Problem 39 - Chemistry: The Central Science in SI Units, Global Edition

Question

The standard reduction potentials of the following half-reactions are given in Appendix E: $$ \begin{array}{l} \mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) \\ \mathrm{Cd}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{Cd}(s) \\\ \mathrm{Sn}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{Sn}(s) \\\ \mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{Ag}(s) \end{array} $$ (a) Determine which combination of these half-cell reactions leads to the cell reaction with the largest positive cell potential and calculate the value. (b) Determine which combination of these half-cell reactions leads to the cell reaction with the smallest positive cell potential and calculate the value.

Step-by-Step Solution

Answer

The combination of half-cell reactions that leads to the largest positive cell potential is Fe(s) with Ag(s), and the value can be calculated using \( E°_{Ag} - E°_{Fe} \). The combination with the smallest positive cell potential is Fe(s) with Cd(s), and the value can be calculated using \( E°_{Cd} - E°_{Fe} \).

Step 1: List all possible half-cell reaction combinations

To find the combination of half-cell reactions that produce the largest and smallest positive cell potentials, start by listing all the possible combinations. Ignore same element combinations as they result in no reaction. There are four different elements, and hence, six possible combinations: 1. Fe(s) with Cd(s) 2. Fe(s) with Sn(s) 3. Fe(s) with Ag(s) 4. Cd(s) with Sn(s) 5. Cd(s) with Ag(s) 6. Sn(s) with Ag(s)

Step 2: Find the standard cell potentials for each combination

Using the standard reduction potentials from Appendix E, calculate the standard cell potential E° of each combination by subtracting the lower potential from the higher potential, ensuring that the resultant potential is positive. (Note: You can find the standard reduction potentials of these half-cell reactions in the AIME table). 1. Fe(s) with Cd(s): \( E°_{Cd} - E°_{Fe} \) 2. Fe(s) with Sn(s): \( E°_{Sn} - E°_{Fe} \) 3. Fe(s) with Ag(s): \( E°_{Ag} - E°_{Fe} \) 4. Cd(s) with Sn(s): \( E°_{Sn} - E°_{Cd} \) 5. Cd(s) with Ag(s): \( E°_{Ag} - E°_{Cd} \) 6. Sn(s) with Ag(s): \( E°_{Ag} - E°_{Sn} \)

Step 3: Determine the combination with the largest positive cell potential

After calculating the standard cell potentials for each combination, you'll find that Fe(s) with Ag(s) has the largest positive cell potential. 3. Fe(s) with Ag(s)

Step 4: Determine the combination with the smallest positive cell potential

Similar to step 3, you'll find that Fe(s) with Cd(s) has the smallest positive cell potential. 1. Fe(s) with Cd(s)

Step 5: Calculate the values of the largest and smallest positive cell potentials

By comparing the calculated E° values, you'll be able to find the largest and smallest positive cell potentials: (a) The largest positive cell potential value occurs when Fe(s) and Ag(s) react together, and its E° value can be calculated as follows: \( E°_{Ag} - E°_{Fe} \) (b) The smallest positive cell potential value occurs when Fe(s) and Cd(s) react together, and its E° value can be calculated as follows: \( E°_{Cd} - E°_{Fe} \)

Key Concepts

Standard Reduction Potentials

Understanding standard reduction potentials is key in electrochemistry. These are the voltages associated with reduction reactions, measured under standard conditions (1 M concentration, 1 atm pressure, and 25°C). They tell us how easily a species gains electrons to become reduced. The higher the reduction potential, the more likely the substance will gain electrons and get reduced. For instance, in our exercise, you'd look up values from the table and see that silver ions (\(\text{Ag}^+ + \text{e}^- \rightarrow \text{Ag}(s)\)) might have a higher potential than, say, cadmium ions.
This means in a reaction, silver ions are more likely to get reduced than cadmium ions. Using these values, comparisons between half-cells can be made to determine which reactions will occur spontaneously.

Cell Potential

Cell potential, also known as electromotive force (EMF), expresses the energy difference between two half-cells in an electrochemical cell. It's calculated using the formula:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]
This equation involves subtracting the anode's potential from the cathode's potential. A positive value means the reaction is spontaneous, indicating which half-reaction will proceed.
Calculating these in our exercise allows us to determine the largest and smallest positive cell potentials by using the given standard reduction potentials.

• The combination with the highest positive potential usually involves the strongest oxidizing and reducing agents.
• The combination resulting in the lowest positive potential usually involves weaker agents.

Half-Reactions

In electrochemistry, a full reaction is broken down into two half-reactions: one for oxidation and one for reduction. These reactions display how electrons transfer between substances.
The importance of half-reactions lies in their ability to provide a clear view of the electron movement. For example, in the exercise, iron might act in one half-reaction:\( \text{Fe}^{2+} + 2\text{e}^- \rightarrow \text{Fe}(s) \), indicating reduction. Whereas another substance may undergo oxidation in its half-reaction.
Finding these pairs is essential to determine the overall reaction direction. The half-reactions are summed to yield the full electrochemical reaction that the exercise aims to explore.

Electrochemical Cells

Electrochemical cells are devices that convert chemical energy into electrical energy. They consist of two half-cells, each containing an electrode and electrolyte. The cells can generate electricity through spontaneous chemical reactions.
In our exercise, identifying the right combination of half-reactions constructs a cell. For example, combining silver and iron reactions in a cell sets up a flow of electrons. These electrons move through an external circuit, driven by the difference in standard reduction potentials.

• The anode is where oxidation occurs, and it loses electrons.
• The cathode gains electrons, where reduction happens.

The overall potential of the cell reflects its ability to perform work, making it critical for applications like batteries and galvanic cells.