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Chapter 17: Problem 78

Consider the galvanic cell based on the following halfreactions: $$ \begin{array}{ll} \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ}=1.50 \mathrm{V} \\ \mathrm{Tl}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl} & \mathscr{E}^{\circ}=-0.34 \mathrm{V} \end{array} $$ a. Determine the overall cell reaction and calculate \(\mathscr{C}_{\text {cell. }}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\) c. Calculate \(\mathscr{E}_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} \mathrm{M}\) and \(\left[\mathrm{Tl}^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)

Short Answer

Expert verified
The overall cell reaction is Au³⁺ + 3Tl⁺ → Au + 3Tl, with a cell potential (E°cell) of 1.84 V. The standard Gibbs free energy change (ΔG°) for this reaction is -532,724 J/mol, and the equilibrium constant (K) at 25°C is 6.28 x 10^47. At the given concentrations of Au³⁺ and Tl⁺, the cell potential (E_cell) at 25°C is 1.79 V.

Step by step solution

01

Determine the overall cell reaction

First, we should balance the electrons in both half reactions. Since there is a difference in the number of electrons in both reactions (3e⁻ in the first reaction and 1e⁻ in the second), we should multiply the second reaction by 3 to balance the number of electrons. Then, we will add both reactions to obtain the overall cell reaction: Au³⁺ + 3e⁻ → Au (E° = 1.50 V) 3(Tl⁺ + e⁻ → Tl) (E° = -0.34 V) Now, add the reactions: Au³⁺ + 3Tl⁺ + 3e⁻ → Au + 3Tl
02

Calculate the cell potential

The cell potential (E°cell) is the difference between both half-reactions potentials. Since the Au³⁺/Au half-reaction has a higher potential value, it will act as the cathode, while the other half-reaction will act as the anode. Thus: E°cell = E°cathode - E°anode E°cell = (1.50 V) - (-0.34 V) E°cell = 1.84 V
03

Calculate the standard Gibbs free energy change

Now, we can calculate the standard Gibbs free energy change (ΔG°) using the formula: ΔG° = -nFE°cell Where n = number of moles of electrons transferred (in this case, 3), F = Faraday's constant (96,485 C/mol), and E°cell is the cell potential obtained in Step 2. ΔG° = -(3)(96485 C/mol)(1.84 V) ΔG° = -532,724 J/mol
04

Calculate the equilibrium constant

We can calculate the equilibrium constant (K) for the cell reaction at 25°C using the standard Gibbs free energy change and the equation: ΔG° = -RT ln K Where R = gas constant (8.314 J/mol K), T = temperature in Kelvin (25°C = 298.15 K), and ΔG° is the standard Gibbs free energy change calculated in Step 3. Rearranging the equation: ln K = -(ΔG°)/(RT) K = exp((532,724 J/mol)/((8.314 J/mol K)(298.15 K))) K = 6.28 x 10^47
05

Calculate the cell potential under given concentrations

We will now calculate the cell potential (E_cell) at 25°C, taking into account the given concentrations of both Au³⁺ ([Au³⁺] = 1.0 x 10⁻² M) and Tl⁺ ([Tl⁺] = 1.0 x 10⁻⁴ M) using the Nernst equation: E_cell = E°cell - (RT/nF) ln Q Where Q is the reaction quotient. For the balanced cell reaction, Q = [Au³⁺]/[Tl⁺]³. Now we can plug in the values: E_cell = 1.84 V - ((8.314 J/mol K)(298.15 K))/((3)(96485 C/mol)) ln (1.0 x 10⁻²)/(1.0 x 10⁻⁴)³ E_cell = 1.84 V - 0.0496 V E_cell = 1.79 V So, under the given concentration conditions, the cell potential at 25°C will be 1.79 V.

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Most popular questions from this chapter

When jump-starting a car with a dead battery, the ground jumper should be attached to a remote part of the engine block. Why?

Consider the following galvanic cell: a. Label the reducing agent and the oxidizing agent, and describe the direction of the electron flow. b. Determine the standard cell potential. c. Which electrode increases in mass as the reaction proceeds, and which electrode decreases in mass?

What is electrochemistry? What are redox reactions? Explain the difference between a galvanic and an electrolytic cell.

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.

Consider a galvanic cell based on the following half-reactions: $$\begin{array}{ll} & \mathscr{E}^{\circ}(\mathrm{V}) \\ \mathrm{La}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{La} & -2.37 \\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} & -0.44 \\ \hline \end{array}$$ a. What is the expected cell potential with all components in their standard states? b. What is the oxidizing agent in the overall cell reaction? c. What substances make up the anode compartment? d. In the standard cell, in which direction do the electrons flow? e. How many electrons are transferred per unit of cell reaction? f. If this cell is set up at \(25^{\circ} \mathrm{C}\) with \(\left[\mathrm{Fe}^{2+}\right]=2.00 \times 10^{-4} \mathrm{M}\) and \(\left[\mathrm{La}^{3+}\right]=3.00 \times 10^{-3} \mathrm{M},\) what is the expected cell potential?
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