Self Studies
Self Studies
Self Studies
Self Studies

Electrochemistry Test - 5

Result Self Studies

Electrochemistry Test - 5
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    4 / -1

     

    Which cell will measure the standard electrode potential of zinc electrode?

     

    Solution

     

    Standard electrode potential is the potential difference under standard state, i.e.
    When [Mn+] = 1M
    It can be measured by coupling it with SHE electrode
    Pt (s) | H2(gr. 1 bar) | H+ (1.0 M)

     

  • Question 2
    4 / -1

    For the following cell with hydrogen electrodes at two different  pressure p1 and p2 

     emf is given by

    Solution

    For SHE E°SHE = 0.00 V
    Oxidation at anode (left)

    Reduction at cathode (right) 
    Net

    This is the type of the cell in which electrodes at different pressures are dipped in same electrolyte and connectivity is made by a salt-bridge.

    Reaction Quotient (Q) 

    ∵ 

  • Question 3
    4 / -1

    For the cell,

    Thus (x/y) is

    Solution

    This is a type of concentration cell using hydrogen electrode as anode and cathode.





  • Question 4
    4 / -1

     

    A solution of Fe2+  is titrated potentiometrically using Ce4+ solution.

    Fe2+ → Fe3+ + e- , E0 = -0.77 V

    emf of the Pt | Fe2+ , Fe3+ pair at 50% and 90% titration of Fe2+ are   

     

    Solution

     

    When Fe2+ is 50% titrated

    =

    where Fe2+ is 90% titrated

     

     

  • Question 5
    4 / -1

     

    Two half-cells are given

    Ag | AgCl | KCl (0.2 M), Ag | AgBr | KBr (0.001 M),

    Ksp(AgCl) = 2.8 x 10-10, Ksp (AgBr) = 3.3 x 10 -13

    For a spontaneous cell reaction, cell set up is    

     

    Solution

     

    Ksp (AgCI) = 2.8 x 10-10
    [Ag+] [Cl-] = 2 .8 x 10-10
     

    ∴ [Ag+]left = 
    Ksp (AgBr) = 3.3 x 10-12 
    [Ag+][Br-] = 3.3 x 10-13

    ∴ [Ag+]left =  

    Net

    = -0.037 V
    Thus, cell reaction is non-spontaneous
    (b) Cell is reversed of (a), thus spontaneous.

     

  • Question 6
    4 / -1

    In the following cell at 298 K,two weak acids (HA) and (HB) with pKa (HA) = 3 and pKa (HB) = 5 of equal molarity have been used as shown.

    Thus, emf of the cell is   

    Solution


    For weak acid by Ostwald's dilution law




    = 0 - 0.0591 [log [H+]L - log [H+]R]
    = 0.0591 [-log (H+)L - (-log (H+)R]
    = 0.0591 [(pH)HA - (pH)HB]
    = + 0.0591 [1.5 - 2.5]= - 0.0591

  • Question 7
    4 / -1

    A concentration cell reversible to anion (Cl-) is set up

       

    cell reaction is spontaneous ,if 

    Solution

    This is a type of concentration cell with gas electrodes at the same pressure (1 bar) but dipped in aqueous solution of different concentration. Hence, a potential difference is set up.

    At anode 

    At cathode



    = 


    To make cell reaction spontaneous, Ecell > 0, hence C, > C2

  • Question 8
    4 / -1

    Which has the maximum potential at 298 K (numerical value) for the half-cell reaction?

    2H+ + 2e-  → H2 (1 bar)   

    Solution

    2H+ + 2e- → H2
    This represents reduction half-cell reaction 


    (a) [H+] = 1 M, E = 0
    (b) pH = 4, [H+] = 10-4M
    ∴ E = 0.0591 log 10-4
    = - 4 x 0.0591 = - 0.2364 V
    (c) Pure water, [H+] = 10-7M
    ∴ E = 0.0591 log 10-7
    = - 7 x 0,0591 = - 0.4137 V
    (d) 1.0 M NaOH
    [OH-] = 1 M

    ∴ E = 0.0591 log 1 x 10-14
    = -14 x 0.0591
    = - 0.8274 V (maximum)

  • Question 9
    4 / -1

    For the cell,  and for the cell Pt(H2) | H+ (1M)| Ag,  

    Thus Ecell for the

    Ag|Ag+ (0.1M) || Zn2+ (0.1M) | Zn is  ....................and cell reaction is...............

    Solution








    Ecell < 0, hence reaction is non-spontaneous.

  • Question 10
    4 / -1

    Consider the following cell reaction,

    2Fe(s) + O2(g) + 4H+ (aq) →2Fe2+ (aq) + 2H2O(l) , E°= 1.67 V

    At [Fe2+] = 1 x 10-3 M.  and pH = 3,the cell potential at 298 K is 

    Solution

    pH = 3, [H+] = 10-3M

    Electrons involved n = 4 

      

  • Question 11
    4 / -1

    For the following cell with gas electrodes at p1 and p2 as shown:

    Cell reaction is spontaneous , if  

    Solution

    This is a type of concentration cell in which two electrodes are at different pressure but dipped in same electrolyte. Salt-bridge is used to make connectivity and liquid junction potential is minimised and
    E°cell = 0.00V



  • Question 12
    4 / -1

    For the half-cell, Cl-|Hg2Cl2, Hg(l), E = 0.280V at 298K electrode potential has maximum value when KCl used is 

    Solution

    Cl- / Hg2CI2, Hg (/)
    This is reduction half-cell
    Hg2CI2(s) + 2e- → 2Hg (/) + 2Cl- 
    Reaction quotient (Q) = [Cl-]2

    Thus, larger the value of [Cl-], smaller the value of Ecaiomel.
     


  • Question 13
    4 / -1

    Given at 298 K standard oxidation potential of quinhydrone electrode = -0.699 V Standard oxidation potential of calomel electrode = -0.268 V

    Thus, emf of the cell at 298 K is 

    Solution








  • Question 14
    4 / -1

    E°red (standard reduction electrode potentials) of different half-cell are given

       

    In which cell , is ΔG° most negative?

    Solution






    In (b) E°cell is most positive, then ΔG° is most negative.

  • Question 15
    4 / -1

    Given the following half-cell reactions and corresponding reduction potentials: 

    Which combination of two half-cell would result in a cell with largest (E°cell > 0)?   

    Solution

    (a)

    Oxidation

    C3- → C- + 2e-        E0 = 1.25 V

    Reduction 

  • Question 16
    4 / -1

    For

     

    Then E° for the reaction    

    Solution



  • Question 17
    4 / -1

    Following half-cell, Pt (H2)|H2O behaves as SHE at a pressure of

    Solution

    Pt(H2)| H2O
    This is oxidation half-cell. Instead of
    [H+] = 1 M and pH2 = 1 bar
    H2O has been taken as a source of [H+]




  • Question 18
    4 / -1

    Electrode potential of the following half-cell is dependent on
    Hg, HgO |OH-(aq)

    Solution




    Thus, electrode potential is dependent on (i) pH, (ii) temperature.

  • Question 19
    4 / -1

     

    The aqueous solution of which of the following compounds is the best conductor of electric current?

     

  • Question 20
    4 / -1

     

    An increase in the conductivity equivalent of a solid electrolyte with dilution is primarily due to

     

  • Question 21
    4 / -1

     

    The equivalent conductance of Ba2+ and Cl– are respectively 127 and 76 ohm-1 cm-1 eq-1 at infinite dilution. The equivalent conductance of BaCl2 at infinite dilution will be

     

  • Question 22
    4 / -1

     

    When heating one end of a metal plate, the other end gets hot because of

     

  • Question 23
    4 / -1

    Comprehension Type

    This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)

    Passage I

    1.05 g of lead ore containing impurity of Ag was dissolved in HNO3 and the volume was made 350 mL. A silver electrode was dipped in the solution and Ecell of Pt(H2) | H+ (1M)|| Ag+| Ag was 0.500 V at 298 K. E°Ag+/Ag = 0.80 V

    Q.

    Pure [Ag+] in the ore is    

    Solution

    Cell Reaction

    ∴ x = 8.4 x 10-6 M

    [Ag+] = 8.4 x 10-6 mol L-1

  • Question 24
    4 / -1

    Passage I

    1.05 g of lead ore containing impurity of Ag was dissolved in HNO3 and the volume was made 350 mL. A silver electrode was dipped in the solution and Ecell of Pt(H2) | H+ (1M)|| Ag+| Ag was 0.500 V at 298 K. E°Ag+/Ag = 0.80 V

    Q.

    Percentage of silver in the sample is      

    Solution

    Cell Reaction

    ∴ x = 8.4 x 10-6 M

    [Ag+] = 8.4 x 10-6 mol L-1

    = 8.4 x 10-6 x 0.350 mol in 350 mL

    = 8.4 x 10-6 x 0.350 x 108 g in 350 mL

    = 3.1752 x 10-4 g in 1.05 g sample

  • Question 25
    4 / -1

    Passage II

    For the following,

    Q.

    pH of the solution in the half-cell containing 0.02 HA is(HA is a weak monobasic acid)   

    Solution

    Anodic

    Cathodic

    Reaction quotient (Q) 





  • Question 26
    4 / -1

    Passage II

    For the following,

    Q.

    pKa of the weak monobasic acid is  

    Solution

    Anodic

    Cathodic

    Reaction quotient (Q) 



  • Question 27
    4 / -1

     

    The weight of silver displaced by a quantity of electricity which displaces 5600ml of O2 at STP will be

     

  • Question 28
    4 / -1

     

    Once a current of 1.0 ampere was passed through one litre of CuCl2 solution for 16 min and 5 sec, all of the solution’s copper was deposited at the cathode. The strength of solution CuCl2 was (Molar mass of Cu = 63.5; Faraday constant = 96,500 C/mol)                                                                      

     

  • Question 29
    4 / -1

     

    On electrolysis of dilute sulphuric acid using platinum electrodes, the product obtained at the anode will be

     

Self Studies
User
Question Analysis

  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now