Answer : The cell potential for this reaction is 0.50 V
Explanation :
The given cell reactions is:

The half-cell reactions are:
Oxidation half reaction (anode): 
Reduction half reaction (cathode): 
First we have to calculate the cell potential for this reaction.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B2.303RT%7D%7BnF%7D%5Clog%20%5Cfrac%7B%5BZn%5E%7B2%2B%7D%5D%7D%7B%5BPb%5E%7B2%2B%7D%5D%7D)
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = 
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = +0.63 V
= cell potential for the reaction = ?
= 3.5 M
= 
Now put all the given values in the above equation, we get:


Therefore, the cell potential for this reaction is 0.50 V
The changes in the energy law of conservation of energy is Potential energy is converted to kinetic energy. Kinetic energy is converted into potential energy.
<h3>What is the law of conservation of energy?</h3>
Law of conservation of energy says that energy can neither be created nor destroyed, it just transformed from one form to another.
The energies are kinetic, potential, mechanical, gravitational, electrical, etc.
Thus, the changes in the energy law of conservation of energy is Potential energy is converted to kinetic energy. Kinetic energy is converted into potential energy.
Learn more about law of conservation of energy
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