Problem 65 - Chemistry: Principles and Reactions

Question

Consider a cell in which the reaction is $$ 2 \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow 2 \mathrm{Ag}^{+}(a q)+\mathrm{Cu}(s) $$ (a) Calculate \(E^{\circ}\) for this cell. (b) Chloride ions are added to the \(\mathrm{Ag} \mid \mathrm{Ag}^{+}\) half- cell to precipitate \(\mathrm{AgCl}\). The measured voltage is \(+0.060 \mathrm{~V}\). Taking \(\left[\mathrm{Cu}^{2+}\right]=1.0 \mathrm{M}\), calculate \(\left[\mathrm{Ag}^{+}\right]\). (c) Taking \(\left[\mathrm{Cl}^{-}\right]\) in \((\mathrm{b})\) to be \(0.10 M\), calculate \(K_{\text {sp }}\) of \(\mathrm{AgCl}\).

Step-by-Step Solution

Answer

Question: Calculate the standard cell potential, concentration of Ag+, and solubility product of AgCl for the given redox reaction. Answer: a) The standard cell potential, \(E^{\circ}\), can be calculated using the equation \(E^{\circ}_{\text{cell}} = E^{\circ}_{\mathrm{Ag/A{g}^{+}}} - E^{\circ}_{\mathrm{Cu/Cu^{2+}}} = 0.800\mathrm{V} - 0.337\mathrm{V}\). b) To find the concentration of \(\mathrm{Ag}^{+}\), use the Nernst equation and given values: \(0.060\mathrm{~V} = (0.800\mathrm{V} - 0.337\mathrm{V}) - \frac{0.0592\mathrm{V}}{2} * \log_{10}\left(\frac{[\mathrm{Ag}^{+}]^2}{[\mathrm{Cu}^{2+}]}\right)\). c) The solubility product (\(K_{\text{sp}}\)) of AgCl can be calculated using the equation \(K_{\text{sp}} = [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}]\) with the obtained value for \([\mathrm{Ag}^{+}]\) and given concentration of \([\mathrm{Cl}^{-}] = 0.10\,\mathrm{M}\).

Step 1: a) Standard Cell Potential, \(E^{\circ}\)

We first need to determine the reduction potentials of the species involved. We can find these values in a table of standard reduction potentials. For Ag: \(\mathrm{Ag}^{+}(aq) + e^{-} \longrightarrow \mathrm{Ag}(s) \qquad E^{\circ}_{\mathrm{Ag/A{g}^{+}}}=0.800\mathrm{V}\) For Cu: \(\mathrm{Cu}^{2+}(aq) + 2e^{-} \longrightarrow \mathrm{Cu}(s) \qquad E^{\circ}_{\mathrm{Cu/Cu^{2+}}}=0.337\mathrm{V}\) Now we can calculate the standard cell potential, \(E^{\circ}\), for the overall reaction. It is given by: \(E^{\circ}_{cell} = E^{\circ}_{\text{reduction}} - E^{\circ}_{\text{oxidation}}\) In this case, the Ag gets reduced and the Cu gets oxidized. So, we obtain: \(E^{\circ}_{\text{cell}} = E^{\circ}_{\mathrm{Ag/A{g}^{+}}} - E^{\circ}_{\mathrm{Cu/Cu^{2+}}} = 0.800\mathrm{V} - 0.337\mathrm{V}\)

Step 2: b) Concentration of \(\mathrm{Ag}^{+}\)

Since the measured voltage is \(+0.060\mathrm{~V}\) and the concentration of \(\mathrm{Cu}^{2+}\) is \(1.0\mathrm{M}\), we can use the Nernst equation to calculate the concentration of \(\mathrm{Ag}^{+}\): \(E_{cell} = E^{\circ}_{cell} - \frac{0.0592\mathrm{V}}{n} * \log_{10}\left(\frac{[\mathrm{Ag}^{+}]^2}{[\mathrm{Cu}^{2+}]}\right)\) We have the cell potential \(E_{cell} = 0.060\mathrm{~V}\), \(n=2\) since there are two electrons transferred in the reaction, and \([\mathrm{Cu}^{2+}]=1.0\mathrm{M}\). We can solve for \([\mathrm{Ag}^{+}]\): \(0.060\mathrm{~V} = (0.800\mathrm{V} - 0.337\mathrm{V}) - \frac{0.0592\mathrm{V}}{2} * \log_{10}\left(\frac{[\mathrm{Ag}^{+}]^2}{[\mathrm{Cu}^{2+}]}\right)\)

Step 3: c) Solubility Product \(K_{\text{sp}}\) of \(\mathrm{AgCl}\)

With the concentrations of \(\mathrm{Ag}^{+}\) and \(\mathrm{Cl}^{-}\), we can now calculate the solubility product constant \(K_{\text{sp}}\) for \(\mathrm{AgCl}\): \(K_{\text{sp}} = [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}]\) Since \([\mathrm{Cl}^{-}] = 0.10\,\mathrm{M}\), we can calculate \(K_{\text{sp}}\) using the calculated \([\mathrm{Ag}^{+}]\) from the previous step.

Key Concepts

Cell Potential

In electrochemistry, the cell potential, also known as electromotive force (EMF), is a measure of the voltage, or electric power, generated by a galvanic cell under standard conditions. A galvanic cell consists of two half-cells, each containing an electrode and an electrolyte, where a chemical reaction occurs. The cell potential arises from the difference in reduction potentials between the cathode and anode.

The standard cell potential, denoted as \(E^{\circ}_{\text{cell}}\), is calculated using the standard reduction potentials for the two half-reactions involved. The equation is:

• \(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\)

In this problem, \(\mathrm{Ag}^{+}\/\mathrm{Ag}\) has a standard reduction potential of 0.800 V, and \(\mathrm{Cu}^{2+}\/\mathrm{Cu}\) has a value of 0.337 V.

Here, silver acts as the cathode (reduction occurs), while copper is the anode (oxidation occurs). Therefore, the calculation becomes:

• \(E^{\circ}_{\text{cell}} = 0.800\ \mathrm{V} - 0.337\ \mathrm{V} = 0.463\ \mathrm{V}\)

This positive cell potential indicates the reaction is spontaneous under standard conditions.

Nernst Equation

The Nernst equation is a fundamental tool in electrochemistry used to calculate the cell potential of a galvanic cell under non-standard conditions. It accounts for variations in concentration and temperature, which alter the cell potential from its standard value.

The general form of the Nernst equation is:

• \(E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln{Q}\)

where:

• \(E_{\text{cell}}\) is the cell potential under non-standard conditions,
• \(R\) is the universal gas constant \(8.314\ \mathrm{J}(\mathrm{mol}^{-1}\ \mathrm{K}^{-1})\),
• \(T\) is the temperature in Kelvin,
• \(n\) is the number of moles of electrons transferred,
• \(F\) is the Faraday constant \(96485\ \mathrm{C}\ \mathrm{mol}^{-1}\),
• \(Q\) is the reaction quotient.

For this exercise, the simplified form using logs is used:

• \(E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0592\ \mathrm{V}}{n} \log_{10}{Q}\)

In our example, \(Q\) is based on the concentrations of Ag⁺ and Cu²⁺. Given an \(E_{\text{cell}} = +0.060\ \mathrm{V}\) and considering \(\left[\mathrm{Cu}^{2+}\right] = 1.0\ \mathrm{M}\), we calculate the concentration of \([\mathrm{Ag}^{+}]\) and adapt the Nernst equation accordingly:

• \(Q = \frac{[\mathrm{Ag}^{+}]^2}{[\mathrm{Cu}^{2+}]} = \frac{[\mathrm{Ag}^{+}]^2}{1.0}\)

Plugging in the known values allows us to solve for \([\mathrm{Ag}^{+}]\).

Solubility Product Constant

The solubility product constant, \(K_{\text{sp}}\), is essential in understanding the solubility of sparingly soluble ionic compounds. It's a specific equilibrium constant for the dissolution of such compounds in water.

For any ionic compound that dissolves as: \(AB \rightarrow A^+ + B^-\), the \(K_{\text{sp}}\) expression is:

• \(K_{\text{sp}} = [A^+][B^-]\)

In the context of this exercise, the compound is \(\text{AgCl}\), which dissolves as:

• \(\text{AgCl} \rightarrow \mathrm{Ag}^+ + \mathrm{Cl}^−\)

Given the concentrations:\( [\mathrm{Cl}^-] = 0.10\ \mathrm{M}\) and \([\mathrm{Ag}^+]\) (calculated using Nernst equation), we determine \(K_{\text{sp}}\) for \(\text{AgCl}\).

Calculating \([\mathrm{Ag}^+][\mathrm{Cl}^-]\) provides the necessary \(K_{\text{sp}}\) value, reflecting the solubility dynamics of silver chloride in the solution.