<h2>
Answer:</h2><h3>22226.026 g</h3><h2>
Explanation:</h2>
To get an approximate result, multiply the mass value by 454.
<em>hope</em><em> </em><em>this</em><em> </em><em>help</em><em>!</em>
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
Equal. Good luck with other questions!