A cell is setup between copper and silver electrodes as cu|cu2+||ag+|ag. if two half cwll work under standard condotions calculate the emf of the xell
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2
a
)
E
0
Cu
2
+
/
Cu
<
E
0
Ag
+
/
Ag
Therefore
copper
electorode
isthe
anode
and
silver
electrode
act
as
a
cathode
.
The
half
reaction
are
:
anode
:
Cu
(
s
)
→
Cu
2
+
+
2
e
−
Cathode
:
2
Ag
+
+
2
e
−
→
2
Ag
b
)
net
reaction
:
Cu
(
s
)
+
2
Ag
+
(
aq
)
→
Cu
2
+
(
aq
)
+
2
Ag
(
s
)
c
)
The
potential
of
the
cell
is
given
by
E
=
E
0
cell
−
0
.
059
n
log
[
Cu
2
+
]
[
Ag
+
]
2
0
=
(
+
0
.
80
−
0
.
34
)
−
0
.
059
2
log
0
.
01
[
Ag
+
]
2
log
[
0
.
01
(
Ag
+
)
2
+
]
=
2
x
0
.
46
0
.
059
=
15
.
593
taking
antilog
0
.
01
[
Ag
+
]
2
=
antilog
15
.
593
=
3
.
919
x
10
15
[
Ag
+
]
2
=
0
.
01
3
.
919
x
10
15
=
2
.
55
x
10
−
18
[
Ag
+
]
=
1
.
59
x
10
−
9
M
)
E
0
Cu
2
+
/
Cu
<
E
0
Ag
+
/
Ag
Therefore
copper
electorode
isthe
anode
and
silver
electrode
act
as
a
cathode
.
The
half
reaction
are
:
anode
:
Cu
(
s
)
→
Cu
2
+
+
2
e
−
Cathode
:
2
Ag
+
+
2
e
−
→
2
Ag
b
)
net
reaction
:
Cu
(
s
)
+
2
Ag
+
(
aq
)
→
Cu
2
+
(
aq
)
+
2
Ag
(
s
)
c
)
The
potential
of
the
cell
is
given
by
E
=
E
0
cell
−
0
.
059
n
log
[
Cu
2
+
]
[
Ag
+
]
2
0
=
(
+
0
.
80
−
0
.
34
)
−
0
.
059
2
log
0
.
01
[
Ag
+
]
2
log
[
0
.
01
(
Ag
+
)
2
+
]
=
2
x
0
.
46
0
.
059
=
15
.
593
taking
antilog
0
.
01
[
Ag
+
]
2
=
antilog
15
.
593
=
3
.
919
x
10
15
[
Ag
+
]
2
=
0
.
01
3
.
919
x
10
15
=
2
.
55
x
10
−
18
[
Ag
+
]
=
1
.
59
x
10
−
9
M
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