Calculate the EMF of the cell, where the cell reation is
Cu(s) + 2Ag (aq) → Cu²+ (aq) + 2Ag(s) [C₁²] = 0.130 M and [Ag] = 102 M. The standard reduction potentials of copper and silver electrodes are 0.34V and 0.80V respectively.
Answers
Answer:
Correct option is
A
0.010 V
Note that the given cell will not work as electrochemical cells since,
E
OP
Cu
o
>E
OP
Ag
o
The equation for electrochemical cells will be: (i)
Cu→Cu
2+
+2e
2Ag
+
+2e→2Ag
Thus, e.m.f. of cell Cu∣Cu
2+
∣∣Ag
+
∣ Ag will be :
E
cell
=E
OP
Cu
o
+E
RP
Ag
o
+
2
0.059
log
[Cu
2+
]
[Ag
+
]
2
∴[Ag
+
]=1M and [Cu
2+
]=1MCu→Cu
2+
+2e
2Ag
+
+2e→2Ag
Thus, e.m.f. of cell Cu∣Cu
2+
∣∣Ag
+
∣ Ag will be :
E
cell
=E
OP
Cu
o
+E
RP
Ag
o
+
2
0.059
log
[Cu
2+
]
[Ag
+
]
2
∴[Ag
+
]=1M and [Cu
2+
]=1M
∴E
cell
=E
OP
Cu
o
+E
RP
Ag
o
(E
cell
o
=E
OP
Cu
o
+E
RP
Ag
o
)
E
cell
=E
cell
o
After the passage of 9.65 ampere for 1 hr, i.e., 9.65×60×60 coulomb change, during which the cell reactions are reversed, thus, Ag metal passes in solution state and Cu
2+
ions are discharged.
The reactions during this passage of current are:
2Ag→2Ag
+
+2e
Cu
2+
+2e→Cu
Thus, Ag
+
formed =
96500
9.65×60×60
=0.36eq.=0.36 mole
Cu
2+
discharged =
96500
9.65×60⋊60
=0.36eq.=0.18 mole
Thus, [Ag
+
]left=1+0.36=1.36 mole
[Cu
2+
] left=1−0.18=0.82 mole
Now, e.m.f. can be given as
E
cell
=E
cell
o
+
2
0.059
log
0.82
(1.36)
2
=E
cell
o
+0.010V
Thus, E
cell
increases by 0.010V.
Answer:
there you go
Explanation:
Correct answer is 0.010 V
Note that the given cell will not work as electrochemical cells since,
E
OP
Cu
o
>E
OP
Ag
o
The equation for electrochemical cells will be: (i)
Cu→Cu
2+
+2e
2Ag
+
+2e→2Ag
Thus, e.m.f. of cell Cu∣Cu
2+
∣∣Ag
+
∣ Ag will be :
E
cell
=E
OP
Cu
o
+E
RP
Ag
o
+
2
0.059
log
[Cu
2+
]
[Ag
+
]
2
∴[Ag
+
]=1M and [Cu
2+
]=1M
∴E
cell
=E
OP
Cu
o
+E
RP
Ag
o
(E
cell
o
=E
OP
Cu
o
+E
RP
Ag
o
)
E
cell
=E
cell
o
After the passage of 9.65 ampere for 1 hr, i.e., 9.65×60×60 coulomb change, during which the cell reactions are reversed, thus, Ag metal passes in solution state and Cu
2+
ions are discharged.
The reactions during this passage of current are:
2Ag→2Ag
+
+2e
Cu
2+
+2e→Cu
Thus, Ag
+
formed =
96500
9.65×60×60
=0.36eq.=0.36 mole
Cu
2+
discharged =
96500
9.65×60⋊60
=0.36eq.=0.18 mole
Thus, [Ag
+
]left=1+0.36=1.36 mole
[Cu
2+
] left=1−0.18=0.82 mole
Now, e.m.f. can be given as
E
cell
=E
cell
o
+
2
0.059
log
0.82
(1.36)
2
=E
cell
o
+0.010V
Thus, E
cell
increases by 0.010V