NEET-XII-Chemistry

exam-1 year:2016

with Solutions - page 2
  • #75
    The pressure of ``H_2`` required to make the potential
    of ``H_2``-electrode zero in pure water at 298 K is :-
    (1) ``10^{-14}`` atm
    (2) ``10^{-12}`` atm
    (3) ``10^{-10}`` atm
    (4) ``10^{-4}`` atm
    digAnsr:   1
    Ans : (1)
    Sol. 2H+(aq) + 2e-  H2(g)
     +
    ∴ =  2
    H0
    2
    P0.0591
    E E log
    2 H
    0 = 0- 0.0295 log
     
    2H
    27
    P
    10
     
    2H
    27
    P
    10
    =1
    =
    2
    14
    HP 10 atm