Question

Low concentrations of hydrogen sulfide in air react with silver to form Ag₂S, more familiar to us as tarnish. Silver polish contains alumium metal powder as an alkaline suspension.

Calculate E for the reaction in part A. (Hint: Derive E values for the half-reactions in which Ag₂S is reduced to Ag metal and Al(OH)₃ is reduced to Al metal. You can do this by combining the standard potentials for the reduction of Ag+ and Al3+ in Appendix 6 with the Ksp expressions and values of Ag₂S (Ksp= 1.6E-49) and Al(OH)₃ (Ksp= 1.9E-33).

Answer 1

To calculate the cell potential (E) for the tarnish reaction and cleaning, we must determine the reduction potentials for the reduction of Ag2S to Ag metal and Al(OH)3

Step-by-step explanation:

To determine the cell potential (E) for the reaction involving the tarnishing of silver and its cleaning process with an aluminum-based silver polish, we utilize the Nernst equation and the known standard reduction potentials.

Tarnish formation is depicted by the reaction 2Ag + H₂S → Ag₂S + H₂.

To calculate E, we need the half-reactions for the reduction of Ag₂S to Ag and the reduction of Al(OH)₃ to Al metal.

The standard reduction potential for Ag+ + e- → Ag is +0.80 V. However, we are dealing with Ag₂S, so we need to incorporate the solubility product (Ksp = 1.6E-49 for Ag₂S).

Using the relation E₀ = -RT/nF ln(Ksp), we can calculate the reduction potential for the process involving Ag₂S being reduced to Ag metal.

Similarly, for the Al(OH)₃, we combine the standard reduction potential for Al³+ + 3e- → Al, which is -1.66 V, with the Ksp of Al(OH)₃ (1.9E-33) following a similar approach.

After calculating individual E₀ values considering Ksp, we use them to calculate the overall cell potential (E) for the full reaction. This process incorporates electrochemistry principles and requires careful calculations to ensure accuracy.