Chemistry, 25.10.2019 22:43 michellemitchel

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?

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Answer from: demeckoferguson

(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

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

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

Answer from: Quest

electron microscopy and scanning tunneling microscopes gives us a greater understanding of atoms by being able to see the atom itself, and lets us see some atomic reactions.

explanation:

Answer from: Quest

For cyclohexane compounds drawn in a chair conformation: -axial bonds are vertical -equatorial bonds are parallel to two bonds within the ring. thus, as i assume the chair conformation is drawn: -the x group is in an axial position, as are the five other vertical bonds. -three axial bonds are on the top face of the ring -the other three axial bonds are on the bottom face the position of the second substituent on each ring carbon is equatorial. gauche groups must be on an adjacent carbon such as carbon 6 and carbon 2, and also gauche groups are rotated by 60 degrees relative to each other. drawing the newman projections for c1–c2 and c1–c6 aids in visualizing the groups that are gauche to the y group on c1. click this link and watch some academic videos that may you more understand the topic