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

Expert verified
The ion-product \(K_w\) for water at 25°C is \(10^{-14}\).

Step by step solution

01

Formulating the overall reaction

Combine the two half-cell reactions to get the net ionic equation. The net ionic equation - where water dissociates into protons \(H^{+}\) and hydroxide ions \(OH^{-}\) - is obtained by multiplying the first equation by 2 and adding it to the reversed version of the second equation. The result is: \n\[2H_2O(l) \rightarrow 2H^+(aq) + 2OH^-(aq)\]
02

Calculate the cell potential (E°)

Next, calculate the cell potential (\(E°\)) for this reaction by subtracting the reduction potential of the second reaction from the first one (since the second reaction was reversed). \nThe standard potential for this reaction, \(E°_{\text{cell}}\), is thus equal to \(E°_{H2/H+}\) - \(E°_{H2O/H2 + 2OH-}\), which equals to \(0 - (- 0.83)\) V = 0.83 V.
03

Apply the Nernst equation

Now, we will have to apply the Nernst equation to relate the standard cell potential to the ion concentrations in water at equilibrium. The Nernst equation is given by: \n\[E = E° - \(\frac{0.0592}{n}\) \log Q\] Where \(E\) is the cell potential, \(E°\) is the standard cell potential, \(n\) is the number of electrons transferred in the cell reaction (which is 2 in this case), and \(Q\) is the reaction quotient. For water at equilibrium \(E\) equals to 0. So, \n\[0 = 0.83 V - \(\frac{0.0592}{2}\) \log Q\]. Solving for \(Q\) yields the equilibrium constant \(K_w\) which equals \(10^{-14}\).
04

Convert Q into \(K_w\)

Since \(Q\) represents the ion products at equilibrium, it can be considered as the ion-product of water or the auto-ionization constant for water, which is defined as \(K_w\) = [\(H^{+}\)][\(OH^{-}\)].

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Key Concepts

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

Standard Reduction Potentials
Standard reduction potentials are a measurement of the tendency of a chemical species to acquire electrons and thereby be reduced. They are typically measured in volts (V) and are determined under standard conditions, which includes a temperature of 25°C, a 1 molar concentration for each ion participating in the reaction, and a partial pressure of 1 atmosphere for any gases involved.

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
The Nernst equation is fundamental in the field of electrochemistry, providing a relation between the concentration of chemical species and the cell potential in an electrochemical cell.

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
The equilibrium constant, represented as \( K \), is a dimensionless value that quantifies the ratio of the concentrations of products to reactants at equilibrium. Each chemical reaction has its own unique equilibrium constant. For the reaction: \( aA + bB \rightarrow cC + dD \), the equilibrium constant \( K \) is expressed as: \[ K = \frac{[C]^c \times [D]^d}{[A]^a \times [B]^b} \], where the square brackets indicate the concentration of each species.

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
The auto-ionization of water is an important concept in chemistry, which describes the process where water molecules spontaneously dissociate into hydrogen ions (\( H^+ \)) and hydroxide ions (\( OH^- \)). The reaction can be represented as: \[ 2H_2O(l) \rightarrow H_3O^+(aq) + OH^-(aq) \] or simply \[ H_2O(l) \rightarrow H^+(aq) + OH^-(aq) \].

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.

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Most popular questions from this chapter

Calculate the standard emf of a cell that uses the \(\mathrm{Mg} / \mathrm{Mg}^{2+}\) and \(\mathrm{Cu} / \mathrm{Cu}^{2+}\) half-cell reactions at \(25^{\circ} \mathrm{C} .\) Write the equation for the cell reaction that occurs under standard-state conditions.

Industrially, copper is purified by electrolysis. The impure copper acts as the anode, and the cathode is made of pure copper. The electrodes are immersed in a \(\mathrm{CuSO}_{4}\) solution. During electrolysis, copper at the anode enters the solution as \(\mathrm{Cu}^{2+}\) while \(\mathrm{Cu}^{2+}\) ions are reduced at the cathode. (a) Write half-cell reactions and the overall reaction for the electrolytic process. (b) Suppose the anode was contaminated with \(\mathrm{Zn}\) and \(\mathrm{Ag} .\) Explain what happens to these impurities during electrolysis. (c) How many hours will it take to obtain \(1.00 \mathrm{~kg}\) of \(\mathrm{Cu}\) at a current of \(18.9 \mathrm{~A} ?\)

Under standard-state conditions, what spontaneous reaction will occur in aqueous solution among the ions \(\mathrm{Ce}^{4+}, \mathrm{Ce}^{3+}, \mathrm{Fe}^{3+},\) and \(\mathrm{Fe}^{2+}\) ? Calculate \(\Delta G^{\circ}\) and \(K_{\mathrm{c}}\) for the reaction.

Explain why most useful galvanic cells give voltages of no more than 1.5 to \(2.5 \mathrm{~V}\). What are the prospects for developing practical galvanic cells with voltages of \(5 \mathrm{~V}\) or more?

What is the emf of a cell consisting of a \(\mathrm{Pb}^{2+} / \mathrm{Pb}\) half-cell and a \(\mathrm{Pt} / \mathrm{H}^{+} / \mathrm{H}_{2}\) half-cell if \(\left[\mathrm{Pb}^{2+}\right]=0.10 \mathrm{M}\) \(\left[\mathrm{H}^{+}\right]=0.050 \mathrm{M},\) and \(\mathrm{P}_{\mathrm{H}_{2}}=1.0 \mathrm{~atm} ?\)

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