Cytochrome, a complicated molecule that we will represent as \(\mathrm{CyFe}^{2+}\), reacts with the air we breathe to supply energy required to synthesize adenosine triphosphate (ATP). The body uses ATP as an energy source to drive other reactions (Section 19.7). At \(\mathrm{pH} 7.0\) the following reduction potentials pertain to this oxidation of \(\mathrm{CyFe}^{2+}\) $$ \begin{aligned} \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & & E_{\mathrm{red}}^{\circ}=+0.82 \mathrm{~V} \\\ \mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-} & \longrightarrow \mathrm{CyFe}^{2+}(a q) & E_{\mathrm{red}}^{\circ} &=+0.22 \mathrm{~V} \end{aligned} $$ (a) What is \(\Delta G\) for the oxidation of \(\mathrm{CyFe}^{2+}\) by air? \((\mathbf{b})\) If the synthesis of \(1.00 \mathrm{~mol}\) of ATP from adenosine diphosphate (ADP) requires a \(\Delta G\) of \(37.7 \mathrm{~kJ},\) how many moles of ATP are synthesized per mole of \(\mathrm{O}_{2} ?\)

Step by step solution

01

Write the balanced redox reaction between \(\mathrm{CyFe}^{2+}\) and \(\mathrm{O}_2\)

First, we need to write the balanced redox reaction by combining the given half-reactions: $$ \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \\ \mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{CyFe}^{2+}(a q) $$ To balance the number of electrons being transferred, we multiply the second half-reaction by 4: $$ 4[\mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{CyFe}^{2+}(a q)] $$ Now, we can add the two half-reactions: $$ \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{CyFe}^{3+}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) + 4 \mathrm{CyFe}^{2+}(a q) $$ This is the balanced redox reaction we will use for the calculations.

02

Calculate the overall cell potential, \(E_{cell}\), for the redox reaction

To find the overall cell potential, we subtract the reduction potential of the cathode from the reduction potential of the anode: $$ E_{cell} = E_{red}^{\circ{\text{(cathode)}}} - E_{red}^{\circ{\text{(anode)}}} $$ In this case, the oxygen reduction half-reaction acts as the cathode and the \(\mathrm{CyFe}^{2+}\) oxidation half-reaction acts as the anode: $$ E_{cell} = (+0.82\,\text{V}) - (+0.22\,\text{V}) = +0.60\,\text{V} $$ So the overall cell potential is \(+0.60\,\text{V}\).

03

Calculate the Gibbs free energy change, \(\Delta G\), using the relationship between \(\Delta G\) and \(E_{cell}\)

To find \(\Delta G\), we use the following equation: $$ \Delta G = -nFE_{cell} $$ where \(n\) is the number of moles of electrons transferred, \(F\) is the Faraday constant \((96,485\,\text{C/mol})\), and \(E_{cell}\) is the overall cell potential. In the balanced redox reaction, \(4\) moles of electrons are being transferred: $$ \Delta G = -(4\,\text{mol})(96,485\,\text{C/mol})(+0.60\,\text{V}) = -231,564\,\text{J/mol} = -231.6\,\text{kJ/mol} $$ So, \(\Delta G = -231.6\,\text{kJ/mol}\) for the oxidation of \(\mathrm{CyFe}^{2+}\) by air.

04

Calculate the moles of ATP synthesized per mole of \(\mathrm{O}_{2}\) using the given information

We are given that the synthesis of \(1.00\,\text{mol}\) of ATP requires a \(\Delta G\) of \(37.7\,\text{kJ}\). We can use the stoichiometry of the balanced redox reaction to find the number of moles of ATP synthesized per mole of \(\mathrm{O}_2\): $$ \frac{-\Delta G_\text{redox}}{-\Delta G_\text{ATP}} = \frac{231.6\,\text{kJ/mol}}{37.7\,\text{kJ/mol}} = 6.14\,\text{mol} $$ Thus, approximately \(6.14\,\text{mol}\) of ATP are synthesized per mole of \(\mathrm{O}_{2}\).

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