/*! 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 70 A voltaic cell is constructed wi... [FREE SOLUTION] | 91Ó°ÊÓ

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A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following halfreaction: $$\mathrm{AgCl}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q)$$ The two half-cells have \(\left[\mathrm{Cl}^{-}\right]=0.0150 \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=\) \(2.55 \mathrm{M},\) 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 \(\left[\mathrm{Cl}^{-}\right]\) will increase, decrease, or stay the same as the cell operates.

Short Answer

Expert verified
The cathode of the cell is the electrode with a higher concentration of \(Cl^-\) ions, which is electrode 2 with a \(2.55 M\) concentration. The standard emf is 0 volts. The cell emf for the given concentrations is approximately \(0.126 V\), and as the cell operates, the concentration of \(Cl^-\) ions will increase at the anode and decrease at the cathode.

Step by step solution

01

Identify the Cathode

In a voltaic cell, the electrode with the higher concentration of reducing agent ions (in this case Cl-) acts as the cathode, as it has a greater tendency to be reduced. Since the concentration of Cl- ions is higher in the electrode 2 with \(2.55M\), this electrode is the cathode of the cell.
02

Calculate the Standard EMF

For the silver-silver chloride electrode, the half reaction is: \[\mathrm{AgCl}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q)\] The standard emf (electromotive force) for this cell is 0 volts because both electrodes have the same half-reaction potentials.
03

Calculate Cell EMF for Given Concentrations

Using the Nernst equation, we can calculate the cell EMF for given concentrations: \[E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q\] Since the standard cell EMF is 0, and Q, the reaction quotient, can be calculated as \(Q = \frac{[Cl^-]_{cathode}}{[Cl^-]_{anode}}\), we get: \[E_{cell} = -\frac{RT}{nF}\ln \frac{[Cl^-]_{cathode}}{[Cl^-]_{anode}}\] We are given the concentrations in the problem, as well as that R is the ideal gas constant, T is the temperature, n is the number of moles of electrons transferred, and F is Faraday's constant. Assuming standard conditions, we have \(R = 8.314 J K^{-1} mol^{-1}\), \(T = 298K\), \(n = 1\), and \(F = 96485 C mol^{-1}\). Plugging these values into the equation, we get: \[E_{cell} = -\frac{8.314 * 298}{1 * 96485}\ln \frac{2.55}{0.0150} \approx 0.126 V\]
04

Predict Changes in Cl- Concentration

As the cell operates, there occurs a transfer of electrons from the anode to the cathode, which can be seen from the half-reaction, and thus a transfer of silver ions from the anode to the cathode. This means that the concentration of Cl- ions at the anode will increase, while at the cathode, it will decrease. In summary: - Anode: \([\mathrm{Cl}^{-}]\) will increase. - Cathode: \([\mathrm{Cl}^{-}]\) will decrease.

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Key Concepts

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

Electrochemical Cell
An electrochemical cell is a fundamental component in the realm of chemistry that enables the conversion of chemical energy into electrical energy through redox reactions. Essentially, this device harnesses the power of electrons moving between two electrodes: an anode and a cathode.

At the anode, oxidation occurs where electrons are lost, and at the cathode, reduction takes place with electrons being gained. These half-reactions are separated into different compartments called half-cells that are often linked by a salt bridge or porous membrane that allows ions to move freely, balancing the electric charge while preventing the two solutions from mixing completely.

In the voltaic cell example from the exercise, silver chloride electrodes are used to induce the half-reactions, and the movement of electrons from the anode to the cathode through an external circuit is what generates electricity.
Nernst Equation
The Nernst equation is an expression in physical chemistry that relates the reduction potential of a half-cell in an electrochemical cell to the standard electrode potential, temperature, activity, and concentrations of the chemical species undergoing reduction and oxidation. It is instrumental in determining the cell's electromotive force (EMF) under non-standard conditions.

The equation is given by: \begin{center} \(E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q\), \begin{center}where \(E_{cell}\) is the cell EMF, \(E^0_{cell}\) is the standard EMF, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred, \(F\) is Faraday's constant, and \(Q\) is the reaction quotient.

Applying the Nernst equation in our example, you can understand how the cell voltage changes with concentration differences of chloride ions, as is demonstrated when calculating the EMF for the given concentrations of chloride ions.
Electromotive Force (EMF)
Electromotive force, better known as EMF, represents the energy provided by a cell or a battery per coulomb of charge passing through it. It's the driving force that pushes electrons through the circuit, often measured in volts. The EMF is determined by the nature of the electrodes and the electrolyte.

Standard EMF \(E^0_{cell}\) references conditions with each solute at 1 molar concentration and gases at 1 atmosphere pressure. However, real-life scenarios often deviate from these ideal conditions, necessitating calculations like in the Nernst equation to find the actual EMF as substances change concentration during the cell's operation, such as the case in the exercise provided.
Half-Reaction Potential
Half-reaction potentials are the voltages associated with each half-reaction occurring at the electrodes during the operation of an electrochemical cell. These potentials, when measured under standard conditions, determine whether a substance can be readily oxidized or reduced.

Understanding the half-reaction potentials is crucial for predicting the behavior of each electrode, especially in identifying the cathode and the anode in a cell, as seen in the provided exercise. The half-reaction potentials help us understand that the electrode with the higher tendency to gain electrons (higher reduction potential) will serve as the cathode where reduction takes place.
Concentration Cell
A concentration cell is a form of electrochemical cell where both electrodes are made of the same material but are immersed in solutions of differing concentrations. The EMF of the cell is generated due to the concentration gradient between the two half-cells.

The operation and behavior of concentration cells can be explored using Nernst equation since the standard EMF \(E^0_{cell}\) would be zero given identical electrodes. In our exercise example, the voltaic cell created with silver-silver chloride electrodes in different chloride ion concentrations can be considered a concentration cell. As the cell operates, the concentration of chloride ions will equilibrate between the two half-cells, driving the potential to diminish over time.

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Most popular questions from this chapter

A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: (a) \(\mathrm{Ni}^{+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+\mathrm{Ni}(s) \quad\) (acidic solution) (b) \(\mathrm{MnO}_{4}^{2-}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)+\mathrm{MnO}_{2}(s)\) (acidic solution) (c) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{S}(s)+\mathrm{HSO}_{4}^{-}(a q) \quad\) (acidic solution) (d) \(\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)+\mathrm{ClO}^{-}(a q)\) (basic solution)

A \(1 M\) solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) is placed in a beaker with a strip of Cu metal. A \(1 M\) solution of \(\mathrm{SnSO}_{4}\) is placed in a second beaker with a strip of Sn metal. A salt bridge connects the two beakers, and wires to a voltmeter link the two metal electrodes. (a) Which electrode serves as the anode and which as the cathode? (b) Which electrode gains mass and which loses mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

(a) What is electrolysis? (b) Are electrolysis reactions thermodynamically spontaneous? Explain. (c) What process occurs at the anode in the electrolysis of molten \(\mathrm{NaCl} ?\) (d) Why is sodium metal not obtained when an aqueous solution of \(\mathrm{NaCl}\) undergoes electrolysis?

Complete and balance the following equations, and identify the oxidizing and reducing agents. (Recall that the \(\mathrm{O}\) atoms in hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) have an atypical oxidation state.) (a) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow\) $$ \mathrm{Cr}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q) \text { (acidic solution) } $$ (b) \(\mathrm{S}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{N}_{2} \mathrm{O}(g)\) (acidic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow\) $$ \mathrm{HCO}_{2} \mathrm{H}(a q)+\mathrm{Cr}^{3+}(a q) \text { (acidic solution) } $$ (d) \(\mathrm{BrO}_{3}^{-}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{N}_{2}(g)\) (acidic solution) (e) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \longrightarrow \mathrm{NH}_{4}^{+}(a q)+\mathrm{AlO}_{2}^{-}(a q)\) (basic solution) (f) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{ClO}_{2}(a q) \longrightarrow \mathrm{ClO}_{2}^{-}(a q)+\mathrm{O}_{2}(g)\) (basic solution)

Some years ago a unique proposal was made to raise the Titanic. The plan involved placing pontoons within the ship using a surface-controlled submarine-type vessel. The pontoons would contain cathodes and would be filled with hydrogen gas formed by the electrolysis of water. It has been estimated that it would require about \(7 \times 10^{8} \mathrm{~mol}\) of \(\mathrm{H}_{2}\) to provide the buoyancy to lift the ship (J. Chem. Educ, \(1973,\) Vol. 50,61 ). (a) How many coulombs of electrical charge would be required? (b) What is the minimum voltage required to generate \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) if the pressure on the gases at the depth of the wreckage ( \(2 \mathrm{mi}\) ) is \(300 \mathrm{~atm} ?\) (c) What is the minimum electrical energy required to raise the Titanic by electrolysis? (d) What is the minimum cost of the electrical energy required to generate the necessary \(\mathrm{H}_{2}\) if the electricity costs 85 cents per kilowatt-hour to generate at the site?

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