The cell potential of the following electrochemical cell is determined using an unspecified concentration of acid.What is the pH of the acid solution given that the measured cell potential is -0.558 V and the anode reduction potential (E^$\circ$is 0.222 V at 25^$\circ$C?
Ag(s) | AgCl(s) | Cl^-(aq,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 potential, E° is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred, F is the Faraday constant, and Q is the reaction quotient.In this case, the reaction quotient is:Q = [H+]/[Cl-]Substituting the given values into the Nernst equation, we get:-0.558 V = 0.222 V - (0.0257 V/K * 298 K / 1 mol * 96485 C/mol)ln([H+]/1.0 M)Solving for [H+], we get:[H+] = 2.0 * 10^-6 MThe pH of the solution is:pH = -log[H+] = -log(2.0 * 10^-6 M) = 5.68

The Cell Potential of the Following Electrochemical Cell Is Determined ∘\circ∘

The Cell Potential of the Following Electrochemical Cell Is Determined $\circ$