Question

Consider the oxidation-reduction reaction \(\mathrm{Al}(s)+\mathrm{Ni}^{2+}(a q) \rightarrow \mathrm{Al}^{3+}(a q)+\mathrm{Ni}(s)\) Sketch a galvanic cell that makes use of this reaction. Which metal ion is reduced? Which metal is oxidized? What half-reaction takes place at the anode in the cell? What half-reaction takes place at the cathode?

Short answer

In the given reaction, Aluminum (Al) is oxidized, and Nickel ion (\(Ni^{2+}\)) is reduced. The anode half-reaction is \(Al(s) \rightarrow Al^{3+}(aq) + 3e^-\), and the cathode half-reaction is \(Ni^{2+}(aq) + 2e^- \rightarrow Ni(s)\). The galvanic cell consists of two half-cells: anode half-cell with an Aluminum electrode in a \(Al^{3+}\) solution, and cathode half-cell with a Nickel electrode in a \(Ni^{2+}\) solution. The half-cells are connected by a salt bridge and an external wire for electron flow.

Step-by-step solution

Identify the metals and their corresponding ions

In the given reaction, there are two metals involved: 1. Aluminum (Al) and its ion: \(Al^{3+}\) 2. Nickel (Ni) and its ion: \(Ni^{2+}\)

Determine the oxidation and reduction processes

Recall the mnemonic OIL RIG, which stands for Oxidation Is Loss (of electrons) and Reduction Is Gain (of electrons). Let's analyze the changes in oxidation states in the given reaction: 1. \(Al(s)\rightarrow Al^{3+}(aq)\): Since Aluminum loses 3 electrons to form its ion, this is the oxidation process. 2. \(Ni^{2+}(aq) \rightarrow Ni(s)\): Since the Nickel ion gains 2 electrons to form metallic Nickel, this is the reduction process.

Identify the anode and cathode by their half-reactions

The anode is where oxidation occurs, and the cathode is where reduction occurs. Since the Aluminum is undergoing oxidation, the half-reaction occurring at the anode is: \(Al(s) \rightarrow Al^{3+}(aq) + 3e^-\) As the Nickel ion is undergoing reduction, the half-reaction occurring at the cathode is: \(Ni^{2+}(aq) + 2e^- \rightarrow Ni(s)\)

Sketch the galvanic cell based on the identified anode, cathode, and their respective half-reactions

To sketch the galvanic cell, we need to set up two half-cells: 1. Anode half-cell: This consists of an Aluminum electrode (Al) immersed in a solution of Aluminum ions (\(Al^{3+}\)). The oxidation half-reaction that takes place here is: \(Al(s) \rightarrow Al^{3+}(aq) + 3e^-\) 2. Cathode half-cell: This consists of a Nickel electrode (Ni) immersed in a solution of Nickel ions (\(Ni^{2+}\)). The reduction half-reaction that takes place here is: \(Ni^{2+}(aq) + 2e^- \rightarrow Ni(s)\) Connect the two half-cells with a salt bridge to maintain the neutrality of the solution, and an external wire to allow electron flow from anode to cathode.

Key concepts

Understanding Oxidation-Reduction Reactions

Oxidation-reduction reactions, commonly known as redox reactions, are the essence of energy flow in chemistry. They involve the transfer of electrons between substances. In essence, oxidation refers to the loss of electrons, while reduction means the gain of electrons. This exchange is fundamental to the operation of batteries and other electrochemical devices, including biological systems.

Take the reaction mentioned in the exercise, \( \mathrm{Al}(s) + \mathrm{Ni}^{2+}(aq) \rightarrow \mathrm{Al}^{3+}(aq) + \mathrm{Ni}(s) \). Aluminium (Al) loses three electrons (oxidation) to become \( Al^{3+} \) and nickel ions (\( Ni^{2+} \) gain two electrons (reduction) to become nickel metal (Ni).A solid understanding of these processes is crucial for students of chemistry, especially when dissecting the intricacies of electrochemical reactions like those in a galvanic cell.

Decoding the Half-Reaction Method

The half-reaction method is a way of handling the individual oxidation and reduction occurrences within a redox reaction. This approach simplifies complex reactions by breaking them down into two separate equations, each representing half of the overall process.

For example, in the exercise, the half-reaction at the anode, where oxidation takes place, is \( Al(s) \rightarrow Al^{3+}(aq) + 3e^- \), and the half-reaction at the cathode, where reduction occurs, is \( Ni^{2+}(aq) + 2e^- \rightarrow Ni(s) \). By isolating these events, it becomes easier to analyze and balance redox reactions, anticipate voltage outputs, and comprehend the electron flow during these chemical transformations.

Principles of Electrochemistry in a Galvanic Cell

Electrochemistry explores the relationship between electricity and chemical reactions. Key principles include understanding how chemical energy is converted into electrical energy and vice versa. A galvanic cell, like the one described in the exercise, is the classic representation of these electrochemical principles at work.

In such a cell, spontaneous oxidation-reduction reactions occur, driving electrons from the anode to the cathode through an external circuit. This electron flow generates an electric current that can do work. The two half-cells are connected by a salt bridge, completing the circuit and allowing ions to flow, which balances the charge changes due to the electron transfer. The correctness and understandability of the electrode design, the nature of the electrolyte, and the direction of the electron flow are important when depicting a galvanic cell, ensuring students can grasp both the theoretical and practical aspects of electrochemistry.