Three electrolytic cells A, B and C containing solutions of ZnSO4

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsLong Answer Type

Advertisement

141. Three electrolytic cells A, B and C containing solutions of ZnSO4(zinc sulphate), AgNO(silver nitrate) and CuSO4  (copper sulphate),  respectively are connected in series. A steady current of 1.5 ampere was passed through them until 1.45 g of silver is deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited? (Atomic masses: Ag = 108, Zn = 65.4, Cu = 63.5, all in amu). 


Ag++e-  Ag  1 mol            1 mol ( = 108 g)              (Cathode reaction in cell B)
108 g of silver is deposited at cathode when 1 mol of electrons are passed or 108 g of silver deposit needs = 1 Faraday = 96,500 coulombs. Therefore, 1.45 g of silver needs
                     = 96,500×1.45108  = 1295.6 columbs
       But quantity of electricity passed
                    = current x time
                    = 1295.6 C  = 1.5 A x time (in sec.)
       or time for which current is passed
               = 1295.61.5  seconds = 863.7 s = 14.40 min.
          The cathode reaction in copper sulphate cell is
                         Cu2++2e-    Cu     2 mol                   1 mol (63.5 g)    (2×96500 c)
              2 x 96500 coulombs gives a deposit of 63.5 g of Cu.
          Therefore, 1295.6 coulombs will deposit of 
                       = 63.5 × 1295.62×96, 500 = 0.4263 g
Similarly,
                        Zn2++2e-  Zn     2 mol              1 mol (65.4 g)   (2×96500)
               (cathode reaction in zinc sulphate cell)
Mass of zinc deposited
                         =65.4 x 1295.62×96, 500 g = 0.44 g.
341 Views

Advertisement

 Multiple Choice QuestionsShort Answer Type

142. Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
 2Cr(s) + 3Cd2+ (aq) → 2Cr3+ (aq) + 3Cd
(Calculate the ΔrG° and equilibrium constant of the reaction.)

257 Views

143. Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
 Fe2+ (aq) + Ag+ (aq) → Fe3+(aq) + Ag(s)
Calculate the ΔrG° and equilibrium constant of the reaction.

146 Views

 Multiple Choice QuestionsLong Answer Type

144. In the button cell widely used in watches and other devices the following reaction takes place:
Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s) + 2OH (aq)
Determine ΔrG° and E° for the reaction.

550 Views

Advertisement

 Multiple Choice QuestionsShort Answer Type

145. The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500Ω. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146 x 10–3 s cm–1
219 Views

 Multiple Choice QuestionsLong Answer Type

146. Conductivity of 0.00241 M acetic acid solution is 7.896 x 10–5 S cm-1. Calculate its molar conductivity in this solution. If Λ°m for acetic acid be 390.5 S cm2 mol–1, what would be its dissociation constant?
183 Views

 Multiple Choice QuestionsShort Answer Type

147. Calculate the emf of the cell Zn/Zn2+ (0.1 M) || Cd2+ (0.01 M) | Cd at 298 k. (given)
Zn2+/Zn = – 0.76 V and E°Cd2+/Cd = – 0.40 V).

1204 Views

 Multiple Choice QuestionsLong Answer Type

148.

Cu2+ + 2e → Cu E° = + 0.34 V
Ag+ + 1e → Ag E° = + 0.80 V
(i) Construct a galvanic cell using the above data.
(ii) For what concentration of Ag+ ions will the emf of the cell be zero at 25°C, if the concentration of Cu2+ is 0.01 M ? [log 3.919 = 0.593]

357 Views

Advertisement
149. Silver is electrodeposited on a metallic vessel of total surface area 900 cm2 by passing a current of 0.5 amp for two hours. Calculate the thickness of silver deposited. [Given : Density of silver = 10.5 g cm–3, Atomic mass of silver = 108 amu, F = 96,500 C mol–1]
603 Views

 Multiple Choice QuestionsShort Answer Type

150. Calculate the number of coulombs required for the oxidation of 1 mole of water to oxygen as per equation:
2H2O → 4H+ + O2 + 4e
[Given: 1 F = 96,500 C mol–1]
190 Views

Advertisement