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In the mitochondrial electron transfer chain, one of the steps that mediates transfer of electrons from NADH to O2 is the transfer of electrons from the reduced (ferrous) heme of cytochrome b to the oxidized (ferric) heme of cytochrome c1:Fe II (cyt b) + Fe III (cyt c1) → Fe III (cyt b) + Fe II (cyt c1) The standard state reduction potentials of Fe III (cyt c1) and Fe III (cyt b) are 0.25 V and 0.08 V, respectively, relative to the standard hydrogen electrode.(a) Express the above redox reaction as two half reactions, one as an oxidation and one as a reduction. (b) Determine the equilibrium constant for the reaction as written. Is this reaction spontaneous, favoring products?

Respuesta :

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Answer:

(a)  Oxidation:  Fe(II)(cyt b) ⇌ Fe(III)(cyt b) + e⁻

     Reduction: Fe(III)(cyt c1) + e⁻ ⇌ Fe(II)(cyt c1)

(b) K = 750; spontaneous

Explanation:

(a) Half-reactions

Oxidation:  Fe(II)(cyt b) ⇌ Fe(III)(cyt b) + e⁻

Reduction: Fe(III)(cyt c1) + e⁻ ⇌ Fe(II)(cyt c1)

Overall:      Fe(II)(cyt b) + Fe(III)(cyt c1) ⇌ Fe(III)(cyt b) + Fe(II)(cyt c1)

(b) Equilibrium constant

(i) Standard cell potential

The standard reduction potentials are

                                                       E°/V

Fe(III)(cyt c1)  + e⁻ ⇌ Fe(II)(cyt c1); 0.25

Fe(III)(cyt b) + e- ⇌ Fe(II)(cyt b);    0.08

For the reaction in Part (a),                                                                    

                                                                                           E°/V

Fe(II)(cyt b) ⇌ Fe(III)(cyt b)                                                -0.08

Fe(III)(cyt c1) ⇌ Fe(II)(cyt c1)                                               0.25

Fe(II)(cyt b) + Fe(III)(cyt c1) ⇌ Fe(III)(cyt b) + Fe(II)(cyt c1)  0.17

The standard cell potential is 0.17 V.

(ii) Equilibrium constant

[tex]\begin{array}{rcl}E^{\circ}& = & \dfrac{RT}{nF}\ln K \\\\0.17 & = &\dfrac{8.314 \times 298.15}{1\times 96485}\ln K \\\\& = & 0.0257 \ln K\\\ln K & = & \dfrac{0.17}{0.257}\\\\& = & 6.62\\K& = & e^{6.62}\\& = & \mathbf{750}\\\end{array}[/tex]

E° and K are positive, so the reaction is spontaneous.

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