Problem 49 - Chemistry: An Atoms-Focused Approach

Question

Using the appropriate standard potentials from Appendix 6, determine the equilibrium constant for the following reaction at \(298 \mathrm{K}:\) $$ \mathrm{Fe}^{3+}(a q)+\mathrm{Cr}^{2+}(a q) \rightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Cr}^{3+}(a q) $$

Step-by-Step Solution

Answer

Answer: The equilibrium constant (K) for the given redox reaction at 298 K is approximately 36,315.

Step 1: Write the half-reactions

First, we need to identify the half-reactions for the given redox reaction. The two half-reactions are: Reduction: Fe³⁺(aq) + 1e⁻ → Fe²⁺(aq) Oxidation: Cr²⁺(aq) → Cr³⁺(aq) + 1e⁻

Step 2: Look up standard reduction potentials

Consult Appendix 6 to find the standard reduction potentials (E⁰) for the two half-reactions: Reduction: \(E_0(\mathrm{Fe}^3+ \to \mathrm{Fe}^2+)\) = +0.77 V Oxidation: \(E_0(\mathrm{Cr}^3+ \to \mathrm{Cr}^2+)\) = -0.41 V Note that the oxidation half-reaction potential must be reversed since Cr²⁺ is getting oxidized to Cr³⁺.

Step 3: Calculate the overall cell potential

To calculate the overall cell potential (E) for the given redox reaction, we add the standard reduction potentials of the two half-reactions: \(E = E_\textrm{Reduction} - E_\textrm{Oxidation} = 0.77 - (-0.41) = 1.18 \textrm{ V}\)

Step 4: Use the Nernst equation to find the equilibrium constant at 298 K

At equilibrium, the cell potential is 0. We can use the Nernst equation to relate standard cell potential and equilibrium constant: \(0 = E - \dfrac{RT}{nF} \ln (K)\) Where R is the gas constant (8.314 J/mol·K), T is the temperature (298 K), n is the number of electrons involved (1, in this case), and F is Faraday's constant (96485 C/mol). Rearrange the equation and solve for K: \(K = e^{\frac{nFE}{RT}} = e^{\frac{1 \times 96485 \times 1.18}{8.314 \times 298}}\)

Step 5: Calculate K

Now, we can plug in the values we found earlier into the equation to calculate K: \(K = e^{10.5} = 36315\) The equilibrium constant for the given redox reaction at 298 K is approximately 36,315.

Key Concepts

Standard Reduction Potentials

Standard reduction potentials are crucial for understanding and predicting redox reactions. These potentials, often represented by the symbol \(E^0\), give us an idea about how likely a species is to gain electrons, also known as reduction. Each half-reaction has its own \(E^0\) value.

• If the \(E^0\) value is positive, the reaction readily occurs as reduction.
• If the \(E^0\) value is negative, the reaction is less favorable for reduction and favors oxidation instead.

Knowing these values is essential when calculating the overall cell potential of a battery or any electrochemical cell. By looking at the standard reduction potentials, we can determine which element is more likely to be reduced or oxidized. Remember that you usually have to reverse the signs of the \(E^0\) values when a species is oxidized. This flip helps to calculate the correct overall voltage of the redox reaction. For example, in our problem, we used the standard potentials of iron (Fe) and chromium (Cr) to compute the overall cell potential by adding the respective \(E^0\) values of reduction and oxidation.

Nernst Equation

The Nernst equation is a powerful tool in electrochemistry. It links the cell potential, \(E\), with the concentrations of the involved species, thus bridging the gap between chemistry under standard conditions and real-world scenarios. The equation is:\[E = E^0 - \frac{RT}{nF} \ln Q\]Where:

• \(E\) is the cell potential under non-standard conditions,
• \(E^0\) is the standard cell potential,
• \(R\) is the universal gas constant \(8.314 \, \mathrm{J/mol \, K}\),
• \(T\) is the temperature in Kelvin,
• \(n\) is the number of moles of electrons transferred in the reaction,
• \(F\) is Faraday's constant \(96485 \, \mathrm{C/mol}\),
• \(Q\) is the reaction quotient, which depends on the concentrations of the reactants and products.

At equilibrium, \(E = 0\) because there's no net driving force, indicating the potential difference between two half-cells equals zero. This simplifies the process for calculating the equilibrium constant \(K\) in redox reactions by rearranging the equation. For instance, we used the Nernst equation in the original exercise to find the equilibrium constant for the reaction between \( ext{Fe}^{3+} \) and \( ext{Cr}^{2+} \) at 298 K.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are chemical processes where electrons are transferred between two substances. These reactions play a vital role in countless everyday processes, including energy metabolism in our bodies and battery operations.

• **Reduction** is the gain of electrons by a molecule, atom, or ion.
• **Oxidation** is the loss of electrons by a molecule, atom, or ion.

In every redox reaction, one substance is reduced, and another is oxidized. Together, these two make up the half-reactions that are combined to complete the redox process. Understanding the balance between oxidation and reduction is essential for solving problems involving redox equations, such as determining equilibrium constants as seen in the original exercise.
When balancing redox reactions, it's also crucial to ensure the number of electrons lost in oxidation equals the number gained in reduction. This balance ensures mass and charge balance across the reaction and allows us to use mathematical tools like the Nernst equation to analyze the system further.