Problem 35 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The EMF for the cell: \(\mathrm{Ag}(\mathrm{s}) \mid \mathrm{AgCl}(\mathrm{s})\) \(\mathrm{KCl}(0.2 \mathrm{M}) \| \mathrm{KBr}(0.001 \mathrm{M}) \mid \mathrm{AgBr}(\mathrm{s})\) \(\mathrm{Ag}(\mathrm{s})\) at \(25^{\circ} \mathrm{C}\) is \(\left(K_{\mathrm{sp}}(\mathrm{AgCl})=2.0 \times 10^{-10}\right.\) \(K_{\mathrm{sp}}(\mathrm{AgBr})=4.0 \times 10^{-13}, 2.303 R T / F=0.06\) \(\log 2=0.3\) ) (a) \(0.024 \mathrm{~V}\) (b) \(-0.024 \mathrm{~V}\) (c) \(-0.24 \mathrm{~V}\) (d) \(-0.012 \mathrm{~V}\)

Step-by-Step Solution

Answer

-0.102 V (None of the given options is correct; the closest one is option (c) -0.24 V, but it is still not accurate based on calculations.)

Step 1: Write the Nernst Equation for the Half-Cells

For the given cell, the half-cell reactions are based on the solubility product constants of AgCl and AgBr. We need to use the Nernst equation to relate the EMF to the ion concentrations in solution: For AgCl: \(E_1 = E^\text{º}_{\text{AgCl/Ag}} - \frac{0.06}{1} \log \frac{1}{[\text{Cl}^{-}]}\)For AgBr: \(E_2 = E^\text{º}_{\text{AgBr/Ag}} - \frac{0.06}{1} \log \frac{1}{[\text{Br}^{-}]}\)Since both electrodes are silver, their standard potentials cancel each other when computing the overall cell potential. Therefore, we only need to focus on the concentration terms.

Step 2: Calculate the Chloride and Bromide Ion Concentrations

Using the given solubility product constants (Ksp), we can find the concentration of Cl- and Br- ions that would be in equilibrium with solid AgCl and AgBr respectively. For AgCl and AgBr, the reactions are takes the form \(\text{AgCl} (s) \leftrightarrows \text{Ag}^+ (aq) + \text{Cl}^- (aq)\)and\(\text{AgBr} (s) \leftrightarrows \text{Ag}^+ (aq) + \text{Br}^- (aq)\)Given that the reaction quotient Q matches Ksp at equilibrium, we can assume\([\text{Ag}^+] = [\text{Cl}^-]\) for AgCl and \([\text{Ag}^+] = [\text{Br}^-]\) for AgBr. Thus:\([\text{Cl}^-] = \sqrt{K_{\text{sp} (\text{AgCl)}}} = \sqrt{2.0 \times 10^{-10}}\)\([\text{Br}^-] = \sqrt{K_{\text{sp} (\text{AgBr)}}} = \sqrt{4.0 \times 10^{-13}}\)

Step 3: Determine the Cell EMF

Now that we have the concentrations, we can calculate the cell EMF using the difference in potentials (E2 - E1) from the Nernst equation we wrote earlier:\(\text{EMF} = E_2 - E_1\)\(\text{EMF} = \left(- \frac{0.06}{1} \log \frac{1}{[\text{Br}^{-}]}\right) - \left(- \frac{0.06}{1} \log \frac{1}{[\text{Cl}^{-}]}\right)\)Simplifying, we obtain:\(\text{EMF} = 0.06 \log \frac{[\text{Cl}^{-}]}{[\text{Br}^{-}]}\)Now plug in the concentrations we calculated in step 2:\(\text{EMF} = 0.06 \log \frac{\sqrt{2.0 \times 10^{-10}}}{\sqrt{4.0 \times 10^{-13}}}\)\(\text{EMF} = 0.06 \log \frac{10^{-5}}{2 \times 10^{-7}}\)\(\text{EMF} = 0.06 \log \frac{1}{2 \times 10^2}\)\(\text{EMF} = 0.06 \log (2 \times 10^{-2})\)\(\text{EMF} = 0.06 (\log 2 + \log 10^{-2})\)Using the given \(\log 2 = 0.3\), we have:\(\text{EMF} = 0.06 (0.3 - 2)\)\(\text{EMF} = 0.06 (-1.7)\)\(\text{EMF} = -0.102 \text{V}\)The EMF is closest to the option (c) -0.24 V, but there seems to be an error as the precise calculation gives -0.102 V.

Key Concepts

Nernst Equation

The Nernst equation is a fundamental principle in electrochemistry that relates the electromotive force (EMF) of an electrochemical cell to the concentration of ions in solution. At its core, it is derived from the Gibbs free energy change for the redox reaction.To put it simply, as ions' concentrations change, so does the cell's ability to do electrical work. The Nernst equation is expressed as:\[ E = E^\text{°} - \frac{0.0592}{n} \log Q \]Where:

• \(E\) is the electrode potential,
• \(E^\text{°}\) is the standard electrode potential,
• \(n\) is the number of moles of electrons exchanged per mole of reactant,
• \(Q\) is the reaction quotient, which is the ratio of the products divided by reactants, each raised to the power of their stoichiometric coefficients.

The '0.0592' value is derived from constants including temperature, and it changes with different temperatures. For simplification, at room temperature (298 K), '0.0592' is often approximated to '0.06'. As seen in the exercise, when ions are at equilibrium with their solids, Q can be substituted by the solubility product \(K_{\text{sp}}\), thus linking the Nernst equation with solubility.

EMF Calculation

EMF, or electromotive force, is essentially the voltage generated by an electrochemical cell. The calculation of EMF allows us to predict the direction of spontaneous reaction and its ability to perform electrical work. The steps to compute EMF often involve understanding the chemistry of the cell and applying the Nernst equation, as demonstrated in the example exercise.Calculating EMF includes:

• Determining the standard reduction potentials of the half-reactions,
• Calculating individual cell potentials using the Nernst equation,
• Combining half-cell potentials to find the overall EMF of the cell.

Bringing together these components can be simple for standard conditions but requires careful attention when dealing with non-standard conditions, such as different concentrations, temperatures, or pressures. Therefore, a systematic approach, starting from the chemical nature of the electrodes and the specifics of the cell's contents, is essential to obtaining an accurate EMF value.

Solubility Product

The solubility product, denoted as \(K_{\text{sp}}\), plays a crucial role in predicting the solubility of ionic compounds in solution. It is a constant that represents the maximum product of the ions' concentrations in a saturated solution at equilibrium. In the context of electrochemistry problems, \(K_{\text{sp}}\) allows us to calculate the ion concentrations necessary for the Nernst equation.Consider a general salt, AB, dissolving as:\[ AB_{(s)} \leftrightarrows A^+_{(aq)} + B^-_{(aq)} \]The \(K_{\text{sp}}\) expression for this would be:\[ K_{\text{sp}} = [A^+] \times [B^-] \]For salts that dissolve to give one mole of each ion, the concentration of each ion is the square root of \(K_{\text{sp}}\), as seen in the exercise. Understanding solubility product is important, not just for EMF calculations in electrochemistry, but also for understanding phenomena like precipitation, complex ion formation, and ion concentration control in various chemical systems.

Electrochemical Cell

An electrochemical cell is a device capable of either generating electrical energy from chemical reactions or facilitating chemical reactions through the introduction of electrical energy. There are two main types of electrochemical cells: galvanic/voltaic cells, which convert chemical energy to electrical energy, and electrolytic cells, which do the reverse.The cell in our exercise is a galvanic cell consisting of two half-cells, each containing an electrode and an electrolyte. The two half-cells are connected by a salt bridge that allows ions to flow, balancing the charge as electrons flow through the external circuit from the anode to the cathode, generating an electric current.

• The anode is where oxidation occurs, and electrons are lost.
• The cathode is where reduction occurs, and electrons are gained.

Every electrochemical cell has its characterizing EMF, which, as previously discussed, can be calculated using the Nernst equation and the solubility product of the involved substances. Understanding the components and functions of an electrochemical cell is fundamental in the study of electrochemistry, with applications ranging from batteries to sensors and beyond.