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Byju's Answer
Standard XII
Chemistry
EMF
A g + can oxi...
Question
A
g
+
can oxidize
P
b
to
P
b
2
+
under standard state conditions.
E
o
A
g
=
0.799
V
E
o
P
b
=
−
0.126
V
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Solution
For this cell must be like this oxidation half cell reaction:
P
b
⟶
P
b
2
+
2
e
−
(
a
t
a
n
o
d
e
)
E
o
P
b
=
−
0.126
V
reduction half cell reaction
2
A
g
+
+
2
e
−
⟶
2
A
g
(
a
t
c
a
t
h
o
d
e
)
Overall cell
P
b
|
P
b
2
+
|
|
A
g
+
|
A
g
E
o
c
e
l
l
=
E
o
c
a
t
h
o
d
e
−
E
o
a
n
o
d
e
=
E
o
A
g
−
E
o
P
b
=
0.799
−
(
0.126
)
=
0.925
For working of all
E
o
c
e
l
l
must be +ve.
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1
Similar questions
Q.
Set up the cell constiting of
H
+
(
a
q
)
|
H
2
g
)
and
P
b
2
+
(
a
q
)
|
P
b
(
s
)
electrodes. Calculate the emf at
25
℃
of the cell if
[
P
b
2
+
]
= 0.1 M,
[
H
+
]
=
0.5
M
and hydrogen gas is at 2 atm pressure,
E
o
P
b
=
−
0.126
V
.
Q.
E
o
F
e
3
+
/
F
e
2
+
=
0.771
V
and
E
o
A
g
+
/
A
g
=
0.799
V
The minimum conc. of Ag
+
that would remain unreduced by a standard
F
e
3
+
/
F
e
2
+
electrode at equilibrium is:
Q.
A silver wire dipped in
0.1
M
H
C
l
solution saturated with
A
g
C
l
develops oxidation potential of
−
0.25
V
.
If
E
o
A
g
/
A
g
+
=
−
0.799
V
,
the
K
s
p
of
A
g
C
l
in pure water will be :
Q.
In simple electrochemical cell, which is in its standard state, the
half cell reactions with their appropriate reduction potential are?
P
b
2
+
+
2
e
→
P
b
(
E
o
=
−
0.13
v
)
A
g
+
+
e
→
A
g
(
E
o
=
+
0.80
v
)
Q.
Consider the cell;
V
2
+
+
V
O
2
+
+
2
H
+
→
2
V
3
+
+
H
2
O
;
E
o
=
0.616
V
V
3
+
+
A
g
+
+
H
2
O
→
V
O
2
+
+
2
H
+
+
A
g
;
E
o
=
0.439
V
Potential of
E
o
V
3
+
/
V
2
+
is :
(Given that
E
o
A
g
+
/
A
g
=
0.799
V
)
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