Question 22:
"Zn/Zn++ || Cu++/Cu"cell, during its working produces which kind of energy
Answers
Explanation:
Given Zn∣Zn 2+
(0.001M)∣∣Cu 2+
(0.1M)∣Cu
Overall cell reaction:
Zn⟶Zn2+ +2e −
Cu 2+ +2e − ⟶Cu
Zn+Cu 2+ ⟶Zn 2+ +Cu
E cell0
=standard reduction potential of cathode + standard oxidation potential of anode
Ecell 0 =0.34 to 0.76 V
E ocell =1.1 V
KC = [Cu 2+ ][Zn 2+ ] = 10 −110 −3
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 cello
=standard reduction potential of cathode + standard oxidation potential of anode
E ocell =0.34 to 0.76 V
E cello
=1.1 VK C = [Cu 2+ ][Zn 2+ ]
= 10 −1 10 −3 =10 −2
EMF of the cell at any electrode concentration is:
E=E o − n0.059
log(K C )=1.1− 2 0.059
log(10 −2 )=1.1− 0.059 ×(2)=1.1−0.059=1.041V
=10 −2
EMF of the cell at any electrode concentration is:
E=E
o − n0.059
log(K C )=1.1− 2 0.059
log(10 −2 )=1.1− 20.059 ×(2)=1.1−0.059=1.041V
Answer:
zn+zn=cu for energy working process