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Standard X
Chemistry
Ionization Enthalpy
10. Calculate...
Question
10. Calculate half cell potential at 298k for the reaction Zn\pm +2 ===Zn if [Zn]=2M, Enot Zn\pm /Zn = -0.76V
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Q.
Calculate the half-cell potential at
298
K
for the reaction,
Z
n
2
+
+
2
e
−
→
Z
n
if
[
Z
n
2
+
]
=
0.1
M
and
E
∘
=
−
0.76
v
o
l
t
.
Q.
Reduction potential for the following half/cell reaction are
Z
n
→
Z
n
2
+
+
2
e
−
(
E
0
(
Z
n
2
+
/
Z
n
)
=
−
0.76
V
)
F
e
→
F
e
2
+
+
2
e
−
E
0
=
0.41
V
The EMF for the cell reaction
F
e
2
+
+
Z
n
→
Z
n
2
+
+
F
e
will
Q.
The standard reduction potential
E
o
for the half reactions are as
Z
n
⟶
Z
n
2
+
+
2
e
−
;
E
o
=
+
0.76
V
F
e
⟶
F
e
2
+
+
2
e
−
;
E
o
=
+
0.41
V
The emf for cell reaction,
F
e
2
+
+
Z
n
⟶
Z
n
2
+
+
F
e
, is
Q.
The standard reduction potential
E
o
for half reaction are:
Z
n
→
Z
n
2
+
+
2
e
−
;
E
o
=
+
0.76
V
F
e
→
F
e
2
+
+
2
e
−
;
E
o
=
+
0.41
V
The EMF of the cell reaction is:
F
e
2
+
+
Z
n
→
Z
n
2
+
+
F
e
Q.
Calculate using Nernst equation cell potential of the following electrochemical cell at
298
K
.
Z
n
(
s
)
|
Z
n
2
+
(
e
q
)
(
0.04
M
)
|
|
S
n
2
+
(
e
q
)
|
S
n
(
s
)
(
0.03
M
)
(Given
E
∘
=
−
0.76
V
,
E
∘
=
−
0.14
V
)
.
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