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A voltaic cell utilizes the following reaction: $$ \mathrm{Al}(s)+3 \mathrm{Ag}^{+}(a q)-\infty \mathrm{Al}^{3+}(a q)+3 \mathrm{Ag}(s) $$ What is the effect on the cell emf of each of the following changes? (a) Water is added to the anode compartment, diluting the solution. (b) The size of the aluminum electrode is increased. (c) A solution of \(\mathrm{AgNO}_{3}\) is added to the cathode compartment, increasing the quantity of \(\mathrm{Ag}^{+}\) but not changing its concentration. (d) \(\mathrm{HCl}\) is added to the \(\mathrm{AgNO}_{3}\) solution, precipitating some of the \(\mathrm{Ag}^{+}\) as \(\mathrm{AgCl}\)

Short Answer

Expert verified
In summary, (a) adding water to the anode compartment will increase the cell emf; (b) increasing the size of the aluminum electrode has no impact on cell emf; (c) adding a solution of AgNO鈧 to the cathode compartment without changing its concentration has no effect on cell emf; and (d) adding HCl to the AgNO鈧 solution, consequently precipitating some Ag鈦 ions as AgCl, will decrease the cell emf.

Step by step solution

01

(a) Water added to the anode compartment

When water is added to the anode compartment, it will dilute the solution. Consequently, the concentration of the Al鲁鈦 ions will decrease. According to the Nernst Equation, if the concentration of Al鲁鈦 ions decreases, the cell emf will increase. Therefore, adding water to the anode compartment will increase the cell emf.
02

(b) Increase the size of the aluminum electrode

Increasing the size of the aluminum electrode does not affect the concentration of any species involved in the redox reaction. Therefore, there will be no impact on the cell emf as the Nernst Equation depends only on the concentrations of the reactants and products.
03

(c) Solution of AgNO鈧 added to the cathode compartment

Adding a solution of AgNO鈧 increases the quantity of Ag鈦 ions in the cathode compartment, but does not change its concentration. Since the Nernst Equation is affected by changes in concentration and not quantity, there will be no impact on the cell emf.
04

(d) HCl added to the AgNO鈧 solution

Adding HCl to the AgNO鈧 solution will cause the precipitation of some Ag鈦 ions as AgCl. This will lead to a decrease in the concentration of Ag鈦 ions. According to the Nernst Equation, as the concentration of Ag鈦 ions decreases, the cell emf will decrease. Thus, adding HCl to the AgNO鈧 solution will decrease the cell emf.

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Key Concepts

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

Cell EMF
The electromotive force (EMF) of a voltaic cell, often called cell emf, represents the cell's ability to do work. It is the voltage difference between the two electrodes and drives the flow of electrons through the circuit. In a voltaic cell, this is generated when a redox reaction occurs, where one element loses electrons while another gains them.
Several factors can influence the cell emf, such as the concentration of ions in the solution, temperature, and the nature of the electrodes. Variations in these factors can increase or decrease the emf, affecting the efficiency of the cell.
  • Increasing the concentration of reactants generally increases the emf.
  • Decreasing the concentration can lower the emf.
  • The intrinsic properties of the substances in the reaction also play a crucial role in determining the magnitude of emf.
Understanding how these factors interact allows us to control and optimize the performance of voltaic cells in practical applications.
Nernst Equation
The Nernst equation is a key tool in electrochemistry that relates the concentration of the ions involved in the redox reactions to the cell potential or cell emf. It is defined as:
\[E_{cell} = E^{ heta}_{cell} - \frac{RT}{nF} \ln Q\]
where:
  • \(E_{cell}\) is the emf of the cell under non-standard conditions.
  • \(E^{\theta}_{cell}\) is the standard cell potential.
  • \(R\) is the gas constant \( (8.314 \, J \cdot mol^{-1} \cdot K^{-1}) \).
  • \(T\) is the temperature in Kelvin.
  • \(n\) is the number of moles of electrons transferred in the reaction.
  • \(F\) is Faraday's constant \( (96485 \, C \cdot mol^{-1}) \).
  • \(Q\) is the reaction quotient, a measure of the concentrations of reactants and products.
This equation helps predict how changes in concentration and temperature affect the cell potential. For example, if the concentration of ions decreases, the emf typically increases to counterbalance the reduction in ion availability.
Redox Reaction
Redox reactions are the cornerstone of the functioning of a voltaic cell. They involve the transfer of electrons between two substances. The term 'redox' stands for reduction-oxidation. Oxidation is losing electrons, while reduction is gaining electrons. These reactions allow the cell to convert chemical energy into electrical energy.
In the given exercise, aluminum (\( \mathrm{Al} \)) is oxidized by losing electrons, while silver ions (\( \mathrm{Ag}^+ \)) are reduced by gaining electrons.
  • Aluminum serves as the anode where oxidation occurs: \( \mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(aq) + 3e^- \).
  • Silver acts as the cathode where reduction happens: \(\mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag}(s) \) .
These simultaneous reactions are interconnected, allowing current to flow through an external circuit, hence generating electrical energy.
Concentration
Concentration plays a fundamental role in determining the cell potential in a voltaic cell. According to the Nernst equation, a change in the concentration of ions directly impacts the cell emf.
  • Increasing the concentration of ions like \( \mathrm{Ag}^+ \) increases the voltage of the cell.
  • Conversely, decreasing the concentration, such as through dilution, usually reduces emf, although certain scenarios, like dilution at the anode, can increase emf.
Each change in concentration needs to be evaluated based on how it affects the reaction quotient (\(Q\)) in the Nernst equation since even minor modifications can lead to significant variations in the overall cell performance. Careful control of concentrations allows for precise tuning of the voltaic cell's potential output.
Precipitation
Precipitation is the process by which dissolved ions in a solution form a solid. In the context of voltaic cells, precipitation can significantly affect cell potential.
When chemical reactions cause soluble ions like \( \mathrm{Ag}^+ \) to combine with others to form a solid, such as \( \mathrm{AgCl} \), the concentration of the dissolved ions decreases. This directly impacts the Nernst equation as it reduces the concentration of reactive ions.
  • For instance, the addition of \( \mathrm{HCl} \) to an \( \mathrm{AgNO}_3 \) solution results in the precipitation of \( \mathrm{Ag}^+ \) as \( \mathrm{AgCl} \), decreasing its concentration and thus the cell emf.
Understanding this principle is critical for predicting changes in cell behavior and for designing cells with stable and predictable performance.

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Most popular questions from this chapter

(a) What is an electrolytic cell? (b) The negative terminal of a voltage source is connected to an electrode of an electrolytic cell. Is the electrode the anode or the cathode of the cell? Explain. (c) The electrolysis of water is often done with a small amount of sulfuric acid added to the water. What is the role of the sulfuric acid?

The standard reduction potentials of the following halfreactions are given in Appendix E: $$ \begin{gathered} \mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) \\ \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) \\ \mathrm{Ni}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ni}(s) \\ \mathrm{Cr}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Cr}(s) \end{gathered} $$ (a) Determine which combination of these half-cell reactions leads to the cell reaction with the largest positive cell emf, and calculate the value. (b) Determine which combination of these half-cell reactions leads to the cell reaction with the smallest positive cell emf, and calculate the value.

A voltaic cell similar to that shown in Figure \(20.5\) is constructed. One electrode compartment consists of an aluminum strip placed in a solution of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\), and the other has a nickel strip placed in a solution of \(\mathrm{NiSO}_{4}\). The overall cell reaction is $$ 2 \mathrm{Al}(s)+3 \mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Ni}(s) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two electrode compartments. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the aluminum electrode to the nickel electrode, or from the nickel to the aluminum? (f) In which directions do the cations and anions migrate through the solution? Assume the Al is not coated with its oxide.

The Haber process is the principal industrial route for converting nitrogen into ammonia: $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ (a) What is being oxidized, and what is being reduced? (b) Using the thermodynamic data in Appendix \(\mathrm{C}\), calculate the equilibrium constant for the process at room temperature. (c) Calculate the standard emf of the Haber process at room temperature.

A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: $$ \mathrm{AgCl}(s)+\mathrm{e}^{-\longrightarrow} \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) $$ The two cell compartments have \(\left[\mathrm{Cl}^{-}\right]=0.0150 \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=2.55 M\), respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether [Cl \(^{-}\) ] will increase, decrease, or stay the same as the cell operates.

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