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Class 12
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Nernst Equation
Nernst Equation
Revise with Concepts
Nernst Equation and Calculation of Emf
Example
Definitions
Formulaes
Emf, Gibbs Energy and Equilibrium Constant
Example
Definitions
Formulaes
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Important Questions
Calculate the emf of the cell.
Z
n
∣
Z
n
2
+
(
0
.
0
0
1
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.
Medium
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>
The equilibrium constant of the reaction;
C
u
(
s
)
+
2
A
g
+
(
a
q
)
⟶
C
u
2
+
(
a
q
)
+
2
A
g
(
s
)
E
o
=
0
.
4
6
V
at
2
9
8
K
Medium
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>
Write the Nernst equation and emf of the following cells at
2
9
8
K
:
(i)
M
g
(
s
)
∣
M
g
+
2
(
0
.
0
0
1
M
)
∥
C
u
+
2
(
0
.
0
0
0
1
M
)
∣
C
u
(
s
)
.
(ii)
F
e
(
s
)
∣
F
e
+
2
(
0
.
0
0
1
M
)
∥
H
+
(
1
M
)
∣
H
2
(
g
)
(
1
b
a
r
)
∣
P
t
(
s
)
(iii)
S
n
(
s
)
∣
S
n
2
+
(
0
.
0
5
0
M
)
∥
H
+
(
0
.
0
2
0
M
)
∣
H
2
(
g
)
(
1
b
a
r
)
∣
P
t
(
s
)
(iv)
P
t
(
s
)
∣
B
r
2
(
l
)
∣
B
r
−
(
0
.
0
1
0
M
)
∥
H
+
(
0
.
0
3
0
M
)
∣
H
2
(
g
)
(
1
b
a
r
)
∣
P
t
(
s
)
.
Hard
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>
Represent the cell in which the following reaction takes place
M
g
(
s
)
+
2
A
g
+
(
0
.
0
0
0
1
M
)
→
M
g
2
+
(
0
.
1
3
0
M
)
+
2
A
g
(
s
)
Calculate its
E
(
c
e
l
l
)
if
E
(
c
e
l
l
)
o
=
3
.
1
7
V
Medium
View solution
>
Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
(i)
2
C
r
(
s
)
+
3
C
d
2
+
(
a
q
)
→
2
C
r
3
+
(
a
q
)
+
3
C
d
(ii)
F
e
2
+
(
a
q
)
+
A
g
+
(
a
q
)
→
F
e
3
+
(
a
q
)
+
A
g
(
s
)
Calculate the
Δ
r
G
⊖
and equilibrium constant of the reactions
Hard
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>
Calculate the emf of the following cell at
2
9
8
K
?
F
e
(
s
)
∣
F
e
2
+
(
0
.
0
0
1
M
)
∣
∣
H
+
(
1
M
)
∣
H
2
(
g
)
(
1
b
a
r
)
,
P
t
(
s
)
(Given:
E
c
e
l
l
∘
=
+
0
.
4
4
V
)
Medium
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>
Given the standard electrode potentials,
K
+
/
K
=
−
2
.
9
3
V
,
A
g
+
/
A
g
=
0
.
8
0
V
,
H
g
2
+
/
H
g
=
0
.
7
9
V
,
M
g
2
+
/
M
g
=
−
2
.
3
7
V
,
C
r
3
+
/
C
r
=
−
0
.
7
4
V
Arrange these metals in their increasing order of reducing power.
Hard
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>
The Gibbs energy for the decomposition of
A
l
2
O
3
at
5
0
0
o
C
is as follows:
2
/
3
A
l
2
O
3
→
4
/
3
A
l
+
O
2
;
Δ
r
G
=
+
9
6
6
kJ/mol.
The potential difference needed for electrolytic reduction of
A
l
2
O
3
at
5
0
0
o
C is at least:
Hard
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>
The standard reduction potentials
E
⊝
for the half reactions are follows :
Z
n
→
Z
n
2
+
+
2
e
−
;
E
⊝
=
+
0
.
7
6
V
F
e
→
F
e
2
+
+
2
e
−
;
E
⊝
=
0
.
4
1
V
The EMF for the cell reaction
F
e
2
+
Z
n
→
Z
n
2
+
+
F
e
is:
Hard
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>
Calculate the emf of the cell
C
r
∣
C
r
3
+
(
0
.
1
M
)
∣
∣
F
e
2
+
(
0
.
0
1
M
)
∣
F
e
(Given:
E
C
r
3
+
/
C
r
∘
=
−
0
.
7
5
v
o
l
t
;
E
F
e
2
+
/
F
e
∘
=
−
0
.
4
5
v
o
l
t
)
.
Medium
View solution
>