Problem 96 - Chemistry

Question

A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate \(\left(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\), at \(25^{\circ} \mathrm{C}\). The measured potential difference between the rod and the SHE is \(0.589 \mathrm{~V},\) the rod being positive. Calculate the solubility product constant for silver oxalate.

Step-by-Step Solution

Answer

The solubility product constant, \(K_{sp}\), is determined from the concentrations of ions using the Nernst equation and the measured potential difference.

Step 1: Understand the Reaction

The dissolution of silver oxalate involves the dissociation of silver oxalate into its ions in solution:\[\text{Ag}_2 \text{C}_2 \text{O}_4 (s) \rightleftharpoons 2 \text{Ag}^+ (aq) + \text{C}_2 \text{O}_4^{2-} (aq)\] Our goal is to find the solubility product \(K_{sp}\) for this equilibrium.

Step 2: Electrochemical Cell Setup and Nernst Equation

The standard half-cell potentials are:- SHE: \(E^0_{\text{SHE}} = 0.00\, \text{V}\)- Ag+/Ag: \(E^0_{\text{Ag}^+/\text{Ag}} = 0.7996\, \text{V}\)The Nernst equation relates the electrode potential to concentrations:\[E = E^0 - \frac{RT}{nF} \ln Q\]where:- \(E\) is the electrode potential, \(+0.589\, \text{V}\) for our setup since Ag rod is positive.- \(E^0\) is the standard potential for the silver cell.- \(n\) is the number of electrons transferred (2 for silver ionization).- \(Q\) is the reaction quotient.

Step 3: Calculate Reaction Quotient (Q)

From the measured potential difference and knowing that the rod is positive:\[E = E_{\text{Ag}^+/\text{Ag}} - E_{\text{SHE}} \Rightarrow 0.589 \text{ V} = 0.7996 \text{ V} - \frac{RT}{2F} \ln \left( \frac{1}{a_{\text{Ag}^+}^2} \right)\]Rearranging gives:\[\ln a_{\text{Ag}^+}^2 = \frac{2F}{RT} (0.7996 - 0.589)\]Substitute the constants \(R = 8.314 \text{ J/mol}\, \cdot\, \text{K}, T = 298 \text{ K}, F = 96485 \text{ C/mol}\):\[\ln a_{\text{Ag}^+}^2 = \frac{2 \times 96485 \cdot (0.2106)}{8.314 \times 298}\]Solve for \(a_{\text{Ag}^+}^2\).

Step 4: Solve for Silver Ion Concentration

Calculate the numerical value from Step 3:\[\ln a_{\text{Ag}^+}^2 \approx 16.916\]So:\[a_{\text{Ag}^+}^2 \approx e^{16.916}\]Calculate \(a_{\text{Ag}^+}\):\[a_{\text{Ag}^+} \approx e^{8.458}\] Determine its numerical value to find the solubility of silver ions.

Step 5: Calculate Solubility Product \(K_{sp}\)

Knowing \( \text{[Ag}^+] = a_{\text{Ag}^+} \) from previous step, and assuming complete dissociation of oxalate, \( \text{[C}_2\text{O}_4^{2-}] = \frac{a_{\text{Ag}^+}}{2} \):\[K_{sp} = [\text{Ag}^+]^2[\text{C}_2\text{O}_4^{2-}] = (a_{\text{Ag}^+})^2 \cdot \frac{a_{\text{Ag}^+}}{2}\]Calculate this value to find \(K_{sp}\).

Key Concepts

Nernst Equation

Electrochemistry often involves evaluating how electrochemical cells work under various conditions. One key tool for such analysis is the Nernst Equation. It helps predict the potential of an electrochemical cell under non-standard conditions. This is crucial in systems like the one involving the silver rod and Standard Hydrogen Electrode (SHE).

The Nernst Equation is given by:\[E = E^0 - \frac{RT}{nF} \ln Q\]

Where:

• \(E\) is the actual electrode potential.
• \(E^0\) is the standard electrode potential.
• \(R\) is the universal gas constant \(8.314 \text{ J/mol}\cdot\text{K}\).
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of electrons exchanged in the reaction.
• \(F\) is Faraday's constant \(96485 \text{ C/mol}\).
• \(Q\) is the reaction quotient representing the ratio of product over reactant activities.

In the given system with silver, the measured potential is influenced by the concentration of silver ions \(\text{Ag}^+\) in the solution. By rearranging the Nernst Equation, the concentration can be determined through measurable variables such as cell potential \(E\) and standard cell potential \(E^0\).

This connection through the Nernst Equation is fundamental in the calculation of solubility product constants, as it allows us to find activity coefficients necessary for determining equilibria.

Solubility Product Constant

The solubility product constant \(K_{sp}\) is an essential concept in understanding how sparingly soluble compounds behave in solution. It quantifies the level of solubility of those compounds, providing a snapshot of equilibrium between the solid and its respective ions in solution.

For silver oxalate \(\text{Ag}_2\text{C}_2\text{O}_4\), the dissociation in a saturated solution can be described by:\[\text{Ag}_2 \text{C}_2 \text{O}_4 (s) \rightleftharpoons 2 \text{Ag}^+ (aq) + \text{C}_2 \text{O}_4^{2-} (aq)\]

The expression for the solubility product \(K_{sp}\) is:\[K_{sp} = [\text{Ag}^+]^2[\text{C}_2\text{O}_4^{2-}]\]

This constant helps in calculating the saturation point of silver oxalate in a solution. If the ion product exceeds \(K_{sp}\), precipitation occurs until equilibrium is re-established. Hence, knowing \(K_{sp}\) value is crucial in predicting whether a precipitate will form in given conditions.

In practice, \(K_{sp}\) is derived from electrical measurements, relating it back to ion concentrations using the Nernst Equation. This determines how the values interact in quantifying a compound's solubility under specific environmental conditions.

Half-cell Potentials

Half-cell potentials are the building blocks of electrochemical cells. Each half-cell has a potential, typically recorded against a standard reference, like the SHE. Understanding these potentials is crucial to calculate overall cell potentials and interpret electrochemical reactions.

For silver, the standard half-cell potential \(E^0_{\text{Ag}^+/\text{Ag}}\) is approximately \(0.7996\, \text{V}\). When paired with SHE, which has a potential of \(0.00\, \text{V}\), the potential difference can be measured to provide insights into reaction tendencies and equilibria.

The combination of half-cells in an electrochemical setup defines the feasibility of electron flow, thus dictating the direction of reaction. The silver half-cell’s tendency to gain or lose electrons emphasizes its behavior of either reduced or oxidized state relative to the SHE.

This is critical when calculating the solubility product constant for a dissolution reaction, as the potential difference directly informs the ion activity via the Nernst Equation. Without recognizing individual half-cell contributions, determining such dissolution properties becomes difficult, underscoring the significance of half-cell potentials in electrochemical applications.