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JEE Main Chemistry Class 11 Electrochemistry Part 1 Questions
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© examsnet.com
Question : 20
Total: 100
250
m
L
of a waste solution obtained from the workshop of a goldsmith contains
0.1
M
A
g
N
O
3
and
0.1
M
A
u
C
l
. The solution was electrolyzed at
2
V
by passing a current of 1 A for 15 minutes. The metal\/metals electrodeposited will be:
[Sep. 04, 2020(II)]
(
E
A
g
+
∕
A
g
0
=
0.80
V
,
E
A
u
+
∕
A
u
0
=
1.69
V
)
only gold
silver and gold in proportion to their atomic weights
only silver
silver and gold in equal mass proportion
Validate
Solution:
Millimoles of
A
u
+
=
0.1
×
250
=
25
Mole of
A
u
+
=
25
1000
=
1
40
=
0.025
Similarly, moles of
A
g
+
=
0.025
Charge passed
=
I
×
t
=
1
×
15
×
60
=
900
C
Moles of e- passed
=
900
96500
=
0.0093
m
o
l
.
Species withhigher value of SRP will get deposited first at cathode.
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