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JEE Main Chemistry Class 11 Electrochemistry Part 2 Questions
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© examsnet.com
Question : 27
Total: 79
The potential for the given half cell at
298
K
is
×
10
−
2
V
.
2
H
+
(
aq
)
+
2
e
−
⟶
H
2
(
g
)
[
H
+
]
=
1
M
,
P
H
2
=
2
atm
(Given:
2.303
RT
∕
F
=
0.06
V
,
log
2
=
0.3
)
[1-Feb-2024 Shift 1]
Your Answer:
Validate
Solution:
E
=
E
H
+
∕
H
2
o
−
0.06
2
log
P
H
2
[
H
+
]
2
E
=
0.00
−
0.06
2
log
2
[
1
]
2
E
=
−
0.03
×
0.3
=
−
0.9
×
10
−
2
V
© examsnet.com
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