Question

Given the following standard reduction potentials, calculate the ion-product, \(K_{\mathrm{w}},\) for water at \(25^{\circ} \mathrm{C}:\) $$ \begin{array}{ll} 2 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{H}_{2}(\mathrm{~g}) & E^{\circ}=0.00 \mathrm{~V} \\ 2 \mathrm{H}_{2} \mathrm{O}(l)+2 e^{-} \longrightarrow \mathrm{H}_{2}(g)+2 \mathrm{OH}^{-}(a q) & E^{\circ}=-0.83 \mathrm{~V} \end{array} $$

Short answer

First apply Nernst Equation, then calculate using E = E_1 - E_2.

Step-by-step solution

Finding the Nernst Equation for Water

The Nernst equation relates the Gibbs free energy change and the potential difference ( (E^∘ )). First, remember that the Gibbs free energy is related to the equilibrium constant by ΔG^∘=−RTlnK . We can use ΔG^∘=−nFE^∘ , where n is the number of moles of electrons, F is Faraday's constant, and E^∘ is the standard electrode potential to find it. We have two reduction reactions for water: 1. for H⁺ to H₂, E^∘=0.00 V 2. for 2H₂O to OH⁻, E^∘=-0.83 V

Key concepts

Standard Reduction Potential

The standard reduction potential, denoted as \(E^{\circ}\), measures a substance’s tendency to gain electrons. It’s crucial in determining cell voltage in electrochemical cells. In our exercise, we have two reactions, each with different \(E^{\circ}\) values:

• The reaction of \(2\ \text{H}^+(aq) + 2\ e^- \rightarrow \text{H}_2(g)\) has an \(E^{\circ}\) of \(0.00\ \text{V}\)
• The second reaction, \(2\ \text{H}_2\text{O}(l) + 2\ e^- \rightarrow \text{H}_2(g) + 2\ \text{OH}^-(aq)\), has an \(E^{\circ}\) of \(-0.83\ \text{V}\)

These values indicate how likely each reaction is to occur under standard conditions (1 M concentration and 1 atm pressure).
An important point to note is that more positive \(E^{\circ}\) values suggest a greater tendency to be reduced. Since the first reaction has a higher \(E^{\circ}\), it is more favorable in reduction potential terms.

Ion-Product Constant

The ion-product constant for water, \(K_{\mathrm{w}}\), is a fundamental concept in chemistry that expresses the product of the concentrations of hydrogen ions and hydroxide ions in water. At \(25^{\circ}C\), \(K_{\mathrm{w}}\) is typically \(1.0 \times 10^{-14}\). It links with the reduction potentials to show the balance of ions at equilibrium.
The relationship \(K_{\mathrm{w}} = [\text{H}^+][\text{OH}^-]\) suggests that understanding \(K_{\mathrm{w}}\) helps in predicting how any chemical reaction will shift to maintain equilibrium. In the context of the exercise, calculating \(K_{\mathrm{w}}\) involves using the standard electrochemical methods to ultimately find the Gibbs free energy, revealing how likely water autoionizes under given conditions.

Nernst Equation

The Nernst equation is a fundamental formula used in electrochemistry. It helps link the cell potential to temperature, pressure, and concentration of chemicals involved. The equation \(E = E^{\circ} - \frac{RT}{nF} \ln Q\) explains how a change in concentration or pressure alters the potential of an electrochemical cell.
For this exercise, understanding the Nernst equation helps relate the standard reduction potentials to real-world conditions, allowing us to calculate the ion-product constant. Here, it serves to connect \(E^{\circ}\) and the equilibrium constant through Gibbs free energy. It translates the theoretical standard conditions to equilibrium situations.

Gibbs Free Energy

Gibbs free energy, \(\Delta G^{\circ}\), represents the energy available to do work during a chemical reaction at constant temperature and pressure. Its relation to electrochemistry is seen in the equation \(\Delta G^{\circ} = -nFE^{\circ}\), where \(n\) is the number of moles of electrons and \(F\) is Faraday's constant.
In this context, \(\Delta G^{\circ}\) links to the equilibrium constant by \(\Delta G^{\circ} = -RT \ln K\). By combining these equations, we can find the equilibrium constant \(K\), which then helps us to identify properties such as \(K_{\mathrm{w}}\).

• If \(\Delta G^{\circ}\) is negative, the reaction is spontaneous
• If \(\Delta G^{\circ}\) is positive, the reaction is non-spontaneous

This helps us determine how the reaction will proceed and is vital for solving problems related to ion-product constants in water.