The following cell has a potential of 0.45 V at 25°C. Pt(s) ∣ H₂(1 atm) |H⁺(? M) ∣∣ Cl^-(1 M) ∣ Hg₂Cl₂(s) |Hg(l)
The standard half-cell potential for the half-reaction Hg₂Cl₂(s) + 2 e^- → 2 Hg(l) + 2 Cl^-(aq) is 0.28 V.What is the pH in the anode compartment?

Question

Quizplus · Accepted Answer

The Nernst equation can be used to calculate the cell potential at non-standard conditions: Ecell = E°cell - (RT/nF)lnQ where E°cell 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 in the reaction, F is Faraday's constant, and Q is the reaction quotient. In this case, the reaction is: Hg₂Cl₂(s) + 2 e^- → 2 Hg(l) + 2 Cl^-(aq) The reaction quotient can be expressed as: Q = [Hg2+][Cl-]^2 / [HgS][Cl-] = [Hg2+][Cl-] / [HgS] Since the solid HgS does not appear in the reaction quotient expression, its activity is taken to be 1. Therefore, the reaction quotient simplifies to: Q = [Hg2+][Cl-]^2 At equilibrium, the cell potential is equal to the standard cell potential. Therefore: 0.45 V = 0.28 V - (0.0257 V/K)(log([Hg2+][Cl-]^2) - log(1)) Simplifying: log([Hg2+][Cl-]^2) = 4.08 [Hg2+][Cl-]^2 = 10^4.08 [Hg2+] = sqrt(10^4.08 / [Cl-]^2) = 704.4 M Now we need to find the pH of the anode compartment, which is where the HgS and Hg2+ are located. We can use the following half-reaction to find the relationship between [Hg2+] and pH: 2 Hg2+ + 2 e^- + 2 H2O⁺ → Hg22+(aq) + 4 HSO4-^- + 4 H+ The standard reduction potential for this half-reaction is 1.23 V, which is sufficiently far from the potential of the cell in order to assume that this reaction is at equilibrium. Therefore, the Nernst equation can be used to find the relationship between [Hg2+] and pH: E° = 1.23 V E = E° - (RT/nF)ln[Hg2+][H2O]^2 / [HSO4-]^4 0.45 V = 1.23 V - (0.0257 V/K)(ln[Hg2+] + 2ln[H2O] - 4ln[HSO4-]) ln[Hg2+] = 28.53 - 44.50pH [Hg2+] = e^(28.53 - 44.50pH) Substituting the value for [Hg2+] obtained earlier: 704.4 M = e^(28.53 - 44.50pH) ln 704.4 = 28.53 - 44.50pH pH = (28.53 - ln 704.4) / 44.50 = 2.9

The Following Cell Has a Potential of 0