Problem 8 - Chemistry and Chemical Reactivity

Question

voltaic cell is constructed using the reaction \(\mathrm{Mg}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Mg}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})\) (a) Write equations for the oxidation and reduction halfreactions. (b) Which half-reaction occurs in the anode compartment and which occurs in the cathode compartment? (c) Complete the following sentences: Electrons in the external circuit flow from the ___ electrode to the ___ electrode. Negative ions move in the salt bridge from the ___ half-cell to the ___ half-cell. The half-reaction at the anode is ___ and that at the cathode is ___.

Step-by-Step Solution

Answer

a) Oxidation at anode: \( \mathrm{Mg} \rightarrow \mathrm{Mg}^{2+} + 2e^- \); Reduction at cathode: \( 2\mathrm{H}^{+} + 2e^- \rightarrow \mathrm{H}_{2} \). b) Anode: Mg oxidation, Cathode: H⁺ reduction. c) Electrons flow from anode to cathode; negative ions move from cathode to anode; oxidation at anode, reduction at cathode.

Step 1: Analyze the overall cell reaction

The given cell reaction is \( \mathrm{Mg}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Mg}^{2+}(\mathrm{aq}) + \mathrm{H}_{2}(\mathrm{g}) \). We need to separate it into two half-reactions, one for oxidation and one for reduction.

Step 2: Identify the oxidation half-reaction

Oxidation is the loss of electrons. In the given reaction, magnesium (\( \mathrm{Mg}(s) \)) is oxidized to \( \mathrm{Mg}^{2+}(aq) \), losing two electrons in the process. The oxidation half-reaction is: \( \mathrm{Mg}(s) \longrightarrow \mathrm{Mg}^{2+}(aq) + 2e^- \).

Step 3: Identify the reduction half-reaction

Reduction is the gain of electrons. Here, hydrogen ions (\( \mathrm{H}^{+}(aq) \)) gain electrons to form hydrogen gas (\( \mathrm{H}_{2}(g) \)). The reduction half-reaction is: \( 2\mathrm{H}^{+}(aq) + 2e^- \longrightarrow \mathrm{H}_{2}(g) \).

Step 4: Determine the anode and cathode compartments

The anode is where oxidation occurs, and the cathode is where reduction occurs. From steps 2 and 3, we see that the oxidation half-reaction (Mg to Mg²⁺) occurs at the anode, and the reduction half-reaction (H⁺ to H₂) occurs at the cathode.

Step 5: Describe electron and ion flow

Electrons flow from the anode to the cathode via the external circuit. Negative ions in the salt bridge generally move to balance charge: they would move from the cathode compartment to the anode compartment to balance the movement of cations.

Step 6: Complete the sentences

Based on Steps 4 and 5: Electrons in the external circuit flow from the anode electrode to the cathode electrode. Negative ions move in the salt bridge from the cathode half-cell to the anode half-cell. The half-reaction at the anode is oxidation and that at the cathode is reduction.

Key Concepts

Oxidation and Reduction

In a voltaic cell, chemical reactions are split into two types: oxidation and reduction. These processes involve the transfer of electrons. Oxidation refers to the loss of electrons. For example, when magnesium (\( \mathrm{Mg}(s) \)) loses two electrons, it becomes \( \mathrm{Mg}^{2+}(aq) \). This is the oxidation half-reaction:

• \( \mathrm{Mg}(s) \rightarrow \mathrm{Mg}^{2+}(aq) + 2e^- \)

Reduction, on the other hand, is the gain of electrons. In our case, hydrogen ions (\( \mathrm{H}^{+}(aq) \)) gain electrons to produce hydrogen gas (\( \mathrm{H}_{2}(g) \)). Here is the reduction half-reaction:

• \( 2\mathrm{H}^{+}(aq) + 2e^- \rightarrow \mathrm{H}_{2}(g) \)

In every redox reaction (a combination of reduction and oxidation), the total number of electrons lost in oxidation equals the number gained in reduction.

Anode and Cathode

In the realm of electrochemistry, the anode and cathode are the two essential components where these half-reactions occur. The anode is where oxidation takes place. In our example, it is the site where magnesium (\( \mathrm{Mg}(s)\)) is converted into \( \mathrm{Mg}^{2+}(aq) \) by losing electrons.

• The anode reaction: \( \mathrm{Mg}(s) \rightarrow \mathrm{Mg}^{2+}(aq) + 2e^- \)

Conversely, the cathode is where reduction occurs. In our voltaic cell, this is the location where hydrogen ions (\( \mathrm{H}^{+}(aq) \)) gain electrons to form hydrogen gas (\( \mathrm{H}_{2}(g)\)).

• The cathode reaction: \( 2\mathrm{H}^{+}(aq) + 2e^- \rightarrow \mathrm{H}_{2}(g) \)

Knowing where these reactions occur is vital as it helps us understand the direction and flow of electrons in the cell.

Electron Flow

In a voltaic cell, electron flow is key to generating electricity. Electrons travel from the anode, where oxidation occurs, to the cathode, where reduction takes place. This flow happens through an external circuit, providing electrical energy useful for powering devices.

• Electrons flow from the anode to the cathode.

Additionally, to maintain charge balance in the system, ions must move through a salt bridge. Negative ions flow from the cathode side to the anode side. This movement helps neutralize the charge buildup created by the flow of electrons. It ensures that the voltaic cell continues to operate smoothly, without interruption.

• Negative ions move from the cathode to the anode in the salt bridge to balance the system.