Given the following electrochemical cell, calculate the potential for the cell

Question

Given the following electrochemical cell, calculate the potential for the cell in which the concentration of Ag⁺ is 0.0595 M, the pH of the H⁺ cell is 1.400, and the pressure for H₂ is held constant at 1 atm. The temperature is held constant at 55°C.

Standard reduction potentials:

2H⁺ +2e^-H₂  E^o= 0 V

Ag⁺ +1e^-Ag  E^o=0.800 V

V =

Answer 1

2Ag⁺ +2e- ==> 2Ag(s) .. reduction reaction (cathode) Eº = 0.800

H₂ ==> 2H⁺ + 2e- .. oxidation reaction (anode) Eº = 0

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2Ag⁺ + H₂ ==> 2H⁺ + Ag .. cell reaction

Eºcell = 0.800 V - 0 V = 0.800 V

Nernst equation:

Ecell = Eºcell - RT/nF ln Q

Ecell = ?

Eºcell = 0.800 V

R = gas constant = 8.314 J/Kmol

T = Temp in Kelvin = 55 + 273 = 328K

n = mols electrons = 2 mols e-

F = Faraday constant = 96500 C/mol e-

Q = reaction quotient = [H⁺]² / [Ag⁺]²

[H⁺] = 1x10^-1.400 = 0.0398 M

[Ag⁺] = 0.0595 M

Plug in the values:

Ecell = 0.800 - (8.314)(328) / (2)(96500) ln (0.0398)²/(0.0595)²

Solve for Ecell