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Byju's Answer
Standard XII
Chemistry
Nernst Equation
The EMF of th...
Question
The EMF of the cell :
P
t
,
H
2
(
1
a
t
m
)
|
H
⊕
(
a
q
)
∥
A
g
C
l
|
A
g
is 0.27 and 0.26 V at
25
∘
C
and
35
∘
C
, respectively. The posituve value of heat of the reaction occurring inside the cell at
25
∘
C
(
i
n
k
J
K
−
1
) is:
( write your answer to nearest integer)
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Solution
(
∂
E
∂
T
)
p
=
(
0.26
−
0.27
35
−
25
)
=
−
0.07
10
V
K
−
1
[
n
c
e
l
l
=
1
]
Δ
H
=
−
n
F
[
E
c
e
l
l
(
a
t
25
∘
C
)
−
T
(
∂
E
∂
T
)
p
]
Δ
H
=
−
1
×
96500
[
0.27
−
298
(
−
0.01
10
)
]
Δ
H
=
−
1
×
96500
C
×
0.568
V
K
−
1
=
−
54812
J
K
−
1
=
−
54.8
k
J
K
−
1
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Similar questions
Q.
Emf of the cell
P
t
.
H
2
(
1
a
t
m
)
|
H
+
(
a
q
.
)
|
|
A
g
C
l
|
A
g
is
0.27
V
and
0.26
V
at
25
∘
C
and
35
∘
C
. Heat of reaction occuring inside the cell at
25
∘
C
is:
Q.
Calculate the cell EMF in mV for :
P
t
|
H
2
(
1
a
t
m
)
|
H
C
l
(
0.01
M
)
|
A
g
C
l
(
s
)
|
A
g
(
s
)
a
t
298
K
if
Δ
G
o
t
values are at 25
o
C
−
109.56
k
J
m
o
l
for AgCl(s) and
−
130.79
k
J
m
o
l
for (H
+
+ Cl
−
) (aq)
Q.
The cell Pt,
H
2
(
1
a
t
m
)
|
H
+
(
p
H
=
x
)
|
|
Normal calomel Electrode has an EMF of 0.67 V at
25
∘
C
. The pH of the solution is:
The oxidation potential of the calomel electrode on hydrogen scale is
−
0.28
V
.
(write the value to the nearest integer)
Q.
The standard potential of the following cells is 0.23 V at 15
o
C and 0.21 V at 35
o
C.
P
t
|
H
2
(
g
)
|
H
C
l
(
a
q
)
|
A
g
C
l
(
s
)
|
A
g
(
s
)
The solubility of
A
g
C
l
in water 25
o
C.
[Note : The standard reduction potential of the
A
g
+
(
a
q
)
/
A
g
(
s
)
couple is 0.80 V at 25
o
C]
Q.
The EMF of the cell,
C
d
(
s
)
|
C
d
C
l
2
(
a
q
,
0.1
M
)
|
|
A
g
C
l
(
s
)
|
A
g
(
s
)
in which the cell reaction is
C
d
(
s
)
+
2
A
g
C
l
(
s
)
→
2
A
g
(
s
)
+
C
d
2
+
(
a
q
)
+
2
C
l
–
(
a
q
)
is
0.6915
V
at 0°C and
0.6753
V
at 25°C. The
Δ
S
and
Δ
H
of the reaction at 25°C is:
Take,
F
=
96485
C
m
o
l
−
1
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