/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 Tarnished silver contains \(\mat... [FREE SOLUTION] | 91Ó°ÊÓ

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Tarnished silver contains \(\mathrm{Ag}_{2} \mathrm{~S}\). The tarnish can be removed by placing silverware in an aluminum pan containing an inert electrolyte solution, such as \(\mathrm{NaCl}\). Explain the electrochemical principle for this procedure. [The standard reduction potential for the half- cell reaction \(\mathrm{Ag}_{2} \mathrm{~S}(s)+2 e^{-} \rightarrow 2 \mathrm{Ag}(s)+\mathrm{S}^{2-}(a q)\) is \(-0.71 \mathrm{~V} .]\)

Short Answer

Expert verified
The cleaning of tarnished silverware in an aluminum pan with an electrolyte solution works on the principle of electrochemistry. The aluminum acts as a sacrificial anode, oxidizing to release electrons. These electrons reduce the tarnish (Ag2S) on the silverware back to silver metal, effectively cleaning it. The direction of electron flow is determined by the standard reduction potentials of the reactions involved.

Step by step solution

01

Identify the Redox Reactions

In the aluminium pan with a NaCl solution, two half-cell reactions are occurring. On the silverware (anode), a reduction reaction occurs: \[ \mathrm{Ag}_{2} \mathrm{~S}(s)+2 e^{-} \rightarrow 2 \mathrm{Ag}(s)+\mathrm{S}^{2-}(a q) \] Meanwhile, the aluminum acts as the cathode where an oxidation reaction takes place. This has to be determined first.
02

Determine the Half-cell Reaction for Aluminum

Knowing that aluminum is oxidized in this process, we can write the half-cell reaction for aluminum: \[ \mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(a q)+3 e^{-} \]
03

Use the Electrochemical Series

The electrochemical series puts the electrode reactions in order of their standard electrode potentials. The series shows that aluminum has a more negative standard electrode potential than silver and thus, is more likely to lose electrons (oxidize).
04

Explain the Cleaning Process

The electrolyte solution enables ions to travel. As a result, the Al is oxidized, allowing electrons to be released and move to the Ag2S on the silverware. Here, the electrons are gained by the Ag2S, reducing it back to silver metal and thus, cleaning the tarnish.
05

Importance of Standard Reduction Potential

The negative potential given (-0.71V for the reduction of Ag2S to Ag) indicates that the reaction is non-spontaneous and requires an external source of electrons. Here, the aluminium provides the necessary electrons to reduce Ag2S on the silverware.

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Key Concepts

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

Redox Reactions
Redox reactions are fundamental chemical reactions involving the transfer of electrons between two substances. They play a significant role in the process of cleaning tarnished silver. In our scenario, the tarnish on the silver, composed of silver sulfide (\( \mathrm{Ag}_{2} \mathrm{~S} \)), interacts in a redox reaction involving aluminum (Al) in the presence of an electrolyte solution.
To clarify, **redox** is a portmanteau of reduction and oxidation:
  • **Reduction** is the gain of electrons by a molecule, atom, or ion, and for \( \mathrm{Ag}_{2} \mathrm{~S} \), **silver ions gain electrons** to become elemental silver.
  • **Oxidation** is the loss of electrons, which happens to the aluminum as it interacts with the electrolyte solution, transforming from solid aluminum to Al ions.
In the cleaning process, the transfer of electrons facilitates the transformation of tarnished \( \mathrm{Ag}_{2} \mathrm{~S} \) back to shiny silver.
Standard Reduction Potential
Standard reduction potentials are essential in understanding electrochemical reactions. They measure the tendency of a chemical species to be reduced, expressed in volts (V). Each half-cell reaction has a specific reduction potential, which indicates how likely it is to gain electrons.
For example, the standard reduction potential provided is \(-0.71 \mathrm{~V}\) for the conversion of \( \mathrm{Ag}_{2} \mathrm{~S} \) to silver. This negative value suggests that the reaction is not spontaneous and requires external influence, like electrons from aluminum, to proceed.
  • The more positive a reduction potential, the greater the tendency of the species to gain electrons (reduce).
  • Conversely, the more negative the reduction potential, the less favorable the reduction is without outside assistance.
In oxidizing aluminum, the electrons are liberated and used to spontaneously reduce the silver sulfide tarnish because aluminum's reduction potential is more negative than silver's.
Electrochemical Series
The electrochemical series is a ranking of elements and their ions based on standard electrode potentials. This series is crucial for predicting the direction of electron flow in electrochemical reactions.
In our exercise, aluminum's more negative electrode potential compared to silver's indicates that aluminum is more willing to lose electrons. Placed above silver in the electrochemical series, aluminum acts as an effective source of electrons in the cleaning process.
The series can help us predict:
  • Which substances will more readily undergo oxidation or reduction
  • The flow direction of electrons in a redox process
Using the electrochemical series, the direction of the electronic transfer, from aluminum to \( \mathrm{Ag}_{2} \mathrm{~S} \), becomes apparent, explaining why aluminum must be used in cleaning tarnish from silver.
Electrolyte Solution
An electrolyte solution, such as \( \mathrm{NaCl} \), is vital to the function of the redox reaction in cleaning silverware. The electrolyte serves as a medium that allows ions to move freely, facilitating electron transfer between reactants.
Here’s how an electrolyte solution aids in the process:
  • As aluminum loses electrons (oxidizes), these electrons need a path to travel from aluminum to the silver sulfide tarnish. The electrolyte solution provides this path.
  • It ensures that ions and electrons are efficiently exchanged, speeding up the tarnish removal process.
Without an electrolyte, the electron transfer would be inefficient, and the cleaning reaction would not proceed effectively. Thus, the electrolyte acts as a conduit that is critical for completing the redox loop.

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Most popular questions from this chapter

Oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is present in many plants and vegetables. (a) Balance the following equation in acid solution: $$ \mathrm{MnO}_{4}^{-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_{2} $$ (b) If a 1.00 -g sample of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) requires \(24.0 \mathrm{~mL}\) of \(0.0100 M \mathrm{KMnO}_{4}\) solution to reach the equivalence point, what is the percent by mass of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in the sample?

The concentration of a hydrogen peroxide solution can be conveniently determined by titration against a standardized potassium permanganate solution in an acidic medium according to the following unbalanced equation: $$ \mathrm{MnO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{O}_{2}+\mathrm{Mn}^{2+} $$ (a) Balance the above equation. (b) If \(36.44 \mathrm{~mL}\) of a \(0.01652 \mathrm{M} \mathrm{KMnO}_{4}\) solution are required to completely oxidize \(25.00 \mathrm{~mL}\) of a \(\mathrm{H}_{2} \mathrm{O}_{2}\) solution, calculate the molarity of the \(\mathrm{H}_{2} \mathrm{O}_{2}\) solution.

Industrially, copper is purified by electrolysis. The impure copper acts as the anode, and the cathode is made of pure copper. The electrodes are immersed in a \(\mathrm{CuSO}_{4}\) solution. During electrolysis, copper at the anode enters the solution as \(\mathrm{Cu}^{2+}\) while \(\mathrm{Cu}^{2+}\) ions are reduced at the cathode. (a) Write half-cell reactions and the overall reaction for the electrolytic process. (b) Suppose the anode was contaminated with \(\mathrm{Zn}\) and \(\mathrm{Ag} .\) Explain what happens to these impurities during electrolysis. (c) How many hours will it take to obtain \(1.00 \mathrm{~kg}\) of \(\mathrm{Cu}\) at a current of \(18.9 \mathrm{~A} ?\)

An acidified solution was electrolyzed using copper electrodes. A constant current of 1.18 A caused the anode to lose \(0.584 \mathrm{~g}\) after \(1.52 \times 10^{3} \mathrm{~s}\). (a) What is the gas produced at the cathode and what is its volume at STP? (b) Given that the charge of an electron is \(1.6022 \times 10^{-19} \mathrm{C}\), calculate Avogadro's number. Assume that copper is oxidized to \(\mathrm{Cu}^{2+}\) ions.

The zinc-air battery shows much promise for electric cars because it is lightweight and rechargeable: The net transformation is \(\mathrm{Zn}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{ZnO}(s)\) (a) Write the half-reactions at the zinc-air electrodes and calculate the standard emf of the battery at \(25^{\circ} \mathrm{C}\). (b) Calculate the emf under actual operating conditions when the partial pressure of oxygen is 0.21 atm. (c) What is the energy density (measured as the energy in kilojoules that can be obtained from \(1 \mathrm{~kg}\) of the metal) of the zinc electrode? (d) If a current of \(2.1 \times 10^{5} \mathrm{~A}\) is to be drawn from a zinc-air battery system, what volume of air (in liters) would need to be supplied to the battery every second? Assume that the temperature is \(25^{\circ} \mathrm{C}\) and the partial pressure of oxygen is \(0.21 \mathrm{~atm} .\)

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