Answer
In the given case, the $\Delta{E}$ is negative and the $\Delta{G}$ is positive. Therefore, the given reaction is not spontaneous under standard conditions.
Work Step by Step
Recall that 'The Nernst Equation Describes Oxidation–Reduction Reactions' (page 463).
The Redox-reactions based on the oxidation and reduction potentials of the given reaction under standard conditions are as follows:
Oxidation: Cyto a (Fe2+) <-> Cyto a (Fe3+) + e- (Anode) (0.29V)
Reduction: Cyto b (Fe3+) + e- <-> Cyto b (Fe2+) (Cathode)(0.077V)
Please note that these values were obtained from 'TABLE 14-4 Standard Reduction Potentials of Some Biochemically
Important Half-Reactions' on page 466.
The formula to calculate the standard reduction potential is as follows:
$$ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$$
where:
$ E^{\circ}_{cathode} $ = Standard reduction potential at the cathode
$ E^{\circ}_{anode} $ = Standard reduction potential at anode
Plugging in the values from table 14:
$$ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$$
$$ E^{\circ}_{cell} = 0.077 - 0.29$$
$$ E^{\circ}_{cell} = -0.213 V $$
Recall the formula for $\Delta{G}^{{\circ}}$:
$$\Delta{G}^{{\circ}} = -nFE^{\circ}_{cell} $$
where $n$ = number of moles of
electrons transferred per mole of reactant converted and
$F$ = faraday constant = 96485 J/Vmol. (See page 464 for more details)
The calculation for $\Delta{G}^{{\circ}}$ is:
$$\Delta{G}^{{\circ}} = -nFE^{\circ}_{cell} $$
$$\Delta{G}^{{\circ}} = -(1)(96485 J/Vmol)(-0.213V) $$
$$\Delta{G}^{{\circ}} = 20.5 kJ/mol $$
In the given case, the $\Delta{E}$ is negative and the $\Delta{G}$ is positive. Therefore, the given reaction is not spontaneous under standard conditions.
