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

a) Oxidation and Reduction

In order to determine what is being oxidized and what is being reduced, we need to look at the overall cell reaction and determine what species have their oxidation states changed. In the given reaction: \[2 \mathrm{Al}(s)+3 \mathrm{Fe}^{2+}(a q) \longrightarrow 3 \mathrm{Fe}(s)+2 \mathrm{Al}^{3+}(a q)\] We can see that aluminum (Al) goes from the elemental state (oxidation state of 0) to \(\mathrm{Al}^{3+}\) (oxidation state of +3). This means that aluminum is being oxidized. On the other hand, iron (Fe) goes from \(\mathrm{Fe}^{2+}\) (oxidation state +2) to the elemental state (oxidation state 0), so it's being reduced.

02

b) Half-Reactions

For the half-reactions at each half-cell, we can see that the following processes are taking place: Oxidation half-reaction (Aluminum): \[2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Al}^{3+}(a q) + 6\mathrm{e}^{-}\] Reduction half-reaction (Iron): \[3 \mathrm{Fe}^{2+}(a q) + 6\mathrm{e}^{-} \longrightarrow 3 \mathrm{Fe}(s)\]

03

c) Anode and Cathode

The anode is the electrode where oxidation occurs. In this case, since aluminum is being oxidized, the anode is the aluminum electrode. The cathode is the electrode where reduction occurs. In this case, as iron is being reduced, the iron electrode acts as the cathode.

04

d) Electrode signs

In a voltaic cell, the anode has a negative (-) potential, and the cathode has a positive (+) potential. Since the aluminum electrode is the anode, it has a negative sign, and the iron electrode, which is the cathode, has a positive sign.

05

e) Electron Flow

In a voltaic cell, the electrons flow from the anode to the cathode through the external circuit. Therefore, electrons flow from the aluminum electrode (anode) to the iron electrode (cathode).

06

f) Cation and Anion Migration

In a voltaic cell, the cations migrate towards the cathode, and the anions migrate towards the anode to balance the charges in the half-cells. In this case, \(\mathrm{Fe}^{2+}\) ions migrate towards the iron cathode, where reduction occurs, and \(\mathrm{Al}^{3+}\) ions migrate away from the aluminum anode, where oxidation occurs. The \(\mathrm{SO}_{4}^{2-}\) anions will also migrate towards the aluminum anode to balance the charges in the half-cells.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!