NEET-XII-Chemistry

Previous Year Paper year:2017

with Solutions - page 6
  • #82
    For a cell involving one electron ``E°_{cell}``= 0.59 V
    at 298 K, the equilibrium constant for the cell
    reaction is :[Given that ``\frac{2.303 RT}{F}``=0.059V at T = 298 K.
    (1) ``1.0 × 10^2``
    (2) ``1.0 × 10^5``
    (3) ``1.0 × 10^{10}``
    (4) ``1.0 × 10^{30}``
    digAnsr:   3
    Ans : ( 3 )
    Sol.
    E
    cell
    = E°
    cell
    -
    0.059
    log Q
    n
    ...(i)
    (At equilibrium, Q = K
    eq
    and E
    cell
    = 0)
    0 = E°
    cell
    - eq
    0.059
    logK
    1
    (from equation (i))
    cell
    eq
    E° 0.59
    logK = = = 10
    0.059 0.059
    K
    eq
    = 1010 = 1 × 1010