Question

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: $$\mathrm{M}^{a+}+a \mathrm{e}^{-} \longrightarrow \mathrm{M} \quad \mathscr{E}^{\circ}=0.400 \mathrm{V}$$ $$\mathrm{N}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{N} \quad \mathscr{E}^{\circ}=0.240 \mathrm{V}$$ The cell contains 0.10\(M \mathrm{N}^{2+}\) 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 \(w_{\text { max }}\) for this electrochemical cell.

Short answer

To find the concentration of \(\mathrm{M}^{a+}\), we apply the Nernst equation and solve for \(\left[\text{M}^{a+}\right]\): \[ \left[\text{M}^{a+}\right] = \frac{(0.10\text{M})^a (9.32 \times 10^{-3})}{(0.10\text{M})^{2a-1}} \\ \] To calculate the maximum work (\(w_{\text{max}}\)) for this electrochemical cell, we use the equation: \[ w_{\text{max}} = -(2a)(96485\mathrm{C/mol})(0.18\mathrm{V}) \\ \]

Step-by-step solution

Write balanced cell reaction

To find the balanced cell reaction, we must balance the individual half-reactions: \[ \text{M}^{a+} + a \text{e}^- \longrightarrow \text{M} \quad\quad (1)\\ \] and \[ \text{N}^{2+} + 2\text{e}^- \longrightarrow \text{N} \quad\quad (2)\\ \] Multiplying half-reaction (1) by 2 and half-reaction (2) by \(a\), we then add them together to form the balanced cell reaction: \[ 2\text{N}^{2+}(aq) + 2a\text{M}(s) \longrightarrow 2a\text{N}(s) + 2\text{M}^{a+}(aq)\\ \]

Calculate the standard cell voltage (\(\mathscr{E}^{\circ}_{cell}\))

To find the standard cell voltage, we can subtract the standard reduction potential of the less positive half-reaction from the more positive half-reaction: \[ \mathscr{E}^{\circ}_{cell} = \mathscr{E}^{\circ}(\text{M}^{a+}/\text{M}) - \mathscr{E}^{\circ}(\text{N}^{2+}/\text{N})\\ \] \[ \mathscr{E}^{\circ}_{cell} = 0.4 \mathrm{V} - 0.24 \mathrm{V} = 0.16 \mathrm{V}\\ \]

Apply the Nernst equation to find the concentration of \(\text{M}^{a+}\)

The Nernst equation is given by: \[ E_{cell} = \mathscr{E}^{\circ}_{cell} - \frac{RT}{nF} \ln Q \\ \] where \(E_{cell}\) is the cell voltage, \(\mathscr{E}^{\circ}_{cell}\) is the standard cell voltage, \(R\) is the gas constant, \(T\) is the temperature (in Kelvin), \(n\) is the number of moles of electrons transferred, and \(F\) is Faraday's constant. Given \(E_{cell} = 0.18 \mathrm{V}\), \(Q = 9.32 \times 10^{-3}\), and assuming \(T = 298 \mathrm{K}\): \[ 0.18 \mathrm{V} = 0.16 \mathrm{V} - \frac{(8.314\mathrm{J / K \cdot mol})(298\mathrm{K})}{(2a)(96485\mathrm{C/mol})} \ln 9.32 \times 10^{-3} \\ \] Solving for \(\left[\text{M}^{a+}\right]\): \[ \left[\text{M}^{a+}\right] = \frac{(0.10\text{M})^a (9.32 \times 10^{-3})}{(0.10\text{M})^{2a-1}}\\ \]

Calculate the maximum work (\(w_{\text{max}}\)) for this electrochemical cell

To find the maximum work, we can use the following equation: \[ w_{\text{max}} = -nFE_{cell} \\ \] where \(n\) is the number of moles of electrons transferred (in this case, \(2a\)), \(F\) is Faraday's constant, and \(E_{cell}\) is the cell voltage. Plugging in the values: \[ w_{\text{max}} = -(2a)(96485\mathrm{C/mol})(0.18\mathrm{V}) \\ \]

Key concepts

Nernst Equation

In electrochemistry, the Nernst Equation is a pivotal tool for understanding how the voltage of an electrochemical cell is influenced by the concentration of ions involved in the reaction. It calculates the cell potential under non-standard conditions by relating it to the standard cell potential and the reaction quotient (Q). The equation is given by:

• \(E_{cell} = \mathscr{E}^{\circ}_{cell} - \frac{RT}{nF} \ln Q\)

Here:

• \(E_{cell}\) represents the actual cell potential.
• \(\mathscr{E}^{\circ}_{cell}\) is the standard cell potential, calculated from standard reduction potentials.
• \(R\) is the universal gas constant \( (8.314 \text{J/mol K}) \).
• \(T\) is the temperature in Kelvin.
• \(n\) is the moles of electrons involved in the balanced equation.
• \(F\) is Faraday's constant \( (96485 \text{C/mol}) \).
• \(Q\) is the reaction quotient.

When applied to the given problem, we see how changes in concentration affect cell voltage. From the exercise, using the Nernst Equation let us calculate the concentration of \(\text{M}^{a+}\) by substituting known values: actual cell potential \( (0.18 \text{V}) \), and the reaction quotient \( (9.32 \times 10^{-3}) \). This way, the equation balances theory and real-time conditions, demonstrating both consequence and control over reaction factors.

Standard Reduction Potential

The standard reduction potential \((\mathscr{E}^{\circ})\) of a chemical species is a measure of its tendency to gain electrons and thereby be reduced, compared to the standard hydrogen electrode set at 0 volts. These potentials are crucial in determining the direction and spontaneity of redox reactions in electrochemical cells. For any particular half-reaction:

• Positive \(\mathscr{E}^{\circ}\) values signify a higher tendency to gain electrons.
• Negative \(\mathscr{E}^{\circ}\) values indicate a lower tendency to gain electrons.

In this exercise, we used the standard reduction potentials to calculate the standard cell potential \( (\mathscr{E}^{\circ}_{cell}) \) by just subtracting the less positive value from the more positive value. Using the potentials from the exercise:

• \(\mathscr{E}^{\circ}(\text{M}^{a+}/\text{M})\) = 0.400 V
• \(\mathscr{E}^{\circ}(\text{N}^{2+}/\text{N})\) = 0.240 V

The calculation gave us the cell’s standard potential \( (0.16 \text{V}) \), offering insights into the flow of electrons and proving crucial for further calculations like the Nernst Equation. This aspect represents the energy aspect of the electron's journey from one half-cell to another, a key determinant of electrochemical efficiency.

Reaction Quotient (Q)

The reaction quotient \((Q)\) is a dimensionless number that reflects the ratio of concentrations of products to reactants at any point during the reaction, based on their stoichiometric coefficients from the balanced reaction. It has a significant role in determining how far a reaction has proceeded towards equilibrium and helps in calculating real-time cell potential using the Nernst Equation.

For the electrochemical cell:

• If \( Q < K_{eq} \) (equilibrium constant), the reaction will proceed in the forward direction.
• If \( Q = K_{eq} \), the reaction is at equilibrium.
• If \( Q > K_{eq} \), the reaction proceeds in the backward direction.

In the context of this problem, \(Q = 9.32 \times 10^{-3}\), indicating that the reaction is still advancing towards equilibrium. It affects the cell potential directly, since its value gets plugged into the Nernst Equation, predicting how the voltage will alter when concentration changes occur. Understanding \(Q\) enables students to grasp its practical use in predicting reaction directions and shifts, as well as maintaining an electrochemical balance, underscoring its value in wider chemical applications.