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JEE Advanced Model Paper 9 with solutions for online practice
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© examsnet.com
Question : 7
Total: 57
The equilibrium constant for the following reaction is
I
n
2
+
+
C
u
2
+
→
I
n
+
C
u
+
(at 298K)
[Given
E
°
C
u
2
+
∕
C
u
+
= + 0.15V ;
E
°
I
n
3
+
∕
I
n
+
= -0.42V ;
E
°
I
n
2
+
∕
I
n
+
= -0.40V]
10
2
x
The value of ' x ' is _______.
Your Answer:
Validate
Solution:
I
n
3
+
+ 2
e
−
→
I
n
+
; E° = - 0.42V
I
n
3
+
+
e
−
→
I
n
+
; E° = - 0.40V
∴ For
I
n
3
+
+
e
−
→
I
n
2
+
; E° = - 0.44V
E
°
c
e
l
l
= 0.15 - (- 0.44) = 0.59V
E
°
c
e
l
l
=
0.059
n
log k
0.59 =
0.059
1
log k ⇒ k =
10
10
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