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. Which of the following occurs at the cathode? (A) \(\mathrm{Cr}^{3+}\) is reduced to \(\mathrm{Cr}(\mathrm{s})\) (B) \(\mathrm{Pb}^{2+}\) is reduced to \(\mathrm{Pb}(\mathrm{s})\) (C) \(\mathrm{Cr}(s)\) is oxidized to \(\mathrm{Cr}^{3+}\) (D) \(\quad \mathrm{Pb}(s)\) is oxidized to \(\mathrm{Pb}^{2+}\)

Step by step solution

01

Identify the given reduction half-reactions

Two reduction half-reactions are given in this problem: \(Cr^{3+}+3e^- \rightarrow Cr(s)\) with \(E^° = -0.41V\) and \(Pb^{2+} + 2e^- \rightarrow Pb(s)\) with \(E^° = -0.12V\). These half-reactions are written in the reduction form, which means electrons are on the left side.

02

Determine the half-reaction that is more likely to occur

In electrochemical cells, the cathode is where the reduction occurs. In this case, the reduction process with the higher standard potential (more positive value) will occur. \nPb^2+ to Pb has higher standard potential (\(E^°\)) than \( Cr^3+ \) to Cr. Thus, the reaction where \(Pb^{2+}\) is reduced to \(Pb(s)\) will occur at the cathode.

03

Select the correct answer

From the options given, only (B) \(\mathrm{Pb}^{2+}\) is reduced to \(\mathrm{Pb}(\mathrm{s})\) accurately describes what happens at the cathode. Thus, (B) is the correct answer.

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Key Concepts

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

Electrochemical Cells

An electrochemical cell consists of two half-cells connected by a salt bridge or porous partition. The primary purpose of an electrochemical cell is to convert chemical energy into electrical energy through redox (reduction-oxidation) reactions.
Each half-cell contains an electrode and an electrolyte solution where either oxidation or reduction reactions occur.
Voltaic or galvanic cells are specific types of electrochemical cells where these spontaneous reactions generate a current.

• The anode is where oxidation occurs, releasing electrons.
• The cathode is where reduction happens, gaining electrons.
• A connection wire allows the flow of electrons from the anode to the cathode.

Understanding these components helps visualize how electrons transfer during these processes.

Cathode Reactions

In an electrochemical cell, the cathode is the electrode where reduction occurs. Since reduction involves gaining electrons, substances at the cathode accept these electrons during the reaction.
The potential of each cathode reaction determines which process is favorable in a cell. For example, in our given voltaic cell, the reaction with lead ions ( Pb^{2+} ) is more favorable to occur at the cathode than chromium ions ( Cr^{3+} ).
Cathode reactions can be identified by examining their standard reduction potentials. The process with a more positive potential is typically the one that occurs at the cathode in a voltaic cell.

Reduction Potential

The reduction potential, denoted as \(E^°\), measures the tendency of a chemical species to gain electrons and become reduced. It is a crucial concept in understanding electrochemical cells because it allows us to predict which reactions will be spontaneous.

• Reduction potentials are listed in volts (V).
• A more positive \(E^°\) value indicates a greater tendency for reduction.
• In a voltaic cell, the half-reaction with the higher reduction potential will generally occur at the cathode.

In our example, the lead reduction half-reaction has a more positive \(E^°\) than the chromium reduction half-reaction, determining that lead is reduced at the cathode.

Half-Cell Reactions

A half-cell reaction represents either the oxidation or reduction process occurring in each half of an electrochemical cell. These reactions can either lose or gain electrons.
In the case of our voltaic cell, we have two specific half-cell reactions:

• "\(Cr^{3+} + 3e^- \rightarrow Cr(s)\)" with \(E^° = -0.41V\).
• "\(Pb^{2+} + 2e^- \rightarrow Pb(s)\)" with \(E^° = -0.12V\).

Both reactions are written in the form of reduction half-reactions, which show the gain of electrons. Understanding half-cell reactions is important because they allow us to analyze and predict the behavior and outcome of each half of the cell.