An electrochemical cell is set up using the following unbalanced reaction: $$ \mathrm{M}^{a+}(a q)+\mathrm{N}(s) \longrightarrow \mathrm{N}^{2+}(a q)+\mathrm{M}(s) $$ The standard reduction potentials are: $$ \begin{array}{ll} \mathrm{M}^{a+}+a \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=0.400 \mathrm{~V} \\ \mathrm{~N}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{N} & \mathscr{E}^{\circ}=0.240 \mathrm{~V} \end{array} $$ The cell contains \(0.10 \mathrm{M} \mathrm{N}^{2+}\) and produces a voltage of \(0.180 \mathrm{~V}\). If the concentration of \(\mathrm{M}^{a+}\) is such that the value of the reaction quotient \(Q\) is \(9.32 \times 10^{-3}\), calculate \(\left[\mathrm{M}^{a+}\right]\). Calculate \(w_{\max }\) for this electrochemical cell.

Step by step solution

01

Balance the redox reaction

First, we need to balance the unbalanced chemical reaction. We can do this by multiplying the half-reactions in a way that the number of electrons transferred is the same: \(\mathrm{M}^{a+}(a q)+ a\mathrm{N}(s) \longrightarrow a\mathrm{N}^{2+}(a q)+\mathrm{M}(s)\)

02

Calculate the Cell Potential, E°

From the standard reduction potentials equation, we can calculate the cell potential, E°: \(\mathrm{E}°_{cell} = \mathrm{E}°_{cathode} - \mathrm{E}°_{anode}\) In this case, we have: $\begin{array}{ll} \mathrm{M}^{a+} + a \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}_{M}=0.400 \mathrm{~V}\\ \mathrm{N}^{2+} + 2 \mathrm{e}^{-} \longrightarrow \mathrm{N} & \mathscr{E}^{\circ}_{N}=0.240 \mathrm{~V}\\ \end{array}$ M is the cathode (reduction) and N is the anode (oxidation), so: \(\mathrm{E}°_{cell} = 0.400 \mathrm{V} - 0.240 \mathrm{V} = 0.160 \mathrm{~V}\)

03

Use Nernst Equation to calculate the concentration of M^{a+}

Now we can use the Nernst Equation for the electrochemical cell: \(E_{cell} = E°_{cell} - \dfrac{0.0592}{n}\log{\dfrac{[\mathrm{N}^{2+}]}{[\mathrm{M}^{a+}]^a}}\) Given the cell voltage (\(0.180 V\)) and reaction quotient \(Q = 9.32 \times 10^{-3}\), apply the Nernst equation: \(0.180 \mathrm{~V} = 0.160 \mathrm{~V} - \dfrac{0.0592}{2}\log{\dfrac{[\mathrm{N}^{2+}]}{[\mathrm{M}^{a+}]^a}}\) Now, we have \([\mathrm{N}^{2+}] = 0.10 \mathrm{~M}\) and \(Q = 9.32 \times 10^{-3}\). Using the balanced equation, rearrange the Nernst Equation to find the concentration of M^{a+}: \([\mathrm{M}^{a+}]^a = \dfrac{[\mathrm{N}^{2+}]}{Q}\) Replace the values: \([\mathrm{M}^{a+}]^a = \dfrac{0.10 \mathrm{~M}}{9.32\times10^{-3}}\) \([\mathrm{M}^{a+}] = 1.25 \mathrm{~M}\)

04

Calculate the Maximum Work Done, \(w_{\max}\)

To calculate the maximum work done by the electrochemical cell (\(w_{\max}\)), use the Gibb's free energy equation: \(w_{\max} = -nFE_{cell}\) First, find the Faraday constant (F) and the number of electrons transferred (n), and then apply the equation: \(n = 2\) \(F = 96485 \mathrm{\dfrac{C}{mol \cdot e^{-}}}\) \(w_{\max} = -2(96485 \frac{\mathrm{C}}{\mathrm{mol}\cdot\mathrm{e}^{-}})(0.180 \mathrm{~V})\) \(w_{\max} = -34734 \mathrm{\dfrac{J}{mol}}\) So, the concentration of \([\mathrm{M}^{a+}]\) is \(1.25 \mathrm{~M}\), and the maximum work done by the electrochemical cell (\(w_{\max}\)) is \(-34734 \mathrm{\dfrac{J}{mol}}\).

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Key Concepts

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

Nernst Equation

The Nernst Equation is a crucial tool in electrochemistry that relates the electromotive force of an electrochemical cell to the concentrations of the reactants and products. It is described by the formula:
\[ E_{cell} = E^{\circ}_{cell} - \dfrac{0.0592}{n} \log Q \]
where \(E_{cell}\) is the cell potential under non-standard conditions, \(E^{\circ}_{cell}\) is the standard cell potential, \(n\) is the number of moles of electrons transferred in the reaction, and \(Q\) is the reaction quotient, measuring the relative concentrations of products and reactants. In the given exercise, the Nernst Equation was used to determine the concentration of \(\mathrm{M}^{a+}\) ions based on the observed cell voltage and standard cell potential.

Standard Reduction Potential

Standard reduction potential, denoted as \(E^{\circ}\), is a measure of the tendency of a chemical species to be reduced, and it's expressed in volts (V). In the context of the provided exercise, the standard reduction potential helps in calculating the standard cell potential (\(E^{\circ}_{cell}\)), which is the difference between the reduction potentials of the cathode and anode. The cell potential is a driving force behind the movement of electrons and thus the reaction in the cell. The given standard potentials for the reduction of \(\mathrm{M}^{a+}\) and \(\mathrm{N}^{2+}\) are used to figure out which species is reduced or oxidized and to calculate the standard cell potential.

Gibbs Free Energy

Gibbs free energy (\(\Delta G\)) in electrochemistry is related to the work that can be extracted from an electrochemical cell. It's calculated with the equation:
\[ \Delta G = -nFE_{cell} \]
where \(n\) is the number of moles of electrons exchanged during the electrochemical reaction, \(F\) is the Faraday constant representing the charge of one mole of electrons (approximately 96485 coulombs), and \(E_{cell}\) is the cell potential under non-standard conditions. The exercise uses this relationship between Gibbs free energy and the cell potential to calculate the maximum work that can be done by the electrochemical cell, referred to as \(w_{max}\).

Balancing Redox Reactions

Redox reactions involve the transfer of electrons between chemical species. Balancing these reactions is fundamental to understanding and making quantitative calculations in an electrochemical process. The step-by-step solution first balanced the unbalanced chemical reaction given in the exercise. The number of electrons lost in the oxidation half-reaction must equal the number gained in the reduction half-reaction, ensuring that the charge and mass are balanced. Successfully balancing redox reactions allows for the proper use of the Nernst Equation and Gibbs free energy formula as seen in the steps for solving the exercise presented. These concepts work together to provide a comprehensive understanding of an electrochemical cell's behavior.