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KCET 2014 Chemistry Solved Paper
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© examsnet.com
Question : 56
Total: 60
For
C
r
2
O
7
2
−
+
14
H
+
+
6
e
−
→
2
C
r
3
+
+
7
H
2
O
;
E
°
=
1.33
V
At
[
C
r
2
O
7
2
−
]
=
4.5
millimole, [
C
r
3
+
] = 15 millimole, E° is 1.067 V. The pH of the solution is nearly equal to
3
4
2
5
Validate
Solution:
C
r
2
O
7
2
−
+
16
H
+
+
6
e
−
→
2
C
r
3
+
+
7
H
2
O
E
cell
=
E
cell
∘
+
0.0591
6
log
[
C
r
2
O
7
2
−
]
[
H
+
]
4
[
C
r
+
3
]
2
1.067
=
1.33
+
0.0591
6
(
log
[
C
r
2
O
7
2
−
]
[
C
r
3
+
]
2
+
log
[
H
+
]
14
)
1.067
−
1.33
=
0.0591
6
−
(
−
log
[
C
r
2
O
7
2
−
]
[
C
r
+
3
]
2
−
log
[
H
+
]
14
)
−
0.263
=
−
1
100
(
−
log
(
4.5
×
10
−
3
)
(
15
×
10
−
3
)
2
+
14
p
H
)
26.3
=
log
1
20
+
14
p
H
26.3
=
−
1.3010
+
14
p
H
27.6
=
14
p
H
p
H
≈
2
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