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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 Au鲁鈦 in the nonstandard cell at 25掳C is approximately 7.92 脳 10鈦烩伒 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, Au鲁鈦 will be reduced since it has a higher standard reduction potential. Therefore, the Mg虏鈦 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 (Au鲁鈦) and the oxidation half-reaction (Mg). Since the oxidation half-reaction is the reverse of the given Mg虏鈦 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 [Au鲁鈦篯 in the nonstandard cell#
04

Identify the Nernst equation

To find the concentration of Au鲁鈦 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 [Au鲁鈦篯

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

Compute the concentration of [Au鲁鈦篯

Now, compute the concentration of [Au鲁鈦篯: $$\left[\mathrm{Au}^{3+}\right] \approx 7.92 \times 10^{-5} \,\mathrm{M}$$ The concentration of Au鲁鈦 in the nonstandard cell at 25掳C is approximately 7.92 脳 10鈦烩伒 M.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Cell Potential
To understand the concept of standard cell potential, it's essential to know its role in electrochemical cells. The standard cell potential, denoted as \( \mathscr{E}^\circ \), is the potential difference between two electrodes of a cell under standard conditions, which are typically 1 M concentration for aqueous solutions, 1 atm pressure for gases, and a temperature of 25掳C. This potential arises due to the tendency of a redox reaction to occur spontaneously. The higher the potential, the more the reaction will drive the electrons through the external circuit.

For a galvanic cell, which generates electrical energy from spontaneous chemical reactions, the standard cell potential can be calculated by adding the potentials of the cathodic (reduction) and anodic (oxidation) reactions. Usually, this is done by subtracting the standard reduction potential of the anode from that of the cathode. In our given exercise, the standard cell potential is calculated as follows:

- Reduction reaction: \( \mathrm{Au}^{3+} + 3 \mathrm{e}^{-} \rightarrow \mathrm{Au} \) with \( \mathscr{E}^\circ = 1.50 \, \text{V} \)
- Oxidation reaction: Reversing \( \mathrm{Mg}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Mg} \) to \( \mathrm{Mg} \rightarrow \mathrm{Mg}^{2+} + 2 \mathrm{e}^{-} \) which changes \( \mathscr{E}^\circ \) to \(+2.37 \, \text{V} \)
- \( \mathscr{E}^\circ_{\text{cell}} = 1.50 \, \text{V} - (-2.37 \, \text{V}) = 3.87 \, \text{V} \)
Nernst Equation
The Nernst equation plays a crucial role in connecting the electrical potential of a galvanic cell under nonstandard conditions to its concentration, pressure, and temperature. It helps us find the actual cell potential by considering deviations from standard conditions. The formula is:
\[E_\mathrm{cell} = \mathscr{E}^\circ_\mathrm{cell} - \frac{RT}{nF} \ln(Q)\]

Where:
  • \( E_\mathrm{cell} \) is the cell potential at nonstandard conditions
  • \( \mathscr{E}^\circ_\mathrm{cell} \) is the standard cell potential
  • \( R \) is the universal gas constant \((8.314 \, \text{J/mol路K})\)
  • \( T \) is the temperature in Kelvin
  • \( n \) is the number of electrons transferred in the reaction
  • \( F \) is the Faraday constant \((96,485 \, \text{C/mol})\)
  • \( Q \) is the reaction quotient, a measure of the concentrations of chemical species
The Nernst equation can be rearranged to find unknown concentrations in nonstandard conditions, using known values for the cell potential and standard conditions.
Redox Reaction
A redox reaction is a fundamental concept in electrochemistry involving the transfer of electrons between two species. These reactions are composed of two half-reactions: reduction and oxidation. In each galvanic cell, one species donates electrons (oxidation), and another accepts electrons (reduction).

In the classroom example, the reaction involves:
  • The reduction half-reaction: \( \mathrm{Au}^{3+} + 3 \mathrm{e}^{-} \rightarrow \mathrm{Au} \)
  • The oxidation half-reaction: \( \mathrm{Mg} \rightarrow \mathrm{Mg}^{2+} + 2 \mathrm{e}^{-} \) (reverse of the given reduction)
The term 'redox' stands for reduction and oxidation, highlighting the electron flow from one reactant to another. Balancing these reactions requires ensuring that the electrons lost in oxidation match the electrons gained in reduction. In this case, scaling the half-reactions to achieve electron balance is vital, often computed with the Least Common Multiple (LCM) of transferred electrons, resulting in 6 electrons cooperating in this reaction.

Understanding the exchange of electrons allows us to harness these reactions to power electronic devices, battery functioning, and more.
Concentration Calculation
Concentration calculation is essential for understanding how the Nernst equation is applied to real-world scenarios. By rearranging the Nernst equation, you can solve for the unknown concentration of reactants or products in a cell.

In the exercise, we calculate \([\mathrm{Au}^{3+}]\) using:
\[\left[\mathrm{Au}^{3+}\right] = \left[\mathrm{Mg}^{2+}\right]^2 \exp\left( \frac{nF(E_{\text{cell}} - \mathscr{E}^\circ_{\text{cell}})}{RT} \right)\]

This setup involves the concentration of \([\mathrm{Mg}^{2+}]\), the measured cell output \((E_{\text{cell}})\), and constants from the Nernst equation. With experimentally determined values, you can deduce the concentration of the gold ions, which is essential for converting theoretical concepts into practical applications.

Using this equation, students learn to predict and adjust concentrations needed to achieve desired cell potentials, crucial in constructing batteries and electronic devices where precise chemical balances matter.

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Most popular questions from this chapter

A fuel cell designed to react grain alcohol with oxygen has the following net reaction: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ The maximum work that 1 mole of alcohol can do is \(1.32 \times\) \(10^{3} \mathrm{kJ} .\) What is the theoretical maximum voltage this cell can achieve at \(25^{\circ} \mathrm{C} ?\)

Estimate \(\mathscr{E}^{\circ}\) for the half-reaction $$ 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-} $$ given the following values of \(\Delta G_{\mathrm{f}}^{\circ}\) $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}(l) &=-237 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g) &=0.0 \\ \mathrm{OH}^{-}(a q) &=-157 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{e}^{-} &=0.0 \end{aligned} $$ Compare this value of \(\mathscr{E}^{\circ}\) with the value of \(\mathscr{E}^{\circ}\) given in Table 17-1.

Explain the following relationships: \(\Delta G\) and \(w,\) cell potential and \(w,\) cell potential and \(\Delta G,\) cell potential and \(Q .\) Using these relationships, explain how you could make a cell in which both electrodes are the same metal and both solutions contain the same compound, but at different concentrations. Why does such a cell run spontaneously?

Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{C}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are 1.0 atm. a. \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.78 \mathrm{V}\) \(\mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2} \quad \quad \mathscr{E}^{\circ}=0.68 \mathrm{V}\) b. \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} \quad \mathscr{E}^{\circ}=-1.18 \mathrm{V}\) \(\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad \mathscr{E}^{\circ}=-0.036 \mathrm{V}\)

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 \(25^{\circ} \mathrm{C}:\) $$\begin{aligned} &3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\ & \mathscr{E}^{\circ}=0.957 \mathrm{V} \end{aligned}$$ $$\begin{aligned} &\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) &\mathscr{E}^{\circ}=0.775 \mathrm{V} \end{aligned}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) 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.

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