The reduction electrode potential, E of 0.1 M solution of M+ ions

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 Multiple Choice QuestionsMultiple Choice Questions

61.

At 25°C, the molar conductances at infinite dilution for the strong electrolytes NaOH, NaCl and BaCl2 are 248 ×10-4, 126 x 10-4 and 280 x 10-4 Sm2 mol-1 respectively, λm  Ba(OH)2 in Sm2 mol-1 is

  • 52.4 × 10-4

  • 524 × 10-4

  • 402 × 10-4

  • 262 × 10-4


62.

The molar conductivities of KCl, NaCl and KNO3 are 152, 128 and 111 S cm2mol-1 respectively. What is the molar conductivity of NaNO3?

  • 101 S cm2mol-1 

  • 87 S cm2mol-1 

  • -101 S cm2mol-1 

  • -39 S cm2mol-1 


63.

The electrochemical cell stops working after sometime because

  • electrode potential of both the electrodes becomes zero

  • electrode potential of both the electrodes becomes equal

  • one of the electrodes is eaten away

  • the cell reaction gets reversed


64.

The amount of electricity required to produce one mole of copper from copper sulphate solution will be

  • 1 F

  • 2.33 F

  • 2 F

  • 1.33 F


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65.

How long (in hours) must a current of 5.0 A be maintained to electroplate 60 g of calcium from molten CaCl2 ?

  • 27 h

  • 8.3 h

  • 11 h

  • 16 h


66.

For strong electrolytes the plot of molar conductance vs C is

  • parabolic

  • linear

  • sinusoidal

  • circular


67.

If the molar conductance values of Ca2+ and Cl- at infinite dilution are respectively 118.88 X 10-4 m2 Ωho mol-1 and 77.33 X 10-4m2 Ωho mol-1 then that of CaCl2 is ( in m2 Ωho mol-1): 

  • 118.88 X 10-4

  • 154.66  X 10-4

  • 273.54  X 10-4

  • 196.21 X 10-4


68.

The ionic conductance of Ba2+ and Cl- are respectively 127 and 76 ohm-1 cm2 at infinite dilution. The equivalent conductance (in ohm-1cm2) of BaCl2 at infinite dilution will be

  • 139.5

  • 203

  • 279

  • 101.5


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69.

The reduction electrode potential, E of 0.1 M solution of M+ ions (ERP = -2.36 V) is

  • -4.82 V

  • -2.41 V

  • +2.41 V

  • None of these


B.

-2.41 V

E= ERP° +  0.0591nlog[M+]Given, ERP° = -2.36V, [M+] = 0.1 Mn= 1 ( for M+ M)E= ERP° +  0.0591nlog[M+]  = -2.36 +  0.05911log 0.1  =-2.36+ 0.0591 ×(-1)   =-2.36 - 0.0591  =-2.419 V


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70.

The amount of silver deposited on passing 2 F of electricity through aqueous solution of AgNO3

  • 54g

  • 108g

  • 216g

  • 324g


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