IChO 1977 · Theory problem 7

Photosynthesis reversed in a galvanic cell

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Photosynthesis by be summarised by the overall equation:

Figure 1

for which values of \(\Delta H\) and \(\Delta S\) at \(25\,^\circ\mathrm{C}\) are as follows:

\[ \Delta H=2.816\times10^6\ \mathrm{J},\qquad \Delta S=-182\ \mathrm{J\,K^{-1}}\quad\text{or} \]

\(\Delta H=2.816\times10^6\ \mathrm{J\,mol^{-1}},\quad \Delta S=-182\ \mathrm{J\,K^{-1}\,mol^{-1}}\) if \(\Delta H\) and \(\Delta S\) values are related to one mole of reaction changes.

Imagine that there have been devised electrodes that would allow selective reduction of oxygen to water and oxidation of glucose to carbon dioxide in a galvanic cell, i. e a reverse process when compared with that of the photosynthetic reaction.

Problems:

7.1 What will be the electromotive force of the cell in which light energy would be transformed to electric energy by means of the photosynthetic reaction?

Note: In the envelope you can find the relation between the electromotive force and the change of free enthalpy of the reaction. (Attention: If you open the envelope you lose some points.)

7.2 In case we would want to quantify the symbol "light" in the equation of photosynthesis, we would ask: how many moles of photons with wavelength for example \(500\ \mathrm{nm}\) take part in the above reaction? Calculate.

7.3 Calculate, what would be the electric power of a square swimming pool with a side of \(10\ \mathrm{m}\) containing green algae capable of the photosynthetic reaction if under average illumination a current of \(1\ \mathrm{mA}\) can be expected from the area of \(1\ \mathrm{cm^2}\).

Official English text of the problem.

Answers

  • 7.1: \(\boxed{E=1.24\ \mathrm{V}}\).
  • 7.2: \(\boxed{\text{Approximately }12\ \mathrm{mol}\text{ of photons per mole of glucose}}\). The ideal free-energy estimate is \(12.00\ \mathrm{mol}\); an enthalpy-only estimate gives \(11.77\ \mathrm{mol}\).
  • 7.3: \(\boxed{P=1.24\times10^3\ \mathrm{W}=1.24\ \mathrm{kW}}\), assuming the cell supplies the calculated voltage at the stated current.

Solution

The prose identifies the product as glucose. Correcting the apparent errors in the printed scheme, the balanced photosynthetic reaction is

\[ \ce{6 CO2(g) + 6 H2O(l) ->[light][chlorophyll] C6H12O6(s) + 6 O2(g)}. \]

The supplied thermodynamic quantities therefore refer to the formation of one mole of glucose.

The species used below are glucose, represented as open-chain D-glucose by O=C[C@H](O)[C@@H](O)[C@H](O)[C@H](O)CO; carbon dioxide O=C=O; water O; dioxygen O=O; and the hydrogen ion (proton) [H+]. The open-chain glucose SMILES specifies its carbon stereochemistry, but does not describe its cyclic solid-state form or crystal structure.

7.1 — Electromotive force

The galvanic cell carries out the reverse reaction:

\[ \ce{C6H12O6(s) + 6 O2(g) -> 6 CO2(g) + 6 H2O(l)}. \]

Written in acidic form for balancing, the electrode reactions are:

\[ \begin{aligned} \text{Anode:}\quad& \ce{C6H12O6(s) + 6 H2O(l) -> 6 CO2(g) + 24 H+(aq) + 24 e-},\\ \text{Cathode:}\quad& \ce{6 O2(g) + 24 H+(aq) + 24 e- -> 12 H2O(l)}. \end{aligned} \]

Thus \(n=24\) moles of electrons are transferred per mole of glucose. Equivalently, the six carbon atoms increase their mean oxidation number from \(0\) in glucose to \(+4\) in carbon dioxide, releasing \(6\times4=24\) electrons.

At \(T=298.15\ \mathrm{K}\), the Gibbs energy change for photosynthesis is

\[ \begin{aligned} \Delta G_{\mathrm{photo}} &=\Delta H_{\mathrm{photo}}-T\Delta S_{\mathrm{photo}}\\ &=2.816\times10^6 -(298.15)(-182)\\ &=2.8702633\times10^6\ \mathrm{J\,mol^{-1}}. \end{aligned} \]

Reversing the reaction changes its sign:

\[ \Delta G_{\mathrm{cell}} =-2.8702633\times10^6\ \mathrm{J\,mol^{-1}}. \]

The maximum electrical work gives the relation

\[ \Delta G_{\mathrm{cell}}=-nFE. \]

Using \(F=96485\ \mathrm{C\,mol^{-1}}\),

\[ E= \frac{2.8702633\times10^6\ \mathrm{J\,mol^{-1}}} {24(96485\ \mathrm{C\,mol^{-1}})} =1.2395\ \mathrm{J\,C^{-1}}. \]

Therefore,

\[ \boxed{E=1.24\ \mathrm{V}}. \]

7.2 — Amount of photons

The energy of a photon is

\[ \varepsilon_\gamma=\frac{hc}{\lambda}. \]

Using \(h=6.62607\times10^{-34}\ \mathrm{J\,s}\), \(c=2.99792\times10^8\ \mathrm{m\,s^{-1}}\), and \(\lambda=500\times10^{-9}\ \mathrm{m}\),

\[ \varepsilon_\gamma =\frac{(6.62607\times10^{-34})(2.99792\times10^8)} {500\times10^{-9}} =3.97289\times10^{-19}\ \mathrm{J}. \]

With \(N_{\mathrm A}=6.02214\times10^{23}\ \mathrm{mol^{-1}}\), one mole of these photons carries

\[ E_{\gamma,\mathrm m} =N_{\mathrm A}\varepsilon_\gamma =2.39253\times10^5\ \mathrm{J\,mol^{-1}} =239.253\ \mathrm{kJ\,mol^{-1}}. \]

If photon energy is ideally converted into the chemical free energy required for photosynthesis, the amount needed to form one mole of glucose is

\[ n_\gamma =\frac{2.8702633\times10^6\ \mathrm{J}} {2.39253\times10^5\ \mathrm{J\,mol^{-1}}} =11.9968\ \mathrm{mol}. \]

Hence,

\[ \boxed{n_\gamma\approx12.0\ \mathrm{mol} \quad\text{per mole of glucose}}. \]

If the question instead intends a simple comparison with the reaction enthalpy, it gives

\[ n_{\gamma,H} =\frac{2.816\times10^6\ \mathrm{J}} {2.39253\times10^5\ \mathrm{J\,mol^{-1}}} =11.77\ \mathrm{mol}, \]

which also rounds to approximately \(12\ \mathrm{mol}\). The free-energy calculation includes the entropy contribution relevant to reversible electrical work. These energy estimates do not determine the actual biological photon requirement.

7.3 — Electric power of the pool

The illuminated area is

\[ A=(10\ \mathrm{m})^2 =100\ \mathrm{m^2} =1.00\times10^6\ \mathrm{cm^2}. \]

At the specified current density,

\[ I= \left(1.00\times10^{-3}\ \mathrm{A\,cm^{-2}}\right) \left(1.00\times10^6\ \mathrm{cm^2}\right) =1.00\times10^3\ \mathrm{A}. \]

Assuming uniform illumination and neglecting voltage losses,

\[ P=EI =(1.2395\ \mathrm{V})(1.00\times10^3\ \mathrm{A}) =1.2395\times10^3\ \mathrm{W}. \]

Thus,

\[ \boxed{P\approx1.24\ \mathrm{kW}}. \]

The solution was checked against the official solution.