Question

A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2-}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 115 ) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4} .\) The \(8^{\circ}\) value for the following half-reaction is \(0.446 \mathrm{V}\) relative to the standard hydrogen electrode: $$\mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}^{2-}$$ a. Calculate \(\mathscr{C}_{\text {cell }}\) and \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \mathrm{mol} / \mathrm{L}\) b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C}\) ) in which \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \times 10^{-5} \mathrm{M},\) what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is \(0.504 \mathrm{V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}^{2-}\right] .\) What is \(\left[\mathrm{CrO}_{4}^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(17-1,\) calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)

Short answer

In short, here are the answers to each part of the question: a. E°cell = 0.204 V, ΔG° = -39,500 J/mol b. E_cell = E°cell - (0.0592V/n)log(Q) c. E_cell = 0.639 V when [CrO42-] = 1.00 x 10^-5 M d. [CrO42-] = 1.26 x 10^-5 M when E_cell = 0.504 V e. Ksp for Ag2CrO4 = 2.11 x 10^-12

Step-by-step solution

1. Calculate cell potential and Gibbs Free Energy change when Ag2CrO4 is 1.00 mol / L.

The given half-reaction potential (E°) is 0.446 V (relative to the standard hydrogen electrode). We also need the potential of the saturated calomel electrode (SCE). The potential of SCE (E°SCE) is +0.242 V (given in Exercise 115). Now, the overall cell potential (E°cell) is the difference between the SCE potential and the given half-reaction potential: E°cell = E°half-reaction - E°SCE E°cell = 0.446 - 0.242 E°cell = 0.204 V Now we can calculate the Gibbs Free Energy change using the following formula: ΔG° = - nFE°cell where n = number of electrons transferred (which is 2 in this case), F = Faraday's constant (96,485 C/mol), and E°cell = 0.204 V. Now substituting the values, ΔG° = - (2) (96,485) (0.204) ΔG° = - 39,500 J/mol

2. Write the Nernst equation for the cell considering SCE constant.

The Nernst equation is given by: E_cell = E°cell - (RT/nF)ln(Q) For this problem, since the concentrations in the SCE are constant and temperature is 25°C: E_cell = E°cell - (0.0592V/n)log(Q) Where R is the gas constant (8.314 J/mol K), T is the temperature (298 K), n is the number of electrons transferred (2), F is Faraday's constant (96485 C/mol), and Q is the reaction quotient.

3. Calculate the cell potential when the concentration of CrO42- is 1.00 x 10^-5 M.

Now, let us calculate the cell potential for the given concentration of CrO42- (1.00 x 10^-5 M). Rearrange the Nernst equation as follows: log(Q) = (E_cell - E°cell) * (n / 0.0592V) The reaction quotient (Q) is given by: Q = [CrO42-] Now substituting the values, log(1.00 x 10^-5) = (E_cell - 0.204) * (2 / 0.0592) E_cell = 0.204 + 7.39 * 0.0592 E_cell = 0.639 V

4. Determine the concentration of CrO42- when the measured cell potential is 0.504 V.

Given that the measured cell potential (E_cell) is 0.504 V, we can use the Nernst equation to determine the concentration of CrO42-. log(Q) = (E_cell - E°cell) * (n / 0.0592V) log(Q) = (0.504 - 0.204) * (2 / 0.0592) log(Q) = 10.09 Q = 1.26 x 10^-5 M Since Q = [CrO42-], [CrO42-] = 1.26 x 10^-5 M

5. Calculate the solubility product (Ksp) for Ag2CrO4 using data from the problem and Table 17-1.

To calculate the solubility product (Ksp) for Ag2CrO4, we need the potential of the Ag+/Ag couple. From Table 17-1, the standard reduction potential (E°) for Ag+/Ag is 0.7996 V. Now, using the half-cell reaction given in the problem and the reduction potential of Ag+/Ag, we can determine E° for the reduction half-cell reaction of Ag2CrO4: E°(Ag2CrO4 + 2e- -> 2Ag + CrO42-) = E°(Ag+ + e- -> Ag) - E°(CrO42- + 2Ag -> Ag2CrO4) Substituting the values, E° = 0.7996 - 0.446 E° = 0.3536 V Using the Nernst equation, we can calculate the Ksp for Ag2CrO4 at 25°C: Ksp = Q * exp(- nFE° / RT) Since Q = [Ag+]^2[CrO42-], substituting the values, Ksp = [Ag+]^2 * [1.00] * exp(-2 * 96485 * 0.3536 / 8.314 * 298) Ksp = [Ag+]^2 * 1.33 x 10^-10 As [Ag+] = [CrO42-] = 1.26 x 10^-5 M (from step 4), we can calculate K_sp: Ksp = (1.26 x 10^-5)^2 * 1.33 x 10^-10 Ksp = 2.11 x 10^-12

Key concepts

Nernst Equation

The Nernst equation is fundamental in electrochemistry as it relates to the cell potential of electrochemical cells. It essentially accounts for the effect of ion concentration on the potential of a cell.

When ions interact within an electrochemical cell, their concentrations can affect the overall voltage that the cell can produce. The Nernst equation provides a quantifiable relationship between the concentrations of reactants and products and the cell potential. Given by the equation:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \left(\frac{RT}{nF}\right)\ln(Q) \] where \(E_{\text{cell}}\) is the cell potential, \(E^\circ_{\text{cell}}\) is the standard cell potential, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvins, \(n\) is the number of moles of electrons transferred, \(F\) is the Faraday constant, and \(Q\) is the reaction quotient—reflecting the ratio of product activities to reactant activities.

In the given exercise, using the Nernst equation enabled us to predict and calculate the potential of a cell given different concentrations of \(\mathrm{CrO}_{4}^{2-}\). It’s important to note that a solid understanding of this equation allows one to make important electrochemical concentration determinations.

Gibbs Free Energy

Gibbs free energy, denoted as \(\Delta G\), is a vital concept in thermodynamics and electrochemistry that measures the maximum amount of work that a system can perform at constant temperature and pressure. It’s a way to predict whether a process will occur spontaneously.

The relationship between Gibbs free energy and cell potential is given by the formula:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \] where \(n\) represents the number of electrons transferred in the electrochemical reaction, \(F\) is Faraday’s constant, and \(E^\circ_{\text{cell}}\) is the standard cell potential. A negative \(\Delta G\) value suggests that the process is spontaneous.

In our problem, we used this formula to compute how much energy changes when the concentration of \(\mathrm{CrO}_{4}^{2-}\) is 1.00 mol/L. Understanding how to calculate Gibbs free energy is essential for determining the feasibility and spontaneity of electrochemical reactions.

Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is a particular equilibrium constant that applies to the dissolution of sparingly soluble compounds. It quantifies the degree to which a compound will dissolve in solution and thus provides insight into the solubility of substances.

The formula for \(K_{\text{sp}}\) is written in terms of the product of the ions' concentrations raised to the power of their stoichiometric coefficients from the dissolution reaction. It is given as:
\[ K_{\text{sp}} = \prod [\text{ion}]^{\text{coefficient}} \] where the products over which the multiplication is carried out are ions produced when the substance dissolves.

In the exercise, by combining the Nernst equation and the determined concentration of \(\mathrm{CrO}_{4}^{2-}\), we calculated the solubility product constant for \(\mathrm{Ag}_{2}\mathrm{CrO}_{4}\). This value is crucial in understanding the solubility and therefore the concentration at which a substance will precipitate from the solution.

Standard Reduction Potential

The standard reduction potential is an intrinsic property of a chemical species that measures the tendency of the species to gain electrons, commonly symbolized as \(E^\circ\). It is a half-cell potential at standard state conditions, measured relative to the standard hydrogen electrode (SHE).

These values are of paramount importance in predicting the directionality of redox reactions and are typically found in a table like Table 17-1 in our exercise. For an electrochemical cell, the overall cell potential, \(E^\circ_{\text{cell}}\), is the difference between the standard reduction potentials of the cathode and anode.

In our example, we used the standard reduction potentials to calculate the initial cell potential for the half-reaction and then determined alterations in potential when changing chemical environments. This concept helps us understand how electrons will flow in an electrochemical system and fuels our ability to harness this flow for various chemical determinations and applications.