Chapter 19: Problem 113
Given the following standard reduction potentials, calculate the ion-product, \(K_{\mathrm{w}},\) for water at \(25^{\circ} \mathrm{C}\) : \(2 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{H}_{2}(g) \quad E^{\circ}=0.00 \mathrm{~V}\) \(2 \mathrm{H}_{2} \mathrm{O}(l)+2 e^{-} \longrightarrow \mathrm{H}_{2}(g)+2 \mathrm{OH}^{-}(a q)\) $$ E^{\circ}=-0.83 \mathrm{~V} $$
Short Answer
Step by step solution
Formulating the overall reaction
Calculate the cell potential (E°)
Apply the Nernst equation
Convert Q into \(K_w\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Reduction Potentials
In electrochemistry, each half-cell reaction has an associated standard reduction potential, which is the voltage observed when the half-cell is connected to a standard hydrogen electrode, which has a potential defined as 0.00 V. The more positive the potential, the greater the species' affinity for electrons.
In the exercise, the reaction involving the reduction of hydrogen ions (\( H^+ \rightarrow H_2 \text{(g)} \) ) to form hydrogen gas has a potential of 0.00 V, while the reaction where water is reduced to form hydrogen gas and hydroxide ions (\( 2 H_2O (l) + 2 e^- \rightarrow H_2 (g) + 2 OH^- (aq) \) ) has a potential of -0.83 V. The potential for these half-reactions allows us to calculate the overall cell potential when they are combined in a full electrochemical cell.
Nernst Equation
It is expressed as: \[ E = E^{\text{deg}} - (0.0592/n) \times \text{log} Q \], where:
- \( E \) represents the cell potential when non-standard conditions are present.
- \( E^{\text{deg}} \) stands for the standard cell potential, which is the potential of the cell under standard conditions.
- \( n \) is the number of moles of electrons exchanged in the redox reaction.
- \( Q \) is the reaction quotient, a ratio of the concentrations of products to reactants raised to the power of their stoichiometric coefficients.
In the provided exercise, applying the Nernst equation helps us to relate the ion concentrations of \( H^+ \) and \( OH^- \) at equilibrium in water. It is by setting the cell potential \( E \) to zero (since at equilibrium there is no net reaction occurring) that we can solve for \( Q \), which directly gives us the ion-product constant for water, \( K_w \).
Equilibrium Constant
In our context, the ion-product constant for water, \( K_w \), is a type of equilibrium constant that specifically relates to the auto-ionization of water: \[ K_w = [H^+][OH^-] \].
At 25°C, \( K_w \) has a value of approximately \( 1 \times 10^{-14} \), which implies that the product of the hydrogen ion concentration and the hydroxide ion concentration in pure water is a constant at this temperature. By calculating \( K_w \) from standard reduction potentials and applying the Nernst equation, we can determine the extent of water's tendency to undergo auto-ionization under standard conditions.
Auto-ionization of Water
This reversible reaction is significant because it establishes the basis for pH and the concept of acidity and basicity in aqueous solutions. Although the auto-ionization is usually minimal and the concentration of \( H^+ \) and \( OH^- \) is low, it is nonetheless present in all aqueous solutions.
The exercise demonstrates how to calculate \( K_w \), the equilibrium constant for this process, by using the concept of standard reduction potentials and managing it through the Nernst equation. Understanding the auto-ionization of water is crucial, as \( K_w \) provides the fundamental link between hydronium and hydroxide ion concentrations, thus playing a pivotal role in the study of acid-base chemistry.