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Estimate \(\mathscr{E}^{\circ}\) 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 \(\mathscr{E}^{\circ}\) given in Table 17-1.

Short Answer

Expert verified
The standard electrode potential (\(\mathscr{E}^{\circ}\)) for the given half-reaction can be estimated using the provided Gibbs free energies of formation. For the balanced half-reaction, the number of transferred electrons is 2. The standard change of Gibbs free energy is calculated as -314 kJ/mol. Using the relationship between \(\mathscr{E}^{\circ}\) and \(\Delta G_{\mathrm{r}}^{\circ}\), we find that \(\mathscr{E}^{\circ} \approx +1.63\, \mathrm{V}\). This value is higher by 0.4 V than the value given in Table 17-1 (1.23 V), indicating that our estimate might not be as accurate but still provides a reasonable approximation of the standard electrode potential for the given half-reaction.

Step by step solution

01

Identify the number of electrons transferred

The balanced half-reaction is already given: \(2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}\) Here, two electrons are transferred, therefore: $$ n = 2 $$
02

Calculate the standard change of Gibbs free energy

To find the standard change of Gibbs free energy (\(\Delta G_{\mathrm{r}}^{\circ}\)), we will subtract the Gibbs free energy of the reactants from that of the products: $$ \Delta G_{\mathrm{r}}^{\circ} = [\Delta G^{\circ}_{\mathrm{H}_2} + 2\Delta G^{\circ}_{\mathrm{OH}^{-}}] - [2\Delta G^{\circ}_{\mathrm{H}_2 \mathrm{O}} + 0] $$ Substitute the provided values: $$ \Delta G_{\mathrm{r}}^{\circ} = [(0 \,\mathrm{kJ/mol}) + 2(-157 \,\mathrm{kJ/mol})] - [2(-237 \,\mathrm{kJ/mol}) + 0] = -314\,\mathrm{kJ/mol} $$
03

Calculate the standard electrode potential

Now, we will use the relationship between \(\mathscr{E}^{\circ}\) and \(\Delta G_{\mathrm{r}}^{\circ}\), $$ \Delta G_{\mathrm{r}}^{\circ} = -nFE_{\text{cell}}^{\circ} $$ Rearrange the equation to get the standard electrode potential: $$ \mathscr{E}^{\circ} = -\frac{\Delta G_{\mathrm{r}}^{\circ}}{nF} $$ Using the value of Faraday's constant F = 96,485 C/mol, $$ \mathscr{E}^{\circ} = -\frac{-314 \,\mathrm{kJ/mol}}{2 \times 96485 \,\mathrm{C/mol}} \times \frac{1000\,\mathrm{J}}{\mathrm{kJ}} \approx +1.63\, \mathrm{V} $$ Comparing this result with the value of \(\mathscr{E}^{\circ}\) given in Table 17-1 (1.23 V), we see that the calculated value is higher by 0.4 V, indicating that our estimate might not be as accurate but still gives a decent idea of the standard electrode potential for the given half-reaction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gibbs Free Energy
Gibbs free energy, often represented as \( \Delta G \), is a thermodynamic potential that helps predict whether a reaction can occur spontaneously. The change in Gibbs free energy specifically indicates the difference in energy between the products and reactants under constant temperature and pressure. This difference helps determine the favorability of a reaction. A negative \( \Delta G \) implies that the reaction occurs spontaneously.
In our exercise, we calculate the standard change in Gibbs free energy (\( \Delta G_{\mathrm{r}}^{\circ} \)) by considering the formation energies of reactants and products under standard conditions. Given the labeled formation energies (\( \Delta G_{\mathrm{f}}^{\circ} \)), the calculated value for our half-reaction was \(-314\,\text{kJ/mol}\). Since \( \Delta G_{\mathrm{r}}^{\circ} \) is negative, it indicates the reaction is favorable as written.
Half-Reaction
A half-reaction refers to either the oxidation or reduction reactions that occur in a redox process. In these reactions, electrons are transferred between reactants, which is key for understanding the changes in electrode potential.
For our exercise, the given half-reaction is:
  • \( 2 \mathrm{H}_{2} \mathrm{O} + 2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2} + 2 \mathrm{OH}^{-} \)
This half-reaction shows that two electrons are involved, highlighting the importance of balancing the number of electrons for accurate calculations of the Gibbs free energy and, subsequently, the electrode potential. Half-reactions need to correctly account for charges and elements' states to ensure that the overall redox process is balanced.
Faraday's Constant
Faraday's constant is a crucial value in electrochemistry, symbolized by \( F \). It is numerically equivalent to the charge of one mole of electrons, approximately \( 96,485 \text{C/mol} \). This constant is key in calculating the relationship between the Gibbs free energy and the electrode potential in electrochemical cells.
For students, understanding Faraday's constant is essential when relating thermodynamic quantities to electrical properties. In the exercise, it was used to relate \( \Delta G_{\mathrm{r}}^{\circ} \) to \( \mathscr{E}^{\circ} \) via the formula:
  • \( \mathscr{E}^{\circ} = -\frac{\Delta G_{\mathrm{r}}^{\circ}}{nF} \)
Here, \( n \) represents the number of moles of electrons transferred, which is multiplied by Faraday's constant to provide the necessary conversion from energy per mole to potential in volts.
Standard Conditions
Standard conditions in chemistry are often referred to as the conditions under which measurements are made, typically at a temperature of 25°C (298 K) and a pressure of 1 atm for gases. Solutions are often considered to be at a concentration of 1 M.
In electrochemistry, standard conditions are vital because they provide a common reference point to compare electrode potentials across different reactions. When dealing with Gibbs free energy and electrode potentials, like in our example, the "standard" prefix denotes these conditions, hence the notation like \( \Delta G^{\circ} \) and \( E^{\circ} \).
Under these standard conditions, the values of thermodynamic constants and potential differences remain consistent, allowing for reliable calculations and comparisons between theoretical values and those observed in lab settings.

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Most popular questions from this chapter

What is the maximum work that can be obtained from a hydrogen-oxygen fuel cell at standard conditions that produces \(1.00 \mathrm{kg}\) water at \(25^{\circ} \mathrm{C} ?\) Why do we say that this is the maximum work that can be obtained? What are the advantages and disadvantages in using fuel cells rather than the corresponding combustion reactions to produce electricity?

What volume of \(\mathrm{F}_{2}\) gas, at \(25^{\circ} \mathrm{C}\) and 1.00 atm, is produced when molten KF is electrolyzed by a current of 10.0 A for \(2.00 \mathrm{h} ?\) What mass of potassium metal is produced? At which electrode does each reaction occur?

Consider only the species (at standard conditions) $$ \mathrm{Br}^{-}, \quad \mathrm{Br}_{2}, \quad \mathrm{H}^{+}, \quad \mathrm{H}_{2}, \quad \mathrm{La}^{3+}, \quad \mathrm{Ca}, \quad \mathrm{Cd} $$ in answering the following questions. Give reasons for your answers. a. Which is the strongest oxidizing agent? b. Which is the strongest reducing agent? c. Which species can be oxidized by \(\mathrm{MnO}_{4}^{-}\) in acid? d. Which species can be reduced by \(\mathrm{Zn}(s) ?\)

Explain why cell potentials are not multiplied by the coefficients in the balanced redox equation. (Use the relationship between \(\Delta G\) and cell potential to do this.)

The overall reaction in the lead storage battery is $$\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ a. For the cell reaction \(\Delta H^{\circ}=-315.9 \mathrm{kJ}\) and \(\Delta S^{\circ}=\) \(263.5 \mathrm{J} / \mathrm{K} .\) Calculate \(\mathscr{E}^{\circ}\) at \(-20 .^{\circ} \mathrm{C} .\) Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. b. Calculate \(\mathscr{E}\) at \(-20 .^{\circ} \mathrm{C}\) when \(\left[\mathrm{HSO}_{4}^{-}\right]=\left[\mathrm{H}^{+}\right]=4.5 \mathrm{M}\) c. Consider your answer to Exercise \(71 .\) Why does it seem that batteries fail more often on cold days than on warm days?

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