Problem 2

The Hill reaction dissects photosynthesis. (a) Write the overall equation of plant photosynthesis, reducing

\ce{CO2}

\{\ce{CH2O}\}

. (b) Hill found that isolated chloroplasts do not evolve

\ce{O2}

in light even with

\ce{CO2}

, but do so upon adding potassium ferrioxalate

\ce{K3[Fe(C2O4)3]}

(with excess oxalate) without

\ce{CO2}

. Give the oxidant and reducing agent in natural photosynthesis and in the Hill reaction. (c) Hill measured evolved

\ce{O2}

with haemoglobin (Hb binds

\ce{O2}

1:1, initial

[\text{Hb}] = 0.6 \times 10^{-4}

mol dm

^{-3}

); at

[\ce{Fe^{III}}] = 2.0 \times 10^{-4}

mol dm

^{-3}

the HbO

_2

fraction saturates at about 75 %. Estimate the Fe/

\ce{O2}

ratio, write the Hill reaction equation, and calculate its

\Delta G

T = 298

p(\ce{O2}) = 1

mmHg, pH = 8, standard concentrations of other species (

E^{\circ}

\ce{O2 + 4H+ + 4e- -> 2H2O}

+1.23 V;

\ce{[Fe(C2O4)3]^{3-} + e- -> [Fe(C2O4)3]^{4-}}

+0.05 V). Is it spontaneous? (d) Isolated chloroplasts were irradiated 2 h with 672 nm light of 0.503 mJ s

^{-1}

, producing 47.6 mm

^3

\ce{O2}

(10 °C, 740 mmHg). Calculate the quantum requirement (photons per electron transferred). (e) Conclusions: are water oxidation and

\ce{CO2}

reduction spatially separated? Is

\ce{O2}

produced from

\ce{CO2}

? Does water oxidation require light? Do most chlorophylls participate directly? Does each photon transfer one electron?

Step 2 of 5: Hill reaction equation

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\dfrac{n(\mathrm{Fe})}{n(\ce{O2})} = \dfrac{2.0 \times 10^{-4}}{0.75 \times 0.6 \times 10^{-4}} \approx 4.4 \approx 4;\quad \ce{2H2O + 4[Fe(C2O4)3]^{3-} -> O2 + 4[Fe(C2O4)3]^{4-} + 4H+}

Analysis

Saturation at 75 % of Hb means $4.5 \times 10^{-5}$ mol dm $^{-3}$ $\ce{O2}$ from $2.0 \times 10^{-4}$ mol dm $^{-3}$ Fe(III) — a ratio near 4:1, i.e. Fe(III) is reduced to Fe(II), stabilised as $\ce{[Fe(C2O4)3]^{4-}}$ .