Question

In a voltaic cell represented by All \(\mathrm{Al}^{3+} \| \mathrm{Cu}^{2+} | \mathrm{Cu},\) what is oxidized and what is reduced as the cell delivers current?

Short answer

In the given voltaic cell (Al³⁺ ∥ Cu²⁺ | Cu), Aluminum (Al) is oxidized to Aluminum ion (Al³⁺) at the anode, losing 3 electrons, and Copper ion (Cu²⁺) is reduced to Copper metal (Cu) at the cathode, gaining 2 electrons as the cell delivers current.

Step-by-step solution

Identify the anode and cathode

In a voltaic cell, the anode is where oxidation takes place, and the cathode is where reduction occurs. In the given cell notation, we can find the anode and cathode represented by their respective half-reactions. The anode is on the left side of the cell notation (Al³⁺), and the cathode is on the right side of the cell notation (Cu²⁺ | Cu).

Write the half-reactions for each electrode.

Now that we have identified the anode and cathode, we can write the half-reactions for each of them: Anode (oxidation): \[ \mathrm{Al} \rightarrow \mathrm{Al}^{3+} + 3 e^-\] Cathode (reduction): \[ \mathrm{Cu}^{2+} + 2 e^- \rightarrow \mathrm{Cu}\]

Determine what is oxidized and what is reduced

Using the half-reactions, we can now determine which species are oxidized and reduced: Oxidation: Aluminum (Al) is oxidized to Aluminum ion (Al³⁺) at the anode, losing 3 electrons in the process. Reduction: Copper ion (Cu²⁺) is reduced to Copper metal (Cu) at the cathode, gaining 2 electrons in the process. Thus, in the given voltaic cell, Al is oxidized, and Cu²⁺ is reduced as the cell delivers current.

Key concepts

Oxidation and Reduction

To comprehend a voltaic cell's operation, starting with the basics of oxidation and reduction reactions, commonly known as redox reactions, is essential. Oxidation involves the loss of electrons from an element, while reduction involves the gain of electrons.

In any redox reaction, there is always one species that gives up electrons and another that gains them. The substance that loses electrons is said to be oxidized, and the substance that gains electrons is reduced. This can be memorized with the mnemonic 'OIL RIG' - Oxidation Is Loss, Reduction Is Gain of electrons.

Understanding through an Example

Take the reaction in a voltaic cell involving aluminum and copper. Aluminum atoms lose three electrons to form Al³⁺ ions, thus aluminum is oxidized. The electrons that aluminum atoms lose are gained by Cu²⁺ ions, reducing them to copper metal. This transfer of electrons is what drives the electric current in the cell.

Anode and Cathode Identification

Identifying the anode and cathode within a voltaic cell is crucial for understanding how the cell works. The anode is the electrode where oxidation occurs, while the cathode is the electrode where reduction takes place.

In cell notation, the anode is written on the left, and the cathode is on the right, with a double vertical line (||) representing the salt bridge that separates them. It's important to remember that in a voltaic cell, the anode is negatively charged because it releases electrons, and the cathode is positively charged because it accepts electrons.

Recognizing Each Electrode

When presented with a cell notation like Al³⁺ | Cu²⁺ | Cu, the anode is aluminum, producing Al³⁺ ions and releasing electrons, while the cathode is where copper ions Cu²⁺ gain electrons and become copper metal. By understanding this setup, you can predict the flow of electrons and the overall reaction in the cell.

Writing Half-Reactions

Half-reactions are a way of representing the two separate processes that occur at the anode and cathode in a redox reaction. They help to visualize electron transfer and the changes occurring to individual species.

A half-reaction splits the overall redox reaction into two parts: one that accounts for the oxidation and another for the reduction. When writing half-reactions, it's important to balance both the number of electrons lost in oxidation with the number gained in reduction and also the overall charge.

Example of Half-Reactions

For the oxidation at the anode in our voltaic cell example, the half-reaction is written as: Al → Al³⁺ + 3e⁻. Here, aluminum is oxidized, losing three electrons. For the reduction at the cathode: Cu²⁺ + 2e⁻ → Cu, illustrating that the copper ions gain two electrons to form metallic copper. These written half-reactions are pivotal for understanding the individual changes happening within the voltaic cell.