The question is incomplete, here is the complete question:
What is the calculated value of the cell potential at 298 K for an electrochemical cell with the following reaction, when the H₂ pressure is 6.56 x 10⁻² atm, the H⁺ concentration is 1.39 M, and the Sn²⁺ concentration is 9.35 x 10⁻⁴ M?

<u>Answer:</u> The cell potential of the given electrochemical cell is 0.273 V
<u>Explanation:</u>
For the given chemical equation:

The half cell reactions for the given equation follows:
<u>Oxidation half reaction:</u> 
<u>Reduction half 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{[Sn^{2+}]\times p_{H_2}}{[H^+]^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%5Ctimes%20p_%7BH_2%7D%7D%7B%5BH%5E%2B%5D%5E2%7D)
where,
= electrode potential of the cell = ?
= standard electrode potential of the cell = +0.14 V
n = number of electrons exchanged = 2
![[H^{+}]=1.39M](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%3D1.39M)
![[Sn^{2+}]=9.35\times 10^{-4}M](https://tex.z-dn.net/?f=%5BSn%5E%7B2%2B%7D%5D%3D9.35%5Ctimes%2010%5E%7B-4%7DM)

Putting values in above equation, we get:

Hence, the cell potential of the given electrochemical cell is 0.273 V