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 = ?
![[Zn^{2+}]](https://tex.z-dn.net/?f=%5BZn%5E%7B2%2B%7D%5D)
= 3.5 M
![[Pb^{2+}]](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D)
= 
Now put all the given values in the above equation, we get:
Therefore, the cell potential for this reaction is 0.50 V
Answer:
5.15 moles
Explanation:
2zn + o2 = 2zno
5.15 2.57 5.15 moles
nzno=500/(16x2+65)= 5.15 moles
-> nzn = 5.15 x 2 ÷ 2 = 5.15 moles
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Answer:
The question is not complete, the complete question should be "Lipids vesicles are formed containing pure water. If these vesicles are transferred to a solution that contains a rather high concentration of solutes, the solution outside the vesicle is said to be Hypertonic. True or False"
The answer is True
Explanation:
This is because it contains greater concentration of solutes on the outside of the cell than the increase.
In other words hypertonic solutions have more concentrate of solutions on the outside than the inside.
