Question

The standard \(\mathrm{E}^{\circ}\) for \((1 / 2) \mathrm{F}_{2}(\mathrm{~g})+\mathrm{e}^{-} \rightarrow \mathrm{F}^{-}(\mathrm{aq})\), which is \(+2.87 \mathrm{~V}\), applies when the flouride ion concentration is \(1 \mathrm{M}\). What would the corresponding \(\mathrm{E}\) be in \(1 \mathrm{M} \mathrm{H}_{3} \mathrm{O}^{+}\), i.e., for the electrode reaction \((1 / 2) \mathrm{F}_{2}(\mathrm{~g})+\mathrm{e}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \rightarrow \mathrm{HF}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} ?\) (The \(\mathrm{K}_{\text {diss }}\) of HF is \(6.7 \times 10^{-4}\) ).

Short answer

The new electrode potential E for the given reaction \((1/2) F_2(g) + e^- + H_3 O^+ \rightarrow HF(aq) + H_2 O\) in 1 M \(H_3O^+\) is 2.87 V.

Step-by-step solution

Write the Nernst equation

The Nernst equation helps us find the electrode potential(E) at given concentrations of reactants and products. The equation is given as: \[E = E^\circ - \frac{RT}{nF} \ln Q\] Where \(E^\circ\) is the standard electrode potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred, and F is the Faraday constant. Q is the reaction quotient.

Identify the change in concentration of fluoride ions

The given reaction is \((1/2) F_2(g) + e^- + H_3 O^+ \rightarrow HF(aq) + H_2 O\). The change in concentration of fluoride ions is important for the Nernst equation. Since we are given the K\(_{diss}\) of HF, we can find the concentration of fluoride ions in this reaction as follows: \(K_{diss} = \frac{[HF]_e}{[F^-]_e[H_3O^+]_e}\) Since, \([HF]_e = [F^-]_e\) and \([H_3O^+]_e = 1\,\text{M}\), we can substitute and solve for \([F^-]_e\): \(6.7 \times 10^{-4} = \frac{[F^-]_e}{[F^-]_e \times 1}\) This means, \([F^-]_e = 6.7 \times 10^{-4} \,\text{M}\).

Calculate Q for the given reaction

The reaction quotient (Q) is calculated as follows for the given reaction: \(Q = \frac{[HF]_e}{[F^-]_e[H_3O^+]_e}\) Substituting the values of the concentrations, we have: \(Q = \frac{6.7 \times 10^{-4}}{6.7 \times 10^{-4} \times 1} = 1\)

Apply the Nernst equation to calculate E

Use the Nernst equation and the values found above to solve for the new electrode potential E: \(E = E^\circ - \frac{RT}{nF} \ln Q\) Since only one electron is transferred in both reactions \((1/2) F_2(g) + e^- \rightarrow F^-(aq)\) and \((1/2) F_2(g) + e^- + H_3 O^+ \rightarrow HF(aq) + H_2 O\), we have n = 1. Assuming a temperature of 298 K and using the gas constant \(R = 8.314\, J/(K \cdot mol)\) and Faraday constant \(F = 96485\, C/mol\), we get: \(E = 2.87\,V - \frac{8.314 \times 298}{1 \times 96485} \ln 1\) Since \(\ln 1 = 0\), the new E value is equal to the standard E value: \(E = 2.87\,V\)

Conclusion

The new electrode potential E for the given reaction \((1/2) F_2(g) + e^- + H_3 O^+ \rightarrow HF(aq) + H_2 O\) in 1 M \(H_3O^+\) is 2.87 V.

Key concepts

Electrode Potential

Electrode potential is the voltage or electric potential difference between two electrodes when they are connected by a circuit. In simple terms, it’s the ability of a system to pull electrons toward itself. This concept is central to electrochemistry and is the driving force behind batteries and electrochemical cells.

When we discuss the Nernst equation, electrode potential represents the balance point where there is no net flow of electrons. It's influenced by factors like the concentration of ions, temperature, and the inherent capacity of the substance to gain or lose electrons, commonly referred to as its standard electrode potential.

In the exercise, students are asked to calculate the electrode potential of a half-reaction involving fluorine gas and fluoride ions under non-standard conditions, showcasing how electrode potential can change with different concentrations.

Reaction Quotient (Q)

The reaction quotient, denoted as Q, compares the relative amounts of products and reactants present during a reaction at a given point in time. It's expressed in the same form as the equilibrium constant (K), but while K is calculated at equilibrium, Q can be determined at any stage of the reaction.

In the context of the Nernst equation, Q is crucial because it reflects the extent to which a reaction has proceeded. It is essentially a snapshot of the reaction's current state and is calculated by plugging in the current concentrations of the reactants and products into the equilibrium expression. In the exercise, when the reaction quotient is used in the Nernst equation, it helps to adjust the standard electrode potential to the conditions that are present in the electrochemical cell.

Standard Electrode Potential (E°)

Standard electrode potential (E°) represents the voltage of an electrode under standard conditions, which are typically defined as all solutes at a 1 M concentration and gases at 1 atm pressure, with the reaction at 25°C. This value is fundamental in electrochemistry, serving as the baseline to which actual electrode potentials are compared.

The standard electrode potential is useful because it gives us the ability to predict the direction of electron flow. Positive values indicate that a substance is likely to gain electrons (reduction), while negative values suggest a tendency to lose electrons (oxidation). In our exercise, the standard electrode potential for the reaction involving fluoride ions is given as +2.87 V, telling us that the half-reaction under standard conditions is a strong oxidizing process.

Gas Constant (R)

The gas constant, R, is a fundamental constant in the field of chemistry. It connects energy scales to temperature and appears in the ideal gas law as well as the Nernst equation. Its value is approximately 8.314 J/(K⋅mol) and is the same for all gases, which makes it a universal constant.

When we're dealing with the Nernst equation, the gas constant comes into play as part of a term that relates the reaction temperature to the potential energy difference across a cell. It illustrates how temperature can influence the movement of electrons and thus the electrode potential. In our exercise solution, R is used to modify the standard electrode potential based on temperature to find the actual electrode potential.

Faraday Constant (F)

The Faraday constant (F) is the amount of electric charge per one mole of electrons, and it’s a key figure in electrochemistry. It is approximately equal to 96485 C/mol and represents the charge of a mole of electrons. When electrons move in a reaction, they carry charge with them, and the Faraday constant enables us to quantify this charge in moles.

In the Nernst equation, the Faraday constant is part of the calculation of the second term that helps adjust the standard electrode potential to real conditions. It relates the molar charge to the voltage, which is an energy per charge quantity. As illustrated in the exercise, F is essential in determining how changes in the reaction quotient affect the actual electrode potential.