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Consider the galvanic cell based on the following halfreactions: $$ \begin{array}{ll} \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ}=1.50 \mathrm{V} \\ \mathrm{Tl}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl} & \mathscr{E}^{\circ}=-0.34 \mathrm{V} \end{array} $$ a. Determine the overall cell reaction and calculate \(\mathscr{C}_{\text {cell. }}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\) c. Calculate \(\mathscr{E}_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} \mathrm{M}\) and \(\left[\mathrm{Tl}^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)

Short Answer

Expert verified
The overall cell reaction is Au鲁鈦 + 3Tl鈦 鈫 Au + 3Tl, with a cell potential (E掳cell) of 1.84 V. The standard Gibbs free energy change (螖G掳) for this reaction is -532,724 J/mol, and the equilibrium constant (K) at 25掳C is 6.28 x 10^47. At the given concentrations of Au鲁鈦 and Tl鈦, the cell potential (E_cell) at 25掳C is 1.79 V.

Step by step solution

01

Determine the overall cell reaction

First, we should balance the electrons in both half reactions. Since there is a difference in the number of electrons in both reactions (3e鈦 in the first reaction and 1e鈦 in the second), we should multiply the second reaction by 3 to balance the number of electrons. Then, we will add both reactions to obtain the overall cell reaction: Au鲁鈦 + 3e鈦 鈫 Au (E掳 = 1.50 V) 3(Tl鈦 + e鈦 鈫 Tl) (E掳 = -0.34 V) Now, add the reactions: Au鲁鈦 + 3Tl鈦 + 3e鈦 鈫 Au + 3Tl
02

Calculate the cell potential

The cell potential (E掳cell) is the difference between both half-reactions potentials. Since the Au鲁鈦/Au half-reaction has a higher potential value, it will act as the cathode, while the other half-reaction will act as the anode. Thus: E掳cell = E掳cathode - E掳anode E掳cell = (1.50 V) - (-0.34 V) E掳cell = 1.84 V
03

Calculate the standard Gibbs free energy change

Now, we can calculate the standard Gibbs free energy change (螖G掳) using the formula: 螖G掳 = -nFE掳cell Where n = number of moles of electrons transferred (in this case, 3), F = Faraday's constant (96,485 C/mol), and E掳cell is the cell potential obtained in Step 2. 螖G掳 = -(3)(96485 C/mol)(1.84 V) 螖G掳 = -532,724 J/mol
04

Calculate the equilibrium constant

We can calculate the equilibrium constant (K) for the cell reaction at 25掳C using the standard Gibbs free energy change and the equation: 螖G掳 = -RT ln K Where R = gas constant (8.314 J/mol K), T = temperature in Kelvin (25掳C = 298.15 K), and 螖G掳 is the standard Gibbs free energy change calculated in Step 3. Rearranging the equation: ln K = -(螖G掳)/(RT) K = exp((532,724 J/mol)/((8.314 J/mol K)(298.15 K))) K = 6.28 x 10^47
05

Calculate the cell potential under given concentrations

We will now calculate the cell potential (E_cell) at 25掳C, taking into account the given concentrations of both Au鲁鈦 ([Au鲁鈦篯 = 1.0 x 10鈦宦 M) and Tl鈦 ([Tl鈦篯 = 1.0 x 10鈦烩伌 M) using the Nernst equation: E_cell = E掳cell - (RT/nF) ln Q Where Q is the reaction quotient. For the balanced cell reaction, Q = [Au鲁鈦篯/[Tl鈦篯鲁. Now we can plug in the values: E_cell = 1.84 V - ((8.314 J/mol K)(298.15 K))/((3)(96485 C/mol)) ln (1.0 x 10鈦宦)/(1.0 x 10鈦烩伌)鲁 E_cell = 1.84 V - 0.0496 V E_cell = 1.79 V So, under the given concentration conditions, the cell potential at 25掳C will be 1.79 V.

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Key Concepts

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

Electrochemical Cells
Electrochemical cells are devices that convert chemical energy into electrical energy. They function via redox reactions, where one substance is oxidized and another is reduced. In simpler terms, electrons are transferred between substances, generating electricity. There are two main types of electrochemical cells: galvanic (or voltaic) cells and electrolytic cells. The key difference is that galvanic cells generate electricity through spontaneous reactions, while electrolytic cells require an external power source to drive non-spontaneous reactions.
  • **Galvanic Cells**: Utilize spontaneous reactions to produce electric current. The reaction takes place in two separate compartments, known as half-cells. Each half-cell contains an electrode and an electrolyte solution.
  • **Cell Structure**: A typical galvanic cell consists of an anode, where oxidation occurs, and a cathode, where reduction happens. These are connected through a conductive wire, allowing electrons to flow between them, generating electricity.
  • **Salt Bridge**: Often, a salt bridge or porous membrane is used to maintain electrical neutrality by allowing the flow of ions between the half-cells.
Through these components, electrochemical cells play a critical role in batteries and energy storage.
Nernst Equation
The Nernst equation allows for the calculation of cell potentials under non-standard conditions. This is essential because real-life applications rarely operate at standard state conditions, which involve 1 M concentrations and 1 atm pressure.
The equation is given by:
\[E_{ ext{cell}} = E^{ ext{0}}_{ ext{cell}} - \frac{RT}{nF} \ln Q\]
where:
  • \( E_{\text{cell}} \) is the actual cell potential.
  • \( E^{\text{0}}_{\text{cell}} \) is the standard cell potential.
  • \( R \) is the ideal gas constant (8.314 J/mol K).
  • \( T \) is temperature in Kelvin.
  • \( n \) is the number of moles of electrons transferred.
  • \( F \) is Faraday鈥檚 constant (96485 C/mol).
  • \( Q \) is the reaction quotient, calculated similarly to the equilibrium constant but with non-equilibrium concentrations.
The Nernst equation helps us understand how concentration affects cell potential, making it invaluable for predicting battery performance in various conditions.
Gibbs Free Energy
Gibbs Free Energy (\( \Delta G \)) offers critical insight into the feasibility of a reaction. It's a measure of the maximum reversible work a thermodynamic system can do at constant temperature and pressure. For galvanic cells, \( \Delta G \) determines the spontaneity.
  • **Negative \( \Delta G \)**: Indicates a spontaneous reaction, allowing the galvanic cell to generate electrical energy.
  • **Positive \( \Delta G \)**: Implies a non-spontaneous reaction, as seen in electrolytic cells requiring energy input.
  • **Zero \( \Delta G \)**: The system is at equilibrium and can no longer perform work.
For electrochemical cells, the relationship between Gibbs Free Energy and the cell potential is given by the equation:
\[\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}\]
Where \( n \) is the number of moles of electrons exchanged in the reaction and \( F \) is Faraday鈥檚 constant. Thus, a more negative \( \Delta G^{\circ} \) corresponds with a higher \( E^{\circ}_{\text{cell}} \), emphasizing the cell鈥檚 ability to do work.
Equilibrium Constant
The equilibrium constant (\( K \)) for a reaction provides insight into the concentrations of products and reactants at equilibrium. For electrochemical reactions, it connects to cell potentials and Gibbs Free Energy.
  • For reactions that favor products at equilibrium, \( K \) is large.
  • For reactions that favor reactants, \( K \) is small.
The equilibrium constant is linked to Gibbs Free Energy by the equation:
\[\Delta G^{\circ} = -RT \ln K\]
Where \( R \) is the gas constant and \( T \) is the temperature in Kelvin. At equilibrium, \( \Delta G \) is zero, and thus, the reaction's position can be assessed with \( K \). For galvanic cells, a large \( K \) indicates a higher tendency to proceed with the reaction, confirming a strong energy output. Understanding this relation helps in designing cells with optimized efficiencies.

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Most popular questions from this chapter

What is electrochemistry? What are redox reactions? Explain the difference between a galvanic and an electrolytic cell.

Electrolysis of an alkaline earth metal chloride using a current of 5.00 A for 748 s deposits 0.471 g of metal at the cathode. What is the identity of the alkaline earth metal chloride?

When magnesium metal is added to a beaker of \(\mathrm{HCl}(a q),\) a gas is produced. Knowing that magnesium is oxidized and that hydrogen is reduced, write the balanced equation for the reaction. How many electrons are transferred in the balanced equation? What quantity of useful work can be obtained when Mg is added directly to the beaker of HCl? How can you harness this reaction to do useful work?

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 \mathrm{M}\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times\) \(10^{-3} M ?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is \(1.62 \mathrm{V}\). Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume Al is oxidized.)

Which of the following statement(s) is/are true? a. Copper metal can be oxidized by \(\mathrm{Ag}^{+}\) (at standard conditions). b. In a galvanic cell the oxidizing agent in the cell reaction is present at the anode. c. In a cell using the half reactions \(A l^{3+}+3 e^{-} \longrightarrow A l\) and \(\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg},\) aluminum functions as the anode. d. In a concentration cell electrons always flow from the compartment with the lower ion concentration to the compartment with the higher ion concentration. e. In a galvanic cell the negative ions in the salt bridge flow in the same direction as the electrons.

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