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Electrochemistry Test - 17
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Electrochemistry Test - 17
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  • Question 1/30
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    The E° in the given diagram is,

    Solutions

    ∴   ΔG0= ΔG10 +ΔG20 + ΔG30

    -6FE0. = -4F x 0.54 - 1F x 1.07

  • Question 2/30
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    What is cell entropy change of the following cell?

    Pt(s) | H2(g) | CH3COOH, HCl || KCl (aq) |Hg2Cl2| (s) | Hg

    P = 1 atm        0.1M                0.1M

    Emf of the cell is found to be 0.045 V at 298 K and temperature coefficient is 

    3.4 x10–4  VK–1

    Given Ka (CH3COOH) = 10–5 M

    Solutions

     

    = 65.223J/K/mole

  • Question 3/30
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    Following cell has EMF 0.7995 V.

    Pt | H2 (1 atm) | HNO3 (1M) || AgNO3 (1M) | Ag

    If we add enough KCl to the Ag cell so that the final Cl- is 1M. Now the measured emf of the cell is 0.222 V.

    The Ksp of AgCl would be :

    Solutions

  • Question 4/30
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    The solubility of [Co(NH3)4Cl2] CIO4_________ if the  = 50,  = 70, and the measured resistance was 33.5Ω in a cell with cell constant of 0.20 is ____.

    Solutions

    The correct answer is option B
    Given,
    λCo(NH3)4Cl2+=50       λClo-4 =70
    λ∞= λCo(NH3)4Cl2+  + λClo-4
    λ∞= 50             + 70
    λ∞=120
    (x) Cell constant = 1/A
    0.02 = l/A
    Resistance(R)   =33.5Ω
    K =c.x              (x = is cell constant)

    S  =49.7 mol/L

  • Question 5/30
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    We have taken a saturated solution of AgBr.Ksp of AgBr is 12 x 10 – 14 . If 10 – 7 mole of AgNO3 are added to 1 litre of this solution then the conductivity of this solution in terms of  10 – 7 Sm – 1 units will be

    [given  Sm2 mol-1 Smmol-1, 5 x 10-3 Sm2mol-1]

    Solutions

    The solubility of Agbr in presence of 10-7 molar AgNo3 is 3 x 10-7 M. therefore [Br] 3 x 10-4 M3, [Ag+] = 4 x 10-4 m3 and [No3-]= 10-4 m3

    Therefore 

  • Question 6/30
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    At 298K the standard free energy of formation of H2O(L) is – 237.20kJ/mole while that of its ionisation into H+ ion and hydroxyl ions is 80 kJ/mole, then the emf of the following cell at 298 K will be

     H2(g,1 bar) | H+ (1M) || OH (1M) | O2 (g, 1bar)

    Solutions

    Cell reaction

    also we have

    Hence for cell reaction

  • Question 7/30
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    Which of the following cell can produce more electric work.

    Solutions

    For Ecell to be highest [H+]a should be lower and [H+]c should be higher and that why anode compartment should be more basic and cathodic compartment should be acidic. 

  • Question 8/30
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    A hydrogen electrodes is immersed in a solution with pH = 0 (HCl). By how much will the potential (reduction) change if an equivalent amount of NaOH is added to the solution. (Take PH2 = 1 atm) T = 298 K.

    Solutions

    pH changes from 0 to 7

    ∴ [H+] changes from 1 to 10-7 M

    Accordingly Ered. Decrease by 0.059 log 10-7 i.e 0.059 x (-7) = -0.41volt

  • Question 9/30
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    At what  does the following cell have its reaction at equilibrium?

     Ag(s) | Ag2CO3(s) | Na2CO3 (aq) || KBr(aq) | AgBr(s) | Ag(s)

     KSP =  8 x 10 – 12 for  Ag2CO3 and KSP = 4 x 10 – 13 for AgBr

    Solutions

  • Question 10/30
    1 / -0

    Calculate the EMF of the cell at 298 K

    Pt|H2(1atm)|NaOH(xM),NaCl(xM)|AgCl(s)|Ag

    If E°cl-/AgCl/Ag = + 0.222 V

    Solutions

  • Question 11/30
    1 / -0

    A current of 0.1A was passed for 2hr through a solution cuprocyanide and 0.3745g f copper was deposited on the cathode. Calculate the current efficiency for the copper deposition.

    Solutions

  • Question 12/30
    1 / -0

    With t taken in seconds and I taken in Amp, the variation of I follows the equation

     t2 + I2 = 25

    what amount of Ag will be electrodeposited with this current flowing in the interval 0-5 second ? (Ag : 108)

    Solutions

     

    =21.97mg = 22mg

  • Question 13/30
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    A resistance of 50Ω is registered when two electrodes are suspended into a beaker containing a dilute solution of a strong electrolyte such that exactly half of the them are submerged into solution. If the solution is diluted by adding pure water (negligible conductivity) so as to just completely submerge the electrodes, the new resistance offered by the solution would be

    Solutions

    the k is halved while the A is doubled. Hence R remain 50 Ω.

  • Question 14/30
    1 / -0

    The standard reduction potential of a silver chloride electrode is 0.2 V and that of a silver electrode is 0.79 V. The maximum amount of AgCl that can dissolve in 106 L of a 0.1 M AgNO3 solution is          

    Solutions

  • Question 15/30
    1 / -0

    Calculate the cell EMF in mV for

    Pt|H2(1atm) |HCl(0.01M)|AgCl(s)| Ag(s)        at 298 K

     If ΔG°r values are at 25°C

      for AgCl(s) and  for H+ + Cl-) (aq)

    Solutions

  • Question 16/30
    1 / -0

    Adiponitrile is manufactured electrolytically from acrylonitrile

     CH2 = CHCN → CN – (CH2)4 – CN

    How many kg of adiponitrile (molecular mass = 108) is produced in 9.65 hr using a current of 3750 A with 80% efficiency

    Solutions

  • Question 17/30
    1 / -0

    It is observed that the voltage of a galvanic cell using the reaction M(s) + xH+→ Mx+ + X/2H2 varies linearly with the log of the square root of the hydrogen pressure and the cube root of the Mx+ concentration. The value of x is       

    Solutions

  • Question 18/30
    1 / -0

    Acetic acid has Ka = 1.8 X 10 – 5 while formic acid had Ka = 2.1 X 10–4. What would be the magnitude of the emf of the cell

    Solutions

  • Question 19/30
    1 / -0

    Consider the cell Ag(s)|AgBr(s)|Br-(aq)||AgCl(s)|Cl-(aq)|Ag(s) at 25°C. The solubility product constants of AgBr & AgCl are respectively 5 X 10 – 13 & 1 X 10 – 10. For what ratio of the concentrations of Br- & Cl- ions would the emf of the cell be zero ?           

    Solutions

  • Question 20/30
    1 / -0

    Value of  for SrCl2 in water at 25°C from the following data:

    Solutions

  • Question 21/30
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    Calculate the useful work of the reaction Ag(s) + 1/2Cl2(g) → AgCl(s)
    Given  E°cl2/cl- = + 1.36 V,       E°AgCl/Ag,Cl- = 0.22 V

    If         Pcl2 = 1 atm     and T = 298 K

    Solutions

    (1) AgCl (s) + e → Ag(s) + Cl-         E0 = 22 V

    (2) 1/2 Cl2 + e → Cl-             E= 1.36V

    we get

    Ag(s) + 1/2Cl2 (g) →AgCl(s)          E0cell = 1.14V

  • Question 22/30
    1 / -0

    Which of these ions Cu+, Co3+, Fe2+ is stable in aqueous medium.

    Given :

    Cu2+/Cu+ = 0.15 volt ;            E°Cu+/Cu = 0.53 V ;       E°Co3+/Co2+ = 1.82 V ;

    Fe3+/Fe2+ = 0.77 V ;   E°Fe2+Fe = - 0.44 V ;    E°O2,H+/H2O = 1.23 V

    Solutions

     if reduction potential of metal ion is greater then O2/H2O couple, the ion is stable in water. So Co3+ is stable in water

  • Question 23/30
    1 / -0

    Select the correct statement if -

    Mg2+/Mg = - 2.4V,      E°Sn4+/Sn2+ = 0.1 V,     E°MnO4-,H+/Mn2+ = 1.5 V, E° I2/I- = 0.5 V    Here,

    Solutions

    [Hint: Reverse of (B) & (C) is spontaneous; weakest Oxidizing Agent here is Mg2+]

  • Question 24/30
    1 / -0

     What is the value of pKb (CH3COOH-) if λm = 390 & λm = 7.8 for 0.04 of a CH3COOH at 25°C          

    Solutions

  • Question 25/30
    1 / -0

    The temperature coefficient of a standard Cd-cell is – 5.0X10– 5 Vk– 1 whose emf at 25°C is 1.018 V. During the cell operation, the temperature will -           

    Solutions

  • Question 26/30
    1 / -0

     A cell Ag | Ag+ || Cu++ | Cu initially contains 2M Ag+ and 2M Cu++ ions. The charger in cell potential after the passage of 10 amp current for 4825 sec is:          

    Solutions

  • Question 27/30
    1 / -0

    For the cell (at 298 K)

    Ag(s) | AgCl(s) | Cl-(aq) || AgNO3(aq) | Ag(s)

    Which of the following is correct –

    Solutions

    Conductivity is high due to [H+]

  • Question 28/30
    1 / -0

    During an electrolysis of conc. H2SO4, perdissulphuric acid (H2S2O8) and O2 from in equimolar amount. The amount of H2 that will form simultaneously will be (2H2SO4 → H2S2O8 + 2H+ + 2e-)

    Solutions

  • Question 29/30
    1 / -0

    Statement-1: In electrochemical cell, we cannot use KCI in the salt bridge if anodic  or cathodic compartment consists of Ag+ of Pb2+ ion.

    Statement-2: Salt bridge is employed to maintain the electrical neutrality and to minimize the liquid-liquid junction potential.

    Solutions

    Both (S – 1) & (S – 2) are true and (S – 2) is not the correct explanation of    (S – 1)

  • Question 30/30
    1 / -0

    Statement-1: Zinc protect the iron better than tin even after it cracks.

    Statement-2:OPzn < E°OPfe But E°OPSn > E°OPfe

    Solutions

    The correct answer C
    EZn+2∣Zn  < EFe+2∣Fe∣Sn
       SPR ↓  Reactivity ↑

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