Problem 8 - Chemistry and Chemical Reactivity

Question

A voltaic cell is constructed using the reaction \(\mathrm{Mg}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Mg}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})\) (a) Write equations for the oxidation and reduction half-reactions. (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 \(\quad\) electrode. Negative ions move in the salt bridge from the \- half-cell to the \(\quad\) half-cell. The half-reaction at the anode is \(=\) and that at the cathode is

Step-by-Step Solution

Answer

Oxidation: \(\mathrm{Mg} \rightarrow \mathrm{Mg}^{2+} + 2\mathrm{e}^-\); Reduction: \(2\mathrm{H}^+ + 2\mathrm{e}^- \rightarrow \mathrm{H}_2\). Anode: oxidation (Mg); Cathode: reduction (H).

Step 1: Identifying Oxidation Half-Reaction

The oxidation half-reaction involves the loss of electrons. In the given equation, magnesium (Mg) loses electrons and is oxidized to magnesium ions (\(\mathrm{Mg}^{2+}\)). The oxidation half-reaction is:\[ \mathrm{Mg}(\mathrm{s}) \rightarrow \mathrm{Mg}^{2+}(\mathrm{aq}) + 2\mathrm{e}^- \]

Step 2: Identifying Reduction Half-Reaction

The reduction half-reaction involves the gain of electrons. Here, hydrogen ions (\(\mathrm{H}^+\)) gain electrons to form hydrogen gas (\(\mathrm{H}_2\)). The reduction half-reaction is:\[ 2\mathrm{H}^+(\mathrm{aq}) + 2\mathrm{e}^- \rightarrow \mathrm{H}_2(\mathrm{g}) \]

Step 3: Determine Anode and Cathode Reactions

In a voltaic cell, the oxidation reaction occurs at the anode and the reduction reaction occurs at the cathode. Therefore, the reaction \(\mathrm{Mg}(\mathrm{s}) \rightarrow \mathrm{Mg}^{2+}(\mathrm{aq}) + 2\mathrm{e}^-\) occurs at the anode, and the reaction \(2\mathrm{H}^+(\mathrm{aq}) + 2\mathrm{e}^- \rightarrow \mathrm{H}_2(\mathrm{g})\) occurs at the cathode.

Step 4: Electrons Flow in the External Circuit

Electrons flow from the anode to the cathode in the external circuit. Since oxidation occurs at the anode, electrons will move from the magnesium electrode (anode) toward the cathode where reduction occurs.

Step 5: Ion Movement in the Salt Bridge

Negative ions move through the salt bridge to balance the charge. In this reaction, as magnesium cations are produced at the anode (oxidation site), negative ions will move from the cathode half-cell towards the anode half-cell.

Step 6: Completing the Sentences

Electrons in the external circuit flow from the magnesium electrode to the hydrogen electrode. Negative ions move in the salt bridge from the hydrogen half-cell to the magnesium half-cell. The half-reaction at the anode is \(\mathrm{Mg}(\mathrm{s}) \rightarrow \mathrm{Mg}^{2+}(\mathrm{aq}) + 2\mathrm{e}^-\) and that at the cathode is \(2\mathrm{H}^+(\mathrm{aq}) + 2\mathrm{e}^- \rightarrow \mathrm{H}_2(\mathrm{g})\).

Key Concepts

Oxidation Half-Reaction

In the realm of electrochemistry, oxidation involves the loss of electrons. For a voltaic or galvanic cell, this process specifically occurs at the anode. In our exercise, magnesium metal (Mg) is oxidized, shedding electrons to become magnesium ions (Mn^{2+}). This transformation can be expressed as: \[ \mathrm{Mg}(\mathrm{s}) \rightarrow \mathrm{Mg}^{2+}(\mathrm{aq}) + 2\mathrm{e}^- \] This equation shows that two electrons are released for each magnesium atom undergoing oxidation. These electrons will then migrate through the external circuit, contributing to the current flow. Helpful points about oxidation half-reactions include:

• They indicate where electron loss occurs.
• The species being oxidized becomes more positively charged.

Remember, oxidation is often remembered by the phrase “loss of electrons.”
It is a vital process in generating electrical energy in a voltaic cell.

Reduction Half-Reaction

Reduction is the complementary process to oxidation. It involves the gain of electrons and is characterized by a decrease in oxidation state. This gain occurs at the cathode of a voltaic cell. In our case, hydrogen ions (H^+) from the solution gain electrons to form hydrogen gas (H_2). This is represented by the following equation: \[ 2\mathrm{H}^+(\mathrm{aq}) + 2\mathrm{e}^- \rightarrow \mathrm{H}_2(\mathrm{g}) \] Here, two electrons are required for the formation of one molecule of hydrogen gas from hydrogen ions. As electrons continue to flow within the circuit, they reach the cathode, facilitating the reduction of hydrogen ions. Key points to note about reduction include:

• Reduction signifies an electron gain.
• The species undergoing reduction becomes less positively charged.

Remember, this is the process where ‘electron gain’ happens, contributing to the formation of substances with lower oxidation states.

Anode and Cathode Reactions

In electrochemical cells, understanding the roles of the anode and cathode is crucial. - The **anode** is the site of oxidation. Here, the substance loses electrons. In the given exercise, magnesium is oxidized at the anode: \[ \mathrm{Mg}(\mathrm{s}) \rightarrow \mathrm{Mg}^{2+}(\mathrm{aq}) + 2\mathrm{e}^- \] - The **cathode** hosts the reduction reaction, where electrons are gained. Hydrogen ions in our example are reduced: \[ 2\mathrm{H}^+(\mathrm{aq}) + 2\mathrm{e}^- \rightarrow \mathrm{H}_2(\mathrm{g}) \] The flow of electrons between these two electrodes through an external conductor is what forms the electric current. The direction of this electron movement—from anode to cathode—is central to voltaic cell operations. Substances undergoing reactions at each electrode are critical:

• At the anode: Oxidation (electron loss).
• At the cathode: Reduction (electron gain).

In summary, the anode and cathode play pivotal roles in facilitating redox reactions, with electron movement essential for generating electric current in the cell.