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Electrochemistry Test - 6
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Electrochemistry Test - 6
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  • Question 1/10
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    Three Faradays of electricity are passed through molten Al2 O3 , 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-

     

    Solutions

     

     

    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 2/10
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    The quantity of electricity required to liberate 0.01g equivalent of an element at the electrode is-

     

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  • Question 3/10
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    The unit of electrochemical equivalent is-

     

    Solutions

     

     

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

     

     

  • Question 4/10
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    One faraday of electricity will liberate one mole of metal from a solution of-

     

    Solutions

     

     

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

     

     

  • Question 5/10
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    The number of faraday required to generate 1 mole of Mg from MgCl2  is-

     

    Solutions

     

     

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

     

     

  • Question 6/10
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    One gm metal M+2  was discharged by the passage of 1.81 ×1022  electrons. What is the atomic weight of metal?

     

    Solutions

     

     

     

     

  • Question 7/10
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    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

     

    Solutions

     

     

    Molar ratio
    All have the same equivalent

     

     

  • Question 8/10
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    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

     

    Solutions

     

     


    If x = 1 ⇒y = 3, z = 2

     

     

  • Question 9/10
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    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

     

    Solutions

     

     

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


    = 27172 C

     

     

  • Question 10/10
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    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 :

     

    Solutions

     

     

    Equivalent of H2 = Equivalent of O2


    = 4.8 L

     

     

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