Question

What is the equilibrium constant of the reaction: \(2 \mathrm{Fe}^{3+}+\mathrm{Au}^{+} \rightarrow 2 \mathrm{Fe}^{2+}+\mathrm{Au}^{3+} ?\) Given \(E_{\mathrm{Au}^{+} \mid \mathrm{Au}}^{\circ}=1.68 \mathrm{~V}, E_{\mathrm{Au}^{3 \cdot} \mid \mathrm{Au}}^{\circ}=1.50 \mathrm{~V}\), \(E_{\mathrm{Fe}^{3 *} \mid \mathrm{Fe}^{\mathrm{e}}^{\circ}}=0.75 \mathrm{~V}\) and \(2.303 R T / F=0.06\) (a) \(1 \times 10^{22}\) (b) \(1 \times 10^{-22}\) (c) \(1 \times 10^{-11}\) (d) \(1 \times 10^{-72}\)

Short answer

The equilibrium constant (K) for the reaction is approximately \(1 \times 10^{26.5}\). This value does not correspond to any of the given options, which suggests there might be an error in the provided options or in the calculation.

Step-by-step solution

- Write down the Nernst equation for the cell potential

According to the Nernst equation, the cell potential, E, is related to the standard cell potential, E°, and the reaction quotient, Q, by the equation: \[ E = E^\circ - \frac{2.303 RT}{nF} \log Q \]For a standard cell potential (conditions where all reactants and products are at 1 M concentration), Q is equal to 1, hence \[ E = E^\circ \]where - E° is the standard cell potential (V),- R is the universal gas constant,- T is the temperature in Kelvin,- n is the number of moles of electrons transferred in the reaction,- F is the Faraday constant,- Q is the reaction quotient.

- Calculate the standard cell potential (E°)

The standard cell potential (E°) for the overall reaction is calculated using the standard reduction potentials of the half-reactions.The standard reduction potential for the half-reaction \[ \mathrm{Au}^{3+} + 3e^- \rightarrow \mathrm{Au} \]is given as 1.50 V.The standard reduction potential for the half-reaction \[ \mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+} \]is given as 0.75 V.The standard reduction potential for the half-reaction \[ \mathrm{Au}^+ + e^- \rightarrow \mathrm{Au} \]is given as 1.68 V.The cell potential for the oxidation of Au+ to Au3+ can be found using the relation E°(oxidation) = -E°(reduction). Thus, for Au+ to Au3+: \[ E^\circ_\text{Au+|Au3+} = -(1.68 \text{ V}) = -1.68 \text{ V} \]The overall cell potential (E°) is the difference between the cell potentials of the reduction and oxidation reactions:\[ E^\circ = E^\circ_{\text{Au}^{3+}|\text{Au}} - E^\circ_{\text{Au+}|\text{Au3+}} \]\[ E^\circ = 1.50 \text{ V} - (-1.68 \text{ V}) \]\[ E^\circ = 3.18 \text{ V} \]

- Apply the formula relating cell potential and equilibrium constant

The relationship between the standard cell potential (E°) and the equilibrium constant (K) is given by the equation: \[ E^\circ = \frac{2.303 RT}{nF} \log K \]Substitute the values: \[ E^\circ = 3.18 \text{ V}, \quad 2.303 RT/F = 0.06 \text{ V}, \quad n = 2 \text{ electrons} \]into the equation to solve for K: \[ 3.18 \text{ V} = 0.06 \text{ V} \times \log K \]\[ \frac{3.18 \text{ V}}{0.06 \text{ V}} = \log K \]\[ 53 = \log K \]To find the value of K, use the antilogarithm (inverse log): \[ K = 10^{53} \]

- Adjust the equilibrium constant for the stoichiometry of the reaction

Since the reaction involves the transfer of two electrons per iron, but we have two iron ions involved, the total is four electrons – therefore, we have used an incorrect stoichiometric factor in the previous calculation. We need to adjust our calculation of the Nernst equation by the correct number of electrons transferred (n = 4) for the full reaction. We do this by dividing the previously calculated cell potential by two to get the potential for the transfer of one mole of electrons, and then using this in the Nernst equation.\[E^\circ_{corrected} = \frac{E^\circ}{2} = \frac{3.18 \text{ V}}{2} = 1.59 \text{ V}\]We then solve for K using the corrected potential:\[1.59 \text{ V} = 0.06 \text{ V} \times \log K\]\[\frac{1.59 \text{ V}}{0.06 \text{ V}} = \log K\]\[26.5 = \log K\]Using the antilog to find K: \[K = 10^{26.5}\]

- Calculation Check

Ensure that the units and stoichiometry are correct. In this case, the stoichiometry is correct and K is a unitless number, so no further unit adjustments are needed.

Key concepts

Nernst Equation

The Nernst equation is a fundamental principle in electrochemistry that relates the reduction potential of a chemical reaction to the concentration of the reactants and products. It is represented by the formula:
\[ E = E^\circ - \frac{2.303 RT}{nF} \log Q \]
where

• \(E\) is the cell potential under non-standard conditions,
• \(E^\circ\) is the standard cell potential,
• \(R\) is the universal gas constant,
• \(T\) is the temperature in Kelvin,
• \(n\) is the number of moles of electrons transferred,
• \(F\) is the Faraday constant, and
• \(Q\) is the reaction quotient which represents the ratio of the concentrations of the products to the reactants.

For the standard conditions where all reactants and products have 1 M concentration, the reaction quotient \(Q\) is 1, leading to the simplification that \(E = E^\circ\). This is important for calculating the standard cell potential, which is a stepping stone in determining the equilibrium constant for the reaction.

Standard Cell Potential

The standard cell potential \(E^\circ\) is a measure of the driving force behind an electrochemical reaction. It's essentially a voltage that represents the tendency of a chemical species to gain or lose electrons, measured under standard conditions (25°C, 1M concentration of all reactants and products, and a pressure of 1 atm). To calculate the standard cell potential for a full reaction, you use the standard reduction potentials of each half-reaction. Remember, oxidation potentials can be found by taking the negative of the reduction potential. The standard cell potential is then calculated by subtracting the standard oxidation potential from the standard reduction potential of the half-reactions involved in the overall reaction. So for any electrochemical reaction, the accurate determination of \(E^\circ\) is vital since it directly affects the calculation of the equilibrium constant.

Equilibrium Constant

The equilibrium constant, represented by \(K\), is a dimensionless value that indicates the ratio of the concentration of the products to the reactants at equilibrium for a reversible chemical reaction. It gives an idea of how far a reaction will go before reaching equilibrium. There is a direct relationship between the equilibrium constant and the standard cell potential which is described by the formula:
\[ E^\circ = \frac{2.303 RT}{nF} \log K \]
When you know the standard cell potential and the number of electrons transferred in the reaction, this equation allows you to calculate the equilibrium constant. A higher standard cell potential corresponds to a larger equilibrium constant, indicating a greater tendency for the reaction to proceed to completion. It's crucial to take into account the correct stoichiometry of the reaction when calculating \(K\). Incorrect stoichiometry can result in inaccurate values, so any errors must be corrected as demonstrated in the step-by-step solution.

Electrochemical Cell

An electrochemical cell is a device capable of generating electrical energy from chemical reactions or facilitating chemical reactions through the introduction of electrical energy. There are two main types of electrochemical cells: Galvanic (or voltaic) cells, which convert chemical energy into electrical energy, and electrolytic cells, which use electrical energy to drive non-spontaneous chemical reactions. Each cell consists of two half-cells where oxidation and reduction reactions occur. The performance of an electrochemical cell is characterized by its cell potential, which can be calculated using the Nernst equation. Understanding the workings of electrochemical cells is essential in fields such as energy storage (like batteries), industrial chemistry (for electrolysis), and electronics. The principles governing these cells are foundational to the understanding of broader concepts such as the equilibrium constant of a reaction.