Ag⁺(aq)+ e^- → Ag(s)E° = +0.800 V AgBr(s)+ e^- → Ag(s)+ Br^-(aq)E° = +0.071 V Br₂(l)+ 2 e^- → 2 Br^-(aq)E° = +1.066 V Use some of the data above to calculate K_sp at 25°C for AgBr.

Question

Quizplus · Accepted Answer

To calculate the solubility product constant (Ksp) for AgBr at 25°C, we use the Nernst equation and the given standard electrode potentials (E°). The overall reaction for the dissolution of AgBr in water can be obtained by combining the half-reactions for AgBr → Ag⁺ + Br⁻ and Br₂ + 2e⁻ → 2Br⁻. The net cell potential (E°cell) for the dissolution of AgBr is the difference between the reduction potential for the formation of silver from AgBr (E° = +0.071 V) and the reduction potential for bromine to bromide (E° = +1.066 V). Thus, E°cell = 0.071 V - (-1.066 V) = 1.137 V. The relationship between E°cell, Ksp, and the number of electrons transferred (n) in the balanced equation is given by the Nernst equation: E°cell = (RT/nF)ln(Ksp), where R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred, and F is the Faraday constant. Solving for Ksp gives a value that corresponds to the choice provided.

Ag+(aq)+ E- → Ag(s)E° = +0

Ag⁺(aq)+ E^- → Ag(s)E° = +0