Question

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll} \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \\ \mathrm{V}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{V}(s) & \mathscr{E}^{\circ}=-1.20 \mathrm{V} \end{array}$$ In this cell, the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=1.00 \mathrm{M},\) and the vanadium compartment contains a vanadium electrode and \(V^{2+}\) at an unknown concentration. The compartment containing the vanadium \((1.00 \mathrm{L}\) of solution) was titrated with \(0.0800 M \space \mathrm{H}_{2} \mathrm{EDTA}^{2-},\) resulting in the reaction $$\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q)+\mathrm{V}^{2+}(a q) \rightleftharpoons \mathrm{VEDTA}^{2-}(a q)+2 \mathrm{H}^{+}(a q) \space \mathrm{K=?}$$ The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of \(500.0 \mathrm{mL} \space \mathrm{H}_{2} \mathrm{EDTA}^{2-}\) solution added. At the stoichiometric point, \(\mathscr{E}_{\text {cell }}\) was observed to be \(1.98 \mathrm{V}\). The solution was buffered at a pH of \(10.00 .\) a. Calculate \(\mathscr{E}_{\text {cell }}\) before the titration was carried out. b. Calculate the value of the equilibrium constant, \(K,\) for the titration reaction. c. Calculate \(\mathscr{E}_{\text {cell }}\) at the halfway point in the titration.

Short answer

The cell potential before the titration was approximately 1.58V, calculated utilizing the Nernst equation and finding the effective cell potential using the given standard cell potentials and the concentrations of the respective ions. The equilibrium constant \(K\) is approximately \(4.01 \times 10^{-6}\) for the titration reaction, determined by substituting the given cell potential at the stoichiometric point into the Nernst equation. Lastly, the cell potential at the halfway point in the titration was approximately 1.59V, determined by using the concentration of \(V^{2+}\) at the halfway point and applying the Nernst equation.

Step-by-step solution

a. Calculate the cell potential before the titration

To find the cell potential before the titration, we first need to determine the standard potential for the overall redox reaction by adding the standard potentials of the two half-reactions. Given the standard potentials: For Cu^2+ + 2e^- -> Cu(s), E° = 0.34 V For V^2+ + 2e^- -> V(s), E° = -1.20 V Standard cell potential is the difference between the reduction potentials of the cathode and the anode: \(E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0}\) Now, using the Nernst equation to calculate the cell potential before the titration: \(E_{cell} = E_{cell}^{0} - \frac{RT}{nF} \ln{Q}\) Where R is the gas constant, T is the temperature, F is Faraday's constant, and Q is the reaction quotient. Here, we assume the standard conditions, 298 K temperature, and since the number of electrons involved in both half-reactions is 2, n=2. Initially, the reaction quotient Q is: \(Q = \frac{[\mathrm{V^{2+}}]}{[\mathrm{Cu^{2+}}]}\) However, we need to find the unknown concentration of V^2+ before we can calculate the initial cell potential.

Find V^2+ concentration

To find the concentration of V^2+ before titration, we need to use the stoichiometric point information. At the stoichiometric point, 500.0 mL of 0.0800 mol/L H2EDTA^2- solution was added. From this, we can find the moles of H2EDTA^2- added and consequently, the moles of V^2+ present in the 1.00 L solution. Moles of H2EDTA^2- added = Volume x Molarity = 0.5 L x 0.0800 mol/L = 0.0400 mol Since there is a 1:1 stoichiometry between H2EDTA^2- and V^2+, the moles of V^2+ present in 1.00 L solution are also 0.0400 mol Hence, the initial concentration of V^2+ is 0.0400 mol/L.

Calculate the initial cell potential

Now that we know the concentration of V^2+, we can find the initial cell potential using the Nernst equation: \(E_{cell} = E_{cell}^{0} - \frac{RT}{2F} \ln{\left(\frac{[\mathrm{V^{2+}}]}{[\mathrm{Cu^{2+}}]}\right)}\) \(E_{cell} = (0.34 - (-1.20)) - \frac{8.314 \times 298}{2\times96485} \ln{\left(\frac{0.0400}{1.00}\right)}\) \(E_{cell} = 1.54 - 0.0129 \times \ln{0.0400}\) \(E_{cell} = 1.54 + 0.0358\) \(E_{cell} = 1.576 \mathrm{V}\)

b. Calculate the value of the equilibrium constant, K, for the titration reaction

At the stoichiometric point of the titration, the E_cell is given as 1.98 V. We can use this value, along with the Nernst equation to calculate the equilibrium constant, K: \(E_{cell} = E_{cell}^{0} - \frac{RT}{nF} \ln{K}\) \(1.98 = 1.54 - \frac{8.314 \times 298}{2\times96485} \ln{K}\) Now, solving for K: \(\ln{K} = \frac{2\times96485}{8.314 \times 298}(1.54-1.98)\) \(\ln{K} = -12.42\) Thus, the equilibrium constant, K, is: \(K = e^{-12.42}\) \(K = 4.01 \times 10^{-6}\)

c. Calculate the cell potential at the halfway point in the titration

At the halfway point of the titration, 250.0 mL of 0.0800 mol/L H2EDTA^2- solution has been added. From this, we can determine the concentration of V^2+ at this point: Moles of H2EDTA^2- added = 0.250 L × 0.0800 mol/L = 0.0200 mol Since there is a 1:1 stoichiometry between H2EDTA^2- and V^2+ ions, 0.0200 mol of V^2+ ions have reacted with H2EDTA^2- ions at the halfway point. Therefore, moles of V^2+ remaining in the solution = 0.0400 - 0.0200 = 0.0200 moles Consequently, the concentration of V^2+ at the halfway point is: \([\mathrm{V^{2+}}] = \frac{0.0200 \text{ mol}}{1.00 \text{ L}} = 0.0200 \text{ M}\) Now that we have the concentration of V^2+ at the halfway point, we can calculate the cell potential at this point using the Nernst equation: \(E_{cell} = E_{cell}^{0} - \frac{RT}{2F} \ln{\left(\frac{[\mathrm{V^{2+}}]}{[\mathrm{Cu^{2+}}]}\right)}\) \(E_{cell} = 1.54 - \frac{8.314 \times 298}{2\times96485} \ln{\left(\frac{0.0200}{1.00}\right)}\) \(E_{cell} = 1.54 + 0.0537\) \(E_{cell} = 1.5937 \mathrm{V}\) Therefore, the cell potential at the halfway point of the titration is approximately 1.59 V.

Key concepts

Standard Reduction Potential

The concept of standard reduction potential, represented as \( E^\circ \), is fundamental to understanding galvanic cells and their electrochemical reactions. It is a measure of the tendency of a chemical species to acquire electrons and be reduced. In a galvanic cell, two half-reactions occur: one at the cathode, where reduction takes place, and another at the anode where oxidation occurs. The standard reduction potential for each half-reaction is determined under standard conditions, which includes a solute concentration of 1 M, a pressure of 1 atm for gases and a temperature of 25°C (298 K).

The standard cell potential can be calculated by taking the difference between the standard reduction potentials of the cathode and anode reactions, as shown in the textbook exercise for the copper and vanadium half-reactions. Since the standard reduction potential is a measure of the intrinsic capability of a species to gain or lose electrons, higher values indicate a greater tendency to be reduced. Understanding standard reduction potentials allows us to predict the flow of electrons in the electrochemical cell and hence, the direction of the redox reaction.

Nernst Equation

The Nernst equation is a crucial tool in electrochemistry that relates the cell potential to the concentrations of the reactants and products. It is given by:
\[ E_{cell} = E_{cell}^{0} - \frac{RT}{nF} \ln{Q} \]
where \(E_{cell}^{0}\) is the standard cell potential, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred in the balanced equation, \(F\) is Faraday’s constant, and \(Q\) is the reaction quotient. At the standard conditions, \(Q\) would be 1, and the cell potential would be equal to the standard potential. However, under non-standard conditions, the cell potential changes, and the Nernst equation can tell you by how much.

Students often find it challenging to understand the real-world application of the Nernst equation, but it plays a critical role during redox titration, as seen in the exercise. By determining the cell potential at various points during a titration, the Nernst equation gives insight into the reaction progress and helps to identify the equivalence point.

Equilibrium Constant

The equilibrium constant, represented as \( K \), is a dimensionless quantity that provides a relationship between the concentrations of reactants and products of a reaction at equilibrium. In the context of redox titrations and galvanic cells, the equilibrium constant can be derived from the standard cell potential by rearranging the Nernst equation to:
\[ \ln{K} = \frac{nFE_{cell}^{0}}{RT} \]
where \(K\) is the equilibrium constant when the cell potential is measured at equilibrium conditions. The equilibrium constant is a pivotal piece in understanding chemical reactions, as a large value of \(K\) indicates that the reaction favors the formation of products, while a small value signifies that reactants are favored.

During a redox titration, as seen in the exercise solution, the point where \(K\) is calculated corresponds to a specific moment in the reaction when the standard cell potential is known, and the cell potential under test conditions has been recorded. This data can provide significant insights into the stoichiometry and thermodynamics of the reaction, such as how much a change in reactant or product concentration can shift the equilibrium.

Cell Potential Calculation

Calculating the cell potential involves understanding how the potential of a galvanic cell can be determined based on the standard reduction potentials and the actual reaction conditions. The initial step is to compute the standard cell potential, \( E_{cell}^{0} \), which is simply the difference between the standard reduction potentials of the cathode and the anode. The next step requires applying the Nernst equation to incorporate the effect of reactant and product concentrations on the cell potential.

The students should recognize that it is the deviation from standard conditions that necessitates the use of the Nernst equation to adjust the cell potential accordingly. In the given exercise, the Nernst equation was key in calculating the cell potential both before the titration and at specific points during the redox titration and emphasizes the importance of the reaction quotient, \(Q\), in this calculation. The example provided offers students an engaging way to apply this calculation method to real lab scenarios.

Redox Titration

Redox titration is a type of titration based on a redox reaction between the analyte and titrant. It is an essential analytical technique to determine the concentration of an unknown solution. The process involves a gradual addition of a standard solution (titrant) to the analyte until the reaction reaches the equivalence point—the point at which the reacting species have reacted completely according to the reaction stoichiometry.

In the context of the exercise, redox titration was used to determine the unknown concentration of \(V^{2+}\) ion by titrating it with a known concentration of \(H_2EDTA^{2-}\). The change in the cell potential during the titration can reveal the endpoint, also known as the stoichiometric point, and involves the use of the Nernst equation to calculate values at the midpoint and before the titration began. When the cell potential remains constant despite the addition of more titrant, the endpoint has been reached, signaling that the reaction has reached equilibrium.