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To remove the tarnish \(\left(\mathrm{Ag}_{2} \mathrm{~S}\right)\) on a silver spoon, a student carried out the following steps. First, she placed the spoon in a large pan filled with water so the spoon was totally immersed. Next, she added a few tablespoonfuls of baking soda (sodium bicarbonate), which readily dissolved. Finally, she placed some aluminum foil at the bottom of the pan in contact with the spoon and then heated the solution to about \(80^{\circ} \mathrm{C}\). After a few minutes, the spoon was removed and rinsed with cold water. The tarnish was gone, and the spoon regained its original shiny appearance. (a) Describe with equations the electrochemical basis for the procedure. (b) Adding \(\mathrm{NaCl}\) instead of \(\mathrm{NaHCO}_{3}\) would also work because both compounds are strong electrolytes. What is the added advantage of using \(\mathrm{NaHCO}_{3}\) ?

Short Answer

Expert verified
(a) Ag₂S is reduced to Ag by Al; (b) NaHCO₃ neutralizes acids and is gentler than NaCl.

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01

Understanding the Reaction Involved

The tarnish on silver, Ag₂S, interacts with aluminum foil. In this process, aluminum, which is more reactive, donates electrons to the silver ions, reducing them back to metallic silver, while itself becoming oxidized. The reaction is:\[\text{3 Ag}_2\text{S} + 2 \text{Al} \rightarrow 6 \text{Ag} + \text{Al}_2\text{S}_3\]
02

Role of Sodium Bicarbonate

Sodium bicarbonate (\(\text{NaHCO}_3\)) acts as a source of carbonate ions and creates a basic environment in water, which helps to facilitate the reaction rate by increasing ion mobility. It doesn't directly participate in the redox reaction but aids in the completion of the tarnish removal.
03

Role of Electrochemical Reaction

Aluminum acts as the anode (where oxidation occurs), and silver acts as the cathode (where reduction occurs) in this electrolytic cell created in the solution, enabling the transfer of electrons required for the reaction.
04

Advantage of Using NaHCO3

Adding NaHCO₃ not only acts as a weak electrolyte that helps in conducting electricity, but it also neutralizes any mild acids that might have formed during the dissolution of tarnish, offering a gentler cleaning method compared to using NaCl directly.
05

Conclusion

Using aluminum foil and NaHCO₃ creates a redox environment where Al reduces Ag₂S back to Ag, removing tarnish. NaHCO₃ aids the reaction by maintaining a stable and slightly basic solution.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Redox reactions
Redox reactions, short for reduction-oxidation reactions, are essential for many chemical processes. In these reactions, there is a transfer of electrons between substances.
One substance loses electrons (oxidation), while another gains electrons (reduction).
In the tarnish removal process, a redox reaction occurs where aluminum oxidizes by donating electrons, and silver ions in the tarnish are reduced back to metallic silver by gaining these electrons.
  • This reaction is represented by the equation: \[3 \text{Ag}_2\text{S} + 2 \text{Al} \rightarrow 6 \text{Ag} + \text{Al}_2\text{S}_3\]
  • This equation shows aluminum donating electrons (oxidation) and silver sulfide receiving electrons to form silver and aluminum sulfide.
The redox reaction restores the shiny appearance of silver by reversing the tarnishing process.
Strong electrolytes
Strong electrolytes are substances that completely dissociate into ions when dissolved in water. This dissociation allows them to effectively conduct electricity in solution.
In the given scenario, both sodium bicarbonate (\(\text{NaHCO}_3\)) and sodium chloride (\(\text{NaCl}\)) function as strong electrolytes.
Because they produce ions that enhance electrical conductivity, they assist in facilitating the electrochemical reaction needed to remove tarnish.
  • Sodium bicarbonate dissociates into sodium ions (\(\text{Na}^+\)) and bicarbonate ions (\(\text{HCO}_3^−\)).
  • As strong electrolytes, these compounds promote the mobility of ions in the solution, ensuring efficient electron transfer between the aluminum and silver.
This assists in the reaction's progression, making the tarnish removal process more effective.
Tarnish removal
The removal of tarnish involves a chemical reaction that reverses the layer of silver sulfide (\(\text{Ag}_2\text{S}\)) forming on the silver object's surface.
The tarnish removal process uses a setup where the tarnished silver becomes part of an electrochemical reaction with a more reactive metal, aluminum, present in the solution.
This reaction reverts the silver ions back to their metallic form, effectively removing the tarnish.
  • This is achieved by using aluminum foil as part of the reaction, which acts to sacrifice itself (oxidize) to help restore the silver.
  • This method is effective because it directly targets the chemical transformation responsible for tarnishing.
The tarnish is reduced to harmless products and removed without damaging the silver, returning it to its original luster.
Sodium bicarbonate
Sodium bicarbonate, commonly known as baking soda, plays a vital role in the tarnish removal process. While it is not directly involved in the redox reaction, it helps create an ideal environment.
When dissolved in water, sodium bicarbonate increases the pH level, creating a basic solution.
This basic environment enhances the ion mobility necessary for effective electron transfer.
  • Additionally, its buffering capacity prevents the solution from becoming too acidic, which could slow down or inhibit the tarnish removal reaction.
  • Sodium bicarbonate also helps neutralize any mild acids produced during the process, offering a gentle solution compared to harsher alternatives like sodium chloride.
By stabilizing the solution, sodium bicarbonate ensures that the silver retains its integrity throughout the tarnish removal process.
Anode and cathode in electrochemistry
Understanding the roles of anode and cathode in electrochemistry is crucial for grasping electrochemical reactions. Anodes and cathodes represent different electrodes in an electrochemical cell.
The anode is where oxidation occurs, meaning electrons are lost.
In contrast, the cathode is where reduction occurs, with electrons being gained.
  • In the tarnish removal process, aluminum acts as the anode as it donates electrons and gets oxidized.
  • The silver acts as the cathode, as it receives electrons and the tarnish is reduced back to the metal.
The electron flow from aluminum to silver is facilitated by the electrolytic solution, enabling the effective removal of tarnish from the silver.

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Most popular questions from this chapter

When \(25.0 \mathrm{~mL}\) of a solution containing both \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) ions is titrated with \(23.0 \mathrm{~mL}\) of \(0.0200 \mathrm{M} \mathrm{KMnO}_{4}\) (in dilute sulfuric acid), all the \(\mathrm{Fe}^{2+}\) ions are oxidized to \(\mathrm{Fe}^{3+}\) ions. Next, the solution is treated with Zn metal to convert all the \(\mathrm{Fe}^{3+}\) ions to \(\mathrm{Fe}^{2+}\) ions. Finally, \(40.0 \mathrm{~mL}\) of the same \(\mathrm{KMnO}_{4}\) solution is added to the solution to oxidize the \(\mathrm{Fe}^{2+}\) ions to \(\mathrm{Fe}^{3+}\). Calculate the molar concentrations of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) in the original solution.

What is the emf of a cell consisting of a \(\mathrm{Pb}^{2+} / \mathrm{Pb}\) half-cell and a \(\mathrm{Pt} / \mathrm{H}^{+} / \mathrm{H}_{2}\) half-cell if \(\left[\mathrm{Pb}^{2+}\right]=0.10 \mathrm{M},\) \([\mathrm{H}]=0.050 M,\) and \(P_{\mathrm{H}}=1.0 \mathrm{~atm} ?\)

When an aqueous solution containing gold(III) salt is electrolyzed, metallic gold is deposited at the cathode and oxygen gas is generated at the anode. (a) If \(9.26 \mathrm{~g}\) of Au is deposited at the cathode, calculate the volume (in liters) of \(\mathrm{O}_{2}\) generated at \(23^{\circ} \mathrm{C}\) and \(747 \mathrm{mmHg}\). (b) What is the current used if the electrolytic process took \(2.00 \mathrm{~h} ?\)

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Given that: $$ \begin{array}{ll} 2 \mathrm{Hg}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Hg}_{2}^{2+}(a q) & E^{\circ}=0.92 \mathrm{~V} \\\ \mathrm{Hg}_{2}^{2+}(a q)+2 e^{-} \longrightarrow 2 \mathrm{Hg}(l) & E^{\circ}=0.85 \mathrm{~V} \end{array} $$ calculate \(\Delta G^{\circ}\) and \(K\) for the following process at \(25^{\circ} \mathrm{C}:\) $$\mathrm{Hg}_{2}^{2+}(a q) \longrightarrow \mathrm{Hg}^{2+}(a q)+\mathrm{Hg}(l)$$ (The preceding reaction is an example of a disproportionation reaction in which an element in one oxidation state is both oxidized and reduced.)

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