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Byju's Answer
Standard XII
Chemistry
Electrolytes
For Cr 2 O ...
Question
For
C
r
2
O
−
2
7
+
14
H
+
+
6
e
−
→
2
C
r
+
3
+
7
H
2
O
;
E
o
=
1.33
V
.
At
[
C
r
2
O
−
2
7
]
=
4.5
millimole,
[
C
r
+
3
]
=
15
millimole,
E
is
1.067
V
. The pH of the solution is nearly equal to
:
A
3
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B
4
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C
2
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D
5
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Solution
The correct option is
D
2
Using the Nernst equation,
E
=
E
o
−
0.059
n
F
l
o
g
[
C
r
+
3
]
2
[
C
r
2
O
2
−
7
]
[
H
+
]
14
1.067
=
1.33
−
0.06
6
×
96500
l
o
g
(
15
×
10
−
3
)
2
(
4.5
×
10
−
3
)
[
H
+
]
14
[
H
+
]
≃
10
−
2
p
H
=
−
l
o
g
[
H
+
]
=
2
Hence, option C is correct.
Suggest Corrections
0
Similar questions
Q.
In acidic medium
C
r
2
O
−
2
7
is an oxidising agent
C
r
2
O
−
2
7
+
14
H
+
+
6
e
−
→
2
C
r
+
3
+
7
H
2
O
[
E
∘
C
r
2
O
−
2
7
/
C
r
+
3
=
1.33
V
]
The electrode potential of half cell at pH = 1 keeping concentration of other species unity is:
Q.
3
C
H
3
C
H
2
O
H
+
2
K
2
C
r
2
O
7
+
8
H
2
S
O
4
⟶
3
C
H
3
C
O
O
H
+
2
C
r
2
(
S
O
4
)
3
+
2
K
2
S
O
4
+
11
H
2
O
Equivalent mass of
K
2
C
r
2
O
7
for the above reaction is :
Q.
In an oxidation-reduction reaction, dichromate
(
C
r
2
O
−
2
7
)
ion is reduced to
C
r
+
3
ion. The equivalent weight of
K
2
C
r
2
O
7
in this reaction is:
Q.
The percentage composition of
C
r
in
K
2
C
r
2
O
7
is (Given molar mass of
K
2
C
r
2
O
7
=
294.0
g
m
o
l
)
Q.
In acidic medium
C
r
2
O
−
2
7
is an oxiding agent
C
r
2
O
−
2
7
+
14
H
+
+
6
e
−
←
2
C
r
+
3
+
7
H
2
O
[
E
∘
C
r
2
O
−
2
7
/
C
r
+
3
=
1.33
V
]
The electrode potential of half cell at
p
H
=
1
keeping concentration of other species unity is:
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