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Electrochemistry Test - 27

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Electrochemistry Test - 27
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  • Question 1
    1 / -0

    Only One Option Correct Type

    This section contains 15 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct

    Q.

    Temperature coefficient of EMF of a cell in terms of entropy change is

    Solution




     is called temperature coefficient of EMF of a cell. Thus,
    temperature coefficient of

  • Question 2
    1 / -0

    The standard reduction potential at 298 K of the reaction, 

    2H2O + 2e-  H2 + 2OH is 0.8277 V

    Thus,thermodynamic equilibrium constant for the reaction. 

    2H2 H3O+ + OH- at 298 K is  

    Solution



  • Question 3
    1 / -0

    Given ,  

    Thus ,equilibrium constant for the reaction in terms of log k is

    2Fe3+ + 3I-  2Fe2+ + I-3   

    Solution


  • Question 4
    1 / -0

    Given ,

    Thus ,(log Keq) for the reaction   Cu2+ +In2+  Cu+ + In3+ is   

    Solution




  • Question 5
    1 / -0

    An excess of liquid mercury is added to an acidified solution of 1.0 x 10-3 M Fe3+ .Thus  is if 5% of Fe3+ remains at equilibrium at 298 K

      

    Solution

    When equilibrium is set up



  • Question 6
    1 / -0

    EMF of the following cell is 0.2905 V

    Zn/Zn2+ (a = 0.1M)|| Fe2+ (a = 0.01M)| Fe

     The equilibrium constant for the cell reaction is   

    [IIT JEE 2004]

    Solution



  • Question 7
    1 / -0

    The half-cell reactions for rusting of iron are

     

    ΔG° (in kJ) for the reaction is     

    [IIT JEE 2005]

    Solution

    In rusting of iron, Fe2+ is formed. Thus 

  • Question 8
    1 / -0

    The Gibbs free energy for the decomposition of Al2O3 at 800 K is as follows :

    2Al2O→ 4Al + 3O2 , ΔrG = 2898kJ mol-1

    The potential difference needed for electrolytic reduction of Al2O3 is at least  

    Solution


    n= electrons exchanged = 12 in the given reaction 

  • Question 9
    1 / -0

    Given ,  

    The value of standard electrode potential for the half-reaction is

    Fe3+(aq) + e- → Fe2+(aq) 

    Solution


    Since, different number of electrons are involved hence

  • Question 10
    1 / -0

    For the reaction, 2H2(g) + O2(g) → 2H2O(l), E0cell = 1.23 V at 298 K 

    and  ΔH0(H2O) = - 285.8 kJ mol-1 Thus, ΔS° (standard entropy change ) is  

    Solution


    n = 4 (four electrons are involved) as

  • Question 11
    1 / -0

    For the reaction ,

    Thus , for the reaction 

      

    Solution

    Electron involved are different, hence

    Signs are taken as per required reaction

    III is obtained as (I) - (II)

  • Question 12
    1 / -0

    Consider the following equations for a cell reaction. 

    A+B  C+ D, E0 = x volt, Keq = K1

    2A +2B  2C+ 2D, E0 = y volt, Keq = K2

    Then,   

    Solution

    When a chemical reaction is m ultiplied/divided EMF of the changed equation remains constant but equilibrium constant is raised to power of that change.

  • Question 13
    1 / -0

    Which of the following statements about the spontaneous reaction occurring in a galvanic cell is always true?

    Solution

    For a spontaneous reaction,

       ΔG < 0

    Also,  ΔG = -nFEcell
    ∴    -ve = -nFEcell
    ∴ Ecell > 0

    Also, Ecell = 

    if equilibrium is reached, Ecell = 0

    and Q = Keq

    ∴   


    To make       

    Ecell >0

    Ecell > Q

  • Question 14
    1 / -0

    For a (Ag-Zn) button cell ,the net reaction is 

    Zn(s) + Ag2O(s) → ZnO(s) + 2Ag(s)

    ΔG0f(Ag2O) = -11.21kJmol-1,  ΔGf(ZnO) = - 318.3 kJ mol-1

    Hence, E°cell of the button cell is  

    Solution

    ΔG0f(element, as Zn, Ag) = 0

    Thus, ΔG(Button cell) 

    =ΔG(ZnO) - ΔGf (Ag2O)

    =-318.30 - (-11.21)

    =-307.09 kJ mol-1

     ΔG0 = -nFE0cell        (n=2)

    ∴   

  • Question 15
    1 / -0

    For the reaction ,

    Thus  

    Solution

    for the given reaction

    E0cell  = 2.73 V

    n(electrons exchanged) = 12

    (4 Al → Al3+ + 12e-)

    F=96500 C mol-1

    ΔG0 (element) = 0

    ∴ ΔG(reaction) = -nFE0cell

    = -12 x 96500 x 2.73

    =3161340 J = 3161.34 kJ

    thus,

  • Question 16
    1 / -0

    Matching List Type

    Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct

    Q.

    The standard reduction potential data at 298 K is given below:

    Match E° of a redox pair in Column I with the values given in Column II and select the corect answer using the codes given below:

    Solution

    In all cases, we use





    Equal number of electrons are involved. 



  • Question 17
    1 / -0

    The standard potential of the following cell is 0.23 V at 288 K and 0.21 V at 308 K. 

    Match the parameters in Column I with their values in Column II and select the answer from the codes given below the list.

    Solution

    (i) Temperature coefficient of EMF 








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