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 pH 7.0 the following reduction potentials pertain to this oxidation of \(\mathrm{CyFe}^{2+}:\) $$ \begin{aligned} \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \longrightarrow & 2 \mathrm{H}_{2} \mathrm{O}(l) & & E_{\mathrm{red}}^{\circ}=+0.82 \mathrm{~V} \\\ \mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{CyFe}^{2+}(a q) & & E_{\mathrm{red}}^{\mathrm{o}}=+0.22 \mathrm{~V} \end{aligned} $$ (a) What is \(\Delta G\) for the oxidation of \(\mathrm{CyFe}^{2+}\) by air? (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

Half-cell reactions

We are given two half-cell reactions: 1. \(\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2}\mathrm{O}(l)\) with \(E_{\mathrm{red}}^{\circ}=+0.82 \mathrm{V}\) 2. \(\mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{CyFe}^{2+}(a q)\) with \(E_{\mathrm{red}}^{\circ}=+0.22 \mathrm{V}\) The first reaction is a reduction reaction, and the second one is an oxidation reaction. **Step 2: Write the overall balanced redox reaction**

02

Overall balanced redox reaction

Combine the half-cell reactions to obtain the overall balanced redox reaction: \(\mathrm{O}_{2}(g) + 4 \mathrm{H}^{+}(a q) + 4 \mathrm{CyFe}^{2+}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) + 4 \mathrm{CyFe}^{3+}(a q)\) There are four electrons exchanged in this reaction, and the stoichiometry between O₂ and CyFe²⁺ is one-to-four. **Step 3: Calculate E° of the reaction**

03

Calculate E° of the reaction

The standard cell potential E° for this reaction can be computed using the Nernst equation: \(E°_\text{cell} = E°_\text{red} - E°_\text{ox}\) Substitute the given values: \(E°_\text{cell} = (0.82 \, \text{V}) - (0.22 \, \text{V}) = 0.60 \, \text{V}\) **Step 4: Calculate ∆G for the reaction**

04

Calculate ∆G for the reaction

Use the Nernst equation to find ∆G: \(\Delta G = -nFE°_\text{cell}\) where - n = the number of electrons exchanged in the balanced equation, - F = Faraday's constant (96,485 C/mol), and - E° = standard cell potential. Substitute the values: \(\Delta G = - (4\,\text{mol})(96,485 \frac{\text{C}}{\text{mol}})(0.60\text{V}) \Longrightarrow \Delta G = -230,364\text{J}\) So, \(\Delta G = -230.36\text{kJ/mol}\) **Step 5: Calculate the number of moles of ATP synthesized per mole of O₂**

05

Number of moles of ATP synthesized per mole of O₂

The ∆G for the synthesis of 1 mol of ATP is given as 37.7 kJ. To find the number of moles of ATP synthesized per mole of O₂, we'll divide the ∆G for the oxidation of CyFe²⁺ (Step 4) by the ∆G for the synthesis of 1 mol of ATP: \(\text{Moles of ATP synthesized per mole of O₂} = -\frac{\Delta G_\text{oxygen}}{\Delta G_\text{ATP}} \Longrightarrow -\frac{-230.36 \,\text{kJ/mol}}{37.7 \,\text{kJ/mol}} \) \(\text{Moles of ATP synthesized per mole of O₂} \approx 6.11\) So, approximately 6.11 moles of ATP are synthesized per mole of O₂ in this reaction.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!