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Electrochemistry Test - 16

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Electrochemistry Test - 16
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  • Question 1
    1 / -0

    The passage of current through a solution of certain electrolye results in the evolution of H2 at cathode and Cl2 at anode. The electrolytic solution is -

    Solution

    In this Cl–  will oxidise to give Cl2 ,Na+ reduction potential has lower potential than water reduction polenlid so water will reduce to give H2.

  • Question 2
    1 / -0

    In an electrolytic cell current flows from -

    Solution

    In an electrolytic cell, current flows from cathode to anode in outer circuit and in daniell cell, it is the reverse direction of flow of current from anode to cathode in outer circuit.

  • Question 3
    1 / -0

    When an aqueous solution of H2SO4 is electrolysed, the ion discharged at anode is-

    Solution

    Water oxidation at anode can be represented and SO42– can't be discharged
    so
    2H2O → 4H+ + O2 + 4e–
    so ion which will be discharged at anode will be OH–.

  • Question 4
    1 / -0

    1 mole of Al is deposited by X coulomb of electricity passing through aluminium nitrate solution. The number of moles of silver deposited by X coulomb of electricity from silver nitrate solution is-

    Solution

    Since to deposit 1 mole of aluminium 3 columb of electricity is required, as the alency of silver is + 1 so 3 mole of silver will be deposited by 3C of electricity

  • Question 5
    1 / -0

    A solution of Na2SO4 in water is electrolysed using Pt electrodes.The products at the cathode and anode are respectively -

    Solution

    At cathode
    4H2O + 4e– → 4H2 + 4OH–
    At anode
    2H2O → 4H+ + O2 + 4e–
    Thus at cathode we will get H2 whereas at anode we will get O2.

  • Question 6
    1 / -0

    In electrolysis of a fused salt, the weight deposited on an electrode will not depend on-

    Solution

    In electrolysis of a fused salt, the weight deposited on electrode will not depend or temperature.

  • Question 7
    1 / -0

    The electrolysis of a solution resulted in the formation of H2 at the cathode and Cl2 at the anode. The liquid is-

    Solution

    In this Cl– will oxidise to give Cl2 Na+ water reduction potential has higher potential than that of water reduction potential, so water will reduce to give H2.

  • Question 8
    1 / -0

    Which loses charge at cathode-

    Solution

    Cation loses charge at cathode as it gave electron

  • Question 9
    1 / -0

    In the electrolysis of CuSO4, the reaction :

    Cu2+ + 2e¯ → Cu, takes place at :

    Solution

    As reduction will take place at cathode so reaction Cu2+ + 2e → Cu
    Will take place at cathode

  • Question 10
    1 / -0

    If mercury is used as cathode in the electrolysis of aqueous NaCl solution, the ions discharged at cathode are-

    Solution

    If mercury  is used as cathode in the electrolysis of aqueos NaCl solution then the metal is discharged at mercury to from amalgam.

  • Question 11
    1 / -0

    The specific conductance of a solution is 0.3568 ohm-1. When placed in a cell the conductance is 0.0268 ohm-1. The cell constant is-

    Solution

    K = 0.3568 W cm–1
    condutance = 0.0268  Ω –1
    K = G ×1/A
    0.3568 = 0.0268 ×1/9
    13.31 cm–1

  • Question 12
    1 / -0

    A conductance cell was filled with a 0.02 M KCl solution which has a specific conductance of 2.768 × 10-3 ohm-1 cm-1. If its resistance is 82.4 ohm at 25ºC, the cell constant is-

    Solution

    K = G. L/A
    10–3 × 2.768 = 1/R ×L/A
    L/A = 228.08 × 10–3
    = 0.2281 cm–1

  • Question 13
    1 / -0

    The variation of equivalent conductance vs decrease in concentration of a strong electrolyte is correctly given in the plot -

    Solution


    On decreasing the value of M will increase but increase will be hyberbolic.

  • Question 14
    1 / -0

    Which of the following solutions has the highest equivalent conductance ?

    Solution

    Higher the dilution higher will be the equivalent conductance

  • Question 15
    1 / -0

    The resistance of 0.01N solution of an electrolyte AB at 328K is 100 ohm. The specific conductance of solution is (cell constant = 1cm-1) -

    Solution


  • Question 16
    1 / -0

    For an electrolytic solution of 0.05 mol L-1, the conductivity has been found to be 0.0110 Scm-1.The molar conductivity is-

    Solution


  • Question 17
    1 / -0

    Two electrodes are fitted in conductance cell 1.5 cm apart while the area of cross section of each electrode is 0.75 cm2. The cell constant is-

    Solution

    L/A is cell constant.

  • Question 18
    1 / -0

    The best conductor of electricity is in 1M solution of-

    Solution

    That electrolyte will be the best electrolyte which has maximum number of ions.

  • Question 19
    1 / -0

    A certain current liberates 0.504 g of H2 in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in CuSO4 solution -

    Solution

    gm eq.  of H2 = gm eq. of Cu

  • Question 20
    1 / -0

    A currnet of 2.6 ampere is passed through CuSO4 solution for 6 minutes 20 seconds. The amount of Cu deposited is (At. wt. of Cu = 63.5, Faraday = 96500 C)-

    Solution


  • Question 21
    1 / -0

    Three Faradays of electricity are passed through molten Al2O3, aqueous solution of CuSO4 and molten NaCl taken in three different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in the ratio of-

    Solution

    gm equivalent of Al = gm eq. of Cu = gm eq. of Na
    3 = 3 = 3
    3/3/ = 3/2 = 1
    1 : 1.5 : 3

  • Question 22
    1 / -0

    The quantity of electricity required to liberate 0.01g equivalent of an element at the electrode is-

    Solution

  • Question 23
    1 / -0

    The unit of electrochemical equivalent is-

    Solution

    Z = w/it
    w in gm
    it in columb.
    so z = gm/columb

  • Question 24
    1 / -0

    One faraday of electricity will liberate one mole of metal from a solution of-

    Solution

    Since KCl has the n-factor of 1 so 1 faraday of electricity will liberate one mole of metal from a solution.

  • Question 25
    1 / -0

    The number of faraday required to generate 1 mole of Mg from MgCl2 is-

    Solution

    Since magnesium has the n-factor 2 so the number of faraday rquired to genereate 1 mole of Mg will be 2 .

  • Question 26
    1 / -0

    One gm metal M+2 was discharged by the passage of 1.81×1022 electrons. What is the atomic weight of metal?

    Solution

  • Question 27
    1 / -0

    One mole of electron passes through each of the solution of AgNO3, CuSO4 and AlCl3 when Ag, Cu and Al are deposited at cathode. The molar ratio of Ag, Cu and Al deposited are

    Solution

    Molar ratio
    All have the same equivalent

  • Question 28
    1 / -0

    Salts of A (atomic weight = 7), B (atomic weight = 27) and C (atomic weight = 48) were electrolysed under identical conditions using the same quantity of electricity. It was found that when 2.1 g of A was deposited, the weights of B and C deposited were 2.7 and 7.2 g. The valencies of A, B and C respectively are

    Solution


    If x = 1 ⇒ y = 3, z = 2

  • Question 29
    1 / -0

    The density of Cu is 8.94 g cm–3. The quantity of electricity needed to plate an area 10 cm × 10cm to a thickness of 10-2 cm using CuSO4 solution would be

    Solution

    Volume : 10 × 10 × 10–2 = 1 cm3
    mass of Cu = 8.94 g


    = 27172 C

  • Question 30
    1 / -0

    During electrolysis of an aqueous solution of sodium sulphate if 2.4 L of oxygen at STP was liberated at anode. The volume of hydrogen at STP, liberated at cathode would be :

    Solution

    Equivalent of H2 = Equivalent of O2


    = 4.8 L

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