Hi there!I am extremely confused. I cannot seem to figure this out. Can you please me help with this question? I do not understand how to solve this.Thank you so much!Discuss the criteria for a spontaneity of a reaction based on the standard Gibbs free energy (ΔGo and the standard cell potential (Eocell). The “tarnish” on silver is silver sulfide, Ag2S. The tarnish can be removed by placing silverware in an aluminum pan containing an inert electrolyte solution, such as NaCl. The standard reduction potential for the half-cell reaction for tarnish is.Ag2S(s) + 2e --> 2Ag(s) + S2-(aq) Eocell = -0.71V1.) What is the other half of the cell created by placing the tarnished silver in the aluminum pan? Write the overall reaction occurring that removes the tarnish. Calculate the standard cell potential for the overall reaction from (a). Can we tell electricity has passed when the tarnished silver item is in the aluminum pan? What is happening to the tarnish? Where is it going?

For this question, we have to discuss: 1. The criteria for spontaneity based on the standard Gibbs free energy (ΔGo) and the standard cell potential (Eocell) 2. Overall reaction to remove tarnish of silverware in an aluminum panWhen ΔGo is negative, then the reaction is said to be spontaneous. The relationship between Gibbs free energy (ΔGo) and the standard cell potential (Eocell) is given by: ∆G° = -nFE°cell So, when Eocell is positive, ΔGo is negative and the cell reaction is spontaneous.Given reaction: Ag2S(s) + 2e- → 2Ag(s) + S2-(aq) E° = -0.71V The other half-reaction: Al3+ (aq) + 3e- → Al(s) E° = -1.66 V 'Al' is a better reducing agent than silver, hence it provides electrons and reduces Ag+ (tarnish) to Ag (silver). Oxidation at anode: Al(s) →Al3+ (aq) + 3e- ]×2Reduction at cathode: Ag2S(s) + 2e- → 2Ag(s) + S2-(aq) ] ×3Overall reaction: 3 Ag2S(s) + 2Al →3S2-(aq) + 2Al3+ (aq) E°cell = E°red, cathode - E°red, anodeE°cell = -0.71 V - (-1.66 V)E°cell = 0.95 V This reaction will be spontaneous and move in the forward direction. Because the electrons are flowing from the aluminum pan to the tarnish, we can say the electricity has passed when the tarnished silver item is in the aluminum pan. Here, the inert solution of NaCl acts as a salt bridge. A complete electrochemical cell is established here. The tarnish, which is the deposit on silver is getting removed and the silver metal is getting cleared up. After the completion of the reaction, sulfide from the tarnish gets dissolved in the solution in the pan as sulfide ions and what we obtain is metallic silver cleared of the tarnish.

Answered: Discuss the criteria for a spontaneity…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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Hi there!

I am extremely confused. I cannot seem to figure this out. Can you please me help with this question? I do not understand how to solve this.

Thank you so much!

Discuss the criteria for a spontaneity of a reaction based on the standard Gibbs free energy (ΔG^o and the standard cell potential (E^o_cell). The “tarnish” on silver is silver sulfide, Ag₂S. The tarnish can be removed by placing silverware in an aluminum pan containing an inert electrolyte solution, such as NaCl. The standard reduction potential for the half-cell reaction for tarnish is.

Ag₂S(s) + 2e --> 2Ag(s) + S^2-(aq) E^o_cell = -0.71V

1.) What is the other half of the cell created by placing the tarnished silver in the aluminum pan? Write the overall reaction occurring that removes the tarnish. Calculate the standard cell potential for the overall reaction from (a). Can we tell electricity has passed when the tarnished silver item is in the aluminum pan? What is happening to the tarnish? Where is it going?

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