Question

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?

Answer 1

(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

Step-by-step 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

`\begin{array}{rcl}E^(\circ)& = & (RT)/(nF)\ln K \\\\0.17 & = &(8.314 * 298.15)/(1* 96485)\ln K \\\\& = & 0.0257 \ln K\\\ln K & = & (0.17)/(0.257)\\\\& = & 6.62\\K& = & e^(6.62)\\& = & \mathbf{750}\\\end{array}`

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