Problem 71 - Chemistry

Question

From the following information, calculate the solubility product of \(\mathrm{AgBr}\) : $$ \begin{array}{ll} \mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \mathrm{Ag}(s) & E^{\circ}=0.80 \mathrm{~V} \\ \mathrm{AgBr}(s)+e^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Br}^{-}(a q) & E^{\circ}=0.07 \mathrm{~V} \end{array} $$

Step-by-Step Solution

Answer

The solubility product \(K_{sp}\) of \(\mathrm{AgBr}\) is approximately \(5.2 \times 10^{-13}\).

Step 1: Understand the Concepts

The solubility product, denoted as \(K_{sp}\), is an equilibrium constant for a solid substance dissolving in an aqueous solution. In this scenario, we want to determine \(K_{sp}\) for \(\mathrm{AgBr}\). The provided half-reactions and standard electrode potentials \(E^{\circ}\) will help us calculate this value.

Step 2: Write the Relevant Equations

We have two half-reactions: 1. \(\mathrm{Ag}^+(aq) + e^- \rightarrow \mathrm{Ag}(s)\)2. \(\mathrm{AgBr}(s) + e^- \rightarrow \mathrm{Ag}(s) + \mathrm{Br}^-(aq)\). We need the net equation for the dissolution of \(\mathrm{AgBr}\) into \(\mathrm{Ag}^+\) and \(\mathrm{Br}^-\) ions: \(\mathrm{AgBr}(s) \rightleftharpoons \mathrm{Ag}^+(aq) + \mathrm{Br}^-(aq)\).

Step 3: Determine the Cell Potential

To find the \(K_{sp}\), we need to use the Nernst equation, which relates the cell potential to the equilibrium constant (\(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\)). Identify the anode and cathode reactions:- Reaction 1 is the anode with \(E^{\circ} = 0.80\, \mathrm{V}\).- Reaction 2 is the cathode with \(E^{\circ} = 0.07\, \mathrm{V}\).Calculate the overall cell potential: \(E^{\circ}_{\text{cell}} = 0.07 - 0.80 = -0.73\, \mathrm{V}\).

Step 4: Calculate the Solubility Product

Use the equation \(\ln K = \frac{nFE^{\circ}_{\text{cell}}}{RT}\), where \(n = 1\) (the number of moles of electrons), \(F = 96485\, \mathrm{C/mol}\) (Faraday's constant), \(R = 8.314\, \mathrm{J/(mol\cdot K)}\) (universal gas constant), and \(T\) is the temperature (usually 298 K for standard conditions). Rework to find \(K\): \(K = e^{\frac{nFE^{\circ}_{\text{cell}}}{RT}}\). Substitute the known values to get \(K_{sp} = e^{\frac{1 \times 96485 \times -0.73}{8.314 \times 298}}\). Evaluating this gives \(K_{sp}\).

Step 5: Compute and Verify the Value

Calculate the value based on the expression from the previous step: \[K_{sp} = e^{\frac{(1)(96485)(-0.73)}{(8.314)(298)}} = e^{-28.21} \approx 5.2 \times 10^{-13}\]. Recheck calculations to ensure accuracy.

Key Concepts

Electrochemistry

Electrochemistry involves the study of chemical processes that cause electrons to move. This movement is what creates electricity. In an electrochemical reaction, two half-reactions occur; one involves oxidation (loss of electrons) and the other reduction (gain of electrons). These reactions take place in two separate electrodes known as the anode and the cathode. The pushing or pulling of electrons across these electrodes is measured as an electric potential. Electrochemistry is central to processes like battery operation and electroplating.

Understanding electrochemistry helps us connect chemical reactions and the electrical energy they produce or consume. This is vital in various real-world applications, including energy storage and electrolysis.

Nernst Equation

The Nernst equation helps link the electrical potential of a cell to its chemical properties. This vital equation is \[ E = E^{\circ} - \frac{RT}{nF} \ln Q \] where:

• \( E \) is the cell potential under non-standard conditions.
• \( E^{\circ} \) is the standard cell potential.
• \( R \) is the gas constant.
• \( T \) is the temperature in Kelvin.
• \( n \) is the number of moles of electrons.
• \( F \) is Faraday's constant.
• \( Q \) is the reaction quotient.

This equation gives us a way to calculate how the potential changes with varying concentrations or pressures of the products and reactants. It is particularly important for determining equilibrium constants, such as the solubility product \( K_{sp} \) in electrochemical reactions.

Half-Reaction

A half-reaction is a part of the overall redox reaction, focusing on either the oxidation or reduction part. Each half-reaction shows the transfer of electrons. For example:

• Oxidation half-reaction: \( \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \)
• Reduction half-reaction: \( \text{AgBr}(s) + e^- \rightarrow \text{Ag}(s) + \text{Br}^-(aq) \)

These are used in constructing the full redox equation by balancing electrons lost and gained. This ensures the conservation of charge. Each half-reaction has its electrode potential, which helps in calculating the overall potential of the electrochemical cell. This method is key in both studying and applying redox reactions in technology.

Standard Electrode Potential

The standard electrode potential \( E^{\circ} \) is the voltage developed by a cell under standard conditions (1 M concentration, 1 atm pressure, and usually at 25°C). It's a measure of the tendency of a chemical species to be reduced, and it's essential in comparing the reactivity of different electrodes. For example, the standard electrode potential for:

• \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \) is 0.80 V
• \( \text{AgBr}(s) + e^- \rightarrow \text{Ag} + \text{Br}^- \) is 0.07 V

These values guide the determination of which half-reaction will occur at the anode or cathode in an electrochemical cell. By using these potentials, we can calculate the cell potential and further explore how electrical and chemical equilibria relate through the Nernst equation.