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The value of \(E_{\text {cell for the reaction below is } 0.500 \mathrm{V} . \text { What is }}\) the value of \(\Delta G_{\text {cell }}^{\text {? }}\) $$\mathrm{Mn}^{3+}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Mn}^{2+}+\mathrm{MnO}_{2}+4 \mathrm{H}^{+}$$

Short Answer

Expert verified
Answer: The Gibbs free energy change (螖G_cell) for the given reaction is -96,485 J/mol.

Step by step solution

01

Identify the half-reactions

The given reaction can be broken down into two half-reactions: 1. Oxidation half-reaction (Mn虏鈦 being oxidized to Mn鲁鈦): Mn鲁鈦 鈫 Mn虏鈦 + e鈦 2. Reduction half-reaction (MnO鈧 being reduced to Mn虏鈦): MnO鈧 + 4H鈦 + 2e鈦 鈫 Mn虏鈦 + 2H鈧侽
02

Determine the number of moles of electrons transferred (n)

The balanced equation for the given reaction is: Mn鲁鈦 + 2H鈧侽 鈫 Mn虏鈦 + MnO鈧 + 4H鈦 To balance the charges, we will need to transfer 2 moles of electrons (since the reduction half-reaction involves a two-electron transfer). So, n = 2.
03

Calculate 螖G_cell using the equation

Now that we have the values for E_cell (0.500 V) and n (2), we can plug them into the equation: 螖G_cell = -nFE_cell 螖G_cell = -(2)(96,485 C/mol)(0.500 V) 螖G_cell = -96,485 J/mol (remember that 1 V = 1 J/C) The Gibbs free energy change (螖G_cell) for the given reaction is -96,485 J/mol.

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Key Concepts

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

Cell Potential and Its Significance
Cell potential, often represented as \(E_{\text{cell}}\), is a measure of the potential difference between two electrodes of an electrochemical cell. It serves as a measure of how much work a cell can do, reflecting the ability of the cell to drive an electric current through an external circuit.
The cell potential is expressed in volts (V) and is vital in determining the likelihood of a chemical reaction to occur spontaneously. A positive cell potential indicates that the reaction is thermodynamically favorable. In the given exercise, the \(E_{\text{cell}}\) is 0.500 V, signifying a reaction where products are more stable than reactants under standard conditions.
  • This potential is affected by factors such as temperature, pressure, and concentration of solutions.
  • It helps in calculating the Gibbs free energy, \(\Delta G\), which predicts the spontaneity of reactions.
Understanding cell potential also assists in studying battery functions where chemicals convert to electrical energy.
Understanding Half-Reactions
In electrochemistry, reactions are often split into two parts called half-reactions. These help in analyzing the process of electron transfer, highlighting each substance's role in the overall reaction.
For the equation given in the exercise, the breakdown results in:
  • Oxidation half-reaction: where a substance loses electrons. In this case, \(\text{Mn}^{3+}\) changes to \(\text{Mn}^{2+}\).
  • Reduction half-reaction: where a substance gains electrons, shown by \(\text{MnO}_2 + 4\text{H}^{+} + 2e^- \rightarrow \text{Mn}^{2+} + 2\text{H}_2\text{O}\).
The separation into half-reactions allows for a detailed view of oxidation and reduction processes, clarifying the transfer of electrons and balancing of charges. This is essential for calculating aspects such as \(n\), the number of moles of electrons transferred, as seen in the solution, where \(n = 2\).
The Role of Electron Transfer
Electron transfer is a core concept in understanding electrochemical reactions, where electrons shift from one reactant to another, changing their oxidation states.
This transfer is evident in the half-reactions where:
  • Electrons are transferred from the MnO鈧 molecule to the Mn鲁鈦 ions.
  • The net effect results in a balanced flow of electrons, maintaining charge neutrality.
Such electron movements are depicted clearly from the balanced overall equation in the exercise where 2 electrons are transferred, correlating directly to the calculation of Gibbs free energy change using \(n\). This concept not only highlights the technical aspects of reactions but underscores fundamental properties of matter that drive circuit flows in batteries and other applications.
Oxidation and Reduction Reactions
Within any redox reaction, oxidation and reduction must occur simultaneously to conserve charge, adhering to the principle that electrons lost in oxidation are precisely gained in reduction.
This framework is pivotal in understanding the given reaction, where:
  • Oxidation: \(\text{Mn}^{3+}\) is oxidized to \(\text{Mn}^{2+}\), illustrating the loss of an electron by the Mn ion.
  • Reduction: MnO鈧 is reduced to \(\text{Mn}^{2+}\), showing the gain of electrons, creating a reduced element.
Mastering these processes is essential because they are foundational to diverse applications such as energy storage and chemical synthesis and comprehend the calculation of Gibbs free energy change from cell potential, demonstrating the interplay of energy and electron flow in electrochemical systems.

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Most popular questions from this chapter

A voltaic cell is based on the reaction between \(\mathrm{Cu}^{2+}(a q)\) and \(\mathrm{Ni}(s),\) producing \(\mathrm{Cu}(s)\) and \(\mathrm{Ni}^{2+}(a q)\) a. Write the anode and cathode half-reactions. b. Write a balanced cell reaction. c. Draw the cell diagram.

The positive terminal of a voltaic cell is the cathode. However, the cathode of an electrolytic cell is connected to the negative terminal of a power supply. Explain this difference in polarity.

A voltaic cell with a basic aqueous background electrolyte is based on the oxidation of \(\mathrm{Cd}(s)\) to \(\mathrm{Cd}(\mathrm{OH})_{2}(s)\) and the reduction of \(\mathrm{MnO}_{4}^{-}(a q)\) to \(\mathrm{MnO}_{2}(s)\) a. Write half-reactions for the cell's anode and cathode. b. Write a balanced net ionic equation describing the cell reaction. c. Draw the cell diagram.

Oxygen Supply in Submarines Nuclear submarines can stay under water nearly indefinitely because they can produce their own oxygen by the electrolysis of water. a. How many liters of \(\mathrm{O}_{2}\) at \(25^{\circ} \mathrm{C}\) and 1.00 bar are produced in 1 hour in an electrolytic cell operating at a current of \(0.025 \mathrm{A} ?\) b. Could seawater be used as the source of oxygen in this electrolysis? Explain why or why not.

Super Iron Batteries In \(1999,\) scientists in Israel developed a battery based on the following cell reaction with iron(VI), nicknamed "super iron": \(\beth \mathrm{K}_{2} \mathrm{FeO}_{4}(a q)+3 \mathrm{Zn}(s) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{ZnO}(s)+2 \mathrm{K}_{2} \mathrm{ZnO}_{2}(a q)\) a. Determine the number of electrons transferred in the cell reaction. b. What are the oxidation states of the transition metals in the reaction? c. Draw the cell diagram.

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