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Byju's Answer
Standard XII
Chemistry
Equilibrium Constant from Nernst Equation
Hg22+ + 2 e- ...
Question
H
g
2
+
2
+
2
e
−
→
2
H
g
⟹
E
0
=
0.789
V
H
g
2
+
+
2
e
−
→
H
g
⟹
E
0
=
0.854
V
Calculate the equilibrium constant for
H
g
2
+
2
→
H
g
+
H
g
2
+
.
A
3.1
×
10
−
3
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B
3.1
×
10
−
4
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C
6.3
×
10
−
3
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D
6.3
×
10
−
4
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Solution
The correct option is
C
6.3
×
10
−
3
H
g
2
+
2
+
2
e
−
→
2
H
g
⟹
E
0
=
0.789
V
H
g
→
H
g
2
+
+
2
e
−
⟹
E
0
=
−
0.854
V
H
g
2
+
2
→
H
g
+
H
g
2
+
⟹
−
0.065
V
Δ
G
=
−
2
×
(
−
0.065
)
×
96500
=
−
8.314
×
298
l
n
K
e
q
K
e
q
=
6.3
×
10
−
3
.
Hence, option C is correct.
Suggest Corrections
2
Similar questions
Q.
An excess of
H
g
was added to
10
−
3
M
acidified solution of
F
e
3
+
ions. Its was found that only
4.6
% of the ions remained as
F
e
3
+
at equilibrium at
25
∘
C
. Calculate
E
∘
for
2
H
g
/
H
g
2
+
2
at
25
∘
C
for
2
H
g
+
2
F
e
3
+
⇌
H
g
2
+
2
+
2
F
e
2
+
.
Q.
Calculate the standard potential for the reaction,
H
g
2
C
l
2
+
C
l
2
→
2
H
g
+
+
4
C
l
−
Given:
H
g
2
C
l
2
+
2
e
−
→
2
H
g
+
2
C
l
−
;
E
∘
=
0.270
v
o
l
t
H
g
2
+
2
→
2
H
g
2
+
+
2
e
−
;
E
∘
=
−
0.92
v
o
l
t
2
H
g
→
H
g
2
+
2
+
2
e
−
;
E
∘
=
−
0.79
v
o
l
t
C
l
2
+
2
e
−
→
2
C
l
−
;
E
∘
=
1.36
v
o
l
t
.
Q.
Calculate
K
c
for
H
g
2
+
+
H
g
⇌
H
g
2
+
2
.
Give that
E
o
2
H
g
/
H
g
2
+
=
−
0.788
V
and
E
o
H
g
2
+
2
/
2
H
g
2
+
=
−
0.920
V
.
Write answer as nearest integer after dividing by 100.
Q.
An excess of liquid Hg was added to
10
−
3
M acidified solution of
F
e
3
+
ions. It was found that only 4.6% of the ions remained as
F
e
3
+
at equilibrium at
25
o
C. Calculate
E
o
for
2
H
g
/
H
g
2
+
2
at
25
o
C for,
2
H
g
+
2
F
e
3
+
⇌
H
g
2
+
2
+
2
F
e
2
+
and
E
o
F
e
2
+
/
F
e
3
+
=
−
0.7712
V
.
Q.
In which direction can the reaction,
2
H
g
(
l
)
+
2
A
g
+
(
a
q
.
)
⇌
2
A
g
(
s
)
+
H
g
2
+
2
(
a
q
.
)
proceed spontaneously at the following concentrations of the ions participating in the reactions (i) and (ii)?
(i)
[
A
g
+
]
=
10
−
4
m
o
l
L
−
1
and
[
H
g
2
+
2
]
=
10
−
1
m
o
l
L
−
1
(ii)
[
A
g
+
]
=
10
−
1
m
o
l
L
−
1
and
[
H
g
2
+
2
]
=
10
−
4
m
o
l
L
−
1
Given:
E
∘
H
g
2
+
2
/
H
g
=
0.79
V
;
E
∘
A
g
+
/
A
g
=
0.80
V
.
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