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Chapter 18: Problem 135

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

Short Answer

Expert verified
The standard potential for this galvanic cell is 3.87 V. The concentration of Au³⁺ in the nonstandard cell at 25°C with a Mg²⁺ concentration of \(1.00 \times 10^{-5} M\) and a cell potential of 4.01 V is \(2.16 \times 10^{-4} M\).

Step by step solution

01

Calculate Standard Cell Potential

To calculate the standard cell potential (\(E^0_{cell}\)), we need to find the difference in potential (\(\Delta E^0\)) between the two half-reactions. First, identify which half-reaction serves as the anode (oxidation) and which one serves as the cathode (reduction). The reaction with the more positive potential will be the reduction (cathode) reaction, and the reaction with the less positive potential will be the oxidation (anode) reaction. Cathode: Au³⁺ + 3e⁻ → Au (\(E^0_{Au} = 1.50 V\)) Anode: Mg → Mg²⁺ + 2e⁻ (\(E^0_{Mg} = -2.37 V\)) Now calculate the standard cell potential (\(E^0_{cell}\)): \(E^0_{cell} = E^0_{cathode} - E^0_{anode}\)
02

Determine the Standard Cell Potential

Substitute the given values of \(E^0_{Au}\) and \(E^0_{Mg}\) in the formula: \(E^0_{cell} = 1.50 - (-2.37) = 3.87 V\) The standard potential for this cell is 3.87 V.
03

Apply the Nernst Equation

To find the concentration of Au³⁺ in the nonstandard cell, we will use the Nernst equation: \(E_{cell} = E^0_{cell} - \frac{RT}{nF} \cdot ln(Q)\) Where: \(E_{cell}\) = cell potential \(E^0_{cell}\) = standard cell potential \(R\) = gas constant (8.314 J/mol K) \(T\) = temperature in Kelvin \(n\) = number of electrons transferred in the reaction (as calculated from the balanced chemical equation) \(F\) = Faraday’s constant (96485 C/mol) \(Q\) = reaction quotient For our problem, we have: \(E_{cell} = 4.01 V\) \(E^0_{cell} = 3.87 V\) \(T = 298 K\) ( Since \(25^\circ C = 298 K\)) \(n = 6\) (Considering the balanced reaction: 2Au³⁺ + 3Mg → 2Au + 3Mg²⁺)
04

Calculate the Reaction Quotient (Q)

Write the expression for the reaction quotient (Q) using the balanced chemical reaction: \(Q = \frac{([Mg^{2+}]^3)}{([Au^{3+}]^2)}\) We are given \([Mg^{2+}] = 1.00 × 10^{-5} M\), and we want to find \([Au^{3+}]\).
05

Solve for [Au³⁺]

Rearrange the Nernst equation to isolate ln(Q): \(ln(Q) = \frac{nFE_{cell}-nFE^0_{cell}}{RT} \) Now, substitute the values and solve for \(Q\): \(Q = e^{\frac{6 * 96485 (4.01 - 3.87)}{8.314 * 298}} = 103.95\) Next, solve for \([Au^{3+}]\) using the \(Q\) value and the given concentration of \([Mg^{2+}]\): \([Au^{3+}]^2 = \frac{([Mg^{2+}]^3)}{Q}\) \([Au^{3+}] = \sqrt{\frac{(1.00 × 10^{-5})^3}{103.95}} = 2.16 × 10^{-4} M\) The concentration of Au³⁺ in the nonstandard cell is \(2.16 × 10^{-4} M\).

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Most popular questions from this chapter

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium $$2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Consider the following standard reduction potentials at $25^{\circ} \mathrm{C} :$ $$3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.957 \mathrm{V}$$ $$\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.775 \mathrm{V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and NO \(_{2}\) mixture with only 0.20\(\% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.

In the electrolysis of an aqueous solution of $\mathrm{Na}_{2} \mathrm{SO}_{4},$ what reactions occur at the anode and the cathode (assuming standard conditions)? $$\begin{array}{ll} {\text{}} & \quad{ \mathscr{E}^{\circ} } \\ \hline {\mathrm{S}_{2} \mathrm{O}_{8}^{2-}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{SO}_{4}^{2-}} & {2.01 \mathrm{V}} \\ {\mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}} & {1.23 \mathrm{V}} \\ {2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}} & {-0.83 \mathrm{V}} \\\ {\mathrm{Na}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Na}} & {-2.71 \mathrm{V}}\end{array}$$

If the cell potential is proportional to work and the standard reduction potential for the hydrogen ion is zero, does this mean that the reduction of the hydrogen ion requires no work?

One of the few industrial-scale processes that produce organic compounds electrochemically is used by the Monsanto Company to produce 1,4-dicyanobutane. The reduction reaction is $$2 \mathrm{CH}_{2}=\mathrm{CHCN}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}$$ The \(\mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}\) is then chemically reduced using hydrogen gas to $\mathrm{H}_{2} \mathrm{N}-\left(\mathrm{CH}_{2}\right)_{6}-\mathrm{NH}_{2},$ which is used to produce production of nylon. What current must be used to produce \(150 . \mathrm{kg} \mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}\) per hour?

A zinc-copper battery is constructed as follows at \(25^{\circ} \mathrm{C} :\) $$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.10 M)\right|\left|\mathrm{Cu}^{2+}(2.50 M)\right| \mathrm{Cu}$$ The mass of each electrode is 200. g. a. Calculate the cell potential when this battery is first connected. b. Calculate the cell potential after 10.0 A of current has flowed for 10.0 h. (Assume each half-cell contains 1.00 L of solution.) c. Calculate the mass of each electrode after 10.0 h. d. How long can this battery deliver a current of 10.0 A before it goes dead?
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