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Byju's Answer
Standard IV
Mathematics
Rules for Writing Roman Numerals
49. For the d...
Question
49. For the disproportionation of copper 2Cu Cu2+ Cu, E is (Given E for Cu2+/Cu is 0.34 V and Eo for Cu?+/Cut is 0.15 V) (1) 0.49 V (3) 0.38 V the m (2) -0.19 V (4) -0.38 V of Ala 2 50
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Q.
For the disproportionation of copper :
2
C
u
+
→
C
u
+
2
+
C
u
,
E
∘
is:
Given;
E
for
C
u
+
2
/
C
u
is
0.34
V
and
E
for
C
u
+
2
/
C
u
+
is
0.15
V
Q.
Given,
C
u
2
+
+
2
e
−
→
C
u
;
E
o
=
0.337
V
C
u
2
+
+
e
−
→
C
u
+
;
E
o
=
−
0.153
V
E
o
value for the reaction,
C
u
+
+
e
→
C
u
will be:
Q.
E
o
values for the half cell reactions are given:
C
u
2
+
+
e
−
→
C
u
+
;
E
o
=
0.15
V
C
u
2
+
+
2
e
−
→
C
u
;
E
o
=
0.34
V
What will be the
E
o
of the half-cell:
C
u
+
+
e
−
→
C
u
?
Q.
If for the half-cell reactions
C
u
2
+
+
e
−
⟶
C
u
+
E
o
=
0.15
V
C
u
2
+
+
2
e
−
⟶
C
u
E
o
=
0.34
V
Calculate
E
o
of the half cell reaction
C
u
+
+
e
−
⟶
C
u
Also predict whether
C
u
+
undergoes disproportionation or not.
Q.
The
E
⊖
f
o
r
C
u
2
+
/
C
u
⊕
,
C
u
⊕
/
C
u
,
C
u
2
+
/
C
u
,
are 0.15 V, 0.50 V, and 0.325 V, respectively. The redox cell showing redox reaction
2
C
u
+
,
→
C
u
2
+
+
C
u
is made.
E
⊖
of this cell reaction and
Δ
G
⊖
may be :
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