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KEAM 2014 Physics and Chemistry Paper
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© examsnet.com
Question : 119
Total: 120
The change in potential of the half-cell
Cu
2
+
|
Cu
,
when aqueous
Cu
2
+
+ solution is diluted 100 times at 298 K?
(
2.303
‌
RT
F
=
0.06
)
increases by 120 mV
decreases by 120 mV
increases by 60 mV
decreases by 60 mV
no change
Validate
Solution:
Half cell reaction is
C
u
2
+
+
2
e
−
⟶
C
u
;
n
=
2
From Nernst equation,
E
=
E
∘
−
‌
2.303
R
T
n
F
‌
log
‌
1
[
C
u
2
+
]
E
1
=
E
∘
+
‌
0.06
2
‌
log
[
C
u
2
+
]
Let the initial concentration of
C
u
2
+
be 1 .
E
1
=
E
∘
+
‌
0.06
2
‌
log
‌
1
=
E
∘
+
0
‌
‌
∴
E
1
=
E
∘
Further, the
[
C
u
2
+
]
solution is dilued to 100 times.
∴
‌
‌
M
1
V
1
Initial
=
M
2
V
2
After
‌
dilution
1
×
1
=
M
2
×
100
M
2
=
‌
1
100
=
0.01
∴
‌
‌
E
2
=
E
∘
+
‌
0.059
2
‌
log
[
0.01
]
=
E
∘
+
‌
0.059
2
(
−
2
)
=
E
1
−
0.059
V
=
E
1
−
59
m
V
Thus, the potential decreases by
59
(
≈
60
)
m
V
.
© examsnet.com
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