Given the following standard reduction potentials,
Hg₂²⁺(aq)  +  2 e^–    2 Hg (l)                                                          E° = +0.789 V
 
Hg₂Cl₂(s)  +  2 e^–    2 Hg (l)  +  2 Cl^-(aq)                                        E° = +0.271 V
determine K_sp for Hg₂Cl₂(s) at 25 °C.

• A. 3.26×10¹⁷
• B. 3.26×10⁻⁸
• C. 7.52×10¹⁰
• D. 7.52×10⁻¹¹

Answer: (A)

Verified step by step guidance

1. Identify the two half-reactions given in the problem: the reduction of Hg2^2+ to Hg(l) and the reduction of Hg2Cl2(s) to Hg(l) and Cl^-. Note their standard reduction potentials: E° = +0.789 V and E° = +0.271 V, respectively.
2. Write the balanced chemical equation for the dissolution of Hg2Cl2(s) into its ions: Hg2Cl2(s) ⇌ Hg2^2+(aq) + 2 Cl^-(aq). This represents the solubility equilibrium for which we need to find the Ksp.
3. Use the Nernst equation to relate the standard reduction potentials to the equilibrium constant (K) for the overall reaction. The equation is: E°cell = (RT/nF) * ln(K), where E°cell is the difference in standard reduction potentials, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred, and F is Faraday's constant.
4. Calculate the E°cell by subtracting the standard reduction potential of the Hg2Cl2 reaction from that of the Hg2^2+ reaction: E°cell = E°(Hg2^2+/Hg) - E°(Hg2Cl2/Hg, Cl^-).
5. Rearrange the Nernst equation to solve for Ksp: Ksp = e^((nF * E°cell) / (RT)). Substitute the known values (R = 8.314 J/mol·K, T = 298 K, F = 96485 C/mol, and n = 2) to find the Ksp for Hg2Cl2(s).