<u>Answer:</u> The potential of the given cell is 1.551 V
<u>Explanation:</u>
The given chemical cell follows:

<u>Oxidation half reaction:</u> 
<u>Reduction half reaction:</u>
( × 2)
<u>Net cell reaction:</u> 
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the
of the reaction, we use the equation:

Putting values in above equation, we get:

To calculate the EMF of the cell, we use the Nernst equation, which is:
![E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Ag^{+}]^2}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.059%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BZn%5E%7B2%2B%7D%5D%7D%7B%5BAg%5E%7B%2B%7D%5D%5E2%7D)
where,
= electrode potential of the cell = ? V
= standard electrode potential of the cell = +1.561 V
n = number of electrons exchanged = 2
![[Zn^{2+}]=0.125M](https://tex.z-dn.net/?f=%5BZn%5E%7B2%2B%7D%5D%3D0.125M)
![[Ag^{+}]=0.240M](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D%3D0.240M)
Putting values in above equation, we get:


Hence, the potential of the given cell is 1.551 V