Question

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 \(\Delta G_{\mathrm{f}}^{\circ} :\) $$\quad\quad\quad \mathrm{H}_{2} \mathrm{O}(l)=-237 \mathrm{kJ} / \mathrm{mol}$$ $$\mathrm{H}_{2}(g)=0.0$$ $$\quad\quad\quad \mathrm{OH}^{-}(a q)=-157 \mathrm{kJ} / \mathrm{mol}$$ $$\quad \mathrm{e}^{-}=0.0$$ Compare this value of \(\mathscr{E}^{\circ}\) with the value of \(\mathscr{E}^{\circ}\) given in Table 18.1

Short answer

The estimated standard electrode potential (E°) for the given half-reaction \(2H_2O + 2e^- \rightarrow H_2 + 2OH^-\) is \(0.83\,V\). This value is similar to the E° value found in Table 18.1 (\(0.828\,V\)), indicating that our calculated estimate is accurate and reasonable.

Step-by-step solution

Write the reaction equation with the given values of ∆Gf°

Rewrite the half-reaction and substitute the given Gibbs free energy of formation values: \(2H_2O + 2e^- \rightarrow H_2 + 2OH^-\) \(-2(237) + 2(0) \rightarrow 0 + 2(-157)\)

Calculate the total change in Gibbs free energy (∆G°) for the reaction

Determine the total change in Gibbs free energy for the half-reaction using the stoichiometric coefficients. The change in Gibbs free energy for the complete reaction is the sum of ∆Gf° products minus the sum of ∆Gf° reactants: \(\Delta G^{\circ} = [1(0) + 2(-157)] - [2(-237) + 2(0)]\) \(\Delta G^{\circ} = (-314) - (-474)\) \(\Delta G^{\circ} = 160\,kJ/mol\)

Convert ∆G° to standard electrode potential (E°)

Given the relationship between Gibbs free energy and standard electrode potential: \(\Delta G^{\circ} = -nFE^{\circ}\) where: \(\Delta G^{\circ}\) = change in Gibbs free energy (in J/mol) n = number of moles of electrons transferred (mol) F = Faraday constant (\(96,485\,C/mol\)) \(E^{\circ}\) = standard electrode potential (V) In our case, the number of moles of electrons transferred (n) is 2 as we can see from the balanced half-reaction. Also, convert the ΔG° value from kJ/mol to J/mol: \(160 \,kJ/mol = 160,000 \,J/mol\) Now we can solve for \(E^{\circ}\): \(-160,000 \,J/mol = -2*\text{(}96,485 \,C/mol\text{)} E^{\circ}\)

Calculate E° for the reaction

To determine the E° value, we will divide both sides of the equation by -2 multiplied by the Faraday constant: \(E^{\circ} = \frac{-160,000 \,J/mol}{(-2)(96,485 \,C/mol)}\) \(E^{\circ} = 0.83\,V\) Finally, the estimated standard electrode potential (E°) for the given half-reaction is \(0.83\,V\).

Compare to value in Table 18.1

In Table 18.1, the E° value found for the given half-reaction is approximately \(0.828\,V\). This is similar to the E° value calculated in this exercise (\(0.83\,V\)), demonstrating that our calculated estimate is accurate and reasonable.

Key concepts

Half-reaction

A half-reaction is a part of a redox reaction. It involves either the oxidation or reduction process, but not both. When writing half-reactions, we separate the reaction into two parts: one for the gain of electrons (reduction) and one for the loss of electrons (oxidation). In the context of our exercise, the half-reaction is:

• Reacts: 2 \[H_2O + 2e^- \longrightarrow H_2 + 2OH^-\]

Here, two electrons are used, indicating a reduction process. Each component of the half-reaction involves distinct species. Water (\(H_2O\)) and hydroxide (\(OH^-\)) ions are part of the redox transformation, showcasing how electrons are transferred. Seeing half-reactions as part of larger chemical processes helps in understanding redox systems. These reactions are crucial in electrochemistry, helping delineate the steps in energy conversion processes.
Learning to balance and interpret half-reactions is essential for mastering broader chemical reactions.

Gibbs free energy

Gibbs free energy (\(\Delta G^\circ\)) helps predict whether a chemical reaction can occur spontaneously. It embodies the energy difference between reactants and products and influences chemical equilibria. In our example step-by-step solution, we calculate:

• \(\Delta G^{\circ} = 160 \,kJ/mol\)

This calculation stems from subtracting the Gibbs free energy of reactants from that of products. Gibbs free energy determines the feasibility of a reaction under constant pressure and temperature. If \(\Delta G^\circ\) is negative, the reaction proceeds spontaneously. Here, we used it to compute the electrode potential later. Converting values by switching units from kilojoules to joules correctly impacts the subsequent calculation for the electrode potential. This conversion ensures that all necessary units align for proper integration into electrochemical formulas.

Faraday constant

The Faraday constant (\(F\)) plays a significant role in electrochemical calculations. It represents the charge per mole of electrons, valued at approximately \(96,485\) coulombs per mole. In this context, the Faraday constant lets us relate charge transfer to energy changes in electrochemistry.
When finding \(E^\circ\) from \(\Delta G^\circ\), knowing the mole of electrons that participate in the reaction is crucial. For our exercise, the constant is used within the equation:

• \(\Delta G^{\circ} = -nFE^{\circ}\)

Electrons often facilitate reactions in chemistry, and the Faraday constant gives us a bridge to convert these chemical reactions into practical, quantitative data. It is essential not only for electrochemical cell calculations but also for understanding how chemical energy can be converted into electrical energy efficiently.

Standard electrode potential

The standard electrode potential (\(\mathscr{E}^{\circ}\)) measures the ability of a half-cell to be reduced, relative to a standard hydrogen electrode. This is under standard conditions, where concentration, pressure, and temperature are controlled. Calculated electrode potentials give insight into the energy available from a redox reaction. For our half-reaction, we found:

• \(E^{\circ} = 0.83 \,V\)

This value offers an insight into the reaction's tendency to proceed and electron acceptance or donation compared to other processes. We compare it with known values, like in Table 18.1, giving us assurance that our calculations align with established reference material. Understanding standard electrode potential is vital for applications in designing batteries, electrolysis, and evaluating reaction spontaneity. It helps chemists tailor reactions for desired energy output.