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Chapter 17: Problem 137

Consider a galvanic cell based on the following theoretical half-reactions: $$\begin{array}{lr} & \mathscr{E}^{\circ}(\mathrm{V}) \\ \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & 1.50 \\ \mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg} & -2.37 \end{array}$$ a. What is the standard potential for this cell? b. A nonstandard cell is set up at \(25^{\circ} \mathrm{C}\) with \(\left[\mathrm{Mg}^{2+}\right]=\) \(1.00 \times 10^{-5} M .\) The cell potential is observed to be 4.01 V. Calculate \(\left[\mathrm{Au}^{3+}\right]\) in this cell. \mathscr{E}^{\circ}

Short Answer

Expert verified
a. The standard potential for this cell is 3.87 V. b. The concentration of Au³⁺ in the nonstandard cell at 25°C is approximately 7.92 × 10⁻⁵ M.

Step by step solution

01

Write the cell reaction

To find the standard potential for this cell, first, write the overall cell reaction by combining the given half-reactions. The overall cell reaction will have both reduction and oxidation occurring simultaneously. $$\begin{array}{l} \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} \\ \mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg} \end{array}$$
02

Determine the oxidation and reduction reactions

In this cell, Au³⁺ will be reduced since it has a higher standard reduction potential. Therefore, the Mg²⁺ half-reaction must be reversed as the oxidation half-reaction. The reverse half-reaction will be: $$\mathrm{Mg} \longrightarrow \mathrm{Mg}^{2+}+2 \mathrm{e}^{-}$$
03

Calculate the standard cell potential

The standard cell potential is the sum of the standard reduction potential for the reduction half-reaction (Au³⁺) and the oxidation half-reaction (Mg). Since the oxidation half-reaction is the reverse of the given Mg²⁺ half-reaction, the sign of the standard reduction potential must be changed to positive for the oxidation half-reaction. Standard potential = \( \mathscr{E}^{\circ}_(\textrm{reduction}) - \mathscr{E}^{\circ}_(\textrm{oxidation}) \) Standard potential = \(1.50 - (-2.37)\) Standard potential = \(3.87 V\) So, the standard potential for this cell is 3.87 V. #b. Calculate [Au³⁺] in the nonstandard cell#
04

Identify the Nernst equation

To find the concentration of Au³⁺ in the nonstandard cell, we can use the Nernst equation. The Nernst equation is given by: $$E_\mathrm{cell} = \mathscr{E}^\circ_\mathrm{cell} - \frac{RT}{nF} \ln(\mathrm{Q})$$ Where E_cell is the observed cell potential, \( \mathscr{E}^\circ_\mathrm{cell} \) is the standard cell potential, R is the gas constant (8.314 J/mol*K), T is the temperature in Kelvin, n is the number of electrons transferred in the balanced redox reaction (in this case, n = 6), and F is the Faraday constant (96,485 C/mol).
05

Calculate the reaction quotient (Q)

The reaction quotient (Q) for this cell is given by: $$\mathrm{Q}=\frac{\left[\mathrm{Au}^{3+}\right]}{\left[\mathrm{Mg}^{2+}\right]^{2}}$$
06

Solve the Nernst equation for [Au³⁺]

Rearrange the Nernst Equation to solve for the concentration of [Au³⁺]: $$\left[\mathrm{Au}^{3+}\right] = \left[\mathrm{Mg}^{2+}\right]^2 \mathrm{exp}\left( \frac{nF(E_{cell} - \mathscr{E}^\circ_{cell})}{RT} \right)$$ Now, plug in the values for temperature, cell potential, standard cell potential, and the concentration of Mg²⁺: $$\left[\mathrm{Au}^{3+}\right] = \left(1.00 \times 10^{-5}\right)^2 \mathrm{exp}\left( \frac{6 \times 96,485(4.01 - 3.87)}{8.314 \times 298} \right)$$
07

Compute the concentration of [Au³⁺]

Now, compute the concentration of [Au³⁺]: $$\left[\mathrm{Au}^{3+}\right] \approx 7.92 \times 10^{-5} \,\mathrm{M}$$ The concentration of Au³⁺ in the nonstandard cell at 25°C is approximately 7.92 × 10⁻⁵ M.

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Most popular questions from this chapter

Consider the following galvanic cell: A 15.0 -mole sample of \(\mathrm{NH}_{3}\) is added to the Ag compartment (assume \(1.00 \mathrm{L}\) of total solution after the addition). The silver ion reacts with ammonia to form complex ions as shown: $$\begin{aligned} \mathrm{Ag}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{AgNH}_{3}^{+}(a q) \\\& K_{1}=2.1 \times 10^{3} \\ \mathrm{AgNH}_{3}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \\\\\ & K_{2}=8.2 \times 10^{3} \end{aligned}$$ Calculate the cell potential after the addition of 15.0 moles of \(\mathrm{NH}_{3}\)

Define oxidation and reduction in terms of both change in oxidation number and electron loss or gain.

Aluminum is produced commercially by the electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in the presence of a molten salt. If a plant has a continuous capacity of 1.00 million \(A\), what mass of aluminum can be produced in \(2.00 \mathrm{h} ?\)

Look up the reduction potential for \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+} .\) Look up the reduction potential for \(\mathrm{Fe}^{2+}\) to Fe. Finally, look up the reduction potential for \(\mathrm{Fe}^{3+}\) to Fe. You should notice that adding the reduction potentials for the first two does not give the potential for the third. Why not? Show how you can use the first two potentials to calculate the third potential.

Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{Cu}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{NO}(g)\) b. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)\) d. \(\mathrm{Mn}^{2+}(a q)+\mathrm{NaBiO}_{3}(s) \rightarrow \mathrm{Bi}^{3+}(a q)+\mathrm{MnO}_{4}^{-}(a q)\) e. \(\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{Zn}(s) \rightarrow \mathrm{AsH}_{3}(g)+\mathrm{Zn}^{2+}(a q)\)
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