NEET-XII-Chemistry
Previous Year Paper year:2017
- #32In 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 > E2digAnsr: 4Ans : (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






