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Standard VII
Chemistry
Electrolysis
The emf of th...
Question
The emf of the cell in which the following reaction occurs is found to be 0.5105 at 298 K Zn(s) +Ni+(0.1M) -changes to Zn+(1M) +Ni(s). Thw standard emf of the cell is (1)0.4810 V (2)0.5696 V (3)-0.5105 (4)0.5400 V
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Similar questions
Q.
The emf of the cell in which the following reaction,
Z
n
(
s
)
+
N
i
2
+
(
0.1
M
)
→
Z
n
2
+
(
1.0
M
)
+
N
i
(
s
)
occurs, is found to be
0.5105
V
at
298
K
. The standard emf of the cell is:
Q.
The e.m.f. of the cell involving following changes,
Z
n
(
s
)
+
N
i
2
+
(
1
M
)
→
Z
n
2
+
(
1
M
)
+
N
i
(
s
)
is
0.5105
V
.
The standard e.m.f. of the cell is:
Q.
Determine the standard emf of the cell and standard free energy change of the cell reaction
Z
n
|
Z
n
2
+
|
|
N
i
2
+
|
N
i
. The standard reduction potentials of
Z
n
2
+
|
Z
n
and
N
i
2
+
|
N
i
half cells are
−
0.76
V
and
−
0.25
V
respectively.
Q.
Determine the standard emf of the cell and standard free energy change of the cell reaction.
Z
n
/
Z
n
2
+
∥
N
i
2
+
/
N
E
o
Z
n
2
+
/
Z
n
=
−
0.76
V
E
o
N
i
2
+
/
N
i
=
−
0.25
V
Q.
The following electrochemical cell is set up at
298
K
:
Z
n
/
Z
n
2
+
(
a
q
)
(
1
M
)
/
/
C
u
2
+
(
a
q
)
(
1
M
)
/
C
u
Given
→
E
o
Z
n
2
+
/
Z
n
=
−
0.761
V
,
E
o
C
u
2
+
/
C
u
=
+
0.339
V
(1) Write the cell reaction.
(2) Calculate the emf and free energy change at
298
K
.
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