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Chapter 18: Problem 96

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.

Short Answer

Expert verified
Question: Calculate the standard Gibbs free energy change, ΔG°', and the standard reduction voltage, E°', for the overall conversion of acetate ion to ethyl alcohol in the given biological system. Answer: For the overall conversion of acetate ion to ethyl alcohol, the standard Gibbs free energy change, ΔG°', is 150049.2 J/mol, and the standard reduction voltage, E°', is -0.389 V.

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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Most popular questions from this chapter

For the following half-reactions, answer the questions below. $$ \begin{array}{cc} \mathrm{Co}^{3+}(a q)+e^{-} \longrightarrow \mathrm{Co}^{2+}(a q) & E^{\circ}=+1.953 \mathrm{~V} \\ \mathrm{Fe}^{3+}(a q)+e^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) & E^{\circ}=+0.769 \mathrm{~V} \\ \mathrm{I}_{2}(a q)+2 e^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) & E^{o}=+0.534 \mathrm{~V} \\ \mathrm{~Pb}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Pb}(s) & E^{\circ}=-0.127 \mathrm{~V} \\ \mathrm{Cd}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Cd}(s) & E^{\circ}=-0.402 \mathrm{~V} \\ \mathrm{Mn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Mn}(s) & E^{\circ}=-1.182 \mathrm{~V} \end{array} $$ (a) Which is the weakest reducing agent? (b) Which is the strongest reducing agent? (c) Which is the strongest oxidizing agent? (d) Which is the weakest oxidizing agent? (e) Will \(\mathrm{Pb}(s)\) reduce \(\mathrm{Fe}^{3+}(a q)\) to \(\mathrm{Fe}^{2+}(a q) ?\) (f) Will \(\mathrm{I}^{-}(a q)\) reduce \(\mathrm{Pb}^{2+}(a q)\) to \(\mathrm{Pb}(s) ?\) (g) Which ion(s) can be reduced by \(\mathrm{Pb}(s)\) ? (h) Which if any metal(s) can be oxidized by \(\mathrm{Fe}^{3+}(a q)\) ?

Consider three metals, \(\mathrm{X}, \mathrm{Y}\), and \(\mathrm{Z}\), and their salts, \(\mathrm{XA}, \mathrm{YA}\), and \(\mathrm{ZA}\). Three experiments take place with the following results: \- \(\mathrm{X}+\mathrm{hot} \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2}\) bubbles \(\mathrm{X}+\mathrm{YA} \longrightarrow\) no reaction I \(\mathrm{X}+\mathrm{ZA} \longrightarrow \mathrm{X}\) discolored \(+\mathrm{Z}\) Rank metals \(\mathrm{X}, \mathrm{Y}\), and \(\mathrm{Z}\), in order of decreasing strength as reducing agents.

For the following half-reactions, answer the questions below: $$ \begin{array}{cc} \mathrm{Ce}^{4+}(a q)+e^{-} \longrightarrow \mathrm{Ce}^{3+}(a q) & E^{\circ}=+1.61 \mathrm{~V} \\ \mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \mathrm{Ag}(s) & E^{\circ}=+0.80 \mathrm{~V} \\ \mathrm{Hg}_{2}^{2+}(a q)+2 e^{-} \longrightarrow 2 \mathrm{Hg}(l) & E^{\circ}=+0.80 \mathrm{~V} \\ \mathrm{Sn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Sn}(s) & E^{\circ}=-0.14 \mathrm{~V} \\ \mathrm{Ni}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Ni}(s) & E^{\circ}=-0.24 \mathrm{~V} \\ \mathrm{Al}^{3+}(a q)+3 e^{-} \longrightarrow \mathrm{Al}(s) & E^{o}=-1.68 \mathrm{~V} \end{array} $$ (a) Which is the weakest oxidizing agent? (b) Which is the strongest oxidizing agent? (c) Which is the strongest reducing agent? (d) Which is the weakest reducing agent? (e) Will \(\mathrm{Sn}(s)\) reduce \(\mathrm{Ag}^{+}(\mathrm{aq})\) to \(\mathrm{Ag}(s) ?\) (f) Will \(\mathrm{Hg}(l)\) reduce \(\mathrm{Sn}^{2+}(a q)\) to \(\mathrm{Sn}(s) ?\) (g) Which ion(s) can be reduced by \(\operatorname{Sn}(s)\) ? (h) Which metal(s) can be oxidized by \(\mathrm{Ag}^{+}(a q)\) ?

Write the equation for the reaction, if any, that occurs when each of the following experiments is performed under standard conditions. (a) Sulfur is added to mercury. (b) Manganese dioxide in acidic solution is added to liquid mercury. (c) Aluminum metal is added to a solution of potassium ions.

Consider a voltaic cell at \(25^{\circ} \mathrm{C}\) in which the following reaction takes place. $$ 3 \mathrm{O}_{2}(g)+4 \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{NO}_{3}^{-}(a q)+4 \mathrm{H}^{+}(a q) $$ (a) Calculate \(E^{\circ}\) (b) Write the Nernst equation for the cell. (c) Calculate \(E\) under the following conditions: \(\left[\mathrm{NO}_{3}{ }^{-}\right]=0.750 M\), \(P_{\mathrm{NO}}=0.993 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.515 \mathrm{~atm}, \mathrm{pH}=2.85\)
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