chemistry

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