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NEET-XII-Chemistry

03: Electrochemistry

 page 2
Qstn# II-4-i Prvs-QstnNext-Qstn
  • #4-i
    2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd
    () Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)

    Calculate the ΔrGθ and equilibrium constant of the reactions.
    () Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)

    Calculate the ΔrGθ and equilibrium constant of the reactions.
    () Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)

    Calculate the ΔrGθ and equilibrium constant of the reactions.
    Ans :



    The galvanic cell of the given reaction is depicted as:



    Now, the standard cell potential is



    In the given equation,

    n = 6

    F = 96487 C mol-1

    = +0.34 V

    Then, = -6 × 96487 C mol-1 × 0.34 V

    = -196833.48 CV mol-1

    = -196833.48 J mol-1

    = -196.83 kJ mol-1

    Again,

    = -RT ln K



    = 34.496

    K = antilog (34.496)

    = 3.13 × 1034
    ()



    The galvanic cell of the given reaction is depicted as:



    Now, the standard cell potential is



    Here, n = 1.

    Then,

    = -1 × 96487 C mol-1 × 0.03 V

    = -2894.61 J mol-1

    = -2.89 kJ mol-1

    Again,



    = 0.5073

    K = antilog (0.5073)

    = 3.2 (approximately)
    ()



    The galvanic cell of the given reaction is depicted as:



    Now, the standard cell potential is



    Here, n = 1.

    Then,

    = -1 × 96487 C mol-1 × 0.03 V

    = -2894.61 J mol-1

    = -2.89 kJ mol-1

    Again,



    = 0.5073

    K = antilog (0.5073)

    = 3.2 (approximately)
    ()



    The galvanic cell of the given reaction is depicted as:



    Now, the standard cell potential is



    Here, n = 1.

    Then,

    = -1 × 96487 C mol-1 × 0.03 V

    = -2894.61 J mol-1

    = -2.89 kJ mol-1

    Again,



    = 0.5073

    K = antilog (0.5073)

    = 3.2 (approximately)
  • #4-ii
    Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)

    Calculate the ΔrGθ and equilibrium constant of the reactions.
    Ans :



    The galvanic cell of the given reaction is depicted as:



    Now, the standard cell potential is



    Here, n = 1.

    Then,

    = -1 × 96487 C mol-1 × 0.03 V

    = -2894.61 J mol-1

    = -2.89 kJ mol-1

    Again,



    = 0.5073

    K = antilog (0.5073)

    = 3.2 (approximately)