A galvanic cell is based on the following half-reactions: $$\mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) \quad \mathscr{E}^{\circ}=0.34 \mathrm{V}$$ $$\mathrm{V}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{V}(s) \quad \mathscr{E}^{\circ}=-1.20 \mathrm{V}$$ 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 $$\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q)+\mathrm{V}^{2+}(a q) \rightleftharpoons \mathrm{VEDTA}^{2-}(a q)+2 \mathrm{H}^{+}(a q)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad K=?$$ The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of 500.0 $\mathrm{mL} \mathrm{H}_{2} \mathrm{EDTA}^{2-}$ 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.

Step by step solution

01

Identify the overall redox reaction

By looking at the half-reactions provided, we can combine them to form the overall redox reaction: $$\mathrm{Cu}^{2+}(a q)+\mathrm{V}(s) \longrightarrow \mathrm{Cu}(s)+\mathrm{V}^{2+}(a q)$$

02

Use the Nernst equation to calculate the initial cell potential

The Nernst equation for the cell potential is given by: $$\mathscr{E}_{\text {cell}} = \mathscr{E}^{\circ} _{\text {cell}} - \frac{RT}{nF} \ln{Q}$$ For our cell, we have: - \(\mathscr{E}^{\circ} _{\text {cell}} = \mathscr{E}^{\circ} _{\text {Cu}} - \mathscr{E}^{\circ} _{\text {V}} = 0.34V - (-1.20V) = 1.54V\) - \(R = 8.314 \frac{J}{K \cdot mol}\) - \(T = 298 K\) (assuming room temperature) - \(n = 2\) (2 electrons transferred) - \(F = 96,485 \frac{C}{mol}\) - \(Q = \frac{[\mathrm{V}^{2+}]}{[\mathrm{Cu}^{2+}]}\) The initial concentration of \(\mathrm{Cu}^{2+}\) is 1.00 M; and the initial concentration of \(\mathrm{V}^{2+}\) is unknown. However, since we need to find the cell potential before the titration, no \(\mathrm{H}_{2} \mathrm{EDTA}^{2-}\) has been added yet, meaning no \(\mathrm{V}^{2+}\) has been reduced. Therefore, the cell is at its equilibrium at the beginning and the potential is the standard potential.

03

Calculate the cell potential before titration

Before the titration, at equilibrium, we have: $$\mathscr{E}_{\text {cell}} = \mathscr{E}^{\circ} _{\text {cell}} = 1.54V$$ #b. Calculate the value of the equilibrium constant for the titration reaction#

04

Determine the concentration of \(\mathrm{V}^{2+}\) at the stoichiometric point

At the stoichiometric point, 500.0 mL of 0.0800 M \(\mathrm{H}_{2} \mathrm{EDTA}^{2-}\) has been added. So, the moles of \(\mathrm{H}_{2} \mathrm{EDTA}^{2-}\) added is: $$(0.0800 \,\text{mol/L})(0.500\, \text{L}) = 0.0400\text{ mol}$$ Since 1 mol of \(\mathrm{H}_{2} \mathrm{EDTA}^{2-}\) reacts with 1 mol of \(\mathrm{V}^{2+}\), we can calculate the concentration of \(\mathrm{V}^{2+}\) at the stoichiometric point as follows: $$[\mathrm{V}^{2+}] = \frac{0.0400\text{ mol}}{1.00\, \text{L}} = 0.0400\,\text{M}$$

05

Use the Nernst equation at the stoichiometric point to calculate the equilibrium constant

At the stoichiometric point, we are given that the cell potential is 1.98 V. We can use the Nernst equation and the previously found value for the concentration of \(\mathrm{V}^{2+}\) to calculate the reaction quotient \(Q\): $$1.98\,V = 1.54\,V - \frac{8.314\,\frac{J}{K\cdot mol} \cdot 298\,K}{2 \cdot 96,485\,\frac{C}{mol}}\ln{Q}$$ Solve for \(Q\): $$\ln{Q} = \frac{1.98V - 1.54V}{- \frac{8.314\cdot 298}{2 \cdot 96,485}}$$ $$Q = e^{\frac{0.44}{-\frac{8.314\cdot 298}{2\cdot 96,485}}} \approx 0.00161$$ Since we know that \(K = \frac{[\mathrm{VEDTA}^{2-}]}{[\mathrm{H}_{2} \mathrm{EDTA}^{2-}][\mathrm{V}^{2+}]}\), we can use the value of \(Q\) to find the equilibrium constant \(K\): $$K = Q[\mathrm{V}^{2+}][\mathrm{H}_{2} \mathrm{EDTA}^{2-}]$$ $$K = 0.00161 \cdot 0.0400\,M \cdot 0.0800\,M \approx 5.17 \times 10^{-6}$$ #c. Calculate the cell potential at the halfway point in the titration#

06

Determine the concentration of \(\mathrm{V}^{2+}\) at the halfway point

At the halfway point, half of the moles of \(\mathrm{V}^{2+}\) are reduced, so the concentration of \(\mathrm{V}^{2+}\) is halved: $$[\mathrm{V}^{2+}]_{\text{halfway}} = \frac{0.0400\,\text{M}}{2} = 0.0200\,\text{M}$$

07

Use the Nernst equation at the halfway point

We can use the Nernst equation again with this concentration to find the cell potential: $$\mathscr{E}_{\text {cell}}(\text{halfway}) = 1.54\,V - \frac{8.314\,\frac{J}{K\cdot mol} \cdot 298\,K}{2 \cdot 96,485\,\frac{C}{mol}}\ln{Q'}$$ Here, \(Q'\) is the reaction quotient at the halfway point when the concentrations of the reactants and products change. The \(\mathrm{Cu}^{2+}\) concentration remains the same at 1.00 M. So, we have: $$Q' = \frac{[\mathrm{V}^{2+}]_{\text{halfway}}}{[\mathrm{Cu}^{2+}]} = \frac{0.0200\,M}{1.00\,M} = 0.0200$$ Now, we can plug \(Q'\) back into the Nernst equation and solve for the cell potential at the halfway point: $$\mathscr{E}_{\text {cell}}(\text{halfway}) = 1.54\,V - \frac{8.314\cdot 298}{2 \cdot 96,485}\ln{0.0200}$$ $$\mathscr{E}_{\text {cell}}(\text{halfway}) \approx 1.74\,\mathrm{V}$$ To summarize: a. The initial cell potential before the titration was carried out is 1.54 V. b. The equilibrium constant, K, for the titration reaction is approximately 5.17 x 10^-6. c. The cell potential at the halfway point is approximately 1.74 V.

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