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Electrochemistry Test - 13
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Electrochemistry Test - 13
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  • Question 1/14
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    Given the standard electrode potentials  

    I. K+ /K = -2.93V,
    II. Ag+ /Ag = 0.80V,
    III. Hg2+/ Hg = 0.79 V  
    IV. Mg2+ /Mg = -2.37V,
    V. Cr3+ /Cr = - 0.74 V

    These metals are arranged in increasing reducing power as

     

    Solutions

     

     

    Most negative E °red means the Mn+ /M is at the top of ECS and is the best reducing agent.

     

     

  • Question 2/14
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    Statement Type

    Direction : This section is based on Statement I and Statement II. Select the correct answer from the codes given below.

    Statement I  : CuSO4  can be stored in a vessel made of  zinc.

    Statement II :  w.r.t SHE    

     

    Solutions

     

     

    E ° Zn2+ /Zn  = -0.76 V
    E ° Cu2+ /Cu  = + 0.34 V

    In electrochemical series, zinc is above copper and thus zinc is a better reducing agent than copper. When CuS04 is placed in zinc vessel, copper is displaced

    Thus, CuSO4 cannot be stored in a vessel made of zinc.
    Thus, statement I is incorrect and statement II is correct.

     

     

  • Question 3/14
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    Statement I ​:  When AgNO3  solution is stirred with a spoon made of copper,solution turns blue.

    Statement II :  In electrochemical series ,copper is above silver

     

    Solutions

     

     

    In electrochemical series, copper is above silver thus is a better reducing agent. When AgNO3 solution is stirred with copper spoon, Ag is displaced and copper is oxidised to Cu2+ (blue).

    Thus, statement I and II are correct and statement II is the correct explanation of statement I.

     

     

  • Question 4/14
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    The positive value of the standard electrode potential of Cu2+ /Cu,

    (E ° Cu2+ /Cu  = 0.34 V) indicates that

     

    Solutions

     

     


    E °cell  >0 thus spontaneous
    Cu2+ /Cu couple is thus a stronger oxidising agent. Reverse reaction is non-spontaneous.
    Cu + 2H+ →Cu2+ + H2 , E °cell = - 0.34 V
    Thus, copper cannot displace H2 from acid.

     

     

  • Question 5/14
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    Consider the following half-reactions:

    Select the correct statements on the basis of the above data

     

    Solutions

     

     



    Thus, (c) is correct.
    Concentrated H2 SO4  ionises as H+ and HSO4- and then water is not oxidised in concentrated H2 SO4 Thus, (b) is incorrect.
     is not oxidised to in dilute H2 SO4 . Thus, (d) is incorrect.
    (d) In dilute sulphuric acid solution, ions will be oxidised to tetrathionate ion () at anode.

     

     

  • Question 6/14
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    Select the correct statement(s) based on the following half-reaction:

     

    Solutions

     

     

    If dil. H2 SO4 is used, then H+ is the reacting species.
    If cone. H2 SO4  is used, then is the reacting species.

    Thus, copper does not react with H+ (dil. H2 SO4 ) forming H2 .

    Since E °cell  >0, Thus copper reacts with cone. H2 SO4 forming SO2 .
    (c) Zn + +4H+ →Zn2+ + SO2 + 2H2 O
    E °cell  = 1.93 V, E °cell >0, thus zinc also reacts with conc.
    H2 SO4 forming SO2 .
    (d) Zn + 2H+ →Zn2+ + H2 , E °cell = 0.76 V
    E °cell >0, thus zinc reacts with dil. H2 SO4 forming H2 .

     

     

  • Question 7/14
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    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

    Match the half-cell reaction in Column I with their values in Column II and select the answer from the answer codes at the end.

     

    Solutions

     

     




    Thus, (i) →(p)
    (ii) (B) O2 (g) + 2H2 O + 4e-  →40H-  n1  = 4, E1  = 0.401 V
    (C) O2 (g)+4H+ 4e-  →2H2 O  , n2 = 4, E2 = 1.229 V
    (G) 4H+ + 4e-  →2H2 , n3 = 4, E3 = 0.00 V
    (B) + (G) - (C) gives
    4 H2 O  + 4e-  →2H2 + 40H-  
    2 H2 O  + 2e-  →H2 + 2OH-  
    ∴E4 = 0.401 + 0.00- 1.229
    = -0.83 V
    Thus, (ii) →(s)




    Thus, (iv) →(s)

     

     

  • Question 8/14
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    Passage I  

    The most stable oxidised species is

     

    Solutions

     

     




    For (I) and (II), E °>0, but for reduced species.
    For (III), E °>0, thus Cr3+ is the most stable.
    More negative value of standard electrode reduction potential, better the reducing agent. Thus,
    Mn2+ < CI- < Cr3+ < Cr

     

  • Question 9/14
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    Passage I  

    Reducing power of different species is in the order

     

    Solutions

     

     




    For (I) and (II), E °>0, but for reduced species.
    For (III), E °>0, thus Cr3+ is the most stable.
    More negative value of standard electrode reduction potential, better the reducing agent. Thus,
    Mn2+ < CI- < Cr3+ < Cr

     

  • Question 10/14
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    Passage II  

    Accidently chewing of a stray fragment of aluminium foil can cause a sharp tooth pain if the aluminium comes in contact with an amalgam filling. This filling, an alloy of silver, tin and mercury acts as the cathode (of a tiny galvanic cell), the aluminium as anode and saliva (of mouth) serves as electrolyte. When the aluminium and the filling come in contact, an electric current passes from aluminium to the filling, which is sensed by a nerve in the teeth.

                  

    Q.

    Net reaction taking place when amalgam is in contact with aluminium foil is

     

    Solutions

     

     

     

     

  • Question 11/14
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    Passage II  

    Accidently chewing of a stray fragment of aluminium foil can cause a sharp tooth pain if the aluminium comes in contact with an amalgam filling. This filling, an alloy of silver, tin and mercury acts as the cathode (of a tiny galvanic cell), the aluminium as anode and saliva (of mouth) serves as electrolyte. When the aluminium and the filling come in contact, an electric current passes from aluminium to the filling, which is sensed by a nerve in the teeth.

                  

    Q.

    Standard emf experienced by the person with dental filling is   

     

    Solutions

     

     

     

     

  • Question 12/14
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    Passage III

    When suspected drunk drivers are tested with a breathalyser, the alcohol (ethanol) in the exhaled breath is oxidised to ethanoic acid with an acidic solution of potassium dichromate.

     

    Q.

    Under standard state, E°cell   of the breathalyser would be  

     

    Solutions

     

     

    CH3 CH2 OH is oxidised to CH3 COOH and is reduced to Cr3+ .
    E °cell - E °CH3 CH2 OH/CH3 COOH + E ° /Cr3+
    = -0.06+1.33  
    = 1.27 V
    Reaction quotient

     

     

  • Question 13/14
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    Passage III

    When suspected drunk drivers are tested with a breathalyser, the alcohol (ethanol) in the exhaled breath is oxidised to ethanoic acid with an acidic solution of potassium dichromate.

     

    Q.

    Under the condition of unit activity of each species,emf recorded in the breathalyser at pH 4 would be  

     

    Solutions

     

     

    CH3 CH2 OH is oxidised to CH3 COOH and is reduced to Cr3+ .
    E °cell - E °CH3 CH2 OH/CH3 COOH + E ° /Cr3+
    = -0.06+1.33  
    = 1.27 V
    Reaction quotient

     

     

  • Question 14/14
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    E °values of Mn+ + ne-  →M are given below
    Fe2+ /Fe -0.44 V, Mg2+ /Mg -2.37 V, Pt(H2 )/H+ 0.00 V,
    AI3+ /AI -1.66 V , Zn2+ /Zn - 0.76 V, Cu2+ /Cu +0.34 V,
    Ag+ /Ag + 0.80 V, Ni2+ /Ni -0.26 V

    How many of the metals can be used to prevent corrosion of Ni to Ni2+ ?

     

    Solutions

     

     

    Fe, Al, Zn and Mg can be used to protect Ni from being oxidised to Ni2+ since, E °cell  >0 and ΔG °<0.

     

     

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