Redox reactions are at the heart of photosynthesis. Some of them are spontaneous, others are driven — Organic Chemistry Chemistry Question
Redox reactions in photosynthesis
Redox reactions are at the heart of photosynthesis. Some of them are spontaneous, others are driven by light or conjugated chemical reactions. The former are named exergonic (∆G < 0), the latter – endergonic (∆G > 0).
Every redox reaction consists of two conjugated processes (half-reactions) – oxidation and reduction. In photosynthesis, half-reactions are often separated not only in space, but also in time. In living organisms, this is performed by dividing redox reactions into many steps involving bioorganic substances – enzymes, cofactors, etc.
Every half-reaction is characterized by a standard redox potential E° which refers to concentration of 1 mol dm-3 of all substances in solution and 1 bar pressure of all gaseous substances. The values of E° for several reactions involved in photosynthesis are listed in the table. Biochemists usually correct the standard potential to pH 7.0 and designate it as E°’.
Photosynthesis in green plants and algae can be described by an overall equation (see Problem 2):
H2O + CO2 → CH2O + O2
In this process water is oxidized to O2, and carbon dioxide is reduced to carbohydrates. The former reaction occurs under the action of light and consists of the so called light stages, the latter is driven by exergonic chemical reactions and involves the dark stages only.
[VISUAL]
Half-reaction | Standard redox potential, E° (V)
O2 + 4 H+ + 4 e → 2 H2O | 1.23
S + 2 H+ + 2 e → H2S | 0.14
Plastoquinone + 2 H+ + 2 e → Plastoquinone•H2 | 0.52
Cytochrome f(Fe3+) + e → Cytochrome f(Fe2+) | 0.365
NADP+ + H+ + 2 e → NADP•H | –0.11
P680+ + e → P680 | 1.10
Chlorophyll+ + e → Chlorophyll | 0.78
Calculate the standard biochemical redox potential for all half-reactions presented in the table above.
Model Answer
Applying the Nernst equation for a half-reaction
Ox + m H+ + n e → R
and putting [H+] = 10–7, we get a standard biochemical redox potential:
[VISUAL]
E°' = E° + (0.0591 / n) * lg(10^-7)^m = E° - 0.414 * (m / n)
Table of results:
- O2 + 4 H+ + 4 e → 2 H2O: E° = 1.23 V, E°’ = 0.82 V
- S + 2 H+ + 2 e → H2S: E° = 0.14 V, E°’ = -0.27 V
- Plastoquinone + 2 H+ + 2 e → Plastoquinone•H2: E° = 0.52 V, E°’ = 0.11 V
- Cytochrome f(Fe3+) + e → Cytochrome f(Fe2+): E° = 0.365 V, E°’ = 0.365 V
- NADP+ + H+ + 2 e → NADP•H: E° = -0.11 V, E°’ = -0.32 V
- P680+ + e → P680: E° = 1.10 V, E°’ = 1.10 V
- Chlorophyll+ + e → Chlorophyll: E° = 0.78 V, E°’ = 0.78 V
Using the answers obtained in Problem 2, determine E° and E°’ for the half-reaction of CO2 reduction to CH2O.
Model Answer
The standard electromotive force for the reaction
H2O + CO2 → CH2O + O2
is the difference between standard redox potentials for oxidant and reductant:
CO2 + 4 H+ + 4 e → CH2O + H2O (E°1)
O2 + 4 H+ + 4 e → 2 H2O (E°2 = 1.23 V)
For this reaction, the standard Gibbs energy is 480.5 kJ mol-1, and 4 electrons are transferred from H2O to CO2. Hence, the standard emf is:
[VISUAL]
E° = -ΔG° / (n * F) = -480500 / (4 * 96500) = -1.24 V = E°1 - 1.23 V
For CO2 reduction to carbohydrates the standard redox potential is E°1 = -0.01 V.
The standard biochemical potential is:
E°'1 = -0.01 - 0.414 * (4 / 4) = -0.42 V
Write the overall reaction equation of photosynthesis in green sulfur bacteria, which oxidize hydrogen sulfide to elementary sulfur. Separate this equation into the oxidation and reduction steps. Calculate the standard Gibbs energy of the overall reaction at 298 K. Assuming that the reaction is driven by light energy only, determine the minimum number of photons (840 nm) necessary to oxidize one molecule of hydrogen sulfide.
Model Answer
The overall reaction: CO2 + 2 H2S → CH2O + 2 S + H2O
Oxidation: H2S – 2 e → S + 2 H+
Reduction: CO2 + 4 H+ + 4 e → CH2O + H2O
Standard emf: E° = –0.01 – 0.14 = –0.15 V
Standard Gibbs energy: ΔG° = –n F E° = –4 × 96500 × (–0.15) × 10^-3 = 57.9 kJ mol-1.
Energy of light with wavelength 840 nm:
[VISUAL]
E = h c N_A / λ = (6.63 * 10^-34 * 3.00 * 10^8 * 6.02 * 10^23) / (840 * 10^-9) * 10^-3 = 143 kJ mol-1.
One quantum gives enough energy to oxidize two molecules of H2S.
Write the overall reaction of light stages of photosynthesis in green plants.
Model Answer
Both NADP+ reduction and ATP formation require one proton, and during H2O oxidation two protons are released. Hence, the overall reaction equation of light stages is:
H2O + NADP+ + ADP + Pi + hν → ½ O2 + NADP•H + ATP
Calculate the Gibbs energy of the overall reaction describing light stages of photosynthesis given that the standard biochemical Gibbs energy for ATP formation is +30.5 kJ mol-1.
Model Answer
The overall reaction is the sum of two reactions:
H2O + NADP+ + hν → ½ O2 + NADP•H + H+
and
ADP + Pi + H+ → ATP + H2O.
For the latter, the standard biochemical Gibbs energy is known (30.5 kJ mol-1) and for the former it can be determined from the standard biochemical redox potentials:
ΔG°’ = –n F E°’ = –2 × 96500 × (0.82 – (–0.32)) × 10^-3 = 220 kJ mol-1.
The overall light stages reaction contains no protons, hence the standard Gibbs energy is the same as the standard biochemical Gibbs energy:
ΔG° = ΔG°’ = 220 + 30.5 = 250.5 kJ mol-1
Explain this effect qualitatively, considering excitation process as an electronic transition between HOMO and LUMO.
Model Answer
This effect is easily understood using a simple orbital diagram (see Appendix in “Molecular Mechanisms of Photosynthesis” by R.E.Blankenship). In the ground state, a lost electron comes from the low-energy HOMO, while an acquired electron enters the high-energy LUMO. As a result, the molecule is neither a strong oxidant nor a good reductant. In the excited state, the situation is different: a lost electron leaves the high-energy LUMO, and the acquired electron comes to low-energy HOMO: both processes are energetically favorable, and the molecule can act both as a strong oxidant and a powerful reductant.
[VISUAL]
Derive the equation relating the redox potential of the excited state, redox potential of the ground state, and the excitation energy Eex = hν. Using this equation, calculate the standard redox potential for the processes: P680+ + e → P680* (λex = 680 nm) and Chlorophyll+ + e → Chlorophyll* (λex = 680 nm), where asterisk denotes excited state.
Model Answer
Consider two half-reactions:
Ox + e → R (standard redox potential E°_Ox/R)
and
Ox + e → R* (standard redox potential E°_Ox/R*).
The difference in their Gibbs energies is equal to the excitation energy:
-F * (E°_Ox/R - E°_Ox/R*) = E_ex = h * ν = h * c * N_A / λ
whence it follows:
E°_Ox/R* = E°_Ox/R - h * c * N_A / (λ * F)
For P680+:
[VISUAL]
E°_P680+/P680* = 1.10 - (6.63 * 10^-34 * 3.00 * 10^8 * 6.02 * 10^23) / (680 * 10^-9 * 96500) = -0.72 V
For Chlorophyll+:
[VISUAL]
E°_Chl+/Chl* = 0.78 - (6.63 * 10^-34 * 3.00 * 10^8 * 6.02 * 10^23) / (680 * 10^-9 * 96500) = -1.04 V