Question

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: $$2 \mathrm{e}^{-}+\mathrm{Ag}_{2} \mathrm{S}(s) \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{S}^{2-}(a q)$$

Short answer

For the given reaction, the calculated values are: - ΔG° = -410 kJ/mol, - K = 2.5 × 10¹², - ξ° = 0.71 V, - Standard reduction potential for the half-reaction \(2 e⁻ + Ag₂S(s) \rightarrow 2 Ag(s) + S²⁻(aq)\) = 0.25 V.

Step-by-step solution

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.

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.

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.

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.

Key concepts

Gibbs Free Energy

Gibbs free energy (ΔG°) is a vital concept in electrochemistry that helps predict whether a reaction will occur spontaneously at constant temperature and pressure. In our silver cleaner reaction, finding ΔG° tells us if the process of removing tarnish using aluminum is favorable.

To calculate ΔG°, we use the formula: \[ΔG° = \Sigma ΔG°(\text{products}) - \Sigma ΔG°(\text{reactants})\] This involves subtracting the Gibbs free energy of the reactants from that of the products, using values found in reference materials. A negative ΔG° indicates a spontaneous reaction, which in this case means the silver tarnish can be removed effectively using aluminum.

Equilibrium Constant

The equilibrium constant (K) is closely related to ΔG° and provides insight into the position of equilibrium for a reaction. After calculating ΔG°, we can determine K with the relationship:

\[K = e^{-ΔG°/(RT)}\]

Here, R is the ideal gas constant (8.314 \, J/mol\cdot K), and T is the temperature in Kelvin.

• A large value of K indicates that products are favored at equilibrium, meaning the tarnishing reaction proceeds well.
• A small K suggests reactants are favored.

Understanding K helps chemists and students see how conditions like temperature affect the reaction's balance.

Redox Reactions

Redox reactions involve the transfer of electrons between chemical species. Our exercise features a redox reaction where aluminum metal reduces silver sulfide to elemental silver.

In redox reactions:

• Oxidation refers to the loss of electrons.
• Reduction refers to the gain of electrons.

For the silver restoration process:

• Aluminum loses electrons (is oxidized).
• Silver sulfide gains electrons (is reduced).

Tracking these electron movements is essential for understanding the electrochemical process, which ultimately allows silverware to regain its shine.

Standard Reduction Potential

The standard reduction potential (E°) measures the propensity of a chemical species to gain electrons and be reduced. It's a crucial factor in redox reactions, indicating how readily a substance can be reduced compared to the hydrogen standard.

For the silver sulfide reaction, E° is calculated using the equation:\[ΔG° = -nFE°\]where F is Faraday's constant (96,485 \, C/mol) and \(n\) is the number of moles of electrons transferred.To determine E° for the half-reaction of silver sulfide, we compare it to another known redox pair, like aluminum, to see which is more likely to occur.
Understanding E° helps predict the feasibility and direction of redox reactions, vital for practical applications like cleaning silver.