The black silver sulfide discoloration of silverware can be removed by heating the silver article in a sodium carbonate solution in an aluminum pan. The reaction is $$3 \mathrm{Ag}_{2} \mathrm{S}(s)+2 \mathrm{Al}(s) \rightleftharpoons 6 \mathrm{Ag}(s)+3 \mathrm{S}^{2-}(a q)+2 \mathrm{Al}^{3+}(a q)$$ a. Using data in Appendix \(4,\) calculate \(\Delta G^{\circ}, K,\) and \(\mathscr{E}^{\circ}\) for the above reaction at $25^{\circ} \mathrm{C} .\left[\text { For } \mathrm{Al}^{3+}(a q), \Delta G_{\mathrm{f}}^{\circ}=-480 . \mathrm{kJ} / \mathrm{mol.}\right]$ b. Calculate the value of the standard reduction potential for the following half-reaction: $$2 \mathrm{e}^{-}+\mathrm{Ag}_{2} \mathrm{S}(s) \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{S}^{2-}(a q)$$

Step by step solution

01

Find ΔG° for reaction

Using the data from Appendix 4, we find the Gibbs free energy of formation for each compound and follow the formula: ΔG°(reaction) = Σ ΔG°(products) - Σ ΔG°(reactants) For the given reaction, it looks like this: ΔG° = [6(ΔG°(Ag, s)) + 3(ΔG°(S²⁻, aq)) + 2(ΔG°(Al³⁺, aq))] - [3(ΔG°(Ag₂S, s)) + 2(ΔG°(Al, s))] The values of ΔG° from Appendix 4 are: ΔG°(Ag, s) = 0 kJ/mol (since it is in its standard state), ΔG°(S²⁻, aq) = -33 kJ/mol, ΔG°(Al³⁺, aq) = -480 kJ/mol (given), ΔG°(Ag₂S, s) = -32 kJ/mol, ΔG°(Al, s) = 0 kJ/mol (since it is in its standard state) Now plug in these values into the formula and compute the sum.

02

Calculate K

Using the calculated ΔG°, we can now calculate the equilibrium constant K using the following formula: K = e^(−ΔG°/(RT)) Where R is the ideal gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (25°C = 298K). Plug in the values and compute K.

03

Calculate ξ°

To find the standard potential (ξ°) for the given reaction, we use the following equation that relates ΔG°, ξ°, and n (the number of electrons transferred in the reaction): ΔG°=-nFE° To find the number of electrons, n, we multiply the coefficients of the half-reactions by the stoichiometric coefficients from the balanced redox equation. In this case, n = (3 × 2) = 6. Now, we can find ξ° using the equation, where F is the Faraday constant (96,485 C/mol): ξ° = −ΔG°/ (nF) Plug in the values and compute the standard potential.

04

Calculate the standard reduction potential for the half-reaction

We have found the overall standard potential for the reaction (ξ°). Now we need to find the reduction potential of: 2 e⁻ + Ag₂S(s) → 2 Ag(s) + S²⁻(aq) We can use the relationship: ξ°(overall reaction) = ξ°(half-reaction of interest) - ξ°(another half-reaction) In this case, the other half-reaction is: 2 e⁻ + 2 Al³⁺(aq) → 2 Al(s) Using the standard reduction potential for the reaction involving aluminum, which is -1.66 V, we can find the standard reduction potential for the half-reaction of interest using the equation: ξ°(Ag₂S half-reaction) = ξ°(overall reaction) + ξ°(Al half-reaction) Plug in the values and compute the standard reduction potential for the Ag₂S half-reaction.

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