91Ó°ÊÓ

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Based on the given reduction potentials, which of the following would lead to a reaction? (A) Placing some \(\operatorname{Cr}(s)\) in a solution containing \(\mathrm{Pb}^{2+}\) ions (B) Placing some \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) ions (C) Placing some \(\mathrm{Cr}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) ions (D) Placing some \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Pb}^{2+}\) ions

Step by step solution

01

Understand Reduction Potentials

Here we are given the standard reduction potentials of two half-cell reactions. A standard reduction potential indicates the tendency of a species to be reduced. A species with a higher (less negative) reduction potential will tend to gain electrons and be reduced, while the species with a lower (more negative) reduction potential will lose electrons and be oxidized.

02

Identify the Half-Cell Reactions

We have two half-cell reactions: \(\mathrm{Cr}^{3+} + 3e \rightarrow \mathrm{Cr}(s)\) with a reduction potential of -0.41 V and \(\mathrm{Pb}^{2+} + 2e \rightarrow \mathrm{Pb}(s)\) with a reduction potential of -0.12 V. Given the standard reduction potential values, \(\mathrm{Pb}^{2+}\) has a higher tendency to gain electrons than \(\mathrm{Cr}^{3+}\).

03

Analyze the Reaction Scenarios

Looking at the scenarios A to D, we need to find out where a reaction would occur. This means that the solid metal would dissolve into the solution and its corresponding cation would be reduced. Given the reduction potentials, the reaction can only occur if the solid metal has a lower reduction potential (more negative) than the cation in the solution.

04

Determine the Possible Reaction

We consider the proposed scenarios. In scenario (A), \(\mathrm{Cr}(s)\) (reduction potential -0.41V) is placed in a solution of \(\mathrm{Pb}^{2+}\) (reduction potential -0.12V). Since \(\mathrm{Cr}\) has a lower reduction potential than \(\mathrm{Pb}^{2+}\), this scenario is possible and a reaction would occur. Scenario (B) putting \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) won't cause a reaction because \(\mathrm{Pb}^{2+}\) has a higher reduction potential. Scenarios (C) and (D) do not make sense either because they involve placing a metal in a solution with its own ions, which cannot lead to a reaction.

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Key Concepts

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

Reduction Potential

The reduction potential is a key concept in understanding redox reactions. It represents the tendency of a chemical species to gain electrons, thus undergoing reduction. A higher reduction potential means a stronger ability to be reduced. Conversely, a lower (or more negative) reduction potential suggests a lower tendency for reduction and a higher tendency to undergo oxidation.

To decide how two species will react in a redox reaction, we compare their reduction potentials. The species with the higher reduction potential will gain electrons and be reduced, while the species with the lower reduction potential will donate electrons and be oxidized.

• Positive or less negative potential indicates a greater propensity for reduction.
• Negative or more negative potential indicates a lesser propensity for reduction.

This forms the basis for predicting the direction of electron flow in a voltaic cell.

Voltaic Cell

A voltaic cell is a device that uses chemical reactions to generate electrical energy. It consists of two separate half-cells connected through an external circuit and a salt bridge or porous disk which allows ion transfer to balance charges.

Each half-cell in a voltaic cell involves a redox reaction, where one is dedicated to a reduction event and the other to an oxidation event.

• The site of reduction is called the cathode, where a species gains electrons.
• The site of oxidation is the anode, where a species loses electrons.

The potential difference between the two electrodes drives an electric current through the external circuit, powering the connected load.

Half-Cell Reactions

Half-cell reactions are fundamental components of a redox reaction. In a half-cell reaction, either reduction or oxidation takes place, but not both. In a voltaic cell, there are always two half-cell reactions: one representing the reduction process and the other the oxidation process.

For example, in a chromic reaction:\[ \mathrm{Cr}^{3+} + 3e^- \rightarrow \mathrm{Cr}(s) \]This is a reduction half-reaction as chromium ions gain electrons. The potential for such a process is given by its reduction potential, in this case, -0.41 V.

• Reduction half-reactions involve gain of electrons.
• Oxidation half-reactions involve loss of electrons.

Understanding these reactions helps us predict the behavior of redox reactions in voltaic cells.

Oxidation and Reduction

Oxidation and reduction are two sides of a redox reaction and always occur simultaneously. Oxidation is the loss of electrons, while reduction is the gain of electrons. In any redox reaction, one species is oxidized and the other is reduced.

Redox reactions are essential for the functioning of voltaic cells, as they cause the flow of electrons through an external circuit. To simplify, think of OIL RIG: "Oxidation Is Loss, Reduction Is Gain."

• Oxidizing agents are substances that can oxidize other substances, and in the process, they get reduced.
• Reducing agents are substances that can reduce other substances, and in the process, they get oxidized.

Recognizing and identifying these processes and agents helps in predicting the feasibility and direction of redox reactions.