NEET-XII-Chemistry

Previous Year Paper year:2017

with Solutions - page 2
  • #32
    In the electrochemical cell -
    Zn|``ZnSO_4``(0.01M) || ``CuSO_4`` (1.0 M)| Cu,
    the emf of this Daniel cell is ``E_1``. When the
    concentration of ``ZnSO_4`` is changed to 1.0 M
    and that of ``CuSO_4`` changed to 0.01M, the emf
    changes to E2. From the followings, which
    one is the relationship between E1 and E2 ?
    Given, ``\frac {RT}{F}`` = 0.059) .
    (1) E2 = 0 ≠ E1
    (2) E1 = E2
    (3) E1 < E2
    (4) E1 > E2
    digAnsr:   4
    Ans : (4)
    Sol.
    Cu||CuSO||ZnSO|Zn
    )M0.1(
    4
    )M01.0(
    4
    Nernst equation
    ]Cu[
    ]Zn[log
    2
    059.0EEmf 2
    2
    cell +
    +
    ° -=
    In first case
    1
    01.0log
    2
    059.0EE cell1 -=
    °
    In second case
    01.0
    1log
    2
    059.0EE cell2 -=
    °
    So E1 > E2