What is the calculated value of the cell potential at 298K for an
electrochemical cell with the following reaction, when the H2
pressure is 6.56x10- atm, the H concentration is 1.39M, and
the Sn2+ concentration is 9.35x10-4M?
2H+ (aq) + Sn(s)—_H2(g) + Sn2+(aq)

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?

The cell potential of the given electrochemical cell is 0.273 V

Explanation:

For the given chemical equation:

The half cell reactions for the given equation follows:

Oxidation half reaction:

Reduction half reaction:

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:

where,

= electrode potential of the cell = ?

= standard electrode potential of the cell = +0.14 V

n = number of electrons exchanged = 2

Putting values in above equation, we get:

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

iron and sleel

make siroeel

fe 55.85 g/mol and h2o 18.02 g/mol.