Question

Aluminum will displace tin from solution according to the equation $$ 2 \mathrm{Al}(s)+3 \mathrm{Sn}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \operatorname{Sn}(s) $$ What would be the individual half-cell reactions if this were the cell reaction for a galvanic cell? Which metal would be the anode and which the cathode?

Short answer

The oxidation half-cell reaction is \(2\text{Al}(s) \rightarrow 2\text{Al}^{3+}(aq) + 6e^{-}\), while the reduction half-cell reaction is \(3\text{Sn}^{2+}(aq) + 6e^{-} \rightarrow 3\text{Sn}(s)\). Aluminum is the anode, and tin is the cathode.

Step-by-step solution

Write the Oxidation Half-Cell Reaction

Identify the species that loses electrons (is oxidized). Looking at the balanced equation, aluminum (\text{Al}) goes from a neutral state to a positively charged ion (\text{Al}^{3+}), indicating it loses electrons. Write the half-cell reaction for the oxidation process: $$\text{Al}(s) \rightarrow \text{Al}^{3+}(aq) + 3e^{-}.$$

Write the Reduction Half-Cell Reaction

Identify the species that gains electrons (is reduced). Tin ions (\text{Sn}^{2+}) gain electrons to form neutral tin (\text{Sn}). Write the half-cell reaction for the reduction process:$$\text{Sn}^{2+}(aq) + 2e^{-} \rightarrow \text{Sn}(s).$$

Balance the Number of Electrons in Both Half-Cell Reactions

Make sure that the number of electrons lost in the oxidation half-cell equals the number gained in the reduction half-cell. For the given equation, multiply the reduction half-cell by 3 and the oxidation half-cell by 2 to balance the electrons:Oxidation: $$2\text{Al}(s) \rightarrow 2\text{Al}^{3+}(aq) + 6e^{-},$$Reduction: $$3\text{Sn}^{2+}(aq) + 6e^{-} \rightarrow 3\text{Sn}(s).$$

Determine the Anode and Cathode

In a galvanic cell, oxidation occurs at the anode, and reduction occurs at the cathode. Aluminum is oxidized; hence, it constitutes the anode. Tin is reduced; hence, it constitutes the cathode.

Key concepts

Half-Cell Reactions

A half-cell reaction is a reaction that represents either the loss or gain of electrons by species. In a galvanic cell, these reactions occur in separate compartments, known as half-cells. One half-cell undergoes oxidation, where electrons are lost, while the other half-cell undergoes reduction, where electrons are gained. These half-cell reactions are critical for understanding how electrochemical cells work.

For instance, in the reaction between aluminum and tin, two separate half-cell reactions take place. The aluminum atom loses electrons, represented by the reaction \(\text{Al}(s) \rightarrow \text{Al}^{3+}(aq) + 3e^{-}\), meaning aluminum is oxidized. On the other hand, the tin ion gains electrons, represented by \(\text{Sn}^{2+}(aq) + 2e^{-} \rightarrow \text{Sn}(s)\), indicating that tin is reduced.

Oxidation and Reduction

The concepts of oxidation and reduction are fundamental in chemistry, especially when discussing galvanic cells. Oxidation refers to the process of losing electrons, which increases the oxidation state of an atom or ion. Conversely, reduction involves gaining electrons, which decreases the oxidation state.

In every redox reaction, one species is oxidized and another is reduced. These two processes are mutually exclusive and occur simultaneously—a phenomenon known as redox pairing. In our example, the aluminum (\(\text{Al}\)) is oxidized by losing three electrons, becoming \(\text{Al}^{3+}\). Simultaneously, the tin ions (\(\text{Sn}^{2+}\)) are reduced by gaining two electrons each to become neutral tin atoms (\(\text{Sn}\)).

Balancing Electron Transfer

To ensure that a redox reaction adheres to the conservation of charge, it's vital to balance the electron transfer between the oxidation and reduction half-reactions. This means adjusting the coefficients of the reactants and products so that the number of electrons lost is equal to the number of electrons gained.

In our example, we initially have two aluminum atoms losing a total of six electrons and three tin ions each gaining two electrons for a total of six gained. This balancing act confirms that the electron transfer in the entire reaction is equal and maintains charge neutrality, which is a requirement for a valid redox reaction.

Anode and Cathode Identification

Anode and cathode identification is crucial for understanding the flow of electrons in a galvanic cell. The anode is where oxidation occurs—electrons are lost. The cathode is the site of reduction—electrons are gained. Electrons flow from the anode to the cathode through an external circuit.

Remembering that 'AnOx RedCat' (anode oxidation, reduction cathode) can help in identifying them in a cell reaction. In our textbook example, since aluminum (\(\text{Al}\)) loses electrons, it's at the anode, while tin (\(\text{Sn}^{2+}\)) gains electrons at the cathode. The identification of anode and cathode not only helps in understanding the redox process but also in constructing and analyzing the functionality of galvanic cells.