Problem 59 - Chemistry Principles and Practice

Question

The standard potential of the half-reaction $$ 2 \mathrm{D}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{D}_{2}(\mathrm{~g}) $$ (where \(\mathrm{D}=\) deuterium, or \(\left.{ }^{2} \mathrm{H}\right)\) is \(-0.013 \mathrm{~V}\). Determine \(\Delta G^{\circ}\) and \(K_{\mathrm{eq}}\) for the reaction $$ \mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{D}^{+}(\mathrm{aq}) \rightarrow 2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{D}_{2}(\mathrm{~g}) $$ In a mixture of hydrogen and deuterium, which isotope more favors its elemental form under standard conditions?

Step-by-Step Solution

Answer

\(\Delta G^{\circ} = 2.52 \text{ kJ/mol}, K_{eq} = 0.364\); Hydrogen is more favored in the elemental form.

Step 1: Identify the Standard Cell Potential

The standard cell potential for the entire reaction is determined by the given half-reaction potential. The half-reaction potential is given as \(-0.013\, \text{V}\), and since the reaction involves these half-reactions, we will use this value to calculate further parameters.

Step 2: Use the Nernst Equation for Gibbs Free Energy

The relationship between the Gibbs Free Energy change, \(\Delta G^{\circ}\), and the standard cell potential, \(E^{\circ}\), is given by the equation: \( \Delta G^{\circ} = -nFE^{\circ} \), where \(n\) is the number of moles of electrons transferred, and \(F\) is Faraday's constant (approximately \(96485\, \text{C/mol}\)). The number of moles of electrons \(n\) in the reaction is 2.

Step 3: Calculate \(\Delta G^{\circ}\)

Insert the known values into the equation to find \(\Delta G^{\circ}\):\[ \Delta G^{\circ} = -2 \times 96485 \times (-0.013) \]\[ \Delta G^{\circ} = 2516.61 \text{ J/mol} \] or approximately \(2.52 \text{ kJ/mol} \).

Step 4: Calculate the Equilibrium Constant

The relationship between \(\Delta G^{\circ}\) and the equilibrium constant \(K_{eq}\) is given by the equation:\[ \Delta G^{\circ} = -RT \ln K_{eq} \]where \(R\) is the universal gas constant (\(8.314 \text{ J/(mol K)}\)) and \(T\) is the temperature in Kelvin (assuming standard conditions, \(298\, K\)). Solving for \(K_{eq}\):\[ \ln K_{eq} = -\frac{\Delta G^{\circ}}{RT} = -\frac{2516.61}{8.314 \times 298} = -1.01 \]Taking the exponential of both sides, we find:\[ K_{eq} = e^{-1.01} \approx 0.364 \].

Step 5: Determine the Favored Isotope Form

The calculated \(K_{eq} < 1\) suggests that at equilibrium, the reactants (\(\mathrm{H}_2(\mathrm{~g})\) and \(\mathrm{D}_2(\mathrm{~g})\)) are favored. Therefore, under standard conditions, hydrogen in the element form (\(\mathrm{H}_{2}\)) is slightly more favored than deuterium (\(\mathrm{D}_{2}\)).

Key Concepts

Standard Cell Potential

The standard cell potential, often symbolized as \(E^\circ\), represents the voltage potential difference between two half-cells in an electrochemical cell. It is measured under standard conditions where concentrations are 1 M, pressures are 1 atm, and the temperature is 298 K.
When dealing with reactions like the one given in the exercise, the standard cell potential is a key indicator of how spontaneous the reaction is. A negative \(E^\circ\), as seen here with \(-0.013 \, \text{V}\), typically means the cell works in the reverse of what might usually be expected; that is, the reactants are more stable than the products under these conditions.
To find the overall standard cell potential for a reaction, it's simply a matter of summing the potentials of the individual half-reactions. This involves flipping the sign of the potential for oxidation reactions, as reductions are what are provided in standard tables.

Nernst Equation

The Nernst equation provides a relationship between the reaction quotient and the electrode potential of a redox reaction under non-standard conditions. Although primarily used here to estimate Gibbs Free Energy for standard conditions, it aids in understanding how changes in concentration affect cell potential.
The Nernst equation is written as: \[E = E^\circ - \frac{RT}{nF} \ln Q\]Where:

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

In our exercise's context at standard conditions, \(Q = 1\) leading the focus on calculating \(\Delta G^\circ\) utilizing the equation \( \Delta G^\circ = -nFE^\circ \). This establishes that with two electrons transferred and the given \(E^\circ\), the ΔG° is positive, implying non-spontaneity under standard conditions.

Equilibrium Constant

The equilibrium constant \(K_{eq}\), is a fundamental concept reflecting the ratio of products to reactants at equilibrium for a given chemical reaction. It's directed by the relationship with Gibbs Free Energy: \[ \Delta G^\circ = -RT \ln K_{eq} \] This relationship guides us in understanding how \(\Delta G^\circ\) and \(K_{eq}\) mathematically relate to reaction spontaneity.
If \(K_{eq} = 1\), the reaction has no preference between products and reactants. If \(K_{eq} > 1\), products are favored, but if \(K_{eq} < 1\), as in our exercise where \( K_{eq} \approx 0.364\), reactants are favored.
This implies that, at equilibrium, the amount of reactants is greater than the amount of products, reinforcing the finding from \(\Delta G^\circ\) that the reaction is not spontaneous under standard conditions. Therefore, hydrogen and deuterium prefer to exist as \(\mathrm{H}_2\) and \(\mathrm{D}_2\) rather than converting into their ionized forms.