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AIEEE 2008 Solved Paper
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© examsnet.com
Question : 20
Total: 104
Given
E
C
r
3
+
/
C
r
o
=
−
0.72
V
;
E
F
2
+
/
F
e
o
=
−
0.42
V
, The potential for the cell
Cr
∣
Cr
3
+
(
0.1
M
)
|
Fe
2
+
(
0.01
M
)
∣
Fe
is
0.26
V
0.399
V
−
0.399
V
−
0.26
V
Validate
Solution:
From the given representation of the cell,
E
cell
can be found as follows.
E
cell
=
E
F
e
2
+
/
F
e
∘
−
E
C
r
3
+
/
C
r
∘
−
0.059
6
log
[
Cr
3
+
]
2
[
F
e
2
+
]
3
[Nernst - Equ. ]
=
−
0.42
−
(
−
0.72
)
−
0.059
6
log
(
0.1
)
2
(
0.01
)
3
=
−
0.42
+
0.72
−
0.059
6
log
0.1
×
0.1
0.01
×
0.01
×
0.01
=
0.3
−
0.059
6
log
10
−
2
10
−
6
=
0.3
−
0.059
6
×
4
=
0.30
−
0.0393
=
0.26
V
© examsnet.com
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