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JEE Main Chemistry Class 11 Electrochemistry Part 1 Questions
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© examsnet.com
Question : 61
Total: 100
A current of
10.0
A
flows for
2.00
h
through an electrolytic cell containing a molten salt of metal
X
. This results in the decomposition of
0.250
mol of metal
X
at the cathode. The oxidation state of
X
in the molten salt is:
(
F
=
96
,
500
C
)
[Online April 9, 2014]
1
+
2
+
3
+
4
Validate
Solution:
According to Faraday's first law of electrolysis
W
=
E
×
i
×
t
96500
Where
E
=
equivalent weight
=
mol. mass of metal
(
M
)
oxidation state of
m
e
t
a
l
(
x
)
Substituting the value in the formula
W
=
M
x
×
i
×
t
96500
or
x
=
M
W
×
i
×
t
96500
=
10
×
2
×
60
×
60
96500
×
0.250
=
3
[
Given : no. of moles
=
W
M
=
0.250
]
Hence oxidation state of metal is
(
+
3
)
© examsnet.com
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