In biological systems, acetate ion is converted to ethyl alcohol in a two- step process: \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+3 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{CHO}(a q)+\mathrm{H}_{2} \mathrm{O}\) \(E^{o \prime}=-0.581 \mathrm{~V}\) \(\mathrm{CH}_{3} \mathrm{CHO}(a q)+2 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \quad E^{o \prime}=-0.197 \mathrm{~V}\) \(\left(E^{\circ \prime}\right.\) is the standard reduction voltage at \(25^{\circ} \mathrm{C}\) and \(\mathrm{pH}\) of \(7.00 .\) ) (a) Calculate \(\Delta G^{\circ \prime}\) for each step and for the overall conversion. (b) Calculate \(E^{\circ \prime}\) for the overall conversion.

Step by step solution

01

Calculating \(\Delta G^{\circ \prime}\) for each step

To calculate the standard Gibbs free energy change, use the Nernst equation: \(\Delta G^{\circ \prime} = -nFE^{\circ \prime}\) Where: \(ΔG^{\circ \prime}\) - standard Gibbs free energy change \(n\) - number of electrons transferred in the reaction \(F\) - Faraday's constant (96485 C/mol) \(E^{\circ \prime}\) - standard reduction voltage For step 1: \(n_1 = 2 \, e^{-}\) \(E^{\circ \prime}_1 = -0.581 \, V\) \(\Delta G^{\circ \prime}_1 = -2 \times 96485 \times (-0.581)\) \(\Delta G^{\circ \prime}_1 = 112069 \, J/mol\) For step 2: \(n_2 = 2 \, e^{-}\) \(E^{\circ \prime}_2 = -0.197 \, V\) \(\Delta G^{\circ \prime}_2 = -2 \times 96485 \times (-0.197)\) \(\Delta G^{\circ \prime}_2 = 37980.2 \, J/mol\)

02

Calculating the overall reaction

Add the two half-reactions to get the balanced overall reaction: $\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+3 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{CHO}(a q)+\mathrm{H}_{2} \mathrm{O}$ $+\; \;\mathrm{CH}_{3} \mathrm{CHO}(a q)+2 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)$ \(--------------------------------------------------------------\) $\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+5 \mathrm{H}^{+}(a q)+4 e^{-} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+\mathrm{H}_{2} \mathrm{O}$

03

Calculating \(\Delta G^{\circ \prime}\) for the overall conversion

Add the \(\Delta G^{\circ \prime}\) values for each step to calculate the overall standard Gibbs free energy change: \(\Delta G^{\circ \prime}_{\text{overall}} = \Delta G^{\circ \prime}_1 + \Delta G^{\circ \prime}_2\) \(\Delta G^{\circ \prime}_{\text{overall}} = 112069 \, J/mol + 37980.2 \, J/mol\) \(\Delta G^{\circ \prime}_{\text{overall}} = 150049.2\, J/mol\)

04

Calculating \(E^{\circ \prime}\) for the overall conversion

Since we have the total Gibbs free energy change and the number of electrons transferred (4) for the overall reaction, we can calculate the standard reduction voltage for the overall conversion using the Nernst equation: \(\Delta G^{\circ \prime}_{\text{overall}} = -n_{\text{overall}}FE^{\circ \prime}_{\text{overall}}\) \(150049.2 = -4 \times 96485 \times E^{\circ \prime}_{\text{overall}}\) \(E^{\circ \prime}_{\text{overall}} = -\frac{150049.2}{4 \times 96485}\) \(E^{\circ \prime}_{\text{overall}} = -0.389 \, V\)

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