The cell potential of the following cell is determined using an unspecified concentration of acid.What is the pH of the acid solution given that the measured cell potential is -0.508 V and the anode reduction potential (E°) is 0.222 V at 25°C?

Ag(s) | AgCl(s) | Cl−(1.0 M) || H⁺(aq,? M) | H₂(g,1.0 atm) | Pt(s)

Question

Quizplus · Accepted Answer

The Nernst equation is:E = E° - (RT/nF)lnQwhere:E is the cell potentialE° is the standard cell potentialR is the gas constant (8.314 J/mol·K)T is the temperature in Kelvinn is the number of moles of electrons transferredF is the Faraday constant (96,485 C/mol)Q is the reaction quotientIn this case, n = 1 and Q = [H+]/[Cl-]. Substituting the given values into the Nernst equation, we get:-0.508 V = 0.222 V - (0.0257 V/K)(298 K)ln([H+]/1.0 M)Solving for [H+], we get:[H+] = 1.0 x 10^-4.83 MTherefore, the pH of the acid solution is 4.83.

The Cell Potential of the Following Cell Is Determined Using