<u>Answer:</u> The cell voltage of the given chemical reaction is 2.74 V
<u>Explanation:</u>
For the given chemical reaction:

Here, gold is getting reduced because it is gaining electrons and nickel is getting oxidized because it is loosing electrons.
<u>Oxidation half reaction:</u> 
<u>Reduction half reaction:</u> 
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the
of the reaction, we use the equation:


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{[Sn^{2+}]}{[Ba^{2+}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.059%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BSn%5E%7B2%2B%7D%5D%7D%7B%5BBa%5E%7B2%2B%7D%5D%7D)
where,
= electrode potential of the cell = ?V
= standard electrode potential of the cell = +2.76 V
n = number of electrons exchanged = 2
![[Ba^{2+}]=5.15M](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D%3D5.15M)
![[Sn^{2+}]=1.59M](https://tex.z-dn.net/?f=%5BSn%5E%7B2%2B%7D%5D%3D1.59M)
Putting values in above equation, we get:

Hence, the cell voltage of the given chemical reaction is 2.74 V