Question

In a galvanic cell the cathode is an \(\mathrm{Ag}^{+}(1.00 \mathrm{M}) / \mathrm{Ag}(\mathrm{s})\) half-cell. The anode is a standard hydrogen electrode immersed in a buffer solution containing \(0.10 \mathrm{M}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and \(0.050 \mathrm{M}\) sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} \mathrm{Na}^{+}\right)\). The measured cell voltage is \(1.030 \mathrm{~V}\). What is the \(\mathrm{pK}_{a}\) of benzoic acid?

Short answer

The pKa of benzoic acid can be found by first calculating the concentration of \(\mathrm{H}^{+}\) ions using the Nernst equation: \[1.030 \, \mathrm{V} = 0.800 \, \mathrm{V} - \frac{(8.314 \, \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1})(298 \, \mathrm{K})}{(1)(96485 \, \mathrm{C} \cdot \mathrm{mol}^{-1})} \ln \frac{[\mathrm{H}^{+}]}{(0.10 \, \mathrm{M})(0.050 \, \mathrm{M})}\] Next, use the Henderson-Hasselbalch equation with the calculated [\(\mathrm{H}^{+}\)] value, the concentration of sodium benzoate (\(0.050 \, \mathrm{M}\)), and the concentration of benzoic acid (\(0.10 \, \mathrm{M}\)) to find the pKa: \[pH = pK_a + \log \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\] Solve for pKa to find the resulting value.

Step-by-step solution

Write down the Nernst equation

The Nernst equation is given by: \[E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q\] where: - \(E_{cell}\) is the total cell potential - \(E^0_{cell}\) is the standard cell potential - \(R\) is the gas constant (\(8.314 \, \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)) - \(T\) is the temperature in Kelvin (assume room temperature at \(298 \, \mathrm{K}\)) - \(n\) is the number of moles of electrons exchanged in the half-reaction - \(F\) is the Faraday's constant (\(96485 \, \mathrm{C} \cdot \mathrm{mol}^{-1}\)) - \(Q\) is the reaction quotient. The given cell potential (\(E_{cell}\)) is \(1.030 \, \mathrm{V}\). We need to find \(E^0_{cell}\), \(n\), and \(Q\).

Calculate the standard cell potential, E0_cell

In a galvanic cell, the reduction half-cell reactions are as follows: Anode (standard hydrogen electrode): \[\mathrm{2H}^{+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}(\mathrm{g}) \quad E^0_{\mathrm{H}^{+}/\mathrm{H}_{2}} = 0.000 \, \mathrm{V}\] Cathode (silver): \[\mathrm{Ag}^{+} + \mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s}) \quad E^0_{\mathrm{Ag}^{+}/\mathrm{Ag}} = 0.800 \, \mathrm{V}\] Calculate the standard cell potential, \(E^0_{cell}\), using the following formula: \[E^0_{cell} = E^0_{\mathrm{cathode}} - E^0_{\mathrm{anode}}\]

Calculate the number of moles of electrons transferred, n

For both half-cell reactions, we can see that there is 1 mole of electrons involved in each reaction. Therefore, \(n = 1\).

Calculate the reaction quotient, Q

Since the anode half-cell is in equilibrium with the buffer, the reaction quotient, \(Q\), is given by the concentration of \(\mathrm{H}^+\) ions and the ratio of benzoic acid and sodium benzoate: \[Q = \frac{[\mathrm{H}^{+}]}{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COOH}] \cdot [\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COO}^{-}]}\] Now we have all the information needed to use the Nernst equation and solve for the concentration of \(\mathrm{H}^{+}\) ions.

Solve for the concentration of H+ ions using the Nernst equation

Plug all the values into the Nernst equation and solve for [\(\mathrm{H}^{+}\)]: \[1.030 \, \mathrm{V} = 0.800 \, \mathrm{V} - \frac{(8.314 \, \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1})(298 \, \mathrm{K})}{(1)(96485 \, \mathrm{C} \cdot \mathrm{mol}^{-1})} \ln \frac{[\mathrm{H}^{+}]}{(0.10 \, \mathrm{M})(0.050 \, \mathrm{M})}\]

Calculate the pKa of benzoic acid using the Henderson-Hasselbalch equation

Now that we have the concentration of \(\mathrm{H}^{+}\) ions, we can use the Henderson-Hasselbalch equation to find the pKa of benzoic acid: \[pH = pK_a + \log \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\] Where: - [\(\mathrm{A}^{-}\)] is the concentration of sodium benzoate (\(0.050 \, \mathrm{M}\)) - [\(\mathrm{HA}\)] is the concentration of benzoic acid (\(0.10 \, \mathrm{M}\)) Rearrange the equation to find the pKa and substitute the values for pH, [\(\mathrm{A}^{-}\)], and [\(\mathrm{HA}\)]. Calculate the pKa of benzoic acid.

Key concepts

Galvanic Cells

Galvanic cells, also known as voltaic cells, are electrochemical cells that generate electrical energy from spontaneous chemical reactions. These cells are comprised of two distinct half-cells. Each half-cell contains a different chemical reaction. In a typical galvanic cell setup, one half-cell acts as the anode and undergoes oxidation, while the other serves as the cathode and undergoes reduction.
One common example of a galvanic cell involves a silver cathode, as mentioned in the given exercise. Here, the silver ions ext{(Ag}^{+}) are reduced to form solid silver while the anode, which is a standard hydrogen electrode in our case, facilitates the oxidation reaction.
Understanding galvanic cells' workings helps grasp essential concepts in electrochemistry, like electron flow and redox reactions, forming the foundation for interpreting cell potentials and predicting reaction spontaneity.

Nernst Equation

The Nernst equation is a pivotal tool in electrochemistry that allows the calculation of cell potential under non-standard conditions. It is a modified version of the relation between the Gibbs free energy and cell potential. By incorporating real-time conditions, it helps students comprehend how concentrations and temperature affect the system.
The formula is expressed as: \[E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q\] Where:

• \(E_{cell}\) is the cell potential under given conditions.
• \(E^0_{cell}\) is the standard cell potential.
• \(R\) is the universal gas constant (8.314 J/mol·K).
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons transferred.
• \(F\) is the Faraday constant (96485 C/mol).
• \(Q\) is the reaction quotient.

Using the Nernst equation, one can understand how voltage varies with ion concentration changes in electrochemical cells. It further demonstrates that even minor changes in concentrations can significantly affect cell potential, a concept crucial for fields like battery technology and biochemistry.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a straightforward formula used to relate the pH of a solution to its pKa (acid dissociation constant) and the concentrations of an acid and its conjugate base. The equation is expressed as:\[pH = pK_a + \log \frac{[A^-]}{[HA]}\]Where:

• \(pH\) is the measure of the acidity of the solution.
• \(pK_a\) is the negative logarithm of the equilibrium constant for the dissociation reaction.
• \([A^-]\) is the concentration of the conjugate base.
• \([HA]\) is the concentration of the undissociated acid.

In the context of the problem at hand, using this equation allows us to calculate the pKa of benzoic acid, given the concentrations of benzoic acid and sodium benzoate in the buffer solution. Understanding how to apply this equation is fundamental for interpreting buffer systems, crucial in both classroom and laboratory environments.

Standard Cell Potential

The standard cell potential, denoted as \(E^0_{cell}\), is the voltage difference between two half-cells under standard conditions (i.e., 1 M concentrations and 25°C). It serves as a benchmark for predicting the direction of a chemical reaction.
In electrochemistry, the standard potential differences between electrodes:

• Anode reaction is typically denoted as oxidation (e.g., the conversion of hydrogen ions to hydrogen gas in the standard hydrogen electrode), and it often appears first in these equations.
• Cathode reaction denotes reduction (like silver ions converting to solid silver), appearing second.

To find the standard cell potential, use:\[E^0_{cell} = E^0_{cathode} - E^0_{anode}\]
By calculating \(E^0_{cell}\), students can determine whether a reaction is spontaneous. A positive \(E^0_{cell}\) indicates spontaneous reactions under standard conditions, providing insights into the efficiency and feasibility of electrochemical cells.