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Standard XII
Chemistry
Electrochemical Cells, Galvanic Cells
Calculate the...
Question
Calculate the EMF of a Danial cell when the concentration of
Z
n
S
O
4
and
C
u
S
O
4
are
0.001
M and
0.1
M respectively. The standard EMF of the cell is
1.1
V.
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Solution
Let
E
1
=
E
0
c
e
l
l
−
2.303
×
R
T
2
F
×
log
(
0.01
)
1
Therefore the concentration are changed,
E
2
=
E
0
c
e
l
l
−
2.303
R
T
2
F
×
(
0.01
)
1
Hence,
E
1
>
E
2
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Similar questions
Q.
Calculate the EMF of a Daniel cell when the concentration of
Z
n
S
O
4
and
C
u
S
O
4
are
0.001
M and
0.1
M respectively. The standard potential of the cell is
1.1
V.
Q.
The emf of a Daniel cell at 298 K is
E
1
concentration of
Z
n
S
O
4
is 0.1 M and that
C
u
S
O
4
is 0.01 M, the emf changed to
E
2
.What is the relationship between
E
1
and
E
2
Q.
Calculate the emf of the cell.
Z
n
|
Z
n
2
+
(
0.001
M
)
|
|
C
u
2
+
(
0.1
M
)
|
C
u
The standard potential of
C
u
/
C
u
2
+
half-cell is +0.34 and
Z
n
/
Z
n
2
+
is -0.76 V.
Q.
The emf of Daniell cell at
298
K
is
E
1
Z
n
|
Z
n
S
O
4
(
0.01
M
)
|
|
C
u
S
O
4
(
1.0
M
)
|
C
u
When the concentration of
Z
n
S
O
4
is
1.0
M
and that of
C
u
S
O
4
is
0.01
M
, the emf changed to
E
2
What is the relation between
E
1
and
E
2
?
Q.
An electrochemical cell is constructed by immersing a piece of copper wire in 50 ml of 0.1 M
C
u
S
O
4
solution and zinc strip in 50 ml of 0.1 M
Z
n
S
O
4
solution
[
E
0
C
u
2
+
/
C
u
= 0.34 V,
E
0
Z
n
2
+
/
Z
n
= -0.76 V ]
The emf of the cell increases when small amount of concentrated
N
H
3
is added to:
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