Question

Estimate \(\mathscr{C}^{\circ}\) for the half-reaction $$2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}$$ given the following values of \(\Delta G_{\mathrm{f}}^{\circ}\) : $$\begin{aligned}\mathrm{H}_{2} \mathrm{O}(l) &=-237 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g) &=0.0 \\\\\mathrm{OH}^{-}(a q) &=-157 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{e}^{-} &=0.0\end{aligned}$$ Compare this value of \(\mathscr{E}^{\circ}\) with the value of \(\mathscr{b}^{\circ}\) given in Table \(18.1\).

Short answer

The standard cell potential (\(\mathscr{C}^{\circ}\)) for the given half-reaction is calculated to be \(-0.828\ \mathrm{V}\) using the given values of \(\Delta G_{\mathrm{f}}^{\circ}\). To confirm our calculation, compare this value with the one given in Table 18.1 for the same half-reaction.

Step-by-step solution

Calculate the \(\Delta G^{\circ}\) for the reaction

We will use the following relation for calculating the Gibbs free energy change of the reaction: \[\Delta G^{\circ} = \sum \Delta G_{f}^{\circ} (\text{products}) - \sum \Delta G_{f}^{\circ}(\text{reactants})\] For the given reaction: \[\Delta G^{\circ} = [\Delta G_{\mathrm{f}}^{\circ}(H_2) + 2\Delta G_{\mathrm{f}}^{\circ}(OH^{-})] - [2\Delta G_{\mathrm{f}}^{\circ}(H_2O) + 2\Delta G_{\mathrm{f}}^{\circ}(e^{-})]\] Using the given values of \(\Delta G_{f}^{\circ}\): \[\Delta G^{\circ} = [(0.0) + 2(-157)] - [2(-237) + 2(0.0)] = -314 + 474 = 160 \mathrm{kJ/mol}\]

Calculate the \(\mathscr{C}^{\circ}\) for the reaction using the Nernst equation

The relationship between the Gibbs free energy change and standard cell potential is given by the Nernst equation: \[\Delta G^{\circ} = -nFE^{\circ}\] Where \(n\) represents the number of electrons transferred, \(F\) is the Faraday's constant (96,485 C/mol), and \(E^{\circ}\) is the standard cell potential. For the given half-reaction, \(n=2\) (two electrons are transferred): \[160000 = -2(96485)(E^{\circ})\] Now, we need to solve for \(E^{\circ}\): \[E^{\circ} = -\frac{160000}{2(96485)} = -0.828 \mathrm{V}\]

Compare the calculated value with that in Table 18.1

We've calculated the value \(\mathscr{C}^{\circ} = -0.828\). Now, we need to compare it with the value given in Table 18.1. You can consult the table in your textbook and look for the corresponding standard cell potential for the same half-reaction. If the value closely matches the calculated value of \(-0.828 \mathrm{V}\), it confirms the correctness of our calculation.

Key concepts

Gibbs Free Energy

Gibbs free energy is a critical concept in electrochemistry that helps us determine the spontaneity of a chemical reaction. It's represented by the symbol \( \Delta G \). In electrochemical reactions, we calculate the standard Gibbs free energy change (\( \Delta G^{\circ} \)) to predict whether a reaction will proceed without external energy. To compute \( \Delta G^{\circ} \), we use:

\[ \Delta G^{\circ} = \sum \Delta G_{f}^{\circ} (\text{products}) - \sum \Delta G_{f}^{\circ}(\text{reactants}) \]

This equation takes the standard Gibbs free energies of formation (\( \Delta G_{f}^{\circ} \)) for products and reactants. When the result is negative, the reaction is spontaneous under standard conditions.
Positive values imply non-spontaneity. In the exercise, this calculation allows us to understand the potential energy change during the half-reaction of converting water and electrons into hydrogen and hydroxide ions. If we can find a way to balance these energies, we can better comprehend how the elements in the half-reaction are transformed.

Nernst Equation

The Nernst equation is profoundly significant in electrochemistry since it links Gibbs free energy and the electrochemical cell potential. It expresses how the cell potential changes with concentration, temperature, and pressure, which are not under standard conditions. It is given by the equation:

\[ \Delta G^{\circ} = -nFE^{\circ} \]

Here, \( n \) is the number of moles of electrons exchanged in the half-reaction, \( F \) represents Faraday's constant (approximately 96,485 C/mol), and \( E^{\circ} \) symbolizes the standard cell potential. This equation indicates that the Gibbs free energy change and the electrical work done during the reaction are directly connected.
For the given exercise, using the Nernst equation allowed us to calculate the standard cell potential (\( E^{\circ} \)) from an already obtained \( \Delta G^{\circ} \) value. The Nernst equation also serves as a bridge, translating thermodynamic data into electrical parameters that discuss how materials transform during electrochemical reactions.

Standard Cell Potential

The standard cell potential \( E^{\circ} \) is a crucial component in electrochemical reactions, signifying the driving force behind an electrochemical cell operating under standard conditions (concentrations of 1 mol/L, 1 atmosphere, and 25°C). It represents how easily electrons can flow through the circuit and thus how much electrical energy is produced. Standard cell potential is inherently linked to the Gibbs free energy and can be derived using this relationship:

\[ E^{\circ} = -\frac{\Delta G^{\circ}}{nF} \]

In the exercise, we used this formula to convert the Gibbs free energy change into a potential difference, giving us insight into the efficiency and competitiveness of the electrochemical cell.
\( E^{\circ} \) values are critical for predicting the direction of electron flow. Negative values of \( \mathscr{C}^{\circ} \), as calculated in the problem, suggest the reverse reaction might be favorable. Consulting tables, like Table 18.1 in the textbook, allows comparisons to be made for accuracy and confirmation of theoretical calculations.