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Standard XII
Chemistry
Equilibrium Constant from Nernst Equation
Question
For hydrogen-oxygen fuel cell, the cell reaction is:
2
H
2
(
g
)
+
O
2
(
g
)
⟶
2
H
2
O
(
l
)
If
Δ
G
o
f
(
H
2
O
)
=
−
237.2
k
J
m
o
l
−
1
, then emf of this cell is:
+
1.23
V
−
1.23
V
−
2.46
V
+
2.46
V
A
−
2.46
V
B
−
1.23
V
C
+
1.23
V
D
+
2.46
V
Open in App
Solution
Verified by Toppr
The correct option is
C
+
1.23
V
Gibbs free energy,
Δ
G
o
=
−
n
F
E
o
∴
E
o
=
Δ
G
o
n
F
=
237.2
×
1000
J
2
×
96500
=
1.23
V
[
∵
n
=
2
]
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4
Similar Questions
Q1
For hydrogen-oxygen fuel cell, the cell reaction is:
2
H
2
(
g
)
+
O
2
(
g
)
⟶
2
H
2
O
(
l
)
If
Δ
G
o
f
(
H
2
O
)
=
−
237.2
k
J
m
o
l
−
1
, then emf of this cell is:
View Solution
Q2
For a cell reaction,
2
H
2
(
g
)
+
O
2
(
g
)
→
2
H
2
O
(
l
)
△
S
0
298
=
−
0.32
kJK
−
1
. What is the value of
△
f
H
0
298
(
H
2
0
,
l
)
Given :
O
2
(
g
)
+
4
H
+
(
a
q
)
+
4
e
−
→
2
H
2
0
(
l
)
;
E
0
=
1.23
V
View Solution
Q3
For a cell reaction
2
H
2
(
g
)
+
O
2
(
g
)
⟶
2
H
2
O
(
l
)
;
△
r
S
∘
298
=
−
0.32
k
J
/
K
. What is the value of
△
f
H
∘
298
(
H
2
O
,
l
)
?
Given
O
2
(
g
)
+
4
H
+
(
a
q
.
)
+
4
e
−
⟶
2
H
2
O
(
l
)
;
E
∘
=
1.23
V
View Solution
Q4
The value of
Δ
S
⊖
for the fuel cell reaction at 298 K is:
2
H
2
+
O
2
→
2
H
2
O
E
⊖
c
e
l
l
=
1.23
V
Δ
f
H
⊖
(
H
2
O
)
=
−
285.8
K
j
m
o
l
−
1
View Solution
Q5
Hydrazine can be used in fuel cell
N
2
H
4
(
a
q
)
+
O
2
(
g
)
→
N
2
(
g
)
+
2
H
2
O
(
l
)
If
Δ
G
0
for this reaction is -600 kJ, what will be the
E
0
for the cell?
View Solution