Problem 3 - Chemistry Matter and Change

Question

For each of these pairs of half-reactions, write the balanced equation for the overall cell reaction, and calculate the standard cell potential. Describe the reaction using cell notation. Refer to Chapter 19 to review writing and balancing redox equations. $$\mathrm{Hg}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Hg}(\mathrm{l}) \text { and } \mathrm{Cr}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cr}(\mathrm{s})$$

Step-by-Step Solution

Answer

The balanced equation is \(\text{Hg}^{2+} + \text{Cr} \rightarrow \text{Hg} + \text{Cr}^{2+}\) with a cell potential of 1.76 V. Cell notation: \(\text{Cr}(s) | \text{Cr}^{2+}(aq) || \text{Hg}^{2+}(aq) | \text{Hg}(l)\).

Step 1: Write the Half-Reactions

The given half-reactions are already provided:1. \( ext{Hg}^{2+}( ext{aq}) + 2 ext{e}^- \rightarrow ext{Hg}( ext{l})\)2. \( ext{Cr}^{2+}( ext{aq}) + 2 ext{e}^- \rightarrow ext{Cr}( ext{s})\)

Step 2: Identify Oxidation and Reduction

The reaction involving \( ext{Hg}^{2+}\) gaining electrons is a reduction process, while the reaction involving \( ext{Cr}^{2+}\) and gaining electrons should be reversed to represent oxidation. Therefore, the reversed reaction for chromium is:\( ext{Cr}( ext{s}) \rightarrow ext{Cr}^{2+}( ext{aq}) + 2 ext{e}^-\)

Step 3: Combine the Half-Reactions

Add the reduction and oxidation half-reactions:\(\text{Hg}^{2+} ( ext{aq}) + ext{Cr}( ext{s}) \rightarrow ext{Hg} ( ext{l}) + ext{Cr}^{2+}( ext{aq})\). This provides a balanced equation for the redox reaction.

Step 4: Calculate the Standard Cell Potential

Use standard reduction potentials: - For \( ext{Hg}^{2+} + 2 ext{e}^- \rightarrow ext{Hg}\), \( E^0 = 0.85 ext{ V}\).- For \( ext{Cr}^{2+} + 2 ext{e}^- \rightarrow ext{Cr}\), \( E^0 = -0.91 ext{ V}\). When reversed for oxidation, it becomes \(+0.91 ext{ V}\).Calculate the cell potential: \(E_{ ext{cell}}^0 = E_{ ext{cathode}}^0 - E_{ ext{anode}}^0 = 0.85 - (-0.91) = 0.85 + 0.91 = 1.76 ext{ V}\).

Step 5: Write the Cell Notation

Cell notation uses the format: \( ext{Anode} | ext{Anode Ion} || ext{Cathode Ion} | ext{Cathode}\).For this reaction: \( ext{Cr}( ext{s}) | ext{Cr}^{2+}( ext{aq}) || ext{Hg}^{2+}( ext{aq}) | ext{Hg}( ext{l})\).

Key Concepts

Half-Reactions

In electrochemical cells, we break down the overall redox reaction into two simpler components called half-reactions. These half-reactions help us understand which species is oxidized and which is reduced.

A half-reaction shows either the reduction or oxidation process separately. For example, in the provided exercise:

• The reduction half-reaction is: \(\text{Hg}^{2+}(\text{aq}) + 2 \text{e}^- \rightarrow \text{Hg}(\text{l})\)
• The oxidation half-reaction (reversed) is: \(\text{Cr}(\text{s}) \rightarrow \text{Cr}^{2+}(\text{aq}) + 2 \text{e}^-\)

By using these half-reactions, we clarify each step and ensure that the electron flow in the system is balanced. Recognizing this division helps in visualizing the electron transfer and setting the stage for calculating potentials.

Standard Cell Potential

The standard cell potential is the measure of the electromotive force of an electrochemical cell. It signifies the voltage available from the cell when no current is flowing.

To find the standard cell potential \(E_{\text{cell}}^0\), use the formula:
\[E_{\text{cell}}^0 = E_{\text{cathode}}^0 - E_{\text{anode}}^0\]
The values of \(E^0\) are found in standard electrochemical tables. For the problem:

• \(\text{Hg}^{2+} + 2 \text{e}^- \rightarrow \text{Hg}\) has \(E^0 = 0.85\ \text{V}\).
• \(\text{Cr}^{2+} + 2 \text{e}^- \rightarrow \text{Cr}\) has \(E^0 = -0.91\ \text{V}\), and reverses for oxidation to \(+0.91\ \text{V}\).

The equation gives us \(E_{\text{cell}}^0 = 0.85 + 0.91 = 1.76\ \text{V}\). This positive value indicates a spontaneous reaction.

Redox Equations

Redox reactions involve the transfer of electrons between two species. The term "redox" is derived from "reduction" and "oxidation," which occur simultaneously in these processes.

Reduction refers to the gain of electrons. Oxidation, conversely, is the loss of electrons. In redox chemistry, it's vital to balance these reactions to ensure that the number of electrons lost matches those gained.

In the provided exercise, by correctly identifying the reduction and oxidation reactions, we form the overall redox equation with these steps:

• Write each half-reaction.
• Balance electrons between them.
• Combine to find: \( \text{Hg}^{2+} (\text{aq}) + \text{Cr}(\text{s}) \rightarrow \text{Hg} (\text{l}) + \text{Cr}^{2+}(\text{aq})\).

This approach helps solidify the understanding of electron flow and chemical transformations.

Cell Notation

Cell notation is a shorthand way to describe an electrochemical cell and its components.It separates the anode and cathode components using a double vertical line \(||\), indicating the salt bridge or membrane that separates the two cell compartments.

In the given reaction, cell notation is written as follows:

• Anode (oxidation side) goes first: \(\text{Cr}(\text{s})\)
• Anode ion after a single line: \(| \text{Cr}^{2+}(\text{aq})\)
• Double line for separation: \(||\)
• Cathode ion next: \(| \text{Hg}^{2+}(\text{aq})\)
• Cathode (reduction side) at the end: \(| \text{Hg}(\text{l})\)

Thus, the complete cell notation is: \(\text{Cr}(\text{s}) | \text{Cr}^{2+}(\text{aq}) || \text{Hg}^{2+}(\text{aq}) | \text{Hg}(\text{l})\). This format ensures quick comprehension of the processes and substances in each half-cell.