Question

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.

Short answer

In this voltaic cell, aluminum is being oxidized, and iron is being reduced. The half-reactions are: Oxidation: \(2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Al}^{3+}(a q) + 6\mathrm{e}^{-}\) Reduction: \(3 \mathrm{Fe}^{2+}(a q) + 6\mathrm{e}^{-} \longrightarrow 3 \mathrm{Fe}(s)\) The anode is the aluminum electrode (negative sign), and the cathode is the iron electrode (positive sign). Electrons flow from the aluminum anode to the iron cathode. The \(\mathrm{Fe}^{2+}\) ions migrate towards the cathode, while the \(\mathrm{Al}^{3+}\) ions and \(\mathrm{SO}_{4}^{2-}\) anions migrate towards the anode.

Step-by-step solution

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.

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): \[3 \mathrm{Fe}^{2+}(a q) + 6\mathrm{e}^{-} \longrightarrow 3 \mathrm{Fe}(s)\]

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.

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.

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).

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.

Key concepts

Oxidation-Reduction

An oxidation-reduction, or redox, reaction involves the transfer of electrons between two species. In a redox reaction, one species loses electrons and is oxidized, while the other gains electrons and is reduced.
In the voltaic cell discussed, aluminum (Al) undergoes oxidation. This means Al loses electrons, transforming from a neutral atom (oxidation state 0)to an ion with a charge, (Al^{3+}). The loss of electrons can be expressed as: \[2 \ \mathrm{Al}(s) \rightarrow 2 \ \mathrm{Al}^{3+}(aq) + 6e^{-}\]On the other hand, iron (Fe) is reduced. It gains electrons, moving from an ion with a charge (Fe^{2+}) to a neutral atom state. The gain of electrons can be depicted as: \[3 \ \mathrm{Fe}^{2+}(aq) + 6e^{-} \rightarrow 3 \ \mathrm{Fe}(s)\]In this process, the aluminum provides electrons it loses to the iron species.

Half-Reactions

In electrochemistry, a half-reaction is the part of the redox process that involves either oxidation or reduction alone. Each half-reaction shows explicitly the electrons exchanged in the process.- **Oxidation Half-Reaction for Aluminum:** A loss of electrons occurs at the aluminum electrode:\[2 \mathrm{Al}(s) \rightarrow 2 \mathrm{Al}^{3+}(aq) + 6e^{-}\]Here, aluminum atoms lose three electrons each to become aluminum ions.- **Reduction Half-Reaction for Iron:** Electrons are gained by the iron ions, as shown:\[3 \mathrm{Fe}^{2+}(aq) + 6e^{-} \rightarrow 3 \mathrm{Fe}(s)\]Each ion gains electrons to convert back into elemental iron.Together, these half-reactions allow us to detail the full picture of electron movement in the voltaic cell.

Electrode Potential

The electrode potential is a measure of the ability of a chemical species to acquire or lose electrons. In a voltaic cell, there are two electrodes: the anode and the cathode, each with a different potential.
The **anode** is where oxidation occurs. It has a lower electrode potential and a negative sign. In our reaction, the aluminum electrode acts as the anode.
The **cathode** is where reduction takes place, featuring positive electrode potential. In this experiment, it corresponds to the iron electrode. The potential difference drives electrons to flow from negative to positive, thus generating electricity.

Electron Flow

Electron flow is fundamental in a voltaic cell. Electrons are generated at the anode and consumed at the cathode. In our system:
The electrons are released by aluminum at the anode due to oxidation. Following the flow rules, these electrons then travel through the external circuit from the anode to reach the cathode (iron electrode).
Here, at the cathode, they are used to reduce iron ions. This external flow of electrons is essentially what makes the voltaic cell capable of doing work, like powering a device connected in this circuit.

Anion and Cation Migration

Ions flow through the solution in a voltaic cell to maintain charge balance. Cations and anions migrate in opposite directions.
- **Cation Migration:** Cations are positively charged ions. In this setup, ( Fe^{2+} ) ions move towards the iron cathode where electron reduction happens.
- **Anion Migration:** Anions carry a negative charge. ( SO_4^{2-} ) anions migrate towards the aluminum anode to offset the loss of positive ions due to aluminum’s oxidation.
This movement of ions balances out the charges between the solutions in both half-cells ensuring the cell’s operation.