calculate the emf of the cell cd/(cdso4)0.01//(cuso4)0.5/cu at 25 degree celsious if E^0 cell is 0.38v
Answers
Answer:
Given Zn∣Zn
2+
(0.001M)∣∣Cu
2+
(0.1M)∣Cu
Overall cell reaction:
Zn⟶Zn
2+
+2e
−
Cu
2+
+2e
−
⟶Cu
Zn+Cu
2+
⟶Zn
2+
+Cu
E
cell
o
=standard reduction potential of cathode + standard oxidation potential of anode
E
cell
o
=0.34 to 0.76 V
E
cell
o
=1.1 V
K
C
=
[Cu
2+
]
[Zn
2+
]
=
10
−1
10
−3
=10
−2
EMF of the cell at any electrode concentration is:
E=E
o
−
n
0.059
log(K
C
)=1.1−
2
0.059
log(10
−2
)=1.1−
2
0.059
×(2)=1.1−0.059=1.041V
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SIMILAR QUESTIONS
star-struck
The standard reducation potentials of
Zn
2+
∣Zn,Cu
2+
∣Cu and Ag
+
∣Ag
are respectively -0.76, 0.34 and 0.8 V. The following cells were constructed.
a) Zn∣Zn
2+
∣∣Cu
2+
∣Cu
b) Zn∣Zn
2+
∣∣Ag
+
∣Ag
c) Cu∣Cu
2+
∣∣Ag
+
∣Ag
What is the correct order E
0
cell
of these cells?
Hard
View solution
>
The standard reduction potentials of Cu
2+
/Cu and Cu
2+
/Cu
+
are 0.337 and 0.153 volts resectively. The standard electrode potential for Cu
+