Question

Look up the reduction potential for \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+} .\) Look up the reduction potential for \(\mathrm{Fe}^{2+}\) to Fe. Finally, look up the reduction potential for \(\mathrm{Fe}^{3+}\) to Fe. You should notice that adding the reduction potentials for the first two does not give the potential for the third. Why not? Show how you can use the first two potentials to calculate the third potential.

Short answer

The direct addition of potentials does not give the correct result for the third reaction because the stoichiometry of electrons in the reactions is not equal. Instead, we multiply the first reaction by 2 to make the electrons' number the same (3) and add it to the second reaction. Then, we use the Nernst equation, considering the logarithmic term, to accurately calculate the third potential.

Step-by-step solution

Look up the reduction potentials

We look up the reduction potentials for the following half-reactions: 1. \(\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}\) 2. \(\mathrm{Fe}^{2+} + 2e^- \rightarrow \mathrm{Fe}\) 3. \(\mathrm{Fe}^{3+} + 3e^- \rightarrow \mathrm{Fe}\) The corresponding reduction potentials are: 1. \(E_1^\circ = 0.771\, \text{V}\) 2. \(E_2^\circ = -0.440\, \text{V}\) 3. \(E_3^\circ = -0.036\, \text{V}\)

Explain why adding the first two potentials does not give the third potential

Adding the reduction potentials for the first two reactions, \(E_1^\circ\) and \(E_2^\circ\), directly doesn't give the potential for the third reaction, \(E_3^\circ\), because the stoichiometry of the electrons in these reactions is not equal. The first reaction involves 1 electron, while the second reaction involves 2 electrons.

Use the first two potentials to calculate the third potential

To use the first two potentials to calculate the third potential, we need to ensure that the number of electrons in both reactions is the same to satisfy the stoichiometry. We can achieve this by multiplying the first reaction by 2, so the number of electrons in both reactions is the same (3 electrons): 2(\(\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}\)) = \(\mathrm{Fe}^{3+} + 3e^- \rightarrow \mathrm{Fe}^{2+} + 2\mathrm{Fe}^{2+}\) Now, we can add the above reaction and the second reaction: \(\mathrm{Fe}^{3+} + 3e^- \rightarrow \mathrm{Fe}^{2+} + 2\mathrm{Fe}^{2+} + \mathrm{Fe}^{2+} + 2e^- \rightarrow \mathrm{Fe}\) The resulting reaction is: \(\mathrm{Fe}^{3+} + 3e^- \rightarrow \mathrm{Fe}\) For the potential of this reaction, we can use the Nernst equation: \(E^\circ = E_1^\circ - \frac{RT}{nF} \ln{\frac{a_\mathrm{Fe^{2+}}}{a_\mathrm{Fe^{3+}}}}\) Since the Nernst equation has logarithmic terms, we cannot just add the two potentials directly. We need to take the logarithmic term into account to obtain the third potential accurately.

Key concepts

Electrochemistry

Electrochemistry is a branch of chemistry that deals with the relationship between electrical energy and chemical reactions. It primarily focuses on the movement of electrons within a substance, which drives these reactions. This is crucial, as understanding the flow of electrons helps us comprehend how chemical reactions can produce electricity or how electricity can instigate chemical reactions.
One of the fundamental concepts in electrochemistry is the electrode potential, which is a measure of a substance's tendency to gain or lose electrons. Different substances have different electrode potentials, reflecting how easily they are reduced (gain electrons) or oxidized (lose electrons).

• **Reduction involves gaining electrons**. For example, \( \text{Fe}^{3+}\) ion is reduced to \( \text{Fe}^{2+} \) by gaining an electron.
• **Oxidation involves losing electrons**. For example, \( \text{Fe}^{2+}\) can be oxidized to \( \text{Fe}^{3+} + e^- \) when it loses an electron.

Electrochemical reactions usually occur in a cell consisting of two half-cells, each containing an electrode and an electrolyte, and are connected by a salt bridge or a permeable membrane. This setup helps maintain a charge balance by allowing ions to move between the half-cells without mixing the different solutions.

Half-Reaction

A half-reaction is either the oxidation or reduction reaction component of a redox reaction. In electrochemistry, each half-reaction occurs in one of the two half-cells within an electrochemical cell. The overall cell reaction is usually split into two half-reactions to analyze each process separately.
The concept of a half-reaction becomes particularly important when you're dealing with reactions involving electrons that have different tendencies. For instance, in the given problem, we have various half-reactions involving iron:

• A **reduction half-reaction** would be: \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \)
• Another **reduction half-reaction**: \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \)

The half-reaction potential for each reaction signifies how strongly the reaction drives electrons to one side of the reaction. When calculating the potential for a full redox reaction, it is essential to ensure an equal number of electrons are involved in both half-reactions.

To achieve a full reaction from individual half-reactions, sometimes it's necessary to balance the electrons by scaling the reactions appropriately. This is why in the solution, we multiplied the first half-reaction by 2, making the total electron exchange equal, which allowed us to appropriately calculate the overall reduction potential.

Nernst Equation

The Nernst equation is a mathematical formula that relates the reduction potential of a half-reaction or cell to the concentrations of the chemical species involved. This equation becomes crucial when you want to adjust the standard electrode potential for non-standard conditions.
For a generic half-reaction, it can be expressed as:
\[E = E^{\circ} - \frac{RT}{nF} \ln Q\]
Here, \( E^{\circ} \) is the standard reduction potential, \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons exchanged, \( F \) is the Faraday constant, and \( Q \) is the reaction quotient.
The Nernst equation shows us that the effective cell potential \( E \) varies depending on the concentrations of the reactants and products. This means that the cell potential can change from the expected standard potential if the concentration conditions aren't the classic 1 mol/L.
In the exercise provided, we used the Nernst equation to adjust for and understand why the sum of certain half-reaction potentials didn't directly produce the potential for another reaction. By adding the individual logarithmic contributions via the Nernst equation, we were able to achieve the correct overall potential. This showcases the importance of the Nernst equation in understanding and predicting electrochemical reactions under a variety of conditions.