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.

Step by step solution

01

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: \[ 2\text{N}^{2+}(aq) + 2a\text{M}(s) \longrightarrow 2a\text{N}(s) + 2\text{M}^{a+}(aq)\\ \]

02

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: \[ \mathscr{E}^{\circ}_{cell} = \mathscr{E}^{\circ}(\text{M}^{a+}/\text{M}) - \mathscr{E}^{\circ}(\text{N}^{2+}/\text{N})\\ \] \[ \mathscr{E}^{\circ}_{cell} = 0.4 \mathrm{V} - 0.24 \mathrm{V} = 0.16 \mathrm{V}\\ \]

03

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 \(\left[\text{M}^{a+}\right]\): \[ \left[\text{M}^{a+}\right] = \frac{(0.10\text{M})^a (9.32 \times 10^{-3})}{(0.10\text{M})^{2a-1}}\\ \]

04

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: \[ w_{\text{max}} = -(2a)(96485\mathrm{C/mol})(0.18\mathrm{V}) \\ \]

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