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

Step by step solution

01

Identify the Half-reactions

We have two half-reactions: Reaction 1: \(Cu^{2+} + 2e^- \longrightarrow Cu\), \(E^\circ = 0.34V\) Reaction 2: \(V^{2+} + 2e^- \longrightarrow V\), \(E^\circ = -1.20V\) The given cell potential at the stoichiometric point is \(E_{cell} = 1.98V\).

02

Calculate Q_rxn before the titration

Before the titration begins, the concentrations are: \([Cu^{2+}] = 1.00M\) Concentration of \(V^{2+}\) is unknown, we can denote it as x. Next, we calculate \(Q_{rxn}\), the reaction quotient before the titration begins: \(Q_{rxn} = \frac{[Cu^{2+}]}{[V^{2+}]}\)

03

Use the Nernst Equation

We can use the Nernst equation to determine the cell potential: \(E_{cell} = E^\circ_{cell} - \frac{0.0592}{2} * log(Q_{rxn})\) Since the cell reaction involves both half-reactions, we should find \(E^\circ_{cell}\) before applying the Nernst equation : \(E^\circ_{cell} = E^\circ_{1} - E^\circ_{2} = 0.34V + 1.20V = 1.54V\) Now we can plug this value and the given stoichiometric point potential into the Nernst equation: \(E_{stoich} = 1.98V = 1.54V - \frac{0.0592}{2} * log\left(\frac{1.00}{x}\right)\) Next, we solve for the unknown concentration x.

04

Solve for the Unknown Concentration x

Rearranging the equation above: \(\frac{0.0592}{2} * log\left(\frac{1.00}{x}\right) = 1.54V - 1.98V = -0.44V\) Divide both sides by \(\frac{0.0592}{2}\): \(log\left(\frac{1.00}{x}\right) = \frac{-0.44V}{\frac{0.0592}{2}} = -14.87\) \(x = \frac{1.00}{10^{14.87}}\) Now we can rewrite the Nernst equation for the cell potential before the titration began. \(E_{cell} = 1.54V - \frac{0.0592}{2} * log\left(\frac{1.00}{10^{-14.87}}\right)\)

05

Calculate E_cell before the titration began

Now we can calculate the cell potential \(E_{cell}\) before the titration began: \(E_{cell} = 1.54V - \frac{0.0592}{2} * (14.87) = 1.529V\) So, the cell potential before the titration began was \(1.529V\). b. Calculate the equilibrium constant K for the titration reaction.

06

Find EDTA2- concentration at the stoichiometric point

At the stoichiometric point, 500.0mL of 0.0800M H2EDTA2- solution was added. We can calculate the number of moles: mol of H2EDTA2- = 0.5L * 0.0800M = 0.0400mol

07

Calculate the concentration of V2+ at the stoichiometric point

We found in section a that the initial concentration of V2+ was \(x = 10^{-14.87}\). At the stoichiometric point, the concentration of V2+ would be = x - 0.0400M. We can plug in the values: Concentration of V2+ at stoichiometric point = \(10^{-14.87} - 0.0400 = -0.0400M\) c. Calculate E_cell at the halfway point in the titration. We will be using the Nernst equation for this using half way concentrations.

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