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A voltaic cell similar to that shown in Figure 20.5 is constructed. One half- cell consists of an iron strip placed in a solution of \(\mathrm{FeSO}_{4},\) and the other has an aluminum strip placed in a solution of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} .\) The overall cell reaction is $$ 2 \mathrm{Al}(s)+3 \mathrm{Fe}^{2+}(a q) \longrightarrow 3 \mathrm{Fe}(s)+2 \mathrm{Al}^{3+}(a q) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the aluminum electrode to the iron electrode or from the iron to the aluminum? (f) In which directions do the cations and anions migrate through the solution? Assume the \(\mathrm{Al}\) is not coated with its oxide.

Step by step solution

01

Identify Oxidation and Reduction

To determine what is being oxidized and reduced, we need to look at the change in oxidation states. In the overall reaction \(2 \text{Al}(s) + 3 \text{Fe}^{2+}(aq) \rightarrow 3 \text{Fe}(s) + 2 \text{Al}^{3+}(aq)\), aluminum (Al) goes from 0 to +3 oxidation state, meaning it loses electrons and is oxidized. Iron (Fe) goes from +2 in the solution to 0 in the solid form, meaning it gains electrons and is reduced.

02

Write Half-Reactions

The oxidation half-reaction for aluminum is \(\text{Al}(s) \rightarrow \text{Al}^{3+}(aq) + 3e^-\). The reduction half-reaction for iron is \(\text{Fe}^{2+}(aq) + 2e^- \rightarrow \text{Fe}(s)\). These reactions show the loss and gain of electrons for aluminum and iron, respectively.

03

Identify Anode and Cathode

In electrochemical cells, oxidation takes place at the anode and reduction takes place at the cathode. Since aluminum is oxidized, the aluminum electrode is the anode. Since iron is reduced, the iron electrode is the cathode.

04

Assign Electrode Signs

In a voltaic cell, the anode is the negative electrode because electrons are produced there. The cathode is the positive electrode because it consumes electrons. Therefore, the aluminum electrode is negative, and the iron electrode is positive.

05

Determine Electron Flow

Electrons flow from the anode to the cathode in a voltaic cell. Thus, electrons flow from the aluminum electrode to the iron electrode.

06

Ion Migration Directions

Cations migrate toward the cathode where reduction occurs, and anions migrate toward the anode where oxidation occurs. Therefore, \(\text{Al}^{3+}\) cations move toward the cathode (iron side), and any \(\text{SO}_4^{2-}\) anions move toward the anode (aluminum side).

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Key Concepts

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

Understanding Oxidation and Reduction in a Voltaic Cell

In a voltaic cell, chemical reactions involve a transfer of electrons. These reactions can be identified as either oxidation or reduction. Oxidation occurs when a species loses electrons. For instance, in the reaction involving a voltaic cell with aluminum (Al) and iron (Fe), aluminum starts with an oxidation state of 0 and ends up with an oxidation state of +3. This tells us that aluminum is oxidized because it loses electrons.
On the flip side, reduction is when a species gains electrons. Iron goes from an oxidation state of +2 in the aqueous phase to 0 as a solid, meaning it gains electrons during the process. Thus, iron is reduced. You can remember this easily with the mnemonic: **LEO the lion says GER**鈥�**L**ose **E**lectrons in **O**xidation, **G**ain **E**lectrons in **R**eduction.

Half-Reactions in Electrochemical Processes

Breaking down the overall cell reaction into half-reactions helps us understand the individual processes of oxidization and reduction. A half-reaction that shows oxidation might be represented as: - Aluminum: \[ \text{Al}(s) \rightarrow \text{Al}^{3+}(aq) + 3e^- \] This equation indicates aluminum losing three electrons to form aluminum ions.Conversely, the half-reaction for reduction is:- Iron: \[ \text{Fe}^{2+}(aq) + 2e^- \rightarrow \text{Fe}(s) \] This informs us that iron ions in the solution gain two electrons to form solid iron.These half-reactions illustrate how electrons are transferred from one material to another, underpinning the function of the voltaic cell.

Determining Anode and Cathode

Electrochemical cells are divided into two main parts: the anode and the cathode. These two components play distinct roles. The anode is where oxidation happens. In our aluminum-iron voltaic cell, since aluminum is oxidized, the aluminum electrode is the anode. On the other hand, the cathode is the site of reduction. Here, the iron's journey from iron ions in solution to solid iron, due to gaining electrons, makes the iron electrode the cathode. Remember: - **Anode**: Site of oxidation - **Cathode**: Site of reduction

Electrode Signs in Voltaic Cells

In a voltaic cell, each electrode has a designated sign, either positive or negative, based on the reaction occurring at that point. - The anode is the site where electrons are generated, hence it's marked as the negative electrode. Thus, in the aluminum-iron cell, the aluminum, being the anode, is negative. - At the cathode, electrons are consumed during the reduction process, so it is the positive electrode. Accordingly, the iron electrode is positive as it is the cathode. These signs help define the direction and flow of electrons, which is essential for the proper functioning of the cell.

The Path of Electron Flow

Understanding electron flow in a voltaic cell requires recognizing that it moves from the anode to the cathode. In our aluminum-iron cell setup, electrons originate at the aluminum electrode, which serves as the anode. These electrons travel through the external circuit and reach the iron electrode, the cathode, where they facilitate the reduction of iron ions. This flow of electrons is integral for the cell's operation in generating electricity.

Ion Migration in Solutions

Within a voltaic cell, ions in the solution must move to maintain balance and allow the redox reactions to continue. - Cations, or positive ions like \(\text{Al}^{3+}\), migrate towards the cathode. This movement is driven by the cathode's positive charge where reduction occurs. - Anions, such as \(\text{SO}_4^{2-}\), move towards the anode to balance out the charge created by the electrons leaving the anode. This movement ensures that the voltaic cell continues operating smoothly, allowing for constant reaction progress.