe be oxidiesed to Fe2+ by reaction with 1.0 M HCl? Eo for Fe/Fe2+ = +0.44 volt.

Solution:    The reaction will occur if Fr is oxidized to Fe2+.

               Fe + 2HCl --> FeCl2 + H2

               Writing two half reaction,

 Fe --> Fe2+ + 2e-       Oxidation EoFe/Fe2+ = 0.44 volt

2H++ 2e- --> H2         Reduction EoH+/H = 0.0  volt

                                --------------------------------------

                                 Adding,      emf = 0.44 volt

 Since emf is positive, the reaction shall occur.

EAMCET PROBLEMS WITH SOLUTIONS FROM ELECTROCHEMISTRY

 PART-1

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

Solution:             Charge on one ion of N3-

                                = 3 × 1.6 × 10-19 coulomb

                        Thus, charge on one g-ion of N3-

                                = 3 × 1.6 10-19 × 6.02 × 1023

                                = 2.89 × 105 coulomb

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

Solution:             (a) The reduction reaction is

                        Al3+       + 3e- --> Al

                        1 mole       3 mole

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

                        Q = 3 × F

                        = 3 × 96500 = 289500 coulomb

                        (b) The reduction is

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

                        1 mole       5 mole

                                Q = 5 × F

                                 = 5 × 96500 = 48500 coulomb

  

Example 3.        How much electric charge is required to oxidise (a) 1 mole of H2O to O2 and (b)1 mole of FeO to Fe2O3?

Solution:             (a) The oxidation reaction is

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

                        1 mole                       2 mole

                                Q = 2 × F

                                  = 2 × 96500=193000 coulomb

                        (b) The oxidation reaction is

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

                               Q = F = 96500 coulomb

Example 4.        Exactly 0.4 faraday electric charge is passed through three electrolytic cells in series, first containing AgNO3, second CuSO4 and third FeCl3 solution. How many gram of rach metal will be deposited assuming only cathodic reaction in each cell?

Solution:             The cathodic reactions in the cells are respectively.

                         Ag+     + e- --> Ag

                       1 mole    1 mole

                        108 g        1 F

                       Cu2+    + 2e- --> Cu

                        1 mole       2 mole

                       63.5 g       2 F

        and           Fe3+   + 3e- --> Fe

                       1 mole       3 mole

                        56 g          3 F

        Hence,       Ag deposited = 108 × 0.4 = 43.2 g

                        Cu deposited = 63.5/2×0.4 = 12.7 g

        and           Fe deposited = 56/3×0.4 = 7.47 g

Example 5.        An electric current of 100 ampere is passed through a molten liquid of sodium chloride for 5 hours. Calculate the volume of chlorine gas liberated at the electrode at NTP.

Solution:             The reaction taking place at anode is

                        2Cl- -->  Cl2      + 2e- 

                        71.0 g    71.0 g     2 × 96500 coulomb

                                    1 mole

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

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

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

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

Example 6.        A 100 watt, 100 volt incandescent lamp is connected in series with an electrolytic cell containing cadmium sulphate solution. What mass of cadmium will be deposited by the current flowing for 10 hours?

Solution:             We know that

               Watt = ampere × volt

             100 = ampere × 110

             Ampere = 100/110

             Quantity of charge = ampere × second

                                       = 100/110×10×60×60 coulomb

             The cathodic reaction is

            Cd2+       +     2e-    -->    Cd

           112.4 g      2 × 96500 C

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

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

Example 7.        In an electrolysis experiment, a current was passed for 5 hours through two cells connected in series. The first cell contains a solution gold salt and the second cell contains copper sulphate solution. 9.85 g of gold was deposited in the first cell. If the oxidation number of gold is +3, find the amount of copper deposited on the cathode in the second cell. Also calculate the magnitude of the current in ampere.

Solution:             We know that

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

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

                        Mass of copper deposited

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

                        Let Z be the electrochemical equivalent of Cu.

                        E = Z × 96500

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

                Applying W = Z × I × t

                       T = 5 hour = 5 × 3600 second

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

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

Example 8.        How long has a current of 3 ampere to be applied through a solution of silver nitrate to coat a metal surface of 80 cm2 with 0.005 cm thick layer? Density of silver is 10.5 g/cm3.

Solution:             Mass of silver to be deposited

                                 = Volume × density

                                = Area ×thickness × density

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

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

                                                                = 0.42 g

                        Applying to silver E = Z × 96500

                                                Z =  108/96500 g

                        Let the current be passed for r seconds.

                        We know that

                                W = Z × I × t

                        So, 0.42 = 108/96500×3×t

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

Example 9:      1.0 N solution of a salt surrounding two platinum electrodes 2.1 cm apart and 4.2 sq.cm in area was found to offer a resistance of 50 ohm. Calculate the equivalent conductivity of the solution.

Solution:             Given, l = 2.1 cm, a = 4.2 sq xm, R = 5- ohm

                        Specific conductance, k = l/a.1/R

                        or  k = 2.1/4.2×1/50 = 0.01    ohm-1 cm-1

                        Equivalent conductivity = k = V

                        V = the volume containing 1 g equivalent = 1000 mL

                        So Equivalent conductivity    = 0.01 × 1000

                                                                = 10 ohm-1 cm-2 equiv-1

Example 10:      Specific conductance of a decinormal solution of KCl is 0.0112 ohm-1 cm-1. The resistance of a cell containing the solution was found to be 56. What is the cell constant?

Solution:             We know that

                        Sp. conductance = Cell constant × conductance

                        or Cell constant   = (Sp.conductance)/Conductance

                                                = Sp. conductance × Resistance

                                                = 0.0112 × 56

                                                = 0.06272 cm-1